Molecular Physics: educational manual 9786010430006

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AL-FARABI KAKZAH NATIONAL UNIVERSITY

MOLECULAR PHYSICS Educational manual

Almaty «Qazaq university» 2017

UDC 539.2/6 LBC 22.36 я 73 M 79 Recommended for publication by the decision of the Academic Council of the Faculty of Physics and Technology, Editorial and Publishing Council of Al-Farabi Kazakh National University (Protocol №2 dated 03.11.2017)

Reviewers: Doctor of Technical Sciences, Professor B.N. Absadykov Doctor of Technical Sciences, Professor A.B. Ustimenko Doctor of Physical and Mathematical Sciences, Professor М.Е. Abishev Authors: А.S. Askarova, S.А. Bolegenova, М.S. Moldabekova, S.А. Bolegenova, ZH.K. Shortanbayeva

M 79

Molecular Physics: educational manual / А.S. Askarova, S.А. Bolegenova, М.S. Moldabekova [et al]. – Almaty: Qazaq university, 2017. – 284 p. ISBN 978-601-04-3000-6 Main purpose of this educational manual is to acquaint students with fundamental concepts and laws and basic theories of molecular physics, the textbook covers fundamental principles of thermodynamics and matters related to the kinetic theory of gases. The textbook also presents methods of mathematical analysis with special emphasis on application thereof. Content of the educational manual is composed in accordance with the standard «SCES RK Higher Professional Education» for specialities in physics. Published in authorial release.

UDC 539.2/6 LBC 22.36 я 73 © Askarova А.S., Bolegenova S.А., Moldabekova М.S. [et al]., 2017 © Al-Farabi KazNU, 2017

ISBN 978-601-04-3000-6

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PREFACE The educational manual covers basic laws and methods of molecular physics. Particular emphasis is made on their intended use. In molecular physics physical properties of substance in different aggregate states are explained based on the microscopic structure. Such approach does not pay attention to the motion of individual molecules of a substance and requires very complex concepts that describe the motion of large ensemble of molecules controlled in practice using averaged values. In this regard, we tried to explain basic concepts of molecular physics in simple terms. The object of molecular physics study are physical systems that consist of innumerable molecules. In such systems properties are not inherent in individual molecules but in the entire system. These particularities require special methods of study of thermodynamics and of statistical physics. Therefore these theoretical points are included in the textbook. Special attention is paid to discovery of physical phenomena mechanisms, to estimating relationships between such practically measureable values as pressure, volume, temperature, etc., and also to explanation of physical theories use limitation, in explanation of natural phenomena. Contents of the textbook will ensure preparation in the future, that is necessary for excellent understanding of special subjects on physics. Molecular physics is intended for the first year students, thereofore theoretical material is limited to classical approaches. This learning aid is intended for students that study physics in universities. But we believe that this textbook will be usefull for other technical majors as well, since matters related to substance structure and study of its microscopic properties play key role in current natural science. Lectures on molecular physics, which authors over the course of many years delivered to students of faculites of mechanics, mathematics, chemistry, biology, physics and geography of Al Farabi Kazakh National University form the basis of this textbook. Authors of the educational manual would like to thank in advance their readers for their advice and recommendations which they will take into consideration for the following editions. Authors

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INTRODUCTION Subject of molecular physics. It is a branch of physics, studying physical properties of field in the different aggregate states on the basis of microstructure (molecular) calls molecular physics. Study of the properties of substances, dependent on a huge amount of moving molecules is the subject of molecular physics. Bodies or systems, consisting of huge amount of particles (atoms, molecules, or charged particles), are called macroscopics. For example, Avogadro constant is a measure of «huge amount of particles». N A = 6,02 ⋅ 10 mole −1 .

Conformity in the macroscopics systems, stimulated by influence of big amount of particles, are called statistical. Statistically laws are not applicable to the systems with small amount of particles. Consequently, subject of study in the filed of molecular physics is a type of molecular motion, and, accordingly, movement of big ensemble of molecules. Due to such a subject formulization there are two very important problems: 1) studying of particularity of molecular motion type; 2) methods of study of systems, consisting of great many components, and elaboration of the relevant concepts. Problems of molecular physics are solved with a help of methods from statistical physics and thermodynamics, which are connected to the study of motion and contacting of microparticles (atoms, molecules, ions) being a part of physical body. Summary review of historical evolution of molecular physics The first developed branch of molecular physics is kinetic theory of gases. Kinetic theory of gases began upon ancient Greece. In those times Greek philosophers Democritus (470-460 before Christ) and Epicurus (341-270 before Christ) gave a grounding of atomic theory of substance, and later this theory was developed by Lucretius (Titus Lucretius Cams. On the nature of material 95-55 years before Christ). 4

Having substantiated some of the results of kinetic theory, independently of one another, scientific researchers, P. Gassendi (15921655) R. Hooke (1635-1703) and D. Bernoulli (1650-1750), independently of one another in the epoch of establishment of modern science, have once again developed the atomic theory. After a secular crisis, kinetic theory started to develop again. Such scientific workers as Herapath, Waterston, Joule, Clausius, Maxwell, Boltzman) made their contribution in it development. The last three scientific workers (Rudolf Clausius, James Maxwell, Ludwig Boltzman) can be considered as founders of kinetic theory of gases. In 1859 Maxwell established the principle of equal molecules energy distribution in the mixture of gases for molecules with different mass and suggested the law on establishing of distribution of molecules by the velocity in the gas situating in the state of equilibrium, established some kind of temporally homogeneous (time constant), which follows the certain statistical law. Further development on the basis of discussion of this results served to appearance of classic statistical physics (J. U. Gibbs). Views on the mutual effect of molecules got its development in forming of theory of facial phenomenon. These phenomena explored such scientific workers as A. Keri, P. Laplas, C. Poisson, K. Gauss, J.U. Gibbs, I.S. Gromeko and others. It is generally thought that development of theory of real gases started from dissertation of Yan Diderik Van-der-Vaalce « Persistence of liquid and vapor states» written in 1873 y. For the first time ever intermolecular impact is mentioned in this research. In 1910 y. Van-der-Vaalce became Nobel prize laureate in physics. New development of molecular physics started in the beginning of twentieth century. Results of research of Brownian motion demonstrated assurance of molecules existence. A major contribution to the experimental research made Zh.B. Perrin and T.Svedberg, in theory – research of A. Einstein and M.Smolukhovsky. Structure of solid substances, liquids and their characteristics in the time of phase transitions were checked in ХХ century with a help of methods of fluoroscopic analysis of electronography, neutron diffraction. Teaching on intermolecular interaction on the basis of quantum mechanics reflects in the works of such scientific workers as M. Born, F.London, V. Geitner, P. Debye. 5

Molecular physics contains very wide range of problems. For example, compound and structure of substances, changes of their characteristics under the influence of external effects, transfer phenomenon, phase equilibrium and transition curve, critical condition of substance and many other occurrence are observed in that section of physics. On the basis of molecular physics were formed such sections of physics, like physics of metals, polymer, physics of plasma and so on. Among other things, such realm of science received independent development, like statistical physics, physical kinetics, chemical kinetics, physics of solid substance, molecular biology and others. Despite of particularity of objects and methods of rsearch, in all cases, main idea of molecular physics – is dependence of macroscopic characteristic of substance from its microscopic (molecular) structure.

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1 MOLECULAR-KINETIC MODEL OF MATERIAL BODIES. MASS OF ATOMS AND MOLECULES. STRUCTURED ELEMENTS. AMOUNT OF SUBSTANCE – MOLE 1.1. Molecular – kinetic model of material bodies In the heart of molecular physics there is a hypothesis (from Greek word hypothesіs – prognosis) on molecular structure of matter. According to this hypothesіs, substance cannot divided continuously and infinitely, it consists of very small particles, called molecules, which retaining all chemical property. Molecular hypothesis has a great importance in chemistry and physics. Molecular motion is invisible, but today structure of big molecules and atoms in them consist is well seen on the electronic microphoto (Fig. 1.1), and elementary particles in the consist of atom is detected due to the track left in the special experimental camera (Fig. 1.2).

Fig. 1.1. Electronic microphoto of molecules and atoms

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Fig. 1.2. Traces of elementary particles, left in experimental camera

What is material (lat. Materia – material)? Where endless variety of substance comes from? On each of these questions each civilization (on the development stage of culture) answered in the different way. According to one of approach, Allah ordered the chaos and created substance, according to other approach, Allah created substance from nothing. Ancient Greeks thought that substance appeared from a common primitive substance (Latin word substantia – initial foundation of all material, phenomenon) and it can take any shape. According to this theory, the world consists of dense substance. Substance represents an infinite number of small, simple, indivisible, eternal atoms. Atoms have different shapes and in different way and in different order settled in space. They can be connected by indefinitely different methods, this is why there are indefinitely diversity of colors and forms of substance A well-known Greek philosopher and scientist Aristotle (384322 years before Christ) thought, that substance was created from one heavenly and four terrestrial elements. Terrestrial elements: water, fire, earth and air (Fig.1.3) generate the sense of cold, heat, moisture and dryness, and reside in chaos motion. Heavenly element moves circular. In Middle Ages (End of V century – XVI c.) this approach was expanded, the next theory appeared. The most important aspect of civilization is human being – microсosm (from Greek word cosmos – 8

world), and its environment – macrocosm, connected with each other. Microcosm reflects macrocosm, in such a way defining of human destiny (Fig. 1.4).

Fig. 1.3. Fire-Dryness-Earth-Cold-Water-Heat-Air-Moisture

Such a concept became the reason of astrology appearance (Greek star + concept, science), which is still exists in our era of knowledge. It was era of alchemists).

Fig. 1.4. Summer day-Autumn day-Winter day-Spring day

In metaphysical concept gold is very important element, very beautiful and eternally valuable treasure. In present days, gold equi9

valent is used in trade too. Alchemists thought, that if they could turn ordnary metals into gold, they would reach the eternity. For this it is necessary to find « philosopher’s stone», to change the rhythm of time in the space in order to create people and metals. This illusion continuously searched in laboratories in middle ages. In results of that alchemists developed chemical technologies, discovered new materials and researched their properties. Unfortunately, they kept in secrets their discoveries. XVIII century became new stage of material structure knowledge development. The main idea of that century – caloric and phlogiston (from Greek word phlogistos – burning, invisible, weightless, volatile). It was thought, that heat consists from indivisible particles, caloric. Phlogiston is segregated when the substance is burning. French scientist chemists Antoine Lavoisier (1743-1794) is founder of modern chemistry, by experimental way has proved that phlogiston and caloric do not exist. He established, that processes of burning are chemical reactions, which go with heat exchange. British physicist and chemists of XIX century J. Dalton (17661844) considered ancient Greek atomic theory again and developed it. In order to explain interaction of substance, he admitted, that substance consists from atoms, similar to billiard ball, and complex substance – from complex combination of atoms. Applying the law of thermodynamics became a bridge between chemistry and physics, and creation of kinetic theory gave big contribution in scientific explorative succession. First decade of ХХ century in a great measure conduced to understanding of outworld. Quantum theory of atom opened the way to revolution in science. Thus, over 25 centuries the concept of substance had developed from objects, feelable by people to elementary particles as a part of an atom.

1.2. Atomic and molecules mass In accordance to conditional agreement, atomic and molecules mass is defined as atomic mass unit mu and relative molecular mass Mr. Atomic mass unit (a.m.u.) calls relative value of molecular mass. 10

It defines by the following way: atomic particle of mass equal to 1/12 part of carbon 12C. mu =12C mass of carbon mo / 12.

(1.1)

Therefrom, relative molecular mass can be found with a help of the following congruence: Mr = molecular mass muмол / mu = muмол 12 / mo12C.

(1.2)

Amount of relative molecular mass is dimensionless quantity. Atomic mass of chemical elements (except of transuranium elements) indicated in periodic system (Mendeleev’s table). In international system of particles (System International, short-form SI (SI)) particle of mass – kilogram (kg). According to international system of particles, 1 (a.m.u.) = 1,6605655 (86)⋅10-27лп (1984), for example mass of hydrogen atom mo= 1,68⋅10-24g, diameter d=0,5⋅10-8sm. 1.3. Structured elements of substance. Amount of substance – mole Quantity of substance defines by amount of strucuted elements in known mass of substance. Atoms, molecules, ions, and other particles or special group of particles can be associated to the structured elements. Unit of measure of substance quantity in system SI calls mole. Amount of moles indicated by symbol ν (ni). Depending on how many atoms in carbon with mass 0,012 кg 12 С, that much amount of substance is in system with structured elements equal to zero. In such a way, according to specification, in unit of measure of any substance (one mole) there is same amount of structured elements. This amount is called Avogadro constant or Avogadro’s number. This constant is called by the name of Italian scientist A. Avogadro, its symbol is NА. Avogadro constant is one of fundamental physical constant. Its valuation ΝΑ =

0,012кг 10−3 кг mole −1 = 6,022045 ⋅ 1023 mole −1. mole −1 = 12mu mu

11

(1.3)

There is no standard for measuring of mole, molecular amounts are not measured, they are summed up. From example (1.3) it follows: (1.4) mu N A = 10 −3 mole −1 ⋅ кg . Mass of one mole of substance is called molar mass. Molar mass is designated by letter М. By using the formula (1.3), we get, that molar mass is equal to product of mass of one molecule and Avogadro’s number:

M = mo N A .

(1.5)

Substitute the value of relative molecular mass of substance, found in example (1.2), in formula, we get the following:

M = M r mu N A .

(1.6)

If gas molecule N, so amount of moles ν can be found by the following way:

ν=

N . NA

(1.7)

By multiplying of this equation to the molecular mass, we see, that amount of substance with mass m is equal to:

ν=

m . M

(1.8)

Therefore:

N = νN A =

m N A. M

(1.9)

For the carrying out of such calculations relative atomic mass of elements can be taken from periodic system of Mendeleev. 12

1.4. Statistical and thermodynamical methods of molecular systems research. Statistical and dynamical conformities Microscopic approach to the exploration of nature was elaborated before microscopic approach. The goal of such method – explanation of characteristic of substance, which can be felt or observed in real conditions. For the research of condition of macroscopic systems in physics, statistical and thermodynamical research methods are used. Thermodynamical method does not base itself on any model of performance on atomic-molecular structure of substance and by definition is phenomenological method (from Greek word phanomenon – visible, touchable). It means, that task of thermodynamical method is establishing of links between directed observed quantities (pressure, temperature, volume, concentration and others). Thermodynamical method is simple and leads up to the solution of whole range of concrete problems, without expecting of any information on characteristic of atoms or molecules (size of atom and molecules, their mass and etc.). Common feature of thermodynamic is that it can be used for the search of simple mechanisms and human organism. More of that, in the scope of this theory, it is possible to understand some of our everyday experiment. Why does the cup of hot tea is getting cold? Why does the smell of perfume is dispersing in the room? The advantages of thermodynamical method is that it is possible to get information on links between observable characteristics of different substances, by knowing nothing about its internal structure. Information on difficult systems, consisting of huge amount of molecules, can get only on basis of four main thermodynamical laws and several variables. Another one advantage of thermodynamical method is that its results do not depend on concrete character of analyzed system. For example, we can know the temperature of stone, insect or hand of human. Stone, insect body, hand of human – all of them has nature of temperature balance. Special nature characteristics of substance in theory is determined by several parameters, such as, for example, thermal capacity, molar volume. On the basis of this method it is possible to receive information on system characteristics in two different conditions. 13

Balance state in thermodynamic reflects system status in standstill, it is informative can be described by « right here and right now», it does not consider previous or further system status. Statistical method as distinct from thermodynamic approach during learning of macroscopic character of systems bases on atomic and molecular approaches. Purpose of statistic method can be formulated by the following way: for the establishing of law on changes of macroscopic substance characteristic, it is necessary to know the law of particle motion (molecules, atoms, ions and etc.), being part of it. This is why statistical physics proves theoretical laws of thermodynamic from the standpoint of atoms and molecules. Advantages and disadvantages of thermodynamical and statistical methods in researching of physics appearance are indicated in the table 1.1. In general, by combination of that methods, they form composed method of statistical thermodynamic. Table 1.1 Method Thermodynamical

Statistical

Advantage

Disadvantage

Cross functional, simple, does not require information on structure of atoms and molecules of substance Gives opportunity to solve such problems like constitutive equation, theory of heat capacities, and so on, determination of scopes on law appliance, which cannot be solved in the scope of thermodynamic.

Does not reveal develop of internal mechanism of appearances;

Results can be used only in the scope of atomic – molecular model

1.5. Interrelation between structure of substance characteristic and thermal molecules motion. Three phases of substance – gas, liquid and solid body (aggregative states) Kinetic theory is very important step in development of molecular physics. Definition « kinetic» relates to motion, in such a way, for the research of appearance their dynamic characteristics is considered. 14

According to kinetic theory, molecules can move faster or slowly, if they are part of hot or cold body. In reality, thermal energy is a mechanical (kinetic and potential) energy, related to invisible molecular motion inside a body. According to this concept, part of thermal energy forms energy of translational movement of molecules, and also circular kinetic energy of atoms in molecules structure and kinetic and potential energy of vibration. In such a way, term « thermal motion of molecules» appears (Latin termіnus – word, which is used in science and gives exact definition to the intention). Kinetic theory of heat and molecular hypothesis are relating to any condition of the substance. Three aggregative condition of substance (lat. Aggrego – « join») – gaseous, liquid, and substantial – distinguish according to level of aboutness of molecules and intension of motion. Particles in solid body are closely border on each other, this is why particle very rare enters into the place of neighboring particle. When heating up of solid body is taking place, the motion of molecules is enhancing which lead to increasing of body temperature. At a certain point, because of external pressure, closely located molecules can move from one group to another. This equates to substance to get over in liquid condition, further heating allows to break a link between molecules, and molecules start to move freely in space; substance gets over to gaseous state. In estimated valuation of pressure and temperature two states of one substance appears (liquid and gas, solid body and liquid or solid body and gaseous). And only in defined value of pressure and temperature substance can be in all three states. It calls triple point. Gas density in comparison to liquid density or solid substance is less. Consequently, distance between gas molecules is much more than their size. Molecules in motion has little effect on each other, and only in forthcoming of their pathway can deviate sensible from initial trajectory. In this case the question will be on molecules collision. Among other things, molecular-kinetic theory establishes link between state of matter and appearances observed in them with forces, acting between molecules. Nature of intermolecular forces defines by results of experimental research and theoretical forecasts. For example, transformation of gas to liquid proves existence of attractive power between molecules, located on big distance, and very 15

strong resistance of liquid to compression means that on small distance between molecules affects repulsive force, and depending on distance these forces are very affected. Intermolecular forces divided in two groups: distance interaction and close interaction. Forces of close interaction calls valence forces or chemical forces. In such a way, when molecules move closer, to such an extent, that covering of their electronic clouds are taking place, then repulsive forces appear. On the distance that more than its own size molecules attracted to one another, and, therefore, gravity forces appear. Intermolecular theory is studied in a special branch of molecular physics.

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2 MICROPARAMETERS OF EQUILIBRIUM. TEMPERATURE AND PRESSURE 2.1. Thermodynamic Equilibrium. Local thermodynamic equilibrium 2.1.1. Main Terms and Interpretations We already know that in the mechanics real-life objects can be seen as a mass point. Mechanical system is in a state of equilibrium if all the speeds and accelerations of mass points therein equal zero. By definition, the resultant force having an effect on a mass point at any time must be zero. If this condition is not met, the state of mechanical equilibrium of the system is disturbed. On the other hand, the nature of thermodynamic equilibrium is quite different. Therefore, in order to describe the thermodynamic equilibrium, we should agree on a definition of the thermodynamic system. Let’s say that the thermodynamic system is a body or an aggregate of bodies able to interact with one another and with the environment exchanging energy and mass. Any mass consists of a large variety of molecules and microparticles. Such objects are called macroscopic systems. The dimensions of macroscopic systems are always much larger than the size of atoms and molecules. We spoke about it this in the first section above. All macroscopic characteristics which describe the thermodynamic system are called macroscopic parameters. These include, for example, density, volume, pressure, concentration, etc. The macroscopic parameters are subdivided into external and internal. Macroparameters external bodies outside of the system are external parameters. The state of the thermodynamic system is determined by a set of independent macroscopic parameters. The values that fully determine the current state of the system regardless of the previous states are called functions of state. 17

If system parameters do not change over time, this is called stationary state. In addition, the state of a thermodynamic system is determined by a variety of all of its properties. 2.1.2. Thermodynamic Equilibrium Thus, the difference between thermodynamic equilibrium and mechanical equilibrium has to do with chaotic disorderly continuous motion under the influence of unbalanced interacting forces of molecules within the thermodynamic system. In this case, energy or mass exchange occurs under the influence of thermal molecular motion between the system and the external environment. Therefore, in order for the system to be in the state of thermodynamic equilibrium, parameters in the system and in the environment, which are responsible for the exchange, must be the same. For example, say, a vessel, the walls of which conduct heat but not the mass, is a system. In this case, the exchange of heat and energy between the system and the environment depends on their temperature (Fig. 2.1). flow system

Ti

External environment Те

(сі, рі, Ті,…)

Fig. 2.1. Open System

If we leave this system in the vessel, there will be temperature formed, and in each point in the system at any given time

T = Ti = Te ,

(2.1)

here Ti – is the temperature of the system (internal parameter), Te is the ambient temperature around the system (external parameter), and T – the system equilibrium temperature. 18

(2.1) based on the above all points in the system assume a value of T and the system enters equilibrium. The T temperature value remains constant untill the equilibrium state of the external action changes. At the same time, the state of thermodynamic equilibrium of the system is called the preservation of all the system parameters, crtain constant values in the absence of changes in external conditions. Thermodynamic system is a macrosystem, which is in a state of thermodynamic equilibrium. The values that characterize the state of system equilibrium are called thermodynamic parameters. 2.1.3. Local thermodynamics Equilibrium Where thermodynamic system does not interact with the environment, and thus, where there is no exchange of energy and mass, it is called an isolated system. The system may be isolated, but its parts may interact. Consequently, such systems in which there is an exchange of energy and mass between its parts, are open systems (Fig. 2.1). In this case, a very important term of «local equilibrium» is used. If equilibrium conditions are met in only one part of the system, then this state is called «local equilibrium» (Latin: localis – local, characteristic of the site). The term «local macroparameters» may be used with respect to it. They are based on consideration of the part of the system and a specified period of time. Where in various local parts of the system (physical elementary volumes) one or more of macroparameters of the system (density, temperature, etc.) do not coincide, the system is unbalanced and transport processes may occur. Thus, where an isolated system is in equilibrium, pressure and temperature in each part of the system are the same. This condition is observed in a single phase and multiphase isolated system that is in equilibrium. Thus, if a system is divided into two parts with the volumes V1 and V2, the total volume is equal to V=V1+V2, the equilibrium temperature Т1=Т2, the pressure will be equal to р1=р2 . 2.1.4. Process The process (from Latin processus – to pass, to move, to go forward) is a very widely used term. When certain system parameters 19

change over time, the process occurs in this system. In this case a phenomenon is the process of development and change at the same time, you could say. For example, there is a mass shift observed, indicating that there was a diffusion process or heat transfer process into the environment taking place. In the first of the given examples substance concentration changes, while in the second example it is temperature that changes; here, concentration and temperature are macroparameters. The process is the change in the system state. System transition from one state to another is associated with the system coming out of equillibrium. Therefore, when any process occur in the system, there is a sequence of disequilibriums taking place. A process consisting of a sequence of continuous states is called an equilibrium process. In the equilibrium process all macroparameters of the system are very slow, physically infinitely slow, change negatively; parameter values at various points, for example, pressure, will be close to an average 〈р〉. Accordingly, only physically continuously infinitely slow process may be an equilibrium process. If a system that is in the equilibrium state, is disturbed from it, after a while it will come back to the equilibrium state. The process of the system’s transition from a nonequilibrium state to the equilibrium state is called relaxation, and the time of the system’s return to a state of equilibrium is called relaxation time. Process speed in the thermodynamic system is much less than the relaxation rate, while the values of all parameters at any stage of the process are equalized. Thus, such a process consists of a sequence of equilibrium states that are infinitely close to each other. Such rather slow processes are called quasi-static or equilibrium. Graphical method is used to display conditions and processes of the system. Uniform states of the system are marked as points, while processes are marked as simple lines -, P − V -, V-T -, etc. It should be noted that it is imposible to graphically display equilibrium conditions and processes only. Such a state of equilibrium at each moment will be characterized by certain macroparameter values. The concepts equilibrium and equilibrium process are important in molecular physics. Of course, all of the actual processes are non-equilibrium, thereforem they may reach the equilibrium state to a certain degree. Thus, 20

the equilibrium process is reversible over time. In the reverse process the system must undergo a series of the same states as in the forward process. For example, assume that ф system has undergone ABC changes (Fig.2.2). These changes are reversible if: a) if СВА changes in the reverse process correspond to the values in the forward direction; oparametrov. b) where there is a heat or substance exchange with the environment and there is work performed, then parts of СВА of the reverse process generate heat or ubstance and the work is performed. For example, assume that the system on the ABC line has received heat of Q’, then when returning by CBA it should release the same amount of heat Q, namely Q' = Q.

Fig. 2.2

All changes that do not meet these two conditions are called irreversible processes. All processes in nature are irreversible. 2.2. Density Let’s set the value of the thermodynamic system with ∆V volume as ∆Q . q density of ∆V property is defined as the ratio of ∆Q to the value of this ∆V volume:

q=

∆Q . ∆V

(2.2)

∆V is a part of the full system’s volume of V , therefore, the value of V property will be expressed through integration (2.2): 21

Q = ∫ qdV ,

(2.3)

V

where dV is the volume of infinitely small part of the system. Say, it consists of uniform molecules m0 . Number of molecules N . In this case, using the (1.8) example we find the gas mass with a V volume:

m = m0 N .

(2.4)

The density of the gas mass per volume unit is denoted by ρ. Then, using examples (2.2) and (2.4) we can write the following:

ρ=

m , kg/m3. V

(2.5)

The number of gas molecules per unit of volume or number density shall be equal to the following:

n=

N , 1/m3. V

(2.6)

ρ and n are connected, we are going to find them using examples (2.4) and (2.5):

ρ=

m m0 N = = m0 n . V V

(2.7)

Consequently, mass density is determined by multiplying the mass of a single molecule by numerical density, in addition, it is directly proportional to the number of molecules per volume unit. If the density of the substance is known, it is apparent that based on the formula (2.3) its total weight at a V volume equals to: If substance property is determined based on the unit mass, then it is called a specific property. For example, we obtain specific volume using the following equation: 22

m = ∫ ρdV

(2.8)

V

υ=

V , m

(2.9)

where V – gas volume, m - gas mass. Given (2.5), specific volume υ =

1

ρ

is equally proportional to

the density. Substance property per mole is called molar. For example, molar mass or molar volume. Molar volume of ideal gases with equal temperature and pressure will be the same. For example, under certain conditions at a temperature of 0 °С and a pressure of 760 mmHg the volume of an ideal gas is equal to 22,4136 ⋅ 10 −3 m3. 2.3. Concentration In general conditions a substance in any aggregate state may be evenly distributed within one or more substances. Such substances include gas mixtures, liquid solutions, solutions (alloys) of solids and mixed crystals. A mixture of several substances is not a simple distribution of molecular of one substance in the second substance; in fact, such a mixture will depend upon various intermolecular interactions. The substance in the mixture (solution, alloy) composition is called component (from Latin component – an integral part). Composition of compounds mixture is determined in various ways. The amount of components substance in the mixture per mass unit or per volume unit is called concentration. Concentration, as a rule, is a value characterizing the relative amount of the mixture components. Consequently, it can be determined as % percentage or as a mass or molar fraction. Substance properties in a mixture are closely related to the physical properties of the mixture component. It is therefore very important to know substance concentration in a mixture. 23

Molecular physics uses several types of concentration. Based on the above, mass and number density in the examples (2.5) and (2.6) is the concentration in the volume. This concentration has measurement units. For example, the unit of mass density kg/m3, while that of number density is 1/m3. To solve many physical problems, it is more convenient to use the dimensionless concentration. Molar fraction of component i in the mixture or the relative molar concentration xi is the value obtained of the ratio of the molar amount of such component to the molar number of all components in the mixture ν i : k

(ν = ∑ν i = ν 1 + ν 2 + ν 3 + ... + ν k ) , means: i =1

xi =

νi

=

k

∑ν i =1

νi ν

i

or

xi =

Ni

Ni , N

=

k

∑N i

(2.10)

i =1

here N i − i is the amount of component particles, N - total number of molecules, k - number of components. For ideal gases mole fraction of components may be determined using partial pressure of components p1 , p2 , p3 ,..., and the total mixture pressure:

xi =

pi

.

k

∑p i =1

24

i

(2.11)

Gas mixture component pressure that such component would have delivered if it was the only one in the entire volume occupied by the mixture is called partial pressure. Therefore, gas mixture pressure equals the sum of partial pressures of the mixture components: к

p = p1 + p2 + p3 + ... = ∑ pi .

(2.12)

i =1

Thus, the pressure of an ideal gas mixture is determined by formula (2.10), it is called the Dalton law. A mixture of an ideal gas is also a normal ideal gas. Accordingly, partial pressure of any component in an ideal gas mixture is the product of its molar fraction in a mixture by total pressure: к

xi = ∑ pi = xi p .

(2.13)

i =1

Relative molar concentration can be calculated using n i molar amount per unit volume:

xi = here n =

к

∑ ni

ν i N i ni = = , N n ν

(2.14)

.

i =1

Mass fraction of component i or relative mass concentration is determined as follows:

ci =

ρi

n

ρi ∑ i

=

ρi , ρ

(2.15)

=1

here ρ i − i – component density, ρ =

∑i ρi – mixture density

Amount of relative mixture concentration will be equal to one, therefore 25

k

∑xi

= 1 or

k

ci = 1 . ∑ i

(2.16)

=1

i =1

When considering the properties of mixtures, various mixing rules are applied. For example, suppose that there are two components in the mixture, then ρ =

2

∑i ρi

mixture density can be determined using

ρ1 and ρ 2 density of pure components: ρ=

1 x1

ρ1

+

x2

.

(2.17)

ρ2

Molar volume concentration of i component in the mixture is determined as follows:

Ci =

νi , mol/m3, V

(2.18)

here, V – gas volume, ν i – amount of moles of i-component. For ideal gas mixtures volume and relative molar concentrations of the substance are the same.

2.4. Gas pressure. Basic equation of molecular-kinetic theory For a detailed description of gas molecules motion, it is necessary to solve the problem of their collisions with the walls of the vessel. The solution to this problem is our main goal; it can be solved, explaining the connection of gas pressure on the vessel walls and other macroscopic characteristics of the gas, such as pressure. The distance between gas molecules is much larger than their sizes (molecular diameter). Nothing affects molecular motion in be26

tween identical collisions. Thus, each molecule moves freely between collisions. Consider an ideal gas. Molecular forces here are not taken into account; gas molecules are considered to be single mass points. Such molecules are moving freely most of the time, but sometimes collide with each other or with the walls of the vessel that contains the gas. When gas is in equilibrium all the molecules have the same speed, and in Cartesian coordinates 1/6 thereof moves forward along the three axes and 1/6 moves in the opposite direction. Probability of the motion direction is the same. When molecules with absolute elasticity collide with the vessel walls, they impart to the walls the impulse equal to the change in the value of its impulse. The number of such collisions is very large, and they are evenly distributed on the vessel walls. Thus, there is continuous work taking place, affecting the vessel wall. As we know from mechanics, according to Newton's second law, the impulse change associated with time, gives the force affecting dS surface element. The direction and magnitude of this force is equal to the speed of impulse of the molecules colliding against the wall. According to the general definition, pressure is determined by the ratio of the force exerted on the surface S to the normal line component. Hence,

p=

Fn . S

(2.19)

A component of the normal line of force impacting any surface, is positive. The surface of the vessel wall impacts gas molecules with equal and opposite force. Gas pressure exerted on the vessel walls is one of the main characteristics of gas. The presence of pressure indicates the presence of gas in the vessel. Gas pressure exerted on the vessel walls was studied in the 18th century by D. Bernoulli, who defined it as the time-averaged impulse imparted when molecules collide with the walls of the vessel and are reflected from them. The laws of mechanics allow us to find the average force caused by gas molecules, so that gas pressure can be calculated using this approach. 27

Thus, in view of the above, for example, let’s define what pressure equals to exerted perpendicularly to OX-axis on the ds vessel wall (Fig.2.3). Gas mixture in the vessel consists of components i = 1, 2,3,..., r , total weight m and the number of molecules n per unit volume in the vessel is determined as follows: r

m = m1 + m2 + ... + mr = ∑ mi

(2.20)

i =1

r

n = n1 + n2 + ... + nr = ∑ ni . i =1

Here ni is a numerical i component density per unit volume, and mi is the component weight Molecules speed is different in each r component. υi is the average speed of ni molecules in along OX-axis. The abcd vessel wall is perpendicular to OX-axis (Figure 2.3).



υ ix is projection of υi velocity on ОХ axis (Fig. 2.4a.). Here the

dS area is a abcd wall element. Impulse of i-molecule before the collision with the wall is m0 iυix ; after collision with dS h area molecule impulse sign changes.   The reason is that υ i' direction is opposite to that of speed υ i . Fig. 2.4b shows a cylinder with dNi molecules, here molecules pass through dS surface over dτ time. The length of this cylinder is υi dτ ,    while its volume is (n υi )dυi dSdτ . The outer unit normal (vector) n





is directed to the dS surface. Collision is elastic, therefore, υ i = υ i' . Hence, the change in molecule impulse will be equal to:

− m0 iυ ix − m0 iυ ix = −2m0 iυ ix .

(2.21)

During dτ period of time dN i molecules of i-component moving in the ОХ-direction collide with dS element. The number of dN i molecules – the height 28

Fig. 2.3

υ ix dτ base lies in the dS cylinder (Fig. 2.4b). The volume of this cylinder is dV = υ ix d τ d S , and a volume unit has ni molecular, therefore

dN i =

1 niυ ix dτ dS . 2

(2.22)

Here 1 coefficient comes based on the continuance of chaotic 2 motion, therefore, the same amount of molecules that collide with the vessel, return. Multiplying the number of dN i collisions on the change in the impulse of each collision, we obtain the existing dK ix impulse imparted on the dS vessel wall element over dτ time:

1 dK ix = −2m0 iυ ix dN i = 2m0 iυ ix niυ ix d τ dS . 2

(2.23)

On the contrary, the dS surface is directed to the opposite and



provides equal pressure, therefore, there is elastic F force impacting gas from the wall. 29



According to Newton's second law, F∆t force is equal to the impulse change from collisions of molecules against the wall, i.e.,

dK ix∗ = F ∗dt ∗

а)

(2.24)

b) Fig. 2.4

Please note that here dτ is the time during which the molecule achieves the dS surface and dt ∗ is the time of molecules interaction with the wall. In fact, dτ ≠ dt ∗ . However, we could take the average value of force impacting the dS, thus taking into account the impact of the laws of statistics, we assume that

dK ix∗ = dK ix .

(2.25)

We will add dK ix to all i groups: r

r

i =1

i −1

K x = ∑ dKix = m0 ∑ niυ ix2 dSdτ = m0 nυ x2 sτ .

(2.26)

We will assume that m0 i = m0 , according to Newton's second law, we will determine K x impulse imparted along the OX-axis to the S surface over τ time: 30

K x = Fτ . Consequently, the amount of pulse averaged over unit time will be equal to:

(2.27)

τ

F = K x = m0 nυ x2 S .

time or per

(2.28)

Here υ x2 is the square of velocity of the molecules moving in the direction OX. This total impulse with the help of Newton's second law determines the force impacting the wall. Then gas pressure is determined by the ratio of the total K x impulse to S square, therefore,

P=

F m0 nυ x2 S = = m0 nυ x2 . S S

(2.29)

But it should be noted that the number of molecules is countless and speeds therefore may be different. Therefore, this issue is not associated with the movement of individual molecules but with special laws that occur in the interaction of a huge number of particles. These laws that are common to many particles are not typical for individual molecules. Thus, they will be different for complex systems of mathematical research methods. Therefore, theory of chances is the only one that can meet these requirements. In the study of the influence of an enormous number of molecules, there is no need to take into account properties of a single molecule. Based on this, projection of υ x velocities of gas molecules in ex-



amples (2.28) and (2.29) will be identical. Indeed, υ speed of the molecules and projections thereof on the OX-axis will be different; molecules velocity distribution in the system in equilibrium will be discussed in the next section. Generally, molecule velocity is a random variable, therefore, in accordance with the distribution law, υ x2 = υ x2 , i.e. mean square value 31

of the speed is applied. Therefore, we replace υ x2 velocity component in the formula (2.28) by its average value of υ x2 . In connection with this, p gas pressure exerted on the S surface is perpendicular to the OX direction (Example 2.9) is determined as follows:

p = nm0 υ x2 .

(2.30)

Velocity of each molecule can be written by components associated with its coordinate axes, as follows, υ 2 = υ x2 + υ y2 + υ z2 , therefore:

υ 2 = υ x2 + υ y2 + υ z2 = υ x2 + υ y2 + υ z2 .

(2.31)

Molecular motion is chaotic, thus making it impossible to determine the direction of their velocity, so that the average value of squares of their projections will be equal to each other, therefore, υ x2 = υ y2 = υ z2 . Accordingly, the average value of the x component of the velocity can be determined as follows:

1 3

υ x2 = υ 2 .

(2.32)

By replacing the value of υ x2 found in the formula (2.31) with formula (2.30) a very important formula can be obtained: 1 p = nm0 υ 2 . 3

(2.33)

(2.33) This equation is called the fundamental equation of the elementary molecular-kinetic theory. By multiplying and dividing the right part of the equation (2.33), we can write the following: 32

2 m0 υ , n 3 2 2

p=

(2.34)

m0υ 2

= E0 is the average kinetic energy of a single molecule. 2 Then the formula that determines the pressure, can be written as follows (2.34):

here

p=

2 nE 0 . 3

(2.35)

This equation (2.35) is called the Bernoulli equation. Thus, the pressure is equal to two thirds of the kinetic energy of translational movement of molecules per volume unit. If in equation (2.35) n is constant, pressure is proportionate to kinetic energy of translational motion of molecules. Given that

n=

N , this equation can be written as follows: V p=

2 N E0 , 3 V

(2.36)

here N E 0 = E is the total energy of gas molecules N. E Energy-toV volume ratio

E = ε. V

(2.37)

It expresses energy density or gas energy per volume unit. If so, the equation (2.36) can be written as follows:

p=

2E 2 = ε, 3V 3

here ε is energy density. 33

(2.38)

Hence, pressure is equal to two thirds of the energy density of the translational motion of molecules. So, we have come to very important conclusions. Equation (2.33) defines the relationship between the values associated with a single molecule (molecular mass, average molecular square, the number of molecules per volume unit), and pressure of gas as a unified body. Pressure is a directly measurable macroscopic value. Therefore, this equation determines inter-connection between micro and macro; in other words, it serves as a bridge between them. This result confirms that gas pressure is determined by the average kinetic energy of its molecules. Thus, gas pressure can be understood as the value associated with the result of the sum of interaction of an enormous number of molecules and vessel walls and intermolecular interaction. Therefore, one or a small number of molecules cannot exert pressure. In this connection, the characteristics of substances, requiring the presence of interaction sum of multiple particles is called statistical properties of substances. Statistical method does not take into account the motion of individual particles and only uses average values describing microscopic state of multiple particles. Given density determination (2.7), we can write the equation (2.33) as follows: 1 1 (2.39 ) p = nm0 υ 2 = ρ υ 2 . 3 3 This formula makes it possible to determine the mean square speed of gas molecules:

υ = υ2 =

3p

ρ

.

(2.40)

It is possible to calculate the average square velocity of the molecules using practically calculated pressure and gas density. For example, under normal conditions, nitrogen molecules move at a rate of

υ 2 = 454m / s while hydrogen molecules move at a rate of

υ 2 = 1694m / s . 34

2.5. Temperature 2.5.1. Concept of Temperature. Heat Equilibrium According to the basic equation of molecular-kinetic theory (Formula 2.33), the ideal gas pressure is proportional to its number density n and average kinetic energy of translational motion of molecules. If n is constant, if the volume of gas mass does not change, then according to equation (2.34), gas pressure only depends on the average kinetic energy of the molecule. Experience shows that gas pressure at constant volume can only be changed in one way: by heating or cooling. Gas heating increases gas pressure while cooling results in its decrease. This phenomenon can be observed in practice. A special value that characterizes the hot or cold state of any substance is called temperature. It follows that the temperature characterizes thermal properties of an object. It follows from the above that temperature is related to the average kinetic energy of substance molecules. Peculiar features of temperature as a physical value can be considered by a simple experiment. You can put a number of different objects with different temperatures in a row. For example, water vapor is warmer for us than the water itself (the liquid), and is hotter than ice. On one hand, if these substances come in contact with each other, their temperatures begin to level out, i.e. after equalizing their temperature, they reach a state of thermal equilibrium. Zeroth law of thermodynamics is based on this phenomenon. In the future, we are going to come back to the laws of thermodynamics. If two bodies that are independent of each other – A and B – are in thermal equilibrium with the third body C, they are in thermal equilibrium with each other. In other words, thermal equilibrium means that temperature is the same in all points of the system. This law is called the zeroth law of thermodynamics. Figure 2.5 demonstrates the above. A spoon is in equilibrium with hot tea in the cup. If we move the spoonful from the tea into water in a glass and spoon’s temperature does not change, the spoon is in thermal equilibrium with the water in the glass. Thus, the tea in a cup and the water in the glass are in thermal equilibrium, and their temperatures are considered the same (2.5V). 35

Temperature measurement methods are based on this principle. Thermometer is a temperature measuring device. For example, it cold be that thermal balance is not achieved in a body or a body system. If the system is isolated, such a system for some time to come into thermal equilibrium. Practice has shown that in the process of establishing thermal equilibrium, body temperature changes occur, they begin to flatten out, and at some point one body transfers heat to another body, i.e. heat transfer process occurs. In comparison with other values temperature as a physical quantity there has a difference. This difference is that the temperature is not an additive quantity. If we theoretically divide any body into several parts, the total body temperature is not going to be equal to the sum of its parts’ temperatures. For example, body volume can be calculated by adding up the volumes of all parts of the body, thus, the volume is an additive quantity. Temperature measurement methods have been known since ancient times. These methods are based on the fact that in the process of change of a substance temperature its properties change as well. Then the quantities characterizing properties of the substance also change with the temperature change. Changes in the substence temperature affect all of its physical properties: linear dimensions, properties of elasticity, electrical resistance of conductors, etc. Any of these changes can be used to measure temperature. Thus, in order to make a temperature measuring device, i.e. a thermometer, a substance is selected whose property changes continuously and monotonously depending on the temperature, that is, we must choose a thermometric value. For example, mercury is a thermometric substance; therefore, its volume can be adopted as a thermometric magnitude. The volume of mercury is directly proportional to the temperature change so that it changes linearly. Change in the fluid volume depending on the temperature is used in many thermometers. To describe dependance of thermometric value of temperature unit in a numerical value, a temperature measuring unit ust be introduced. A «degree» (Latin gradus – step) is adopted as the temperature unit. 36

It is defined as follows. Take two temperature ranges and divide them into equal parts, called «degrees» and assign a specific numerical value of one of the temperatures. With this we can determine the value of the second temperature and any temperature from the range. This creates a temperature scale. For example, in a well-known Celsius temperature scale ice melting temperature Т 0 and the boiling point of water at atmospheric pressure Т к are adopted as the two temperatures (control points). This temperature range is divided into 100 equal divisions. Thus,

the degree of this scale is defined as follows 1°С =

Т к − Т0 . Using 100

this approach, you can create any thermometer and temperature scale.

а) – cup of tea b) – glass of water с) – cut of tea and glass of water are in equilibrium with each other Fig. 2.4

In the history of science, for example, Reaumur temperature scale (Paris, 1750), Celsius scale (Sweden, 1742.), Fahrenheit (England, 1724) had been used. For example, the Fahrenheit scale in the modern world is used in the United States, 1°F=5/9 °C. 37

Present-day thermometry is based on the idea gas scale. This thermometric scale is determined using a gas thermometer. Gas thermometer is a closed vessel filled with an ideal gas and equipped with a pressure gauge. Ideal gas is thermometric substance in the thermometer, and thermometric value is the gas pressure at constant volume. Dependency of a certain mass pressure on gas temperature is considered to be linear. In gas thermometer hydrogen or helium is used as a thermometric substance for the lower temperatures while nitrogen is used for the upper temperature. According to this hypothesis, the pressure ratio of the corresponding boiling water temperatures and melting ice will be equal to the ratio of the temperatures: Pк Tк , = P0 T0

(2.41)

where Pк is the pressure of an ideal gas at a boiling water tempera-

ture of Т к the in the gas thermometer; P – ideal gas pressure at the 0 ice melting temperature of Т in the gas thermometer. 0

Pк – this ratio is found in practice. The following ratio can be P0

derived from various measurement results: Pк = 1,3661 . Then the P0

temperature ratio will be equal to a value of: Tк = 1,3661 . T0

The degree value can be found by dividing the (Tк − T0 ) difference by a hundred divisions: о Tк − T0 = 100 , Tk − T0 =1 .

(2.42)

100

Examples (2.41) and (2.42) show that ice melting temperature under atmospheric pressure of Т 0 = 273,15 and water boiling point is 373.15 degrees. 38

To measure body temperature using gas thermometer it is necessary to establish contact between the thermometer and the body, after establishing thermal equilibrium, gas pressure should be measured in the thermometer. Thus, body temperature determines the following formula: 273,15 , (2.43) p Т= p0

where p0 is gas pressure thermometer in ice melting. In practice, gas thermometer is used to calibrate all other embodiments of thermometers. According to the ideal gas temperature scale zero temperature value corresponds to gas pressure zero value. If with temperature scale at zero value thermometric value equals zero, this scale is called absolute. The temperature, which is determined by such an absolute scale, is called the absolute temperature and is denoted by a T. Thus, gas thermometer scale is absolute. It is called the Kelvin scale. Unit of measurement in the temperature scale is called Kelvin degree or Kelvin and is denoted by K. It should be noted that the lower temperature value is zero (0 K), and using Тү triple water temperature interval, in 1954 Kelvin measurement unit was defined in the SI system as the main reference point. Triple point of water is water temperature balance in liquid form, in vapor state and ice state (solid state of water). Then it was found that Тү-Т0= Тү-0=273,16. Hence, it appears that 1 К=1/273,16. This scale is called the thermodynamic temperature scale. A degree of a widespread Celsius scale was aligned with Kelvin degree. Therefore, 1К=1 °С, 39

It follows that T К=t °С+273,15К. Here, number 273.15 the zero temperature value on the Celsius scale corresponds to ice melting temperature at atmospheric pressure. On Celsius scale, it looks as t °С. Temperature determination and relationship with pressure. The process of transition to a state of thermal equilibrium of a system is related to temperature equation, and when the system reaches a state of equilibrium, temperature and pressure in all parts of the system are the same. Thus, the value that is equate at equilibrium state in the system, is the temperature. According to equation (2.33) if particle density n in the system is constant, the mass in the gas volume V remains unchanged, then in heating (or cooling) of gas only change average kinetic energy of molecules can change. In the process of heat transfer, temperature equation means equation of the average gas molecules kinetic energy. Thus, in the course of transition to the equilibrium state from one gas particle to another, energy will be transferred, but the energy of the entire gas as a whole body is not going to equalize; only average kinetic energy of each molecule will be equalized. Consequently, we will take the following equation from example (2.34) 2 2  m0 υ n 3  2 

 2  m0 υ 2  = n  3  2 1 

 m υ2   = … 2 n 0  3  2  2

 .  k

(2.44)

Here i = 1, 2,3,..., k – are gas particles. Consequently, it is clear that the change in the average molecule kinetic energy is characterized by a temperature change, in fact, this energy should be considered as temperature. As with temperature, average molecule kinetic energy is not an additive quantity. Based on this, considering the average kinetic energy of translational motion of a molecule as temperature, denoting it as θ, we can write the following: 40

2  m0 υ 3 2 

2

  =θ .  

(2.45)

Taking into account the expression (2.45), the equation (2.34) can be written in the following simple manner:

p = nθ .

(2.46)

In this definition degree should be θ temperature measurement unit. Then you must enter the conversion factor of energy measurement unit from example (2.45) into degree. Conversion factor of energy units is denoted as k , while temperature is denoted as T . Thus, θ = kT , with this in mind, equation (2.45) can be written as follows: 2 2  m0 υ  = kT . 3 2    It follows from here:

 m0 υ 2   2 

 3  = kT .  2 

(2.47)

This equation determines average kinetic energy of molecule translational motion, which is considered as a material point. Its average square speed υ 2 can be divided into three parts υ x2 ,υ y2 ,υ z2 according to space coordinates. Molecular motion is chaotic, so molecule energy is distributed uniformly over the entire speed components, and therefore each molecule receives 1 kT amount of energy. Here, k is the ratio between 2

the energy value J (joule) and the temperature value – K (Kelvin degrees), which represents the Boltzmann constant. Its value is established empirically and in SI system k is equal to the following: 41

k 1,380662 ⋅10−23 J/k. =

(2.48)

According to formula (2.47) when temperature is equal to zero, average random molecular motion kinetic energy is equal to zero, and thus, random motion of molecules stops. This temperature is an absolute zero, i.e. the absolute temperature scale reference point. Therefore, according to (2.47) there is no negative temperature. But to describe some systems, the concept of negative temperature, is used but this is not related to the state of equilibrium and it cannot be said that such a temperature is below the absolute temperature. At the same time, temperature is determined by the average kinetic energy of molecules; like pressure it relates to statistical values. Thus, the molecule temperature concept or the expression "cold" or "hot" molecule will be meaningless.

2.5.2. Ideal gas State. Mendeleev’s-Clapeyron Equation There is a relationship between the parameters characterizing gas state. The main parameters that determine gas state are pressure p , temperature T and gas volume V . The equation that defines the relationship between these parameters is called gas state equation. Total non-specific form of this equation-tion is as follows:

f ( p, V , T ) = 0

(2.49)

or p = f (V, T ) Taking into account the dependence on each other, we can write T = f ( p,V ) , or V = f ( p, T ) . Using a molecular-kinetic equation for determining pressure p = f (V, T ) and determining temperature T = f ( p,V ) , we can determine the following formula:

p=

2 mυ 2 2 3 n = n kT = nkT . 3 2 3 2

(2.50)

If gas mass does not change n is constant, then р pressure is proportional to the T temperature. 42

If we take gas mixture, average velocity of different gas molecules with different masses are different, but the average energy is the same. The mixture pressure will be equal to:

p = (n1 + n2 + ... + n )k kT ,

(2.51)

here n1 , n 2 ,..., ni number of i-component molecules per unit volume or: k

p = n1 kT + n 2 kT + ... = p1 + p 2 + ... + p k = ∑ p i

(2.52)

i =1

here pi is a partial pressure of i – component. Expression (2.52) is called the Dalton law. This law shows that the pressure of an ideal gas mixture equals the sum of partial pressures of components in the mixture. If in V volume the number of molecules is N, then n = N / V , and taking this into account, we write the equation (2.51) as follows:

p=

N kT or pV = NkT . V

(2.53)

This equation (2.53) includes three main states’ parameters, so that it can be called the equation of gas state. However, if this equation should be written for a certain amount of substance, for example, one mole, then N = N A , hence

pV = N AkT ,

(2.54)

where ( N Ak ) , where the product of two fundamental constants is called the universal gas constant and is denoted by letter R, and therefore, R = N Ak . Its value R = 8,31 J/mol⋅К. 43

Given universal gas constant, we can write the equation (2.54) as follows:

pV = RT .

(2.55)

This expression is called Mendeleev-Clapeyron equation or an ideal gas equation in one mole. And for any gas mass, equation of state is determined as follows. N is the number of gas molecules (2.53), using the definition of mole, we will write it as follows:

N=

m N A . Then we can write (2.53) as follows: M pV =

m N kT , M A

(2.56)

where m M = v is the number of moles in the gas mass. Taking into account (2.55), we can write (2.57) -ni as follows:

p = νRT

or

pV =

m RT . M

(2.57)

This example (2.57) of the equation of ideal gas state with any mass m.

2.5.3. Properties of ideal gas as a simple thermodynamic system. Ideal gas laws Properties of an ideal gas are formulated in the form of empirical laws through the generalization of the results of experimental observations of a rather diluted gas. The ideal gas laws may be determined based on elementary kinetic theory of equations. Boyle's law. Boyle's law was formulated in the XVII century on the basis of experimental results. This law describes the isothermal process (from the Greek isos – equal, and thermos – heat), and thus that runs at a constant temperature. 44

According to Boyle's law, gas volume at a given weight at a constant temperature ( Τ = const ) is inversely proportional to gas pressure: (2.58) pV = const .

This law was discovered in 1662 by R. Boyle. and in 1676 was independently formulated by E. Marriott. Equation (2.58) is called the isotherm equation. Isotherm equation can be derived from ideal gas state formula (2.55). If Τ = const then the right member of the equation is constant. pV = RT = const , that is, expression is determined (2.58). According to Boyle's law, increase or decrease of gas at a constant temperature leads to a change in pressure, but pressure and volume of substance remain constant. Temperature-pressure curve p = f (V ) for different temperatures is shown in Figure 2.5. These curves are called isotherms. According to equation (2.58), they are equilateral hyperbola ( T1 < T2 < T3 ). Compressibility of gas at constant temperature is characterized by isothermal coefficient of compressibility χ . It is defined as follows: pV = const , while d ( pV ) = pdV + Vdp = 0 , dT = 0 , consequently, 1  dV  1 (2.59)  = − , χ =  V  dp T p 45

where dV is the change in volume, resulting in a change in pressure dp ; V – initial gas volume; T derivative index represents Τ = const ; dT – A temperature change. According to equation (2.59), isothermal coefficient of compressibility is determined by a relative change in volume χ = dV / V , which results in a change in pressure units (dp = 1). Symbol "-" minus in equation (2.59) determines that the increase in volume occurs towards pressure decrease. Unit of measurement of isothermal compressibility coefficient in the SI system м 2 н . Gay-Lussac's Law. This law was discovered in 1802 by J.L. Gay-Lussac, and proposed independently in 1801 by J. Dalton. GayLussac's Law states that at constant pressure, ideal gas volume varies linearly with temperature:

V = V0 (1 + αt ) , where

V0

(2.60)

is the initial volume of gas; t – difference between the

initial and final temperatures; α – Coefficient of thermal increase or increase in volume. Ideal gas state equation (2.55), at a constant pressure, can be written as: V m R = = const T M p

or V = const . T

(2.61)

V-T ideal gas diagram in the isobaric process. This equation is called isobars, while the process taking place at constant pressure is called isobaric. (2.61) th formula determines Gay-Lussac Law. It is not correct that sometimes this law is called the Charles law. Temperature-volume curve V = f (T ) in the V − T diagram for various pressures is shown in Fig. 2.6. 46

Fig. 2.6

Ideal gas isobars are lines starting at the beginning of coordinate axes in the V − T diagram, angular coefficient f which is equal to R . α We will find coefficient in equation (2.61) based on formula p (2.57):

mR V  dV  = , hence it follows that:   =  dT  P M p T α=

1  dV  1.   = V  dT  P T

(2.62)

Ideal gas volume increase coefficient is inversely proportional to the value of absolute temperature. For example, with 0 0 С coefficient α = 1 273,15 К-1 is the same for all gases. Isochoric process. A process takes place in a system with constant volume, is called called isochoric (from Greek Isos-equal, and chora-space). This process takes place in gases and liquids in closed containers, in the absence of volume change. Ideal gas pressure dependence on temperature p = f (T ) in the isochoric process is called the Charles law. Isochors equation from the ideal gas state equation (2.55) is determined as follows:

p m R = = const T MV 47

or

p = const . T

(2.63)

According to the Charles law (2.63) with changes related to ideal gas pressure dependency on the temperature at a constant volume p T ratio does not change.

p − T diagram (Figure 2.7) shows p = f (T ) dependency cur-

ve for different volumes. Ideal gas isochores are right lines, starting at the beginning of the coordinate axes. Ideal gas pressure thermal coefficient in isochoric process is determined as follows: 1  dp  1 (2.64) γ =   = . p  dT V T

Fig. 2.7. p-Т diagram of ideal gas state in isochoric process

This equation is determined by the equation of state as follows: pV = RT ; Vdp = RdT , V = const , hence,

R p  dp    = = .  dT V V T Temperature scale of ideal gas is based on this law. The results of experiments at high pressure show that real gas properties deviate from the ideal gas laws. Gas laws are based on a synthesis of the results of the kinetic theory in practice, therefore, they serve as a proof of the correctness of theoretical results. These laws operate exactly for ideal gas, thus establishing ideality of gas. Therefore, gas that is subject to the laws of Boyle, Gay-Lussac and Charles is called an ideal gas. 48

2.6. Barometric Formula Chaotic molecular motion causes a uniform distribution of microparticles of gas throughout the vessel volume and to the same number of molecules per unit volume, on average. Pressure and temperature of the gas which is in a state of equilibrium will be the same throughout the entire volume. These provisions are constant, if molecules are not affected by an external force. If there is an external force molecular motion is subject to change. For example, consider a gas (air) under the influence of gravitational field. If there is no thermal motion of air molecules, under the influence of gravity they would have fallen on the surface of the earth, covering it with a thin layer. If there were no gravity, but molecular motion would be retained, air molecules would have departed from the Earth. The reason for the Earth's atmosphere, is the relationship of gravity and thermal molecular motion. Distribution of molecules in the atmosphere occurs in accordance with a certain law. This distribution is determined as follows. Consider a vertical column of air at random height (Figure 2.8). Denote height equal to zero (ground) as x = 0 ; air pressure at this point Р0, and we assume that equal pressure at height х is equal to p. If the height changes to dx, pressure will change to dp. Air pressure at any height is equal to the weight exerted per unit area of the vertical air column.

Fig. 2.8

49

Thus, change in pressure dp is the difference in pressure p and p − dp , and thus, (2.65) ( p − dp) − p = −dp . Dp base area is equal to one, and dx height (Fig. 2.8) is equal to the air mass in the cylinder volume (air column).

dp = − ρgdx ,

(2.66)

here ρ – is air density, g – acceleration of gravity. We know that p = nkT . Hence the number of particles per unit volume of air:

n=

m p p , then air density ρ = 0 . kT kT

(2.67)

T here is the air temperature, k – Boltzmann constant, m0 – particle mass. Considering the example (2.67), we can express (2.66) as follows: mg dp = − 0 pdx . kT Therefore:

mg dp = − 0 dx . p kT

(2.68)

Wherre temperature does not change with altitude, then using integral for example (2.68), we determine pressure as a function of the x height:

ln p = −

m0 g x + ln C kT

or find pressure, using the potential of:

p = Ce

−

50

m0 g x kT ,

(2.69)

where C is constant of integration. We determine the constant C based on the initial conditions, then at the time of x = 0 , p = p0 , substituting these values in the example (2.69), we find the constant C: p0 = C , where p0 is air pressure at h = 0 . Thus, in an isothermal state (at constant temperature) air pressure dependency on the height over ground can be expressed as follows:

p = p0 e

−

m0 g x kT

Where molecular mass equals m 0 =

p = p0 e

−

M g RT

.

(2.70)

M , NA

x

,

(2.71)

here M is molecular air mass, R - universal gas constant. Formulas (2.70) and (2.71) are called a barometric formula. Barometric formula determines the law of decrease in the gravitational field of atmospheric pressure with height. Using this equation, however, height over ground can also be found. Temperature with height are not constant, T = T ( x) and, hence,

acceleration of gravity depends on height g = g ( x) . T = const , therefore, In the midst of the law under pressure dependence law at g = const of height is shown in Figure 2.9. Gas pressure is proportional to the number of molecules in the number of volume units, and hence p ~ n , by replacing p for n, we can determine the law of decrease of the number density of molecules with height:

n = n0 e 51

−

m0 g x kT

,

(2.72)

where n is the number of molecules per unit volume at x height; n0 is the number of molecules per unit volume at a height of zero. The equation (2.72) may be modified by using a molar mass as follows:

Fig. 2.9

n = n 0e

−

Mg x RT .

(2.73)

This formula, like formula (2.72), requires Earth's atmosphere expansion to infinity. However, these theoretical results do not match reality. As indicated above, atmosphere temperature at the surface of the earth is not a constant. For example, at a height of 10-17 km and 80 km temperature has a minimum value, and at a height of 80-250 km temperature rises. Acceleration of gravity is reduced at high altitudes. In fact, according to the law of gravitation acceleration of gravity depends on the distance r from the center of the earth:

g (r ) = γ

Mж Mж , =γ 2 r (r0 + x)2

(2.74)

here γ is gravitational constant, Мж – Earth’s mass, r0 – Earth’s radius. 52

Therefore, we can write example (2.68) as follows:

dp M m dx . =− ж 0γ p kT (r0 + x)2 Taking the temperature as a constant value, using the integral, we obtain the following: ln p = γ

M ж m0 1 + ln C . kT r0 + x

Or by applying the potential, we can write:  M m r2   M m m g r02 . 1   = C expν 2ж 0 0  = C exp 0 p = C exp γ ж 0  r kT r + x  kT r0 + x  kT r0 + x 0  0  

We find constant C, having adopted the following conditions

x = 0 , p = p0 :

 m0 g  . r0   kT 

p0 = C exp

(2.75)

Hence:

C = p0 e

−

m0 g r kT 0

.

Thus, taking the temperature as a constant in changing the acceleration of gravity with a distance, we write the pressure dependence on height as follows:  m0 g

p = p 0 exp − 

kT



r0 1 − 

 .  r0 + x 

r0

(2.76)

The following surprising findings come from this equation: even at an infinite distance from the earth's surface, x → ∞ pressure is not equal to zero:  mg  (2.77) p ∞ = p 0 exp − 0 r0  ≠ 0 .  kT  53

The Earth's atmosphere lasts forever, and nowhere its density equals to zero ρ ≠ 0 . However, this phenomenon is physically impossible since the number of molecules is restricted while the volume of Universe is infinite. Then suppose that Earth’s atmosphere is not in a state of equilibrium. Due to inequality of atmospheric gas, it should be distributed across the Universe. Nevertheless, Earth has beenr retaining its atmosphere over many years. This shows that a very small number of air molecules depart from earth.

2.7. Bolzmann distribution law The above formula (2.73) is applicable for gas that is under the influence of gravity. The m0gx value in the exponent of this expression is the potential molecule energy on the x height. For the general case we denote height х as r. If so, then formula (2.73) defines the n number of molecules, potential energy of which is equal to ϕ (r ) = m0 gr . Denote the number of molecules per unit volume at a height equal to zero (x = 0) as n0. Where gas properties are influenced by forces other than gravity, its molecules have potential energy ϕ (r ) , so the number of molecules is determined as follows:

n = n0 e

−

ϕ (r ) kT

.

From example (2.78), the proportion of molecules n

(2.78)

n0

with the

energy ϕ (r ) depends only on temperature. It shows us the dependence of distribution of molecules on temperature. Since the number of molecules at a temperature T depends on potential energy value ϕ (r ) , in the increase of this energy n share decreases rapidly. If n0 so, then the proportion of molecules with high energy must be very small. With decreasing temperature, the ratio of n share to ϕ (r ) n0 decreases rapidly. 54

This can be understood by analyzing the formula (2.78). With a height at each potential impact ϕ (r ) distribution of molecules can be double-sided. 1) φ (r ) >> 1 . kT In this case, gas must be at a low temperature. Where the value of T is small, the potential energy of impact force must be very high. By increasing impact strength, thermal motion of the molecules is weakened. In this case, ϕ (r ) , has a very high value. The kT → ∞ molecules at zero energy level begin to settle down on the ground. (Figure 2.9). Then, n = ∑ Pi xi ,

(3.31)

i =1

ni determines probability of random numbers that have a n value of xi .

where Pi =

3.5.2.2. Mean value of continuously varying random variables If random number changes continuously and propagation of probability density will be p ( x) , we can determine its mathematical expectation through the following integral: +∞

< x >=

∫ xp(x)dx ,

(3.32)

−∞

where p ( x) is propagation of probability density of x value. It is worth noting that mathematical expectation corresponds to the possible mean values of random variables. For example, function ϕ (t ) continuously changes over a period of time t 0 and t1 . Then the average value of this function on the time interval t 0 and t1 is calculated as follows: t

1 1 < ϕ >t = ϕ (t )dt . t1 − t 0 t∫0 79

(3.33)

Averaging is determined in terms of time, therefore t denotes function ϕ (t ) . For example, we need to find the mathematical expectation χ of a random number uniformly propagating on the interval (a, b): Mχ =

b

1 b2 − a 2 a + b . xdx = = ∫ 2(b − a ) 2 b−a a

(3.34)

From this it follows that mathematical expectation corresponds to the average range of values of random variables, i.e.:

< x >=

a +b . 2

3.5.2.3. Dispersion A measure of the propagation of random variables in the region of average values is called dispersion. Therefore, dispersion of random variables is determined by the mean square of mean values deviation, i.e.: (3.35) σ 2 =< (x− < x > )2 > . Square root of dispersion is called standard or average sqaure deviation. In practice, dispersion is calculated as follows:

σ 2 =< x 2 > −(< x > ) . 2

(3.36)

Dispersion of discrete random variables 2 σ 2 = ∑ (xi − < x > ) .

i

Dispersion of continuously varying random variables 80

(3.37)

∞

2 ∫ (x−  x  ) Ρ(x)dx .

σ = 2

(3.38)

−∞

Or a formula applied in practice

σ2 =

∞

∫

x 2 Ρ( x)dx −

−∞

(∫ xP ( x)dx)2 = x2  −( x ) 2 .

(3.39)

Dispersion has a positive value, so

∫

x 2 Ρ( x)dx ≥

(∫ xP ( x)dx)2 ,

i.е.  x2  > ( x ) 2 . For example, we need to find dispersion of a uniformly distributed random number χ on the interval (a, b). From this example

∫

b

x Ρ( x)dx = ∫ 2

a

x2 b−a

b 3 − a 3 b 2 + ab + a 2 . = 3(b − a ) 3

dx =

The mean determined in the previous 3.5.2.

 x =

a +b 2

.

Therefore,

σ2 =

a 2 + ab + b 2 3

(b − a )2 . a +b −  = 12  2  2

It is seen from the example that dispersion depends on the length of the interval (a, b) and increases according to the length. The 81

greater the spread of random variables, the greater will be the dispersion.

3.5.2.4. Calculation of average values by ensemble It was said above that the values observed in the experiment are calculated from the mean values ensemble. For example, we denote the coordinate of a particle in the statistical ensemble in the i-system through xi , and its square through xi2 . Then, the average in the ensemble is x 2 comprised of Nα systems (3.30) Nα

1

 x a = 2

xi ∑ i

2

Nα

,

(3.40)

=1

where, α – is a figure related to the ensemble If we assume that each system consists of N cells, and the number of systems in the ensemble Nα is much larger than N ( Nα >> N), then Nαj is the number of ensembles of particles in cell j. The probability that there is a particle in cell j equals to:

Pj =

Nα j Nα

,

(3.41)

Therefore, Nα

N

i =1

j =1

∑ xi2 = ∑ Nα j x 2j ,

(3.42)

where x j − j is the cell coordinate; Nαj is number of systems, ensemble with a particle in cell j; N is the number of cells in each statistical ensemble system. Taking into account formulas (3.41) and (3.42), we can write (3.35) as follows:

82

 x2  α =

1

Nα

Nα

xi ∑ i

2

=1

=

N

1

N

Pj x j , ∑ Nα j x j = ∑ N j j 2

=1

2

(3.43)

=1

where Pj − is the probability of existence of a particle in a cell j. Expression (3.43) corresponds to the formula (3.31).

3.5.2.5. Calculation of mean values in time In our example, the change in the state of a particle is observed on a long time interval (even Τ → ∞ ) and from this time we calculate the mean square value of x(t ) coordinates. Then, based on the formula (3.32), the time average x 2 is determined as follows: T

x 2 (t )dt . ∫ Τ→∞ T

 x  t = lim 2

1

(3.44)

0

In this example, the coordinate xi (t ) of a particle changes abruptly when it comes from one cell to another, then taking into account that i − leaping particle exists in the cell ∆t i , we write the following: T

m

0

i =1i

2 2 ∫ x (t )dt = ∑ xi (t )∆ti ,

where m − is the number of leaps, ( i = 1,2,..., m ) during Т time, then m

∑ ∆ti = T . if Τ → ∞ . i =1

Then the particle can hit different cells several times, for examN

ple, in cell j − it will be Τ j = ∑ ∆t i time. Therefore, T = ∑ T j where j =1

N − is the number of ensemble cells in each system ( j = 1,2,..., N ). 83

Considering all of the above, we write down formula (3.44):

 x2  t = lim

Τ→∞

1

N

~

∑ Tj xj = ∑j Pj xj , T j 2

2

(3.45)

=1

where

 Tj  . T →∞ T   

~

P j = lim 

Probability of particles existence in cell j − . For Formulas (3.43) and (3.45),

~

Pj = Pj and

 x 2  a = x 2  t .

(3.46)

Respectively, if the measurement time for equal processes tends to infinity, mean values in time in the phase space will be equal to the average values in ensemble. It was also noted that this is called an ergodic hypothesis. The ergodic hypothesis plays an important role. In order to describe an ensemble of equal systems, the density of points in the phase space must be constant in time. Then the following conditions must be met:

∂ N N N Ρ r , p ,t = 0 , ∂t

(

(

)

(3.47)

)

  where Ρ N r N , Ρ N , t − is probability density function; N − particles of the system or N molecule, they are denoted by points N − ,





coordinates r N -, Ρ N − N molecule pulses. Condition (3.48) is satisfied if it is constant in coordinates and    momenta Ρ N or Ρ N = P N (α ) , where α = α r N , P N is the motion integral. The motion integral is a quantity in an isolated system that does not vary in time.

(

84

)

Energy, momentum, angular momentum refer to the motion

integrals. Therefore, ensemble Ρ N = P N (α ) stands for established equality and distribution. For a system with equality of state, there are two important distributions: microcanonical ensemble and canonical ensemble. A system (from the Greek sistem – a whole, consisting of parts) is a set of related elements that form a certain integrity and unity, are described by interaction with the environment. There are several classification systems in science. We will study physical systems, i.e. a set of physical objects that are in an infinite space. The difference between hierarchical, multilevel and composite systems is related to the property of objects in the system, the exchange of information and control process. In terms of the method of interaction with the environment they are divided into open, closed and isolated systems. An open system exchanges with the external environment through mass, energy and information. A closed system exchanges with the external environment through energy only. An isolated system does not exchange with the external environment through energy, mass and information. Also, the system is characterized by physical, chemical and other properties of its subject. A macroscopic system refers to objects that are composed of multiple particles and which we can see without an instrument. The size of macroscopic systems is always much greater than the size of atoms and molecules in their composition. A macroscopic parameter refers to all macroscopic signs describing the system and the relation with the environment. Macroscopic parameters include such parameters as density, volume, temperature, pressure, elasticity, concentration, etc. In relation to the environment, macroscopic parameters are divided into external and internal. A set of independent macroscopic parameters determines the macro state of the system. Macroscopic laws investigate the properties of objects using macroparameters, and if we apply these laws to individual atoms and molecules, the meaning is lost. The motion of molecules and intermolecular interactions are displayed by microscopic laws, they are formulated through coordinates, molecular pulses or matrix elements and wave functions. 85

Statistical mechanics assigns a connection between macro and microscopic views. Therefore, it is possible to explain a system state through the characteristics of molecules, inside it and the microstate. It can be concluded that one macro state contains several microstates. For this reason, the observed macroproperties of the system are the mean value in time taken along γ − space trajectory. According to the ergodic hypothesis, this mean value in time is equal to the phase average value of the «ergodic surface». An ergodic surface is a surface with constant energy in the γ − space. The energy and volume of te microcanonical ensemble are used in the study of a particular isolated system. The distribution function, which is determined by the uniform distribution of points that represent a system between two neighboring surfaces, whose energy in the phase space is constant is called the microcanonical ensemble. Apart from this zone, there are no other distribution points. For microcanonical ensemble Ρ N = Ρ0 (constant) for all energy values between E and (E + Δ E); Ρ N = 0 , outside this zone. If ΔE → 0, then all systems have the accurate energy E, then the ensemble is surface. According to this ensemble, all microstates over a certain energy surface are equally probable. Canonical ensemble is used in systems with the same temperature and volume. The system is in thermal equilibrium and exchanges energy with this medium, the number of particles and the volume are constant. Energy distribution assigned through the canonical ensemble has an acute maximum. The mean values taken with the help of the canonical ensemble correspond exactly to the values taken with the help of microcanonical ensemble. In order to calculate the probability W of a system being in a certain macrostate, the following two conditions must be met:

∑n j = N , ∑n jE j = E .

(3.48)

j

Where E is the total energy of the system, Ε j − is the energy of a system particle in the cell j − ; n i is the number of points that 86

describe the system particle and that locate in the i − cell of µ − space; n i is a number describing the macrostate. N – the number of cells, according to the system of particles in the µ space. Hence, the amount of the microstate of the system is calculated as follows:

N! , then W = 1 N! , n1!n2!n3!... c ∏ n j!

(3.49)

j

where c is the product of normalization, W (n1, n 2 , n 3,... ) is the probability of the system being in a certain macrostate. In systems with a large number of parts, only the average value of probable macrostates is determined. With an infinitely large value of N , i.e. N → ∞ , fluctuation in the systems state equals zero. When studying properties of an ideal gas, its individual mole-cules are considered as parts of the system, while gas inside the vessel is a system. There are no interactions between the molecules of an ideal gas. The total energy of an ideal gas is found by summing energy Ε j of the molecules in its composition. Ε j is equal to the sum of the kinetic energy of molecules and the energy of the internal degree of freedom.   The Ε j value depends on coordinates q, Ρ of individual mo  lecules of the j − cell and is determined by H q , Ρ Hamiltonian function of individual molecules. Here the generalized coordinate  frame q depicts configuration and arrangement of individual

( )

(



)

molecules, while Ρ depicts a set of momentum. The Hamiltonian of an individual molecule of a monatomic gas is determined by:    P 2 , where m is the molecule mass, Ρ -is the moleH (q , Ρ ) = 0 2m0 cular momentum. If macroscopic systems parameters do not change in time, then its state is stationary. The state of equilibrium of the system is the state under which all the system parameters are stationary in time 87

and there is no flow from the influence of external circumstances. Systems that do not meet these requirements are called non-equilibrium systems. Parameters describing thermodynamic equilibria are called thermodynamic parameters. A thermodynamic process consisting of a continuous chain of equilibrium states is called the equilibrium process. If there are nonequilibrium states inside the chain, then the process is called nonequilibrium process.

3.6. Fluctuaiton The

square root of the root-mean-square deviation ∆x =  ∆x 2  is a measure of deviations from the mean value, we denote it by δ and determine fluctuation of the x value (from the Latin fluactuatіo - to fluctuate): 2

( )

δ =  (∆x 2 )  ,

(3.50)

meaning δ =  (∆x 2 )  = x 2 − (x )2 ,

( )

where  x 2  – is the mean x square value, х 2 – is the square of the mean value of this variable. Fluctuation is a random and erratic deviation of a mean value from some value. According to thermodynamic equality, a random deviation from mean values describing the state of the system is a fluctuation. In macroscopic thermodynamics, changes in the system do not stop at all, even when the system is in the state of equality, continuous fluctuations do not end, thus, minor fluctuations in the system are observed. Therefore, this term is used in the meaning of a collector for different circumstances and one can talk about fluctuations in energy, volume, number of particles, etc. Totality of the theory of processes is explained through a fluctuation. For example, the Brownian motion, the statistical theory of liquids, etc. 88

Let m be the additive value, which describes a system consisting of N particles, meaning

m = ∑ mk ,

(3.51)

k

where mk is a point representing the system in γ space and in the

k cell. The additive property also assumes the mean value m :

m = ∑ mk .

(3.52)

k

Now we need to determine the m value fluctuation. According to equation (3.50), the fluctuation meter is determined from the standard deviation from the mean value, meaning

 (∆m)2  = ∑  (∆mk )2  ,

(3.53)

k

where ∆m = m−  m  ; ∆mk = mk −  mk  . The particles in the system are the same, they are equal, so the mean deviation square for all microstates is the same, i.e.

 (∆m1 )  = (∆m22 )  = ... = (∆mc )  , 2

meaning

(

2

 ∆m

)2  = N  (∆mc )2  ,

(3.54) (3.55)

that the root-mean-square deviation is directly proportional to the number of particles in the system. From this it follows that fluctuation of the m value is directly proportional to the root of the number of particles in the system:

δ =  (∆m)2  ~ N , 89

(3.56)

so if the number of particles increases, the δ value will also increase. However, in order to describe deviations in the system, it will be more convenient to use comparative fluctuations. The ratio of δ -fluctuations to the mean value m is called the comparative fluctuation, and is determined from the following expression:

k=

 (∆m)2  .  m

(3.57)

The mean value m of a variable in the system is directly proportional to the number of particles in its composition, i.e.  m  ~ N . Therefore, N 1 . (3.58) k~ ~ N N And so we came to a very important conclusion: the smaller is the relative value of additive value fluctuation, the more particles there are in this system. Even if N is infinitely many, fluctuation of the system state equals zero. For example, at room temperature and at atmospheric pressure of 1l of air, there are N ~ 10 22 − 10 24 of molecules. Then k = 10 −11 − 10 −12 or 10 −9 − 10 −10 % will be a very small value. For this reason, statistical fluctuations are not observed in a macroscopic system. On the contrary, in systems with a smaller number of particles, there are many more such fluctuations. The theory of fluctuations is considered in a special chapter of statistical physics. We have discussed properties in the state of equality, they are considered to be the most probable values or mean values in the ensemble. In particular, these equality properties obey the laws of thermodynamics. The methods of statistical mechanics allow not only the possibility to determine mean values of the system, but also give methods for calculating fluctuation probabilities in the zone of mean values of any variable. For example, in the study of a critical gas area, special attention is paid to fluctuations in density and energy. Fluctuation in density 90

results in the propagation of light. Fluctuation of internal energy near the critical point is the cause of anomalous growth of speed in a stable volume.

3.7. Correlation of random variables In dynamic illustration of macroscopic systems, the first value depends on the second, and there may also be cases where they may depend on other values or on unknown factors. Between values other than functional, other connections may appear, in such cases the term correlation is used (from the Latin correlatio – the ratio). In probability theory, linear relationship of two Χ i and Χ j random quantities is taken as numerical measure of correlation, which are called correlation coefficient, and correlation function is taken for random processes. If for two random numbers Χ 1 and Χ 2 mathematical expectation a i = MΧ i and dispersion σ i2 = M ( xi − a i )2 are determined by formulas, then correlation coefficient is calculated by the formula:

rij = rij (x1 , x2 ) = M

x1 − a 1 x2 − a 2 , ⋅ σ1 σ2

(3.59)

where rij is correlation coefficient between the two values хi и хj.

The value of the correlation coefficient (− 1,+1) lies in the interval

− 1 ≤ rij ≤ 1 . If random values хi and хj are linearly related to one another, the correlation coefficient will be rij = ±1 , if the correlation coefficient of independent random values хi and хj will be equal to zero ( rij = 0 ),then random variables хi and хj are called noncorrelation values. The correlation coefficient of two X η values can be zero, even if there is a relationship between them. The correlation coefficient does not reflect the entire functional dependence of random variables, it is only a measure of linear connections. 91

Practical importance of correlation functions of random processes is great in physical studies, for example, the correlation function method is widely used in the theory of fluctuations, in Brownian motion and hydrodynamics problems, etc.

3.8. Random variables and distribution function. Binomial distribution. Poisson distribution 3.8.1. Random variables and distribution function From the practice of the history of mankind it is known that events whose probability is closer to a unit are definitely observed. If probability of events is closer to zero, such events are rare. All this determines the criterion for the application of probability theory in practice. In practical actions of an event, the probability of which is closer to zero or to a unit is important. From this one can assign laws of events, the probability of which is closer to a unit, arising from the interaction of random factors. In order to determine probability of various random events, the concept of the random variables distribution function is used. Distribution function is one of the basic concepts of statistical physics, in classical physics it determines distribution of probable densities of a many-particle system or in quantum physics - the quantum mechanical condition. In the kinetic theory of gas, in order to describe unequal sys tems, the distributive function f (r,υi , t ) is applied. Here, r is the radius vector describing the change in the location of the point  (molecule) in space, Descartes coordinates x, y, z ; r , t – this is “at





r point at a t time”, υi - is i velocity of molecules. Distribution  

function f (r ,υ , t ) shows the probable number i of molecules moving     at a velocity interval of υ , dυ in a volume element r , dr at the t time. If there are no gradients of concentration, velocity, and tempe  rature in gas, then f (r ,υi , t ) is Maxwell's distribution function. For systems in the state of equality, this function is independent of the 92

time and space function, it only depends on velocirt, i.e. f (υi ) . In      connection with this, ndr = dr ∫ f (r ,υ , t )dυ , where the number of molecules means

   n = ∫ f (r ,υ , t )dυ .

(3.60)

 

If velocity tends to infinity, then f (r ,υ , t ) function will not be negative, it will tend to zero. For all t times, the distribution function is finite and continuous. In six-dimensional spaces function   f (r ,υ , t ) determines the numerical density of a point, and then all   components of molecules υ and r are the coordinates of the points. For the entire space of velocity and density, the integral (3.60)   will be equal to the numerical density n , which means f (r ,υ , t ) , and this will be the condition for the normalization of the distribution function.

3.8.2. Binomial Distribution If in conducting n independent number of tests an A-event is observed m times, and in other (n − m) times an opposite Α -event is observed, the probability Ρn (m) is called binomial distribution or binomial probability distribution law. Let us consider Jakob Bernoulli’s scheme. In this scheme, the probability of Α is -events is independent of the number s of tests. Where s is the order of tests, s = 1,2,3,..., n . Then Α is is the result of A events during the test. The probability of appearance of the i result in s tests is denoted as Ρis = Ρ Α is , i = 1,2,3,..., k . By the formula (3.29)

∑ Ρi = 1 .

( )

This scheme for k = 2 case was first considered by J.

Bernoulli, hence its name – Bernoulli's scheme. According to Bernoulli's scheme Ρ1 = p , Ρ2 = 1 − Ρ = q , means p + q = 1 . 93

Let's calculate the probability of occurrences of Α s events in m tests, then calculate for n − m . By multiplication theorem of independent probabilities of events (equation 3.19), it is equal to

Ρ m ⋅ Ρ n −m : Probability Ρn (m) is determined by the sum of ( Ρ m ⋅ Ρ n − m ) for all m and ( n − m ). The number of such methods

C nm =

n! , m! (n − m)!

(3.61)

where C nm is binomial coefficient. Thus,

Pn (m) =

n! P mq n − m . m!(n − m)!

(3.62)

Formula (3.61) is called binomial distribution. Here n

∑ Ρn (m) = (P + q )n = 1n = 1 ;

0 ≤ P ≤ 1,

0≤ m≤ n.

m=0

For example, in every test there is a k event observed from all

Α i events. The probability of occurrence of Α i events is Ρi , then in n tests A1 event occurs m1 times, A2 event occurs m2 times and

Α k event occurs mk times. (m1 + m2 + ... + mk = n ) is determined by

the following formula:

Ρn (m1, m2 ,..., mk ) =

n! m1! m2 !,..., mk !

Ρ1m1 Ρ2m2 ,..., Ρkmk .

(3.63)

Equation (3.63) is a general form of binomial distribution, equation (3.62) is a separate form. The formula (3.62) is the

94

probability of occurrence of Ρ1 = Ρ events, whereas for Ρ2 = (1 − Ρ) it is not noticed ( Ρ1 + Ρ2 = 1) . Therefore, during n test this event was observed m times, and was not observed ( n − m ) times, its probability follows from the formula (3.62):

Pn (m, n − m) =

n! n−m Ρ m (1 − Ρ ) , m!(n − m)!

(3.64)

i.е. corresponds to formula (3.61).

3.8.3. Poisson distribution Let us apply the method considered for an ideal gas, where there is no interaction of molecules. Gas is a system in the volume V , its parts are n molecules, n i is the number of molecules in the i cell and in µ space. N is the number of cells where particles can be placed in the µ space. The macro state of the system is described by

n i value. Let us calculate the number of microstates where m particles are located in the V1 volume and which implement the macrostate of gas. If there are m particles in the volume V1 , then the number of microstates in the given macrostate is equal to:

N1! , (N1 − m)!

(3.65)

where N1 is the number of cells in the volume V1 ; m is the number of molecules in the volume V1 ; m  n ; N1  m . In part of the volume (V − V1 ) remained particles (n − m) . For them, the number of microstates in the given macrostate will be equal to: 95

(N − N1 )! . [N − N1 − (n − m)]!

(3.66)

Thus, the total number of microsystems that implement the macrostate is determined by multiplying equations (3.65) and (3.66). And the number of methods is calculated by the following formula:

n! . m!(n − m)!

(3.67)

Then the number of microstates that implement the macrostate is calculated by multiplying formulas (3.65), (3.66), and (3.67):

(N − N1 )! N1! n! . (N1 − m)! [N − N1 − (n − m)]! m! (n − m)!

(3.68)

As was noted above (chapter 3.8.2), the probability of one molecule coming from V volume and n molecules into volume V1 and into cell N1 :

p=

N1 V1 , = N V

(3.69)

and q = 1 − N1 = 1 − p is the probability of a particle in the part N . p (V − V1 ) + q = 1 , i.e. the particle will be in volume V1 or in

volume (V − V1 ) . Then the probability of the system being in a certain macrostate corresponds to the number of microstates that implement it by the total number of microstates.

N !

(N − n )!

.

Will be calculated as follows: PV (m) = 1

n! N1!(N − N1 )!(N − n )! . m!(n − m)!(N1 − m)![N − N1 − (n − m)]!⋅N! 96

(3.70)

Numbers N1 , m , n , N are large. For example, in an air with temperature T = 0 0 C , atmospheric pressure 1 atm, and volume

V = 1 см 3 the number of molecules is equal to n = 2,7 ⋅ 1019 1 / см 3 , and the volume of one molecule is V0 ≈ 10 −24 см 3 ; N=1024; V  N1 = 1024  1  , N1  m . V  Therefore, in order not to make boring calculations and when n number is large, the Stirling formula is applied:

n ~ 2πn ⋅ n n e − n ,

(3.71)

where n! = n (n − 1)(n − 2 )(n − 3)...3 ⋅ 2 ⋅ 1 - is a factorial function. By definition factorial functions of zero is equal to one, 0! = 1 . Logarithm n! of factorial function ln n! =

j =n

∑ ln j .

(3.72)

j =1

Calculated using the Euler-Maclaurin addition formula ln n! ≈ n ln n − n +

1 1 1  ln n + C =  n +  ln n − n + ln (2π ) , 2 2 2 

(3.73)

1 where С= ln 2π . Taking this into account, the Stirling formula is 2 determined. Taking Stirling's formula into account, we write equation (3.70) as follows: m n−m n!  N1   N1  . (3.74) 1 PV1 (m) = −     m!(n − m)!  N   N

In computations (3.73), the following conditions are taken into account: N1  m , N − N1  n − m , N  n and 97

n

x  lim 1 +  = e x . n →∞ n

For example,

(N − n )! =  N − n  

N −n



e

N =  e

N −n

n  1 −  N 

N −n

N =  e

N −n

e −n .

Using the ratio (3.69), we derive the formula (3.74) through (3.74):

PV (m) = 1

n! p m (1 − p )n − m . m! (n − m)!

(3.75)

To solve specific cases for very small p values, you will need

to calculate the probability PV1 (m) . For small values p or q, Poisson proposed an asymptotic equation for calculating probabilities Pn (m) , therefore, formula (3.75) is called the Poisson distribution. If in the n process of tests occurrence is very rare, it means that m → 0 or m → n , then the Pn (m) value will not be large. It is worth noting

that p + q = 1 , n → ∞ . But Pn (m) reaches a large value, when certain m events occur. In order to find this maximum value, you need to solve the differential equation dPn (m) = 0 . dm Consider the case where derivatives V1 and p are very small, and value q is close to zero. To do this, we write the factorials from the equation (3.75) using the Stirling formula as follows:

(m / e) n! nn ≈ ≈ = m! (n − m )! (m / e )n (n − m / e )n − m m m n n − m (1 − m / n )n − m n

 ne   n  (1 − m / n ) =  , =  n m  m  (1 − m / n ) m

98

m

n

n → ∞ , then lim 1 − m  = e −m , therefore n →∞

n

m

 nep  n  ne  m n−m  q . =   p (1 − p ) m    mq   neр   dPn (m)   ≡ 0,  = 0 , m values, ln    dm   mмах q 

Pn (m) ≈ 

m

(3.76)

where these conditions are observed are called probable. The highest value of Pn (m) probability corresponds to the number of observations of m events. If we denote that the number of observations where the probability of events reaches a maximum of m = m мах , we find the following relation:  neр   ≡ 0, ln   mмах q  the value of the divisor is not equal to one, approximately

m мах ≈

nP q

(3.77)

if q ≈ 1 , then mmax ≈ np . By our definition p =

N1 V1 = , this shows the arrangement of N V

particles in the volume V1 . Suppose that there are n particles placed in the V volume, then n / V = n 0 is the numerical particle density or concentration. Its physical value indicates that n particles are evenly distributed in the V volume. Hence, probable concentratoin of particles in V1 volume is

m max V . Then, taking into account p = 1 , we find the folloV1 V wing: n 0 = n мах . From this it follows that the probable concentran max =

99

tion of particles in the V1 volume corresponds to a uniform distribution of particles throughout the V volume. The V1 volume is randomly taken from V , so the equilibrium state of the system corresponds to its most probable state. If p value is very small, then the value of the total number of tests is large, but np = m retains its stable small value n then equation (3.75) written for probabilities Pn (m) I s called the Poisson distribution. When the system is in equilibrium, the average number of particles in the V1 volume will be equal to m = np = const . Then equation (3.75) can be written as follows: m

 m    m   n−m n!  1 −  Pn (m) =  . m!(n − m)!  n   n 

(3.78)

We will transform this equation: Pn (m ) =

n(n − 1)(n − 2)...(n − m + 1)   m     m     1 −  m!! n   n   m

 m m   m   =  1 − m!  n 

n−m

n−m

=

 1  2   m − 1  .  1 − 1 − ...1 − n   n  n  

The number of molecules is very large, values n and m large numbers, if n → ∞ , then in comparison with 1 , 2 ,..., m − 1 , it can n n n be ignored. But brackets with a n − m degree are very large, n − m ≈ n . Therefore

Pn (m) =

m

m m!

  m  1 − n  

n−m

n

≈

mm  m  1 −  . m!  n 

For large values n , you can write it as follows: 100

m −   〈 m〉  〈 m〉  n  Pn (m) =  1 − m!  n    

−  m

m

  m lim1 −  n →∞ n  

−

n 〈 m〉

=e .

In view of the above, we will write th probability Pn (m) with n → ∞ as follows: 〈 m〉 m −  m . (3.79) Pn (m) = e m! In probability theory, this equation is called the Poisson law or Poisson's probability distribution. The difference between this distribution and binomiality is that m =  m  is asymmetric with respect to point. In a general case, the Poisson law is written by the following formula: a m −a e , Pn (m ) = (3.80) m! where n is the number of tests; m is the number of event observations, a n = nPn . (3.81) ∑ Pn (m) = 1 . m

Based on Sterling’s equation: m

m m!≅   , then e

Pn (m) ≅

mm − m m e e = 1. mm

From this it can be concluded that with sufficiently large volumes of gas, in the parts of the volume under consideration, in a state of equilibrium, the number of molecules and the density are the same. 101

Let us investigate probability Ρn (m) through the m function. For this we consider the following relation: Ρn (m) a = , m  a , then Ρn (m)  Ρn (m − 1) ; if m  a , then Ρn (m − 1) m Ρn (m)  Ρn (m − 1) ; if m = a , then Ρn (m) = Ρn (m − 1) . From this Ρn (m) value will increase from m equal to zero, to m0 = a , with

further increase of m , Ρn (m) decreases. If а is a whole number, Ρn (m) will have two maximal numbers

m0 = a and m0 = a − 1 .

For example, the probability that a bullet hits a target with each shot уйгфды 0,001. If the number of shots is equal to 5000, then you need to find the probability of hitting the target from two or more times. Let us denote each shot as a test, and bullet hitting the target – as an event in order to calculate the probability Pn (m ≥ 2 ) , we will apply the Poisson law. In this example:

a n = пр = 5000 − 0.001 = 5 ; Pm (m ≥ 2) = Pn (m) =

5000

∑ Ρn (m) = 1 − Pn (0) − Pn (1) .

m= 2

Based on Poisson formula

50 − 5 P5000 (0) ≈ e ≈ e − 5 ; P5000 (1) ≈ 5e −5 , 0! therefore,

P5000 (m ≥ 2) ≈ 1 − 6e −5 ≈ 0,9596 ;

1 − 6e −5 = 1 − 6 ⋅ 0,00674 = 1 − 0,04044 = 0,95956 . Maximum value of Pn (m) corresponds to m0 = 4 and m0 = 5 . These probabilities are equal, which means that Pn (4) = Pn (5) ≈ 0,179 . 102

The event from the example was used during the Second World War when shooting a plane with a firearm. You can shoot a plane if a bullet hits the engine, the pilot or the petrol tank. With a separate shot, probability of hitting the plane's weak spots is very small, so the shot was fired by a rifle detachment. Then the total number of shots was much larger. As a result, there is a probability of one or two bullets hitting a weak spot of the aircraft. The above was actually observed in reality.

103

4 MAXWELL VELOCITY MOLECULAR DISTRIBUTION 4.1. Ideal gas as molecular system. Molecular collision – the main reason of Maxwell gas distribution in the state of equilibrium, by velocity Let’s look at the monoatomic ideal gas. It is assumed that molecular mass of such gases m0 are the same material points. These particles are not intercommunicate with each other, move continuous and desordered, and also walls of storage reservoir with V volume, in which they are located, stop molecules that impinged on the walls. On molecules, learned in the capacity of material points, non of the laws are influenced, this is why their potential energy of mutual effect equal to zero. In such case full energy of molecular system are enumerated by sum of the kintec particle energy in the molecular composition. Accorfing to our forecast, velocity of all N molecules is equal and they move in three direction of the coordinate systmen of descartes, 1/6 part of particles go to positive direction, 1/6 part of particles go to negative direction. If come molecule impinge on walls of storage reservoir, so the direction is changed. Walls of storage reservoir are in parallel of coordinate system. Thus, by the way of example (model) of gas the ideal gas is taken and its main characteristics formulate following molecular statistical systems: 1) molecules of system are material points, which are not intercommunicate with each other, this is why the volume of particle is not taken into account; 2) molecules of system move continuous and desordered; 3) mutual effect force of molecules are observed during collision, so that each molecule move free before collision, in other words, line is their trajectory; collision of particles between each others and solid walls of vessels; 104

4) all molecules move with the same velocity; in the capacity of molecular velocity it is opportune to apply average arithmetical velocity; 5) all molecules move in parallel with coordinate axix, in particular, 1/6 part in the direction (+х), 1/6 part in the direction (-х), 1/6 part in the direction (+y ) etc. We learn ideal gas in the condition of equilibrium, in other words, pressure P and temperature T, which describe gas in volume V are permanent, no flows are observed. The distance between gaz molecules way more than size of molecules themselves. This is why they influence to each other very faint and move in different direction. In this case it should be considered that collision of molecules took place. As it was noticed above, statistic characteristic of such manyparticle systems, in other words, averaged vaues are enumerated in real gas experiments. This is why, our main task is to understand such valuations as average number of molecules, momentum, energy or average distribution of velocity and etc., as a statistical property. Simplest molecular-kinetic theory explains macroscopical gascharacteristic of gas by the following way: 1) Sum of molecules mass is a visible macroscopical mass of the gas; 2) Thermal energy of littler particle of gas is equal to advancing kinetic energy of molecular motion. Kinetic and molecular energy of protons and electrons being a part of molecules do not include in thermal energy; 3) Pressure, provided to walls of storage reservoir where the gas locates is equal to average velocity of hitting momentum of molecules on that area; hittings are frequents, and provision of momentum is discontinuous, but the influence is continual; 4) average kinetic energy of one molecule is in proportion to thermodynamecal temperature, proportional ratio depends on energy and temperature but independent to the type of gas. In such a way, distribution of molecules on the velocity can explain the processes that is inside and characteristics of the subject. We also marked that molecular physics laws are always prospective. According to forecasts, average squared velocity of gas molecules is the same, because all motion directions are the same 105

presumable, their distribution on the directions is proportional. It is real, velocity of molecules are different, the experiment of Shtern demonstrates it. Disordered motoin and hitting of molecules showes, that distribution on velocity is and fast as well as slow motion inside is observed. Motion of molecules are chaotic, hittings are aselected, this is why, despite on changes of velocity has random character, dirtribution of molecules velocity is subject to estimated laws. Theory as well as practice proves it. English physics, one of the founder of statistical physic Maxwell James Klerk (1831-1879 yrs.) is the first who turns attention to it. In 1859 he suggested laws on velocity distribution between molecules on the convention of British assosiation. Later this law was named by his name.

4.2. Distribution function Determination process of molecules velocity distribution is seems to be similar to process of location of sum molecules with defined velocity. However, there is no physical sence in such setting of the problem, because probability of molecules with such velocity is equal to zero. Because the amount of valuation of different velocity is infinitely many. The amount of molecules is limited. By this reason, amount of molecules, taking up any valuation is equal to zero. From this, our question forms by the following way. How many molecules or parts of molecules taking up th velocity on interval of defined velocity?

4.2.1. Conception of distribution function. For the solution of this problem Maxwell applied theory of chances. As of Maxwell forecasts, distribution of molecules on subcomponent of velocity (element) relating to three direction of system of references, is independent from each other. Therefore, the valuation of component х motion with a probability υ x is not related to other two valuations υ y,υ z . Then we can mark probability of 106

molecules on the axis х , having the component of velocity on the interval υ x + dυ x and υ x , as f (υ x )dυ x . Where f (υ x ) is function depending on velocity . This all can be explained on the basis of following conclusion. As it was marked above (chapter 3.8.1), gas is real as a statistical identity element. All characteristics of statistical group can be determined by distribution function. Molecules velocity continuous changes in the result of collision  and moves out of the volume dr , and others are coming into volume  dr from the outside. By this reason, distribution of molecules points   in the volume dr changes on time. Amount of velocity points ndr is a lot, this is why they have estimated statistical density, as quantitive of n molecules of conventional gas. Density of the velocity points  on form of externals and in proportion to ndr does not depend  element of volume dr .  In general case, spatial coordinate r and t connected with time, and also instead of velocity space, this is why it is denoted by the followig way:     (4.1) f (υ , r , t )dr dυ

Fig. 4.1

By this determination, in the time of t in the elements of      volume r , dr on the intervals of velocity υ , υ + dυ amount of     probabilities of molecules is equal to f (υ , r , t )dr dυ . 107

It does not mean that there is the same amount of molecules in the volume elements in time t on the interval of the velocities υ ,  dυ , but that on interval of time dt all fluctuation of gas collapses and means average amount of molecules.

 

In such a way, function f (υ , r , t ) describes distribution of gas molecules velocity and it calls distribution function in elements of volume    In order to find the amount of molecules dr , it is necessary to integrate fdr dυ in all velocity spaces. By 

definition this numeric is equal to ndr :









 



ndr = dr ∫ fdυ = dr ∫ f (r ,υ , t )dυ .

(4.2)

 It means  that  average amount of molecules, the velocity of which

υ and υ + dυ equal to:

+∞

n=

 



∫ f (r ,υ , t )dυ .

(4.3)

−∞

 

Distribution function f (υ , r , t ) can not be minus number, it will

be tending to zero, if velocity υ is tending to infinity, f (υ ) = 0 , if υ → ∞ . This function is denumerable and continuous for all temp-  oral value. Distribution function f (υ , r , t ) determines numerical  density of molecules in six-dimensional space, valuations υ (υ x,υ y,υ z )



and r ( x, y, z) specifies points of coordinate. Let’s consider monoatomic gas with homogeneous condition. It is possible to define gas characteristics in the condition of equilibrium with help of distribution function. But  in this case, it will not be depending on time t and coordinate r , which determines place in the space. Because of that, distribution function depends only on molecules velocity, in other words f (υ ) . It is necessary to mention



that f (υ ) = f (υ ) , where it showes that there is no special direction of molecules velocity. Gas considered as entire system in latent condition. 108

As direction of molecules motion is equally possible, distribution of molecules by direction is proportional. In such a way, results, taken throught out f (υ ) describes averaged condition or condition of gas probability. In the stable temperature velocity of molecules in the equilibrium condition observed only one stable distribution. It calls maxwell distribution . Let’s outline probability on the interval υ y + dυ y , υ z and υ z + dυ z and projection of molecules velocity, in compliance with coordinate axis υ x and υ x + dυ x , υ y as dP υ x,υ y,υ z .

(

)

Rationing condition of this probability is written this way: +∞

∫∫∫

dP (υ x,υ y,υ z ) = 1 ,

(4.4)

−∞

where

dP (υ ) =

dn(υ ) . n

(4.5)

dn (υ ) – number of molecules on the interval of velocities υ and υ + dυ ; n – molecules amount in the unit of volume. Therefore ,

dn (υ ) – determines posibility, on the interval of velocities n dn (υ ) gives υ and υ + dυ in the unit of volume. In other words, n part of all molecules on the interval of velocities υ and υ + dυ . valuation

4.2.2. Average valuation of function, depending on molecular velocity Let’s suppose, ϕ (υ ) function, depending on molecular velocity. Function ϕ (υ ) can be scalar, vectorial or tensorial. Let’s determine average valuation of this function throught out distribution function f (υ ) : 109

ϕ (υ ) =

1

∞

n ∫0

ϕ (υ ) f (υ )dυ .

(4.6)

For example, average arithmetical function of molecules (4.6) determins:

υ=

1

∞

n ∫0

υf (υ )dυ .

In such a way, knowing about distribution of molecules velocity, can find any average valuation, which is function of velocity

4.3. Conclusion of Maxvell on functions of molecules velocity distribution It is supposed, that components of x velocity of molecules and posibility P (υ x ) on interval υ x and υ x + dυ x do not depend on velocity components υ y and υ z . Valuation of velocity components y and z on the interval υ y , υ y + dυ y and υ z , υ z + dυ z and its possibility P υ y dυ y and P (υ z )dυ z . Accordingly (4.1) :

( )

f (υ x,υ y,υ z )dυ x dυ y dυ z = dn(υ x .υ y , υ z )

(4.7)

velocity components on interval dυ x , dυe , dυ z determine amount of molecules per unit of volume. Taking into account (4.2) – (4.5), distribution function should be writtent by the following way:

f (υ x,υ y,υ z )dυ x dυ y dυ z = nP (υ x )dυ x P (υ y )dυ y P (υ z )dυ z , (4.8)

(

where f υ x,υ y,υ z

)

– distribution function by velocity,

P (υ x ,)P (υ y ), P (υ z ) – posibility of velocity components on interval

dυ x , dυ y , dυ z . 110

In condition of equilibrium there is no priveledged direction of gas velocity, this is why possibilities nP (υ x )P (υ y )P (υ z ) are the same on all directions. Then dependence of distribution function f υ x,υ y,υ z from dυ x , dυ y , dυ z can be defined only by absolute

(

)

value (modul) of velocities υ 2 = υ x2 + υ 2y + υ z2 . These conditions give opportunities to write following ratio:

nP (υ x )P (υ y )P (υ z ) = f (υ x ,υ y ,υ z ) = φ (υ x2 + υ y2 + υ z2 ) .

(4.9)

Solution of this functional equation (4.9): Φ (υ x ) = Χe yυ x . 2

(4.10)

Then taking into account (4.9), can write the formula, calculating distribution fuction by the following way:

(

)

f (υ x ,υ y ,υ z ) = φ υ x2 + υ 2y + υ z2 = nΧ 3 e

y (υ x2 +υ y2 +υ z2 )

.

(4.11)

Where Χ, y – constants. Distribution function on formula (4.11) find by three invariant of  accent ( m0 – molecular mass, m0υ – molecular momentum and

1 m0υ 2 – kinetic energy of molecules: 2

f (υ x ,υ y ,υ z ) = βe

1 −α m0υ 2 2

,

(4.12)

where β , α – constants, they are determined by numerical density  of gas n = ∫ fdυ , average velocity nυ = ∫ υfdυ and temperature 3 1 kT = m0υ 2 . 2 2

111

Formulas (4.10), (4.11), (4.12) were suggested by Maxwell in 1859 in theory. This is why this distribution function was called by his name. So, gas condition, determined by this distribution function is known as maxwell condition. Constant α in formula (4.12) according to corroboration of Maxwell α = 1 / kT , where k – Boltzmann constant, T – temperature. Maxwell suggested following type of distribution function from the formule (4.12):

 m  f (υ ) = n 0   2πkT 

3/ 2

e

−

m0υ 2 2 kT

.

(4.13)

This formula is specified by Maxwell function for distribution of molecules velocity. Index of power of the Maxwell distribution function m0υ 2 / 2kT is kinetic energy, devided kT and describing average molecules energy. In such a way, from definition follows, that only one distribution of molecules velocity is observed, where density, average velocity and temperature are known. 4.4. Special aspects of Maxwell condition. 4.4.1. Disttibution of molecules velocity components Amount of molecules on volume unit in the velocity interval υ

(

)

and υ + dυ equal to f (υ )dυ or f υ x,υ y,υ z dυ x dυ y dυ z . It means that amount of molecules on volume unit with components of velocity υ x and υ x + dυ x , υ y , υ y + dυ y and υ z , υ z + dυ z defines by formula (4.13): 1 m0υ x2 3 / 2 − 2 kT

dn(υ x ,υ y ,υ z ) = n(m0 / 2πkT ) e

dυ x ⋅ e

−

2 1 m0υ y 2 kT

dυ y ⋅ e

−

1 m0υ z2 2 kT

dυ z

, (4.14)

where υ 2 = υ x2 + υ 2y + υ z2 . It is follows from it, that distribution υ x does not depend on valuation υ y and υ z , it means that posibility of 112

location on the estimated interval х of velocity molecules components unrelated with valuation of components υ y , υ z vertical to axis ОХ. If suppose, that υ y ,υ z constants, then dn(υ ) = dn(υ x ) = f (υ x )dυ x . Average amount of molecules on Х component of velocity on the volume unit in the interval υ x and υ x + dυ x of molecules is equal

to dn x (υ x ) . Then condition of rate setting (formula (4.4)) comes to the following mode: +∞

∫

P (υ x ) =

−∞

+∞ dnx (υ x ) − mυ / 2 kT = A ∫−∞ n ∫−∞e x dυ x = 1 . +∞

2

If indicate degree e of exponential in formula m0υ 2 / 2kT = x2 , then integral asquires the other mode: +∞

−m υ ∫e 0

2

/ 2 kT

dυ x =

−∞

2kT

m0

+∞

∫e

− x2

(4.15)

(4.15) as

dx .

−∞

It is famous integral, which equal to: +∞

−x ∫ e dx = π . 2

−∞

Accordingly, +∞

∫e

− m0υ x2 / 2 kT

dυ x =

−∞

2πkT

m0

.

(4.16)

In the capacity of integral υ x possible velocity vallues from (-∞) to (+∞) were taken. Let’s determine probability of achievement of gas molecules velocity by some х componet. Finding A constant from the formula (4.15) by the formula (4.16):

113

1

 m 2 Α= 0  .  2πkT 

Fig. 4.2

Thus, Maxwell distribution function for the х – component of velocity finds like that: 1

 m0  2 −m0υ x2 / 2 kT dn(υ x ) = dn x = n dυ x  e  2πkT 

(4.17)

or 1

 m0  2 −m0υ x2 / 2 kT . f (υ x ) = n  e  2πkT 

(4.18)

Formulas (4.17) and (4.18) determine distribution of molecules velocity components. Last result can be explained like that: +∞

∫ dn(υ x )dυ x = n .

−∞

In other words equal to number density of molecules. It is necessary to take into account, that in a great many educational books (for example, I, E. Yrodov. Objects on physics. – M.: Lab. Whithout knowledges, 2001. – 432 p., A.K. Kykoyn. Molecular physics. – M.: Science, 1976. – 480 p. and so on) distribution function by the formula (4.18) appears as follows: 114

f (υ x ) =

2 1/ 2 − m0υ x m  0  e 2kT .  2πkT   

( x ) = 

dnx υ

ndυ x

(4.19)

Distribution function by the other components of velocity

υ y , υ z must be written according to formulas (4.17) and (4.18): mυ 2 dn υ   m 1 / 2 − 0 y y y  =  0  e 2kT f υy =  2πkT  ndυ y  

( )

2 1 / 2 − m0υ z m   dn (υ ) f (υ z ) = z z =  0  e 2kT .  2πkT  ndυ z  

(4.20)

Distribution of molecules velocity components shown on Figures of diagrams 4.2 и 4.3. On the Figure 4.2 distribution function f (υ x ) is in proportion to e

− S x2

, where

Sx = υ x

m . 2kT

On the Figure 4.2 freehand curve, describing functional connection f (υ x ) from υ x , summetrical to tangential varying (+ υ x ) to (- υ x ) and largest extremum is equal to υ x = 0 , а υ x → ∞ , then

f (υ x ) is tending to zero. But part of molecules

dnx with compon

nent х of velocity is not equal to zero (formula (4.17)). In stable temperature this exponent is equal to previous (Figure 4.3). 1/ 2

 m  A=   2πkT  115

Fig. 4.3

With a rise in temperature, proportion of such molecules will be decrease. From the Figure it is visible, that mean observation of х – component of velocity  υ x  is equal to zero. It is real, as it was said in chapter 4.3.2, according to formula mean observation of х component of velocity determines like that:

 υ x =

+∞

υ x f (υ x ,υ y,υ z )dυ x dυ y dυ z , n −∫∞ 1

(4.21)

∞ 1 +∞  m0  − m 0υ x2 / 2 kT 1 − αx 2 e d m = = = υ υ   x x 0 ∫ xe ∫ n − ∞  2πkT  2α 0 1/ 2

 υ x =

∞ 0  2 2 1 1  m   =  0   ∫ υ x e − m0υ x / 2 kT dυ x + ∫ υ x e − m0υ x / 2 kT dυ x2  = − + = 0 , (4.22)  2πkT  −∞ 0  2α 2α m where α = 0 . 2kT 1/ 2

Value of a quantity, standing ahead of interral has positive sign (+ υ x ), value (- υ x ) negative sign, degree υ x ungerate, this is why mean observation is equal to zero, it means

 υ x  = υ y  = υ z  = 0 . 116

(4.23)



Accordingly, mean observation of velocity vector υ is equal to  zero  υ  = 0 . Is is by reason of isotropism of considered system , hence shows deficiency of any directions. However, mean ovservation of velocity component module is not equal to zero. In order to evaluate arythmetic mean value of module х – component of velocity, it is necessary to apply the following:

 υ x =

1

∞

n ∫0

υ x f (υ x )dυ x .

(4.24)

Averaging out of all velocities υ x determine components for positive value. This is why, by the integration limit of formula (4.24) integral from 0 to ∞ is taken. Enummerate distribution function by formula (4.19): ∞

1/ 2

 m   υ x  = ∫ υ x n 0  n0  2πkT  1

e − m υ x / 2 kT dυ x . 2

0

(4.25)

Consequently, result of calculation of arithmetic mean value of component module 〈 υ x 〉 is equal to:

 υ x =

2kT

πm0

.

(4.26)

4.4.2. Distribution of molecules velocity modules. Find amount of molecules per unit of volume dn (υ ) , the values of which is on interval υ and υ + dυ , velocity of which is equal



to υ ≡ υ

inside of n molecules. Absolute value of molecules

velocity υ does not depend on direction of motion. 117

Suppose, all molecules n on the unit of volume were located on initial point of descartes system of referecne and its velocity equal to

(

υ = υ x2 + υ 2y + υ z2 . Molecules velocity υ x ,υ y ,υ z

)

changes to

dυ x , dυ y , dυ z . Then, in such period of time, they get into volume dυ x dυ y dυ z (Figure 4.4). Inside of it molecules devided by

( )

verisimilar density nP (υ x )P υ y P (υ z ) . Sketch two sides of the sphere

ΑΑ 1 and ΒΒ1 , which walls off parallelipiped dυ x dυ y dυ z on Figure 4.4. Here molecules velocity is limited by internal radius

υ1 = υ x2 + υ 2y + υ z2 , outer radius (υ + dυ ) is limited by sides of the sphere ( ΑΑ 1 ) and ( ΒΒ1 ) . Volume of this class is equal to

4πυ 2 dυ multuplication. Where 4πυ 2 is content of side ( ΑΑ1 ) , dυ – strata.

Thus, the possibility to calculate molecules number in velocity interval υ and υ + dυ appears. For this it is necessary to multiply volume of spherical class 4πυ 2 dυ on the density of molecules inside of it, in other words:

dn (υ ) = 4πυ 2 f (υ )dυ .

(4.27)

By the formula (4.27), in order to enumerate amount of molecules in spherical class υ and (υ + dυ ), it is necessary to replace

f (υ ) by the formula (4.13):  m  dn(υ ) = n 0  π  2kT  4

3/ 2

υ 2 e − m υ x / 2 kT dυ 2

0

or

dn(υ ) 4  m0  3 / 2 2 − m υ x / 2 kT = n dυ .  υ e n π  2kT  

2

0

118

(4.28)

Fig. 4.4

Equation (4.28) calls distribution law of molecules by the velocity or Maxwell law. This formula determines amount of molecules, which velocity is on interval υ and (υ + dυ ) . Molecular fraction dn is on unit of value on the velocity intern vals υ and (υ + dυ ) . Put it in other way, it determines probability of presence of gas molecules velocity on the intervals υ and

(υ + dυ ) . If denote dn  = f (υ ) . ndυ So, we can write the formula (4.28) :

4  m0  dn   = f (υ ) = ndυ π  2kT 

3/ 2

υ 2e − m υ x / 2 kT . 2

0

(4.29)

This formula is one of the type of distribution Maxwell function. It determines fraction of molecules per unit of volume and on 119

  interval of velocity dυ = 1 . Function f (υ ) starts from zero ( υ = 0 ), 

reach maximum valuation in case υ = υ вер (index designates proba-

bility), when go to infinity υ → ∞ , vanishing has asmptotic form (Figure 4.5). In order to draw a curve, we use the following designation on the Figure 4.5. 

S=υ

m0 , S 2 = υ 2 m0 , y = S 2e − S 2 . 2kT 2kT

()

So then function f υ

will be equal to zero, if υ = 0 and

υ → ∞ . This case explains that there are no molecules which moves with high velocity, fail to move at all, and situated in the state of rest. Velocity module, befit to maximun value of distribution function of freehand curve on the Figure 4.5 calls most probable velocity, indicates as υ вер . It is true, biggest fraction of gaz molecules moves with velocity closed to this υ вер velocity. Molecules which velocities is close to υ вер are usually meet in the gas structure, so the probability that molecules velocity will be close to velocity υ вер is mostly high.

Fig. 4.5

120

Molecules fraction per unit of volume and on velocity interval dυ can be find with a help of dependency diagrams f (υ ) from

υ (Figure 4.6). The  dn  value is equal to the area of the shaded part (of unit)  n  on the Figure 4.6. Altitude of the unit f (υ ) , foundation dυ. Area

()

surrounded by freehand curve f υ

on the Figure 4.6 determines

general amount of molecules. By the equation (4.29), the real type of distribution function depends on gas nature and on temperature. Graphic of velocity distribution function of molecules in different temperature shown on Figure 4.7. It can be observes that all freehand curve displaced to the right, in other woeds it displaces to the high velocity.

Fig. 4.6

Fig. 4.7

121

However, freehand curve f (υ ) and area, surrounded velocity axis does not change. By the condition of normalizing (formula (4.4.)): ∞

∫ f (υ )dυ = 1 0

this is why, with increasing of temperature the maximum of freehand curve f (υ ) decreases. The last condition of normalization can be explained by the following way: certainly there is one molecules velocity value on the interval 0 and ∞ , in other words, determine probability of assumption by molecules one of velocity on this interval. Then this is a probability of existance of a real event, that is why it equals to zero. 4.5. Average velocity of molecules Average valuation of molecular velocity of any function φ (υ ) for the gas in the Maxwell state is determined as follows:

nφ = ∫ φf (υ )dυ = φ =

1

n∫

φf (υ )dυ .

(4.30)

4.5.1. Average arithmetical velocity Substitute the calculate value from the formula (4.29) instead of distribution function from the formula (4.30), take into account that φ = υ and integrate it to the possible velocities between 0 and ∞, by this way we find arithmetical mean value of velocity  υ  : ∞  1 4n  m0   υ  = ∫ υf (υ )dυ =   n0 π  2kT 

Where m0 = α , signify 2kT

3/ 2 ∞

∫υe

3 − m0υ 2 / 2 kT

dυ .

0

∞

υ = х , then ∫υ e 0

3 − m0υ 2 / 2 kT

∞

dυ = ∫ x3e−αx dx . 2

0

Find valuation of such integral, by using the following rules: 122

a) If r – even number, then ∞

∫x e r

−αx 2

π 1 3

dx =

α ⋅ ⋅ ⋅⋅⋅ 2 2 2

2

0

r −1

 r +1  −   2 

;

(4.31)

b) If r – odd number, then ∞

1 2

r −αx ∫ x e dx = α 2

0

 r +1  −   2 

 r −1 . ⋅ !  2 

(4.32)

By the formula (4.32) −2

∞

3 −m υ ∫υ e 0

2

/ 2 kT

1  m0   . 2  2kT 

dυ = 

0

(4.33)

If we add to the formula (4.31) valuation of integral from the formula (4.33) and make calculation, so we find the following:  υ =

4  m0    π  2kT 

3/ 2

−2

1 m  4 ⋅  0  = 2  2kT  π

 m0     2kT 

3/ 2

2

1  2kT  2   = 2  m0  π

2kT 8kT , = πm0 m0

it means 8kT

 υ =

πm0

or  υ =

8 N A kT

πN Am0

=

8 RT ,

πM

(4.34)

where M – molal mass, m0 – molecular mass, N A – Avogadro constant, Т – temperature, R – universial gas constant. Equation (4.34) determines an arithmetical mean value of molecule velocity. Conclusion can be made that arithmetical mean value of gas molecule velocity depends only on temperature. 123

If comapre arithmetical mean value of velocity υ x with average valuation of velocity component by the module (formula (4.26) ), then arithmetical mean value will be twice as little from υ , in other words,

υx =

υ

2

.

(4.35)

4.5.2. Mean square velocity of molecules Make calculation similar to evaluation of arithmetical mean value and determine average square velocity 〈υ 2 〉 :

 υ 2 =

∞

1 2 4  m0  υ f (υ )dυ =   ∫ n0 π  2kT 

3/ 2∞

2 4 − m0υ / 2 kT

∫υ e

dυ ,

(4.36)

0

where ∞

∞

4 − mυ ∫υ e 0

2

/ 2 kT

π 1 3  m0  ⋅ ⋅   2 2 2  2kT 

dυ = ∫ x4e −αx dx = 2

0

−5 / 2

.

Because of that average square velocity of molecules will be equal:

4  m0   υ =   π  2kT  2

3/ 2

π 3  2kT   ⋅  2 4  m0 

5/ 2

3 2kT 3kT , = = ⋅ 2 m0 m0

(4.37)

in other words

υ =  υ2  = υ2 =

3kT

m

.

(4.38)

0

This velocity is not equal to arithmetical mean value:

 3π   = 1,086 υ .  8 

υ = υ 2 =υ 

124

(4.39)

4.5.3. Mean kinetic energy of molecules. Definition of temperature Mean kinetic energy of molecules defines with help of velocity. Mean kinetic energy defines this way: m0υ 2

∞

1 mυ2 4  m0  〈 〉 = ∫ 0 f (υ )dυ =   n0 2 2 π  2kT 

3 ∞ 2

∫ 0

m0υ 2 2

υ e 2

−

m0υ 2 2 kT

3 2

dυ = kT .

In other words, 

m0υ 2 3  = kT . 2 2

(4.40)

From the formula (4.40) it can be seen, that temperature defines direct by absolute velocity of molecules and it proportional to average kinetic energy of molecules. Precise meaning of average kinetic energy defines average square velocity. It follows from it that temperature which is macroscopic characteristic of gas designated with regards to its microscopic characteristic. Temperature is connected with medium energy of thermal motion of molecules, in other words it is indicator of average kinetic energy of gas molecules.

4.5.4. The most probable velocity Valuation of molecules ∆N classified alongside of velocity conformed to largest observation of distributional function is the biggest one. It was said above (chapter 4.5.3, Figure 4.3). Then this velocity defines by the biggest valuation of distribution curve of Maxwell and it calls the most probable velocity. In order to find the most probable velocity, we set to zero by velocity first derivative of distribution function and find υвер' : d  4  m0  d 2 −m υ f (υ ) =  υ e   dυ  π  2kT  dυ 3/ 2

0

125

2

/ 2 kT

  = 0. 

This congruence fulfilled, if

(

d 2 −m υ υ e dυ

2

/ 2 kT

)= 0.

2

/ 2 kT

⋅ 2υ

0

(4.41)

Distinguish equation (4.41):

2υe − m0υ

2

/ 2 kT

− υ 2e − m0υ

m0 =0 2kT

or

2υe − m0υ

2

/ 2 kT

 m0υ 2   = 0 . 1 − 2 kT  

(4.42)

Conditions, meeting congruence (4.42) or υ = 0 , or υ = ∞ and  m0υ 2  1 −  = 0 . Two previous conditions do not conform with maxi2 kT   mum of distributional curve. This is why appears valuation of velocity υ prob' , which turns valuation in the brackets to zero:

1−

m0υ 2 = 0 , from this 2kT 2kT υ prob = mo

(4.44)

accordingly m0υ prob

2

2

= kT .

(4.45)

Due to it kinetic molecular energy, relevant to most probable velocity is equal to kT . By comparing formulas ((4.34), (4.38) and 4.44), designating average velocity we find link between this three velocities: υ =υ

3π = 1,086υ = 8

126

3 ⋅υ prob = 1,225υ prob 2

1/ 2

 2kT  3 2kT  ⋅ = 1,2247 2 m0  m0 

υ  = 2

 υ =

2

2kT

π

m0

1/ 2

 2kT   = 1,2183  m0 

.

(4.46)

Average velocity of all velocities is in proportion to (2kT / m0 )1 / 2 , but difference between each others and with most probable velocity are unlike only by approximate numerical multiplier. In calculations, applying mean velocity, taking into account all mentioned above, it is necessary to choose correct average velocity. For example, for the correct determination of average kinetic energy of molecules it is necessary to use average square velocity (equation 4.38), hence and for determination of correct temperature also use υ 2 average velocity.

Numerical valuation of estimated average velocity is remarkable large. For example, in temperature 273 К, arithmetical average velocity of hydrogen molecules  υ  = 1270 м / с , and for oxigen in the same temperature it reach  υ  = 425 м / с . Earlier big valuation of molecules velocity invoked judjement against kinetic theory, as far as observable diffusion velocity is very small. But in this case direction of molecular motion in space on every direction is chaotic and any molecule collides with other molecules continual, by this reason its macroscopic velocity along the gas is very small. It was mentioned, that molecules distribution law by velocity and consequence resulting from it, performs only in cases when gas is in the state of rest. Maxwell law is substantiated by the atomic-molecular example and rest against jungements of theory of probability, in other words use statistical illustration and description method of large particles amount. Then Maxwell law is statistical law, due to it, disposal of molecules, reilized by large amount of microscopic state is mostly probable allocation and strict correspond to the state of equilibrium of gas molecules. This is why, statistical common factors mostly accurate acomplished in the system with big amount of particles. The less amount 127

of particles in the system, the more distribution law inflects from law of probable maxwell distribution. Because of that, results of this theory will be correct for all numbers of particles N → ∞ , volume of system V → ∞ . In this case ratio N / V does not change and stays permanent. Such displacement of limit calls thermodynamical extremum replacement.

4.6. Amount of molecules collided with walls of capacities It is possible to estimate amount of molecules, collided with bonds of the capacity, applying Maxwell distribution function. Molecules moves perpendicularly in three directions ( x, y, z) . Let’s calculate amount of molecules, collided with bonds of x -standards with unit area dS x , in the time of dt . All velocities of molecules x -component coming to element

dS x of the bond x in the time dt

reside in interval υ x and υ x + dυ x (Figure 4.9).   Volume of molecules υ dtn ds coming up to square





υ , in the

time of dt must be inside of parallelepiped. Where υ dt - is length of parallelepiped, ds - square. Amount of molecules, velocity which    is on interval υ and υ + dυ is equal to:

dn (υ x ,υ y ,υ z ) = f (υ x ,υ y ,υ z )dυ x dυ y dυ z . By calculating molecules amount the collision of molecules between each others is not taken into consideration. Thus, general amount of collided molecules is z per unit of time dt = 1 , and per unit of square ds = 1 it can be determined as follows: +∞+∞ +∞

z=

 m0  ∫0 −∫∞−∫∞υ x f (υ x ,υ y ,υ z )dυ xdυ ydυ z = n 2πkT 

3 / 2 +∞+∞ +∞

128

∫ ∫ ∫υ e x

0 − ∞− ∞

(

)

− m0 υ x2 + υ y2 + υ z2 / 2 kT

dυ x dυ y dυ z . (4.47)

where integral limit by υ y and υ z from (− ∞ ) to (+ ∞ ) , for υ x

integral limit containes only between 0 и ∞ .

Fig. 4.9 ∞

∫e

− m0υ y2 / 2 kT

dυ y and

−∞

∞

∫e

− m0υ z2 / 2 kT

dυ z .

−∞

If we specify α = m0 / 2kT , υ y = υ z = y , then

∞

−αy ∫ e dy . Its 2

−∞

valuation is equal to ∞

π . Then intergal is written differently:

∫υ xe

− mυ y2 / 2 kT

1 m   2  2kT 

dυ x = 

0

−1

.

Thus, total amount of collide molecules z determines by the following way:  m  z = n   2πkT 

3/ 2

( )2 ⋅ 12 ⋅ 2mkT =

⋅ π

n 1 2kT n 8kT 1 = = n  υ  . (4.48) π 2 m 4 πm 4 129

In other words

z=

1 n〈υ 〉 . 4

(4.49)

Equation 4.49 determines amount of collision z in one second and one unit of volume. It is also possible to determine amount of collide molecules in time ∆t and on square ∆S by applying average valuation of module x -component (formula 4.35). If there is molecule on the unit of capacity volume, then according to Figure 4.9 base area ∆S , length υ x dt , in other words in parallelepiped with volume υ x dSdt there are υ x dSdt of molecules. Gas is in state of

equilibrium. This is why from left to write in time of ∆t on the area dS falls 1 nυ x dSdt of molucules, it means they are collide. 2 From this, on the unit of time the amount of molecules that fall 1 on unit of area is equal to nυ x . If substitute υ x to average valua2 tion of module υ x = 1 〈υ 〉 , then it is possible to determing average 2 amount of molecules collision per unit of time:

1 2

1 4

z = nυ x = n  υ  .

(4.50)

Thenk equations (4.49) and (4.50) are confotmed. 4.7. Maxwell’s distribution for the relative velocities. Unmeasured type of Maxwell’s distribution For the solution of many tasks, it is more suitable to write the formula (4.28), finding the function of Maxwell distribution with a help of relative velocities. Then velocity of molecules represents throught out relative velocity, in the capacity of velocity unit the most probable velocity υ prob is taken. This is why relative velocity determines this way: 130

u = υ / υ prob ,

(4.51)

where υ – velocity of molecules, and υ prob = 2kT / m 0 – relative velocity in stable terperature T. In the Maxwell formula 4  m0  dn   = f (υ ) = ndυ π  2kT 

3/ 2

υ 2e − m υ x / 2 kT . 2

0

Value is m0 / 2kT , which is two times We will replace this value m0 / 2kT with an equal relation 2 1 / υ prob and considering (4.51), it can be written as follows:

4 −u 2 dn e u . = ndu π 2

(4.52)

This equation is Maxwell’s distribution for the relative velocities, in other words, it gives unmeasured type of Maxwell formula. Ratio (4.52) – is universal, it means scalene. Distribution in this type does not depend on gas nature, and on temperature. For the compose of molecules velocity the similar formula can be written. If take x -component of velocity, then relative velocity is equal to: u x = υ x / υ prob .

(4.53)

Thus, by the formula (4.18) unmeasured type of distribution function for the generated molecules is written by the following way:

dn 1 −u x = e . ndυ π 2

(4.54)

Figure (4.10) shows distribution function for the relative velocities. 131

Fig. 4.10

Valuation of equations (4.52) and (4.54) can be calculated prelimiary for the different valuations u and u x . In the table 4.1 were drawn meanings of these functions. For example, it is necessary to find a fraction dn of nitrogen n molecules in the temperature Т=300 К and with velocities 275 m/s and 276 m/s. In order to solve this task, it is necessary to use equation (4.52). From this we find dn . As a first step we calculate mostly propable

n velocity of nitrogen velocity in the 300К temperature: υвер =

2 N 0 kT 2kT 2 RT 2 ⋅ 8 ⋅ 31 ⋅10330 ⋅10 2 = = = ≈ 394 м/с. m N0 К M 28

Then relative velocity u is equal to: u=

du =

dυ

275 υ = ≈ 0,70 . υ prob 394

, it defines from u =

υ

. In statement of problem υ prob υ prob valuation of velocities interval is very small 132

∆υ = (276 − 275) м / с = 1 м / с , because of that du =

∆υ

υ prob

=

1 ≈ 0,0025 . 394

From the table 4.1 we find valuation of distribution function conformed with relative velocity u = 0,70

dn 4 (0,7 )2 e −(0,7 ) = 0,677 . = ndu π 2

From this

dn = 0,677 ⋅ 0,0025 = 1,7 ⋅ 10 −3 = 1,7 ⋅ 10 −3 ⋅ 10 2 % = 0,17% . n As a resut, we estimate fraction of molecules, where indicated velocity will be more then certain velocities. For the solving of such tasks it is favorable to use formula (4.52). We find fraction of molecules, where velocity is more then υ , it means more then indicated u , integrating equation (4.52) dn 4 −u 2 2 e u du . Bounds of intervals are found from the =

π

ndu

indicated velocity u indefinitely. Then formula defining fraction of molecules

n u is written by the following way: n Table 1

u , ux 1 0,1 0,2 0,3 0,4 0,5

f (u ) =

4 2 −u 2 ue π

2 0,022 0,087 0,185 0,308 0,439

f (u x ) =

1

π

eu

2 −u 2

e

3 0,558 0,542 0,515 0,480 0,439

u , ux 4 1,2 1,3 1,4 1,5 1,6

133

f (u ) =

4 2 −u 2 ue π

5 0,770 0,703 0,623 0,535 0,447

f (u x ) =

1

π

e

u 2e −u

6 0,133 0,104 0,079 0,059 0,044

2

1 0,6 0,7 0,8 0,9 1,0 1,1

2 0,567 0,677 0,761 0,813 0,830 0,814

3 0,393 0,345 0,297 0,251 0,208 0,168

n u 4 = n π

4 1,7 1,8 2,0 2,2 2,4 3,0 ∞

∫z

2 − z2

e

5 0,362 0,286 0,165 0,086 0,041 0,003

6 0,031 0,022 0,010 0,004 0,002 -

dz ,

(4.55)

u

where z = u , dz = du , n  u - amount of molecules, where relative velocity is more then indicated velocity u . For calculating it is necessary to find integral from the formula (4.55). In the table 4.1 valuation of this integral defines for different values. For example, define fraction of molecules, moving with velocity more than mostly probaple velocity, it means υ  υ вер , then

υ = 1 , valuation of integral, conformed to it is 0,5724 or 57,24% . υ ык Amount of gas molecules, moving with velocity υ  υ вер is 57,24% . u=

Table 2

0,1 0,2 0,3 0,4 0,5 0,6 0,7

2 4 z 2e − z dz π ∫u

0,9992 0,9941 0,9807 0,9582 0,9190 0,8685 0,8061

∞

∞

∞

u

u 0,8 0,9 1,0 1,1 1,2 1,3 1,4

2 4 z 2e − z dz ∫ π u

0,7340 0,6550 0,5724 0,4900 0,4105 0,3370 0,2702

u 1,5 1,6 1,7 1,8 1,9 2,0

2 4 z 2e − z dz ∫ π u

0,2123 0,1632 0,1230 0,0905 0,0602 0,0460

It means, it will not be symmetrical to distribution curve of Maxwell, it is apparent from graphics of Figures (Figures 4.6; 4.7; 4.10) 134

4.8. Dimension of average velocity of molecular motion. Experiment of Shtern. First who showed by experiment velocity of gas molecules was O. Shern in 1920. On Figure 4.11 experimental design of Shtern is shown, intended to determination of molecules velocity. In this experiment velocity of atoms of the silver is researched. As the capacity of atoms facility platinum wire covered by silver ouside were used. Wire L was surrounded by two hat orifices; in these backstops two thin gaps S1 and

S2 were made. Wire L and gaps S1 and S2 are located on equal vertical subspace, all locate inside of cylinder D . Opposite gap S2 , on

the internal surface of cylinder D thin object В that is made from brasses was established. Capaccity D can roll on the axis, passing through wire L completeness and inside of it the air is so much thin out. Electricity was transmitted, silver was melted and when heat up to temperature (1235 К) in which it will be dissolved , concentrated atom of silver started to fly on all directions. Directed on the lines, passing through the gaps S1 and S2 flyover very thin groups of atoms, which were not collided with each others. Such directed atoms or molecules flow calls molecular groups.

Fig. 4.11

If cylinder D does not move, then molecular group stops on the point A of the object on the Figure 4.11., on the object remains thin 135

trace as on the Figure of gap S2 . If the capacity D will be rolled, then molecular group will not be able to reach the point A and stop on point A′ , displacement δ = AA′ between points A and A′ can be correlated with average velocity of molecules inside the group. Atoms, that moves with υ velocity pass the r distance from the gap

r . However each point of υ the circumvolved capacity D for the time t displaced on the distance δ = 2π nR . Where n is number of cylinder D rotation per one second, R – radius of cylinder (Figure 4.1). By using of these two S2 up to the object for the equal time t =

equations, find average velocity of atoms of the wire on the estimated temperature: 2πnRr . (4.56) υ=

δ

From the experiment it is possible to determine displacement δ through measurement of traces of gaps S1 and S2 , determine angular velocity through radiuses of capacity R , r and then determine average velocity of molecules. Subsequent to the results of measurements, made for the atoms of silver by method of Shtern, velocity value converge with values, determined with help of kinetic theory of gases. From experiment it was developed that average velocity of atoms of silver is equal to 650 m/s. By changing of current strength, passing through wires and changing temperature, it is possible to search dependence of molecules velocity from temperature. Average velocity of molecules, defined from experiment of Shtern were proportional to square root of temperature, in other words υ ~ Τ . Experiments of Shtern denoted link between average velocity of molecules and temperature and verified that atoms within the group are followed to the law of distribution of maxwell on the velocity. In 1921 M. Bron and E. Berman suggested method of measurement of average length λ of molecules free movement. In this measurement method was used groups of atoms of silver. Method of molecular groups are still applied in research of different properties of microparticles. 136

5 NONIDEAL GASES 5.1. Deviation of gas properties from ideality Boyle-Mariotte’s law, Gay-Lussac’s law, Charles law describe only the condition of fairly sparse and ideal gas. Nonideal gases are not governed by ideal gases law, but under low density and chemical reactions their properties are roughly similar with ideal gases. It is not correct to consider molecules in gases as material points. Intercation between molecules means, that there is an existence of attractive force and repulsive force, therefore the condition of nonideal gas changes into idal gas equation: pV = RT . Deviation of gas properties may be shown through the coefficient of compressibility Z:

Z=

pv , RT

(5.1)

where v – volume mole, Т – temperature, p – pressure, R – universal gas constant. The coefficent of compressibility of ideal gas Z=1. Usually gas coefficient of compressibility is below one under high temperature and pressure. In order to determine the liquid compressibility coefficient it is necessary to use equation (5.1), but their values are significantly lower than one. In the experiment of gas (liquid) compressibility coefficient the specified below temperature τ and stress dependence π are studied:

Z = f (τ ,π ) ,

(5.2)

where, τ = T , π = р , ТК – criticaltemperature, pK – critical pressure, T р к к Т – temperature, р – pressure. Latertheconcept of such critical parametersas ТК and рК will be presented. 137

The production of ideal gas volume and tension under certain temerature does not change due to pressure, it remains constant (figure 5.1 dashed line). p-V-T based on experiment results the dependence of real gas is complicated pV = f ( p ) . Experimental Nitrogen isotherms under temperature of 00С are shown with uninterrupted line. As the temperature rises the deviations are observed.

Fig. 5.1

The deviation appears as the pressure increases. Deviations from idal gas law can be traced in all reall gases, besides that, there are differences peculiar to them because molecules interaction and their sizes are different. Let us review the influence of gas molecules volume in the following example. Let us designate gas molecules diameter as hard sphere σ . The volume V of these billiard-like molecules is located in box. Then the center of molecule, that is in the box, can move 3

1 under volume of V 3 − σ  , because it cannot be closer to the  

vessel wall than 1 σ (fig. 5.2).

2

The movement of the second molecule center, located near the first molecule and comes near to the volume 3  13 3  4 V − σ  − 3 πσ  .  

138

Fig. 5.2

The second molecule cannot approach closer than the dicstance of σ (Fig. 5.2). This process can tbe observed until molecule N is located in volume V. In the event of calculation of average volume of any molecule center, it will be equal to: 3 V 13 − σ  − N 4 πσ 3 ≈ V − 2 πNσ 3 .   2 3 3

(5.3)

The size (volume) of the box, the size of molecule is signgificantly bigger than σ ( 5.3) 2 3

b = πNσ 3 .

(5.4)

Vouminal molecules have «own volume». It is equal to quadruple size of sphere molecules b. It is seen from the equation (5.4), that it does not depend on the temperature b, and that is has constant size and it is determined with the use of gas. Let us calculated the volume of molecules without taking into account any of the mentioned above at all. For example, normally gas volume is equal to 1 cm3 , n = 2,68 ⋅ 1019 cm-3 (Loschmidt 139

number) of molecule location. Diameter of molecule is apporximately equal to 3 ⋅ 10 −8 cm. Then own volume of all molecules will equal to:

V=n

π 6

d3 =

2,68 ⋅ 1019 ⋅ 3,14 ⋅ 33 ⋅ 10 −24 3 = 3,2 ⋅ 10 − 4 сm . 6

Intermolecular forces were not taken into aacount in this example. It is seen from experiment, that own volume of molecules in one thousandth atmospheric pressure is equal to one third of gas volume. Therefore, if gas pressure was a ten fold rate of the atmoshpere, then deviation from Boyle-Mariotte law would have been significant. Relation between real gas parameters do not depend on ideal gas equation under Mendeleev-Clapeyron law. Functional relation between ideal gas equation, pressure and temeprature is more complicated than the equation of Mendeleev-Clapeyron. For the first time, the real gas equation was presented in 1873 by van der Waals the scientist from Netherlands. Suggested equation of van der Waals was supposed to describe gaseousness of object and liquid. Basis foundation for molecular-kinetic state of gaseous and liquid objects starts from the van der Waals equation.

5.2. Experimental Andrew’s isotherms. Analysis of real gases isotherms. Gaseous state is one of the forms of objects aggregative state. Under high pressure and low temperature gas condensates (from Latin language condensatio – thicken, harden, consolidate) and transforms into liquid state. This phase conversion must not be according to ideal gas equation. Liquid can be transformed into gas, and gas into liquid, which means that intertransformation of these two states are known to us from experiments. Gas transforms into liquid state, and liquid may evaporate. Simple similarity in molecule structure is enough. Their molecules are in chaotic thermal motions. Molecules interact with each other, and their interaction forces are 140

equal. The distance of gas molecules are much bigger than their own sizes, and it is otherwise in luquids, the distance is smallm but the sizes of molecules are same. Interaction power of tightly located molecules in liquid is very strong, therefore intermoleculare forces are very improtant in the proccess of liquid phisycal properties study. Interrelation between properties of liquids and gases, for the first time was studied in experiments by English scientist T. Andrews in XIX century. (1813-1885) Van der Waals analysed experimental studies of Andrews and issued theoretical description.

Fig. 5.3

Andrews observed, that the properties of carbon dioxide СО2 are approximately similar to liquid properties under pressure increase and temperature decrease. Andrews’ values for carbon dioxide isotherm СО2 is shown in figure 5.3. If the initial state А of gas in diagram p − V (fig. 5.4) is compressed, the curved line of isotherm in this process that shows gas state rises up to the point B. АВ also as a part, does not subject pressure to change. Beginning from point В, the parallel passes to isotherm axis ОV , which means that deacrease of pressure does not lead to pressure increase. In the 141

part of isotherm DВ pressure remains stable. First product

 ∂p  in  ∂V T

this case is discontinue d in point B.

Fig.5.4

Observation of isothermal horizontal Andrews related with physical processes, taking place in molecular system. The object in part АВ is in homogeneous state; the object that is located at any point of this part remains in gaseous state. Beginning with point В, homogeneity of the object is interrupted, in addition to gaseous state another additional state appears in the form of liquid. At any point of the isotherm horizontal (fig. 5.4) the object meets in two different aggregative states at the same tim or in two different states: liquid and gaseous. These two phases are in balance. So, beginning from the point В gas becomes liquid or condesation begins. One part of the compressed object transforms from gaseous stat into liquid, therefore liquid phase is observed in the system. Further compression leads to the loss of gas masses to liquid masses increase. At any point of isotherm part, for example at point C, which is located between points В and D, the object goes through two phases simultaneously. At point D the process of gas condensation ceases, all gases are condensated, and the object again transforms into homogeneous state, namely, into liquid state phase. Part of isotherm DЕ is similar to liquid state of the object. Further compression of the object will lead to another pressure increase. 142

Thus, isotherm АЕ (fig. 5.4) consists of three different parts, two of them АВ and DE designate one-phase state (АВ – gaseous state, DE – liquid state), and horizontal part BD is similar to two-phase state of the object. In BD part the pressre and volume of the object in corresponding vessel is liquid, and in remaining part there is gas (steam) and the balance between two phases is observed. Gas which is in state of balance with liquid is called saturated steam. Pressure, corresponding to BD horizontal part, is called saturated gas pressure, ususally it is called saturated vapor pressure when it is under certain temperature. If gas again is taken as initial state of the object and its temperature is higher than point A, then isothem will pass through the point A. But, this isotherm will be located higher and part of horizontal will be shorter. With the temperature increase parts of isotherm horizontal shrink and turn into certain point (fig. 5.3), the length of horizontal part will be equal to zero. In such event the difference between liquid and saturated gas disappears, therefore, the difference between gas and liquid disappears. This isotherm is called critical or critical isotherm, and its appropriate temperature, pressure and volume – critical temperature ТК, critical pressure рК, critical volume Vк or critical parameters. Based on the temperature which is higher than critical temperature there is no part of straight line in the middle of isotherm, they cross lines similar to uniform hyperbolae (fig. 5.3). The absence of straight line part in isotherm, located higher than critical temperature mean than these states of the object do not divide into two phases; these states can be observed only in one phase. So, in the event of T < TK , the liquid or gaseous phase of the object is possible. And in the event of T > TK , only one phase of the object is possible. Accordingly, liquid state of the object can be observed only under critical temperature. At first gas should be cooled down from critical to low temperature, then it can be transformed into liquid state. After the mentioned above, critical isotherm can be called as natural border, which divides territory of homogeneous state ( T > TK ) from the territory of two phase state ( T < TK ), in accordance with critical temperature. 143

Based on the results of experiments, at critical point

 ∂2 p   ∂p    = 0 and  2  = 0 .  ∂V T  ∂V T

(5.5)

According to conditions (5.5) critical parameters can be determined from real gas equation.

5.3. Forces and potential of intermolecular interaction Transformation of real gases into liquid assumes, that there is an existence of attractive force between molecules at long distance. Liquids resistance to compression means the existence of repulsive force, which change with distance and has impact when close. Force interaction F between two sphere molecules will constitute intermolecular distance of function r. In the process of intemolecular interaction study it is better to use potential energy of interaction ϕ (r ) or potential. These two F (r ) and ϕ (r ) functions are used in simple relation: ∞ dϕ , (5.6) ϕ (r ) = ∫ F (r )dr . F (r ) = − dr r The nature of intermolecular forces are revealed in results of experimental and theoretical research. Functional type of interaction potential is suggested by theory, and experiment results are used in order to determine parameters of intermolecular interaction potential. For example, viscosity of sparse gases are calculated using intermolecular potentioal ϕ (r ) . Then, viscosity factor may be used in order to determine dependence and parameters of potential from temperature. Intermolecular forces are dividied into two groups: action force from afar and short action. Such grouping is related to estimation of intermolecular forces. Atom or molecule is viewed as stationary core system, surrounded by the cloud with negative charge. Distribution 144

of charges is subjected to quantum mechanics laws. If distribution of charges is known, intermolecular forces are calculated based on classic electrostatics. If two molecules are located at the distance, that do not cross with electric cloud, then the action force from afar or attractive force is observed between them. Attraction forces are inversely proportional to different degree of intermolecular distance and have negative sign: A (5.7) F (r ) = − a , r where А – constant number, r – distance between centers of shere molecules. In the event of molecules approach, their electric clouds touch each other and they may hide each other, in this case short interaction forces or repulsive forces are observed. These repulsive forces are called valence forces or chemical forces. Change of potential interaction energy of two molecules is show in figure 5.5. Along abscissa axis the distance between molecules centers – is r, and along ordinate axis the potential interaction energy – is ϕ (r ) . In line 5,5 line ϕ (r ) at the distance r quickly decreases and at point r = rm there will be minimum.

Fig. 5.5

145

The depth of potential dimple ε in accordance with overstability, determines binding energy of molecules. r = σ – is short distance between stationary centers of molecules. Molecules collide with each other, their intermolecular force is equal to zero. If r > σ , then there is influence of attractive forces between molecules. These forces are weak and quickly weaken with distance. If r < σ , then there is influence of repulsive forces between molecules. In figure 5.5 it is shown how the right side of the line ϕ (r ) at the distance r rapidly increases

∂ϕ

∂r

, consequently, values of pro-

duct r < σ will be large at a distance. Due to these two collisions strong repulsive force is present between molecules. For strong sphere molecules under r < σ , ϕ (r ) tends to infinity. Repulsive forces in various extent of intermolecular distance change in inverse proportion and have positive sign:

F (r )т ебу =

В , rв

(5.8)

but а σ .

(5.10)

5.4.2. Soft spheres (point of replstion center) Hard spheres models have modifications. Gases are well described under high temperature. Graphic form of potential is show in figure 5.7, and mathematical species:

ϕ (r ) = K

δ

σ  ,  δ = ε r r

Fig. 5.7

147

(5.11)

whare δ – repulsion indicator, value is located between 9 and 15; if δ = 4 , then such molecules are called Maxwell molecules.

5.4.3. Square-well potential

Fig. 5.8

Based on this model diameters σ , thickness of hard spherical molecules ε are surrounded by repulsive field; field is stretched along the distance Rσ (fig. 5.8). Potential in addition to repulsive forces takes into account attractive force as well. Mathematical species: ϕ (r ) = ∞ , if r < σ ,

ϕ (r ) = −ε , if σ < r < Rσ , ϕ (r ) = 0 , if r > Rσ .

(5.12)

5.4.4. Sutherland potential This potential describes diameters σ and hard spherical molecules, which change in inverse proportion to attractive force of different degree of distance. (fig. 5.9). Easy to use actual model. Mathematical species: 148

ϕ (r ) = ∞ , if r < σ ,

Fig. 5.9

Fig. 5.10

ϕ (r ) = −cr −γ , if r > σ .

(5.13)

5.4.5. Lennard-Jones Potential Widely used potential for kinetic theory problem solution, also takes into account attractive and repulsive forces (fig. 5.10). Mathematical species:  σ 12  σ 6  (5.14) ϕ (r ) = 4ε   −    ,  r    r  6 where they have are related to sum of repulsive forces  σ  ;  σ  to r  r sum of attractive forces; σ – at distance it changes sign of potential

12

function; ε – the lowest number (minimal) of potential distance.

ϕ (r ) = −ε , if r = r0 ,

ϕ (r ) = ∞ , if r < σ , σ  ϕ (r ) = −ε   , if r  6

r > r0 , ϕ (r ) = 0 , if r → ∞ . 149

1

r = 2 6 σ at (5.15)

5.5. Van der Waals’ equation There are few methods for solution of gas state equation, namely equation which connects state parameters, volume, pressure and temperature with functional dependence. Real gas equation is calculated by making adjustments in ideal gas equation. MendeleevClappeyron

pV = RT

(5.16)

does not take into account forces of influence between molecules volume and molecules themselves. In fact, in this equation when the pressure is very high gas volume may be too little, which means that there is no barriers in unlimited gas compression. But, it is different with real gases because even with little number of gas molecules there will be certain volume anyway. Taking into account repulsive forces between molecules. In formula (5.16) gas volume V if vessel is considered. In this volume of gas molecule move without disturbing each other (considered as material point, molecule of ideal gas do not collide with each other), in volume, namely in volume V molecules move freely. If molecules of real gas are place into this vessel, due to its own molecule size their movement will become not free. Because, there must be some volume which is not located near the gas molecules. This volume is designated as (section 5.1, formula (5.4))) b . Volume of freely moving molecules of real gas will be less than the value of b from toal volume of the vessel V . Therefore, instead of volume V in ideal gas equation (5.16) it is necessary to use value (V − b) , in other words.

p(V − b) = RT .

(5.17)

In equation (5.17) it is suggested that in the process of calculation b pair collision occurs between interacting molecules N. Straightened pressure b in case of endless increase will be equal to 150

final volume of gas. In fact, if equation (5.17) is written in following way:

p=

RT , V −b

(5.18)

then V = b , if p = ∞ . Taking into account the influence of own volume of molecules ((equation 5.17)), it is possible consider their repulsive forces during interaction as well. Even at the distance of repulsive forces of molecules (sections 5.3 and 5.4) r < σ , approaching distance will not be equal to zero. Therefore, consideration of molecules sizes is viewed as description of replulsive forces. Molecules are assumed to be as elastic balls. At the moment of collision balls may deformed, in that event their exterior (or volume) changes due to forces impact. Kinetic energy of collided balls is spent for work, which is carried out against deformation forces and transforms into potential energy of deformed object. After collision the released potential energy transforms into kinetic energy of flied apart molecules. Identifiable way b is located on 5.1. Let us assume that diameter σ two ballshaped molecules tough each other. For example, center of molecule В cannot enter into the filed with radius σ , which draws round line around the middle of molecule A (fig. 5.11). Also, the center of molecule A cannot enter into the field with radius σ , which draws around the center of molecule В (fig. 5.11). Therefore, two molecules with volume which is 8 times larger than the volume of one molecule, and one molecule with volume which is 5 times larger than the volume cannot cut them through. For one it is gas mole

2 3

V = N A πσ 3 = 4 N Ab0 , where b0 =

π 6

(5.19)

σ 3 – own volume of molecule. From this it is seen

that straighten molecule b is equal to four fold of own size. Unit of constant b is m3/mole. 151

Accounting of attractive forces between molecules. Attractive forces between molecules affect the pressure that was put on the walls of gas vessel. Vessel is located close to walls and moving from toward it from neighboring molecules to molecule A making resulting force. This force is directed inside gas. As result of it molecules A collide with vessel walls with weak strength. Because of this with all other being equal, real gas pressure is less than ideal gas to the extent p to the quantity ∆p . Pressure ∆p is directed inside gas, in consequence of which it may be called internal gas pressure. In this case instead of ideal gas shown in table (5.16) it is necessary to put compound p + ∆p :

p=

RT − ∆p V −b

or

p + ∆p =

RT . V −b

(5.20)

Fig. 5.11

Attractive force enables molecules attraction. External forces affect in the same way. Addition ∆p – is a reason of pressure and intermolecular attraction occorrence. Therefore the pressure, that is put on the vessel surface is equal to the force put on all molecules. Value of affecting force result is directed inside the gas. Because of it, this force is in 152

proportion to numeric value n. On the other hand, attractive force and number of affecting molecules are also in proportion to density of n . Therefore additional pressure

∆p ~ n 2 .

(5.21)

Numeric density of molecules V is in inverse proportion to volume mole, in other words

n~

1

V2

.

Thus

∆p ~

1

V

2

, in other words

∆p =

a , V2

(5.22)

Where а – proportional coefficent, numeric value depends on attractive force between molecules. Accordingly, if value that was found through ∆p (5.22) is put in formula (5.20), real gas equation may written in the following way:

α    P + 2 (V − b) = RT . V  

(5.23)

This example connects pressure, volume and temperature and is called real gas equation, determined per one mole. This equation is also called Van der Waals’ equation, where constant coefficients α and b are called Van der Waals’ constants. In Van der Waals equation attractive forces were taken into account (

a – straightening V2

part) as well as repulsive forces ( b – straightener). Relying on molecular-kinetic views, basic properties of gas were taken into account for the first time in this example. Measurement unit of the constant b corresponds to volume, which is m3/mole, and measurement unit α corresponds to pressure unit, therefore in SI system Pa·m6·m-2, in other words Н ⋅ м 4 . For different gases values of constants α and b will be different. 153

Equation (5.23) that was written in accordance with one mole of gas. For any other mass the equation will as follows:  m2 a  m  m  p + 2 2 V − b  = RT , M  M M V  

(5.24)

where m – gas weight, М – molar mass, V – gas volume. 5.6. Van der Waals’ isotherms Having open parantheses of Van der Waals’ equation (5.23), afte simple transformations it is possible to come to the following form:

pV 3 − (bp + RT )V 2 + aV − ab = 0 . From bother sides it has to be divided by p , this equation shall be written like this:

 ab RT  2 a V + V − =0 V 3 −  bp + p p p  

(5.25)

For the volume V this equation is raised to the third power, thefore it has three roots: three real or one real and two imaginary roots. Consequently at certain pressure interval three different values will correspond to on value of pressure. In sutuations without pressure intervals certain value will correspond to one value or pressure. In order to open root values in Van der Waals’ equation, it is necssary to deal with dependence of pressure to the volume in constant temperature under (5.23), in other words isotherm. Then (5.23) it is necessary to write as follows:

p=

RT a − 2. V −b V 154

(5.26)

Based on this equation we see, that there are theoretic minimums of and maxumums of isptherms, when V = b , then, p = ∞ . Therefore volume V cannot be less than b. Volume of vessel where gas is located cannot be less than quadruple volume of molecules.

Fig. 5.12

Based on Van der Waals’ equation in figure 5.12 for CO2 gas theoretical isotherms are shown, calculated for different temperatures (undivided line – real gas; shaded line – ideal gas). Appearance of Van der Waals’ isotherm will not be hyperbola as in the case with ideal gas and it is seen from graphics. With temperatre increase unusual wavelike part of theoretic isotherm bend decreases and disappeas in isothem point Т=Тк. Temperature Тк is critical temperature. Isotherm which is corresponding to criticial temperature and emphasized by sinuous line is called critical isotherm. In respect to volume Vad der Waals’ equation is raised to ther third power, therefore one out of three roots is real and two imaginary (in temperatures that are higher than critical temperature) or under low temperature there will be three real roots. Temperatures T > TK are higher than cricital temperature, isotherms are uninterrupted, homo155

geneous, lines are changeable, have appearance like ideal gas hyperbola. Temperatures T < TK in the middle of isotherm are below critical and have unusual wavelike bends. Van der Waals’ equation is used for description of homogeneous gas and for certain states of liquid. Use of equation become clear by comparison of experimental and theoretical isotherms. Let us compare Van der Waals theoritical isotherm (fig. 5.12) with real gas isotherm, practically determined (fig. 5.13) Territory of isotherm gasous state – ab , territory of liquid state – fg correspond to each other. But, there are bacis differences between them, it is related to parts of experimental isotherm bdf and theoretic isotherm cde. States that are corresponding to cde point are nonexistant in nature, because the dependence of volume to gas exceeds usual one, when the pressure increases, volume does not decrease, volume will also be expanding. There are no objects in nature that in the event of pressure increase expand their volume, in other words there are no objects which decrease density. Even if object that has such remarkable ability existed, their states would have been unstable. These states that correspond to cde part are called unstable (unsteady).

Fig. 5.13

If volume decreases descending along unusual wavelike hillock of isotherm with critical temperature up to the lower temperature (fig. 5.12), then in the beginning pressure will increase to its maximum, and after that will decrease to minimum and again will start inscreasing. 156

∂p  Vс and Ve for volume values   = 0 , and in ther intervals of  ∂V T these volumes this product will be positive. Line P=const passes isotherm (fdв) at three points (fig. 5.12), therefore three values of volume correspond to one value of volume (( three roots 5.25)) (Т=const). Volumes of liquid Vf and gaseous Vb phases correspond to this pressure. Third value of volume  ∂p  belongs to positive  ∂V T isotherm of cde part. In the interval between c and e any states will be unstable (unsteady). States in interval bc and ef are called metastable, they determine the state of homogeneous variability. Let us consider physical value of these particularities. Uninterrupted transformation of the object from liquid state into gaseous state along Van der Waals’ isotherm line is very important. Consequently, part of theoretical isotherm correspond to gaseous state, fg – to liquid state, сde – to absolute equality of liquid and gaseous states, in other words correspond to two phases states. It is discovered from experiment, that in isotherm part bdf volume changed, but pressure remained same (fig. 5.13). In the process of volume decrease part of gas transforms into liquid (become condensed) however the pressure does not change remaining at the same value. And in the process of volume increase part of liquid transforms into gas and the pressure still remains unchanged. Of cours, volume change takes a lot of time, in this moment rate of condensation (or evaporation) and rate of volume change correspond to each other. Segment of isotherm line bdf is perpendicularly to pressure axis, the process is isthermyisobaric. This event is common only for liquid and saturated steam in two phase system. If there is saturated steam on the liquid surface, this similartiy will not be present in two phase system. Part of isotherm fg correspond to liquid state. In the event of little change of volume pressure significantly increases because compression of liquid is too little. If compare unstable parts of Van der Waals’ isotherm bc and ef with states of part cde, then in experiments they may be observed in unusual conditions. States that correspond to part bc are gaseous, but the pressure of this gas is more than the pressure of saturated steam in the same temperature рқан, in other words contradicts with 157

naturalness. Such unusual state of the object is observed during humid air cooling in closed vessel. Such state is called extremely saturated steam. This state is very unstable, it is often called metastable state. In order to receive extremely saturated steam, it is necessary to remove condensation centers. In extremely saturated steam when condensation centers appear extra part of the steam transforms into liquid, steam becomes saturated, states of isotherm bc run into part of points bdf. This property of saturated steam (namely, to transform into liquid at the presence of condensation centers) is widely used in the science and technics. For example, in the process of microparticles study (particles received in the process of nuclear reaction) the state of extremely saturation steam is created in Wilson chamber. Particle entered into chamber plays the role of condensation center. Fog that appeared due to microparticles motions (groupping of extremely saturated steams transformed into liquid) leaves the trace, this trace is called track. From the trace of particles that are noticeable on chamber it is possible to identify weight, charge and energy. In part fe liquid pressure is less than the pressure of saturated gas. This state is also metastable, unstable. If evaporation center appears the liquid begin instensly boiling, steam pressure on liquid surface increases, points of isotherm part fe go into (fig. 5.13) the state of part fd. Liquid with pressure less than the saturated steam pressure is called extremely heated. In nature isothermic transformation of real liquid state into gas occurs intermittently, which is not slowly, but suddenly and always occurs through the territory of two phase state (liquid and saturated steam; isotherm bdf in figure 5.13). This obvious process is known to all. Isothermic conversion into gaseous state may be observed during boiling of water in vessel, where water transforms into steam. Conversion of object from liquid into gaseous state is uninterrupted, that is why in practice cde is not observed. States that correspond to part cde in Van der Waals’ isotherm are not encountered and are not observed in practice. Volume of this part is dependent on pressure, in other words, in the event of unreasonable pressure increase, volume also increases. Even if imagined that in any part of the object such states existed it will be very unstable. Random change of pressure, for instance, its increase will lead to volume increase, and this will 158

lead to further pressure increase, and until object returns to the state at point с (fig.5.13), the process will repeat. And vice-versa, pressure decrease will lead object to the state at point с. How to draw the object in the part of Van der Waals’ isotherm, which transforms from one aggregate state into another? Jumping transformation of physical qualities, for example, water into steam or otherwise, melting of ice is called aggregate state of the object. Based on Maxwell analysis Van der Waals’ isotherms transform into state with equal areas fedf and dcbd, along straight line bdf from the state f into state b (fig. 5.14). There is no difference whether object goes through one phase line fedcb, or two phase line bdf, the conducted work will be the same. In figure 5.14 area, bordering with line fedcb is equal to area under line bdf. Therefore areas of line fedcb hillocks (fig. 5.14, dashed with bent line) will be equal.

Fig. 5.14

Even though Van der Waals’ equation represents two phases: gaseous (part ab ) and liquid (part fg), equation does not define the place of part bdf straight line. During Van der Waals’ equation testing with the use of experiments, it was discovered that relation of real gas parameters between p , V and T is described tentatively. For many gases the value of product, discovered through Van der Waals’ equation ( p ⋅ V ) is more than the one discovered with the use of experiments. Constants 159

a and b depend on temperature. Therefore, Van der Waals’ equation is approximatel discribes the state of real gas. The most important problem arising out of this Van der Waals’ equation – is critical temperature and critical state.

5.7. Critical state of the object. Critical temperature In figure 5.12 one of Van der Waals’ isotherms attracts attention. This isotherm as well as uninterrupted homogeneous hyperbola divides isotherm territory where there is part of wavelike hillocks. As the temperature increases the pressure of saturated steam increases, density increases, and gas volume decreases. With the rise of the temperature point b (fig. 5.12) rises and moves to the left side, liquid volume expands, point f rises up and moves to the right side. With increase of the temperature points b and f they move towared each other and decrease in segment bf. For this subject in certain temperature points b and f get connected at point ТК. This isotherm has indented points, at this point  ∂p   = 0,   ∂V T

 ∂2 p   2  = 0 .  ∂V T

(5.27)

These equations (5.27) are conditions that determine critical state in object state diagramm. In critical state two (or more) thermodynamic equal phases are in absolute equality in their protperties. Point b correspond to gaseous state, and f to liquid state (fig. 5.12). Their conjunction at certain temperature shows the disappearance of particularities beteen liquid and saturated steam. Temperature under which the difference between liquid and saturated steam disappears is called critical. Two phase line of equality (binodal) and line, marking the limits of the centers, contact at critical point (fig. 5.15). Therefore critical point correspond to conditions (5.27). Inside bimodal the object remains in two aggregate states: in liquid and in equality with its saturated steam. 160

Fig. 5.15

It is important to notice that in all text books in Kazakh language the concept of crictical is presented as crisis. (from Greek kritike – selection skill, expression of opinion; figurative meaning rapid change. From Greek language krisis-decision, result, figurative meaning – new stage of development). Term «critical» – international term, therefore we do not use meaning «crisis». Values of critical volume Vк , critical pressure pк , critical temperature T к of the object in critical state is determined with the use of interaction forces of moleculs and structure of molecules. Critical parameters of the object belong to physical fundamental constants. Using critical parameters it is possible to calculate numerous properties of the object (for instance, melting temperature, boiling temperature, phase transition of heat, etc.). With the use of critical parameters the constants a and b are determined in Van der Waals’ equation. In temperatures higher than critical in real gas isotherms there are not any differences, the higher temperature rises, the more ideal gas isotherms approach. Due to loss of differences between liquid in critical state and its saturated steam, the heat of 161

liquid evaporation is equal to zero. Surface tension coefficient of liquid is equal to zero. Numeric density of molecules is significantly deviate from average values. These changes require changes of physical parameters that depend on numeric density. For instance, due to refraction factor change the substance in critical state will be optically dim, the color will be whitish like valuable mineral stone opal. This phenomena is called critical opalescence. This critical phenomena was discovered by Russian scientist D.I. Mendeleev in 1861, he called critical temperature as absolute boiling temperature. Some values of substances critical parameters and constants a and b are given in 5.1. Table 5.1 Substance

Nitrogen Argon Water Hydrogen Carbon dioxide Oxygen Methane

Chemical formula

а, Pa·m6/ mole2

N2 Ar H2O H2 CO2 O2 CH4

b ⋅ 10 6 , 3

Vк , 3

Pк , atm

Tк ,К

m /mole

cm /mole

0,1390 0,1345 0,5464 0,0244 0,3592

39,13 32,2 30,5 26,6 42,7

89,5 56,0 73,4 65,0 94,0

33,5 48,1 217,6 12,8 72,6

126,2 150,8 647,3 33,2 304,2

0,1360 0,2253

31,8 42,8

73,4 99,0

49,4 45,4

154,6 190,6

Find relations in critical parameters of Van der Waals’ equation

Vк , Pк and Tк with constants a and b . If in equation (5.25) Т=ТК

and р=рК , then it is written in the following way:







pk 

(V − Vк )3 = V 3 − 3VK V 2 + 3VK2V − VK3 = V 3 −  b + RTK V 2 +

a ab V− . pK pK (5.28)

In equation (5.28) of degree V must be equal to coefficients from both sides, three equations derive from this: 162

3Vк =

RTк +b, Pк

3Vк2 =

a , Pк

Vк2 =

ab . Pк

(5.29)

Having solved this equation system, it is necessary to determined critical paramters related to constants a and b:

Vк = 3b , Pк =

a 8a , Tк = . 2 27 Rb 27b

(5.30)

These values may be found after differentiation of equation (5.26) and using conditions of (5.27). At critical point compressibility of the object is equal to infinity. In fact, compressibility coeffeicient is calculated as follows:

χ=

1  ∂V  .   V  ∂p T

At critical point ∂p = 0 , then ∂V = ∞ , therefore compressi∂V ∂p bility coefficient is equal to χ infinity.

5.8. Law of corresponding states. Van der Waals’ reduced equation Intrinsic qualities of the object in Van der Waals’ equation penetrate with the use of constants a and b and critical parameters pк , Vк ,T к . Therefore, for instance, if different gas isotherms are brought into the same temperature, their appearance will different. For all gases ideal gases isotherms do not have differences, because they do not depend on gas qualities. With the use of Van der Waals’ equation it is possible to find manifold (univsercal) equation of real gas state, that is not depending from the object nature. For this, proportion of critical parameters p, V, T to critical values pк , Vк ,T к , are calculated, in other words 163

π =

T p V ,ϕ = ,τ = . pK VK TK

(5.31)

Where π ,ϕ ,τ – reduced parameters. These immeasurable values are the values of real gas, which are determined as proportion of critical parameters. In Van der Waals equation

α    P + 2 (V − b) = RT . V   Instead of p, V, T put πpR ,ωVK ,τTK and convert constants a and b according to formula (5.30) with the use of pк , Vк ,T к the following equation is received:

  3  3  1 8  π + 2 (3ϕ − 1) = 8τ or  π + 2  ϕ −  = τ . 3 3 ϕ  ϕ   

(5.32)

This formula unites parameters of any substances, except for constants, characterising individual qualities of substances, therefore equation (5.32) will be universal, which means equal for all substances. Equation (5.32) is called Van der Waals’ reduced equation. π ,ϕ ,τ determine corresponding state. If out of three reduced parameters two in different substances are the same, in the third parameter these substances also will be the same. This law is called law of corresponding states. States that are subjected to Van der Waals’ equation ensure the law of corresponding states, therefore there will be thermodynamic similarity. For thermodynamically similar substances the following ratio is performed:

pKVK 3 = . RTK 8 164

(5.33)

For all substances the value of this relation must be the same, but in reality, in real, dense gases there are deviations from these values. Consequently, for the accomplishment of this similar substances law of corresponding states:

pKVK = const , RTK in other words, this complex must be the same.

165

(5.34)

6 FIRST LAW OF THERMODYNAMICS 6.1. Subject matters of thermodynamics. The basic laws of thermodynamics – the result of practical generalization of phenomena undergoing a transformation of energy. Functions of state Investigation of regularities in equilibrium systems and the thermal motion during the systems transition in a state of equilibrium is the subject of thermodynamics. Initially the main goal of thermodynamics was the creation of the heat engine improving theory and the solution of circulation problem of heat and mechanical operation. However, by using the laws of thermodynamics for studying other properties of the substance it has become possible to get information about complex systems, which consist of countless molecules. For example, to test the effect of temperature, pressure and composition of the substance for the chemical and physical properties and vital processes laws of thermodynamics are applied. As it was mentioned above in Article 1.4 above, thermodynamic methods for studying the properties of the substance and heat (energy) circulation processes are based on the model of the atomic and molecular structure of a substance and do not require information about the structural elements of substance. Following consolidation of information found in the course of experiments on the properties of the phenomena and processes proceeding with the release of energy, the basic laws of thermodynamics were formulated. Information about complex systems consisting of countless molecules, can be obtained only on the basis of four basic laws through the surprisingly large number of mathematical formulas. Another advantage of the thermodynamics is that its results do not depend on the characteristics the system in question. Thermodynamics investigates substances that are in and are not in a state of equilibrium. The approach of description of systems in 166

thermodynamics is based in equilibrium concept («conceptio» Latin – concept, opinion). We consider thermodynamics in the state of equilibrium. Thermodynamics in an equilibrium state represents the system in a state of rest, which can be described as follows: «here and now», it does not take into account the previous and the subsequent state of the system. Theoretical basis of the concepts and definitions were discussed in sections 1.4 and 2.1. Status of the thermodynamic system defines a set of independent macro parameters. They describe measurable material properties (temperature, pressure, density, mass, composition, etc.), and referred to as state functions or status parameters. These parameters describe the transition of a thermodynamic system from one state to another in a particular point in space and time («here» and «now»). As a thermodynamic system, for example, we can take a system with a cylinder and a movable piston (valve) (Figure 6.1). The piston compresses the gas volume of ν moles, but heat decrease through the piston does not accur. Designating this thermodynamic system as a BD, let’s look at four different states. Piston retains gas with a homogeneous composition in the cylinder D with the volume V , pressure p and temperature T.

a)

c)

b)

d)

Fig. 6.1

D of cylinder gas piston V amount of pressure and keep under a temperature T p. BD system is able to have substance and energy exchange with the environment (in the form of heat and operation) by three different ways: 167

а) isolated system, there is no exchange of matter and energy (Figure 6.1а); b) adiabatic process in an isolated system, proceeds without the exchange of substance and energy (6.1b); c) closed-loop system, no exchange of substance will take place, only the exchange of energy (6.1c); d) open system, exchange of substance and energy (6.1d).

6.2. Zeroth law of thermodynamics. Temperature The temperature – is a property inherent in any thermodynamic system (Section 2.5), it is determined by the Zeroth law of thermodynamics. If the two independent of each other bodies A and B are in thermal equilibrium with the third body C, they are in thermal equilibrium with each other. Then the balance of the system is characterized by the equality of all points of the function state (Figure 6.2). It is believed that these bodies A, B and C have the same temperature.

Fig. 6.2

State of the system being in equilibrium, is determined by the parameter T known as the temperature, that describes its internal state. T parameter value is equalized as a result of energy exchange in thermal contact of different systems in an equilibrium state and becomes equal for all of them. A rule that describes the state of 168

equilibrium of the system, that includes a special function state - the temperature is called the zeroth law of thermodynamics.

6.3. The first law of thermodynamics and its physical meaning. Prohibition the idea of perpetual motion of the first order according to the first law of thermodynamics The first law of thermodynamics introduces the familiar and surprising abstract concept of energy. Generally, energy describes the ability of any body to perform work, which means that it is related to application of force. The possibility of heat convesion into mechanical work is shown based on thermodynamics with the help of summary of aggregate experimental data. The equivalncy of heat and work was discovered in the middle of XIX century. Thus, types of energy manifestation were discovered. Mayer (1842) and Helmholtz (1847) summurized threse practical results and came to the following conclusion: All microscopic system have reserve of constant energy and energy may transform from one form into another. Quantity of energy system may change only in the result of its transfer to outside environment or its production from outside environment. This statement determines the meaning of the first law of thermodynamics. “What is energy?” It is not easy to answer this question. We know that the the manifestation of the energy only during mutual transfer. Thus, we may measure only the energies difference and only this difference has physical meaning. According to macroscopic approach, each body is connected with certain quantity of energy. This energy is divided into three types: potential energy, kinetic energy and internal energy. If so, full enerty of the system shall be determined in the following way:

∆E = ∆E кин + ∆E пот + ∆U .

(6.1)

With two previous types of energies we became familiar with in mechanics section. Generally, in the process of thermodynamic systems study the cases, when changes of potential energy do not 169

affect full energy and when kinetic energy is equal to zero, are reviewed. In these cases the change of full energy of system ∆E is equal to change of internal energy ∆U , in other words ∆E = ∆U . In this section we will consider energetic side of ideal gas state change. First of all, change of gas temperature, paying special attention to the process of its status change.

6.3.1. Internal energy of ideal gas Mean energy of ideal gas molecules (article 2.5) is determined in the following way: m0υ 2 3 = kT , 2 2

where k – Boltzmann constant, Τ – gas temperature, m0 – molecules mass, υ 2 – average square molecular velocity. This equation is written for ideal gas, molecules of which are considered as material points, and it is considered that their motions are only translational. If gas molecules consist of more than one molecule, such monoatomic molecules may be in rotational and oscillating motion, then it is necessary to take into account the energy that is related with such motion. Let us first consider monoatomic gas and find formula for in order to determine energy for all molecules of substance when mass is known. If gas consists of N molecule, then the value of full energy will be equal to: m υ2 3 (6.2) U = N 0 = N kT . 2

2

For one mole of gas the example (6.2) may be written in the following way: 3 3 (6.3) U = N A kT = RT , 2 2 where N A k = R – universal gas constant, N A – Avogadro number. 170

Energy that is determined by formula (6.3), is called internal energy of ideal gas (for one mole). Generally, internal energy of body – is “invisible” energy connected with molecules in its composition. Thus, internal energy of body or system of bodies may include kinetic energy of all moving molecules in substance composition, kinetic energy of atoms in molecule (if molecule is monoatomic), and also kinetic and potential energy of microparticles that are included in atoms composition. But the kinetic energy of gas travel is not included here as complete body and its potential energy that is located in external force field. Internal energy of gas under certain mass according to formulas (6.2) and (6.3) depends only on the temperature and does not depend on volume and pressure of gas. For real gases this statement does not work. In order to change the temperature of ideal gas, it is necessary to change its internal energy, which is seen from formulas (6.2) and (6.3). As it is known from mechanics, energy change is related to work performance, its energy changes if the body performs work or if the work is performed toward it, the change of energy is equal to performed work. If so, then it is possible to think that the change of gas temperature or of any body is possible only if mechanical work is performed. For example, in order to heat the body, it is necessary to perform work in respect to it, and in order to cool it, it must perform work by itself. For example, is known since ancient times that the bodies get heated during friction with each other. 6.3.2. Quantity of heat It is well known from practice that any body may be heated or cooled without subjecting it to the work. Such heating or cooling method is possible when the contact of bodies occurs with different temperature or radiation. In the event of division of bodies with different temperature with any environment, it is also possible to observe heat exchange process between them. According to formula (6.2) the change of gas temperature it related to change of its energy. If this is so, in the event of bodies contact occurrence, then due to any reason the energy discharges (heating) or is produced (cooling). 171

This method of energy exchange is connected with exchange during microparticles collisions of two contacting bodies. As a result of heat transfer to particles of cooler body, heated body loses energy, which means that microparticles of one body that are moving chaotically transfer energy to particles of another body. In this case the change of energy does not depend on work performance. The quantity of discharged (produced) energy of such heat exchange is called quantity of heat. It is designated by Latin letter Q (que). In the event of body temperature change (heating or cooling) through radiation or contact with another body, it is said that it was produced or discharged such and such quantity of heat. In fact, in the process of heat exchange the work is performed, but it is done not by microscopic bodies in orderly motion, but by microparticles in its composition that are moving chaotically. It follows that if energy of system changes only in the heat exchange process, then Q = ∆U , in other words this quantity of heat Q becomes the measurement of energy change in the process of heat exchange. Mechanical equivalent of heat. Heat and energy were determined as discharge (production) of energy, namely, change of energy types. Consequently, its unit in SI system is the same, joule (J). In consequence of many experiments in 50’s of 18th century, D. Joule proved that the heat and energy are equivalent. He proved that the heating ability of heat quantity of 1 kilo-calorie requires work performance in the same amount. In this case the work appeared to be equal to the following: 1 kcal=4186,8 J. Quantity of heat, discharged (or produced) during heating (or cooling) of one kilogram of water under atmosphere pressure up to one Kelvin is called on kilocalorie (kcal). Number determining the ratio of mechanical work unit to heat unit is called mechanical equivalent of heat, it is equal to the following:

I = 4186,8 J/kcal = 4,1868 J/cal. 172

Reciprocal of it is called heat equivalent of mechanical work:

I ′ = 2 ⋅ 39 ⋅ 10−4 kcal/J=0,239 cal/J. Thus the quantity of heat, work and energy have the same measuring unit.

6.3.3. The first law of thermodynamics State of body or system changes in the process of work performance. Then the change of system occurs in parallel to its performed work or the work that is directed to it under the external forces influence. It is possible to calculate work with the use of parameters that determine the state of this system. If the stat of the body is determined by two parameters out of p , V and Τ , then in general case the change of any of them requires the performance of external work. For example, gas temperature change, namely, its heating or cooling, is possible as result of mechanical work (heating) or performance of work against external forces (cooling). Under the influence of external forces gas is heated and compressed, it means that the mechanical work is performed against this force. And when gas is cooled down, it increases and performs work by itself. It is possible to change the volume of gas without change of gas temperature, it requires little work. At the same time, it is possible to change gas state (or any other body) having transferred to it certain amount of heat or having taken from it, which means after the contact with hotter or cooler body. Which work is performed under system change in such a way? The answer to this question – energy conservation law. If certain amount of heat dQ is transferred to gas (or to any other body), then, generally, the work is performed dA and its internal energy changes at dU . In accordance with energy conservation law, the work, performed by system is equal to the difference from produced quantity of heat and change of internal energy: 173

dA = dQ − dU or

dQ = dU + dA .

(6.4)

Energy conservation law in respect to mechanical and heating energy is determined by formula (6.4). This equation is the most important law of nature. This law is called the first law of thermodynamics. Prohibition of the idea of the first order perpertual motion machine based on the first law of thermodynamics It is practically confirmed that the change of energy does not depend on the way of how the system transforms from one state into another, it depends only on the properties of initial and final states. The above mentioned may be considered as one of the conclusions of the energy conservation law. If so, then the system will have the same energy in the same state independently of the type of transformation into that state. Thus, when during cyclic process the system returns to initial state, its energy must reach to the original value. In this case mechanical equivalent of all external influence is equal to zero. In order that is would not be equal to zero, it is necessary to change energy of the system. Thus, it is impossible to create machine that would perform continuous work without change in other environment (external environment) or which would perform work for unlimited amount of time without using external energy. Such machine which would perform work during unlimited amount of time without using external energy is called perpetual motion machine (Latin perpetuum mobile – perpetual motion) In order to stop resultless search of perptual motion machine, conducted since 13th century, Paris Academy of Sciences issued special order in 1775. This order was about the impossibility of invention of perpetual motion machine, therefore it is necessary to give up such projects. Because invention of perpetual motion machine contradicts the law of conservation and transformation of energy, namely, contradicts to the first law of thermodynamics. Prohibition of the idea of the first order perpetual motion machine in accordance 174

with the first law of thermodynamics – one of the results of the first law of thermodynamics.

6.3.4. Work during change of gas volume Let us calculate the work performed during the change of gas volume, which means under its increase or decrease. In order to do that, we use thermodynamic system BD, which was mentioned above (article 6.1). For example, in vessel (cylinder), that can be closed by mobile piston with the area of S there is gas (Figure 6.3).

Fig. 6.3

Original volume of gas V , pressure р and temperature Τ . Let us assume that gas volume remains constant during the change of pressure and temperature, in other words, the system will be in the state of balance. Under the influence of external force the piston moves to the distance of dx , gas is compressed. Gas is compressed until the force that influences the piston from gas side is equal to force F . Balancing force is equal to F = pS . Where p - gas pressure. Work for piston movement to dx (known from mechanics)

dA = Fdx = pSdx , 175

(6.5)

where Sdx = −dV , change of compressed gas volume, then the performed work is equal to:

dA = pSdx = − pdV .

(6.6)

With the increase of gas, that is to say increase of gas up to dV , from gas side the work is performed against external forces that is equal to + pdV . If external work in the event of the system state change is performed in connection with volume change, then the first law of thermodynamics shall be written in the following way:

dQ = dU + pdV .

(6.7)

Generally, body state changes can be connected with electrical, magnetic and other properties (parameters). Then additional elements can be included in the right side of the expression (6.7) depending on the types of energy. If body transforms from one state into another, then the performed work A can be found by equation integration (6.6): 2

2

1

1

A = ∫ dA = ∫ pdV .

(6.8)

This integral can be determined with the help of diagram. In fact, on curve p = f (V ) body states are marked by dots (Figure 6.4). If dependence of p = f (V ) can be imaged using 1а2, then diagram form of the integral: 2

∫ рdV . 1

Will be equal to shaded area under the curve. And if the state change is imaged using 1в2, then the work in such transfer will be different. 176

Fig. 6.4

External work that is performed during gas volume change depends on the state circuit between the original and final state. Thus, the work that is connected with pressure change in accordance with difference states is also different (Figure 6.4). Consequently, the work does not depend on the original and final state of the system and depends on process progress. If the process of gas transformation from state p1 ,V1 into state p2 ,V2 is imaged with the use of curve 1а2 , and the curve located above 1в2 is imaging the following process (fig 6.4), values of the performed work in these circumstances will be different. Area of figure of the imaged curve 1в2, located above is not equal to the area under the curve 1а2. Thus, the work – is the function of the process, which is determined only by determination of expansion conditions and gas compression. The presence of work reserve in the form of heat quantity is impossible. Quantity of heat and work is determined based on the process. In case of internal energy, it depends only on the state of the system, its change does not depend on the states in between the transformation of body from one state into another: 2

∆U = ∫ dU = U 2 − U1 ,

(6.9)

1

where internal energy U1 and U 2 – in accordance with the states 1 and 2. 177

For infinitesimal changes of internal energy, the first law of thermodynamics can be written in the following way:

dU = dQ − dA = dQ − pdV .

(6.10)

At this dU – full differential of internal energy which shows the difference of energy between these two states (formula (6.9)). dQ and dA cannot be full differential, because it is infinitesimal quantity of discharged (produced) heat and work of system, which depend from processes. Thus, equation (6.7) can be written in the following way: 2

2

1

1

∫ d Q = U 2 − U1 + ∫ pdV .

(6.11)

Particularly, let us assume that gas returned to its original states after all changes of states, then U1 = U 2 , and ∆U = 0 . In this case, the process of gas state change is called circular or cyclic. This process shall be imaged by closed curve on the diagram (fig.6.5). Work during cyclic process is equal to shaded area. This work is determined by closed integral in the following way:

A = φ рdV . .

(6.12)

Fig. 6.5

If cyclic work is positive, then from gas side in relation to external forces the work was performed, body received the quantity 178

of heat Q from outside, that is equal to work. If the work А is negative, then the external forces perform work in relation to body, then the quantity that was received is equal to work. Thus, cyclic process ∫ Q = ∫ A . The above mentioned could not be possible if the process taking place in the cycle would not be kept. If so, then these processes must be quasi statistic (article 2.1.4). It is necessary during gas volume change that the state of balance would remain. Then this process goes very slow, namely, infinitely slow. In this case, deviations from the balance state quickly disappear, gas goes through the number of balance states. These processes are called quasi statistic, then at any specific time parameters of gas state remain same throughout whole volume. Only such process is imaged in the form of diagrams. Not quasi statistic process cannot be imaged in the form of diagrams.

6.5. Heat capacity of ideal gas Quantity of heat for has heating is connected with the amount of heating substance and with on how many degrees its temperature changed. In order to cool the body, the heat quantity unit must be received from it. Thus, for description of heat properties of gas or a body the special value of heat capacity is used. Heat capacity – is the value of discharged or produced quantity of heat, in order to change body temperature for 1 Kelvin (1 K). Heat quantity is divided into own and molar ones. Heat capacity that belongs to substance mass unit is called own heat capacity. It is designated by lower case letter с (tse). This heat capacity describes substance in body composition. Heat capacity that belongs to one mole of substance is called molar heat capacity. It is designated by upper case letter С (tse). As well as own heat capacity it characterizes the substance. According to heat capacity definition, its measuring unit is J/K, and measuring unit of molar heat and own heat capacity – J kg ⋅ К . capacity is – J mole ⋅ К 179

Ratio between own and molar heat capacity is determined in the following way:

c=

C and С = сМ , М

(6.13)

where М – molar mass. According to definition heat capacity is equal to the following ratio:  dQ  , (6.14) CX =    dT  X where dQ – quantity of heat, discharged (produced) by body, dT – temperature change, x – (index) designation of going process. Since the quantity of heat, spent for body temperature change, depends on the process, then the heat capacity also depends on the conditions of heating or cooling, it means that C is not the same in different in processes х. If this is so, then system is characterized by different heat capacities based on different processes.

6.5.1. Heat capacity during isochoric process For example, gas heating process occurs at constant volume, V = const . Molar heat capacity of gas during isochoric process is

called heat capacity at constant volume or isochoric heat capacity and is designated by CV . Here the value V is written instead of x (example 6.14). According to example (6,14) heat capacity at constant volume is determined in the following way:

 dQ  CV =   .  dT V

(6.15)

If so, then the first law of thermodynamics dQ = dU + pdV for process at constant volume is determined in the following way:

dQ = dU . 180

(6.16)

Work of this process:

dA = pdV = 0 . Since the change of the volume dV = 0 (fig 6.6). Then all quantity of heat transferred to gas is used for change of its internal energy.

Fig. 6.6

Therefore heat capacity CV at constant volume is equal to:  dU  . CV =    dT V

(6.17)

Then internal energy is equal to:

dU = CV dT .

(6.18)

Taking into account formula (6.18), the first law of thermodynamics (formula (6.4)) can be written in the following way:

dQ = CV dT + pdV .

(6.19)

Thus, the quantity of heat dQ , transferred to body is used for change of its temperature up to dT (change of internal energy) and of volume up to dV (performance of external mechanical work). 181

6.5.2. Heat capacity of monoatomic gases Internal energy of one mole of ideal monoatomic gas (expression (6.3)) 3 U = RT . 2 Then according to (6.17) molar heat capacity will be equal to:

CV =

3 R. 2

(6.20)

In the event of relation of this molar heat capacity ratio to Avogadro number, then the average part of each molecule in gas heat capacity will be equal to: 3 3 R (6.21) = k, 2 NA 2 where k – Boltzmann constant. Thus, at the increase of gas temperature on 1 К, the energy of each monoatomic molecule will be growing on

3 k joule. 2

6.5.3. Heat capacity during isobaric process Isobaric process occurs at constant pressure ( p = const ) . Molar heat capacity during isobaric process is designated by CP and is determined by formula (6.14):

 dQ  . CP =    dT  P

(6.22)

Quantity of heat dQ at constant gas pressure will be used for work performance and for change of its internal energy: 182

dQ = CV dT + pdV . Thus, heat capacity of CP power in accordance with formula (6,22) is equal to:

 dV  .  CV dT + pdV    = CV + p dT  dT  P P 

CP = 

(6.23)

Equation of ideal gas state for one mole pV = RT , then during isobaric process volume V and temperature T change, which means that pdV = RdT , because (Vdp = 0 ) , consequently, the second member in the right part of the equation (6.23) is determined in the following way:

 dV   = R.  dT  P

p

(6.24)

Using this formula heat capacity CP for equation (6.23) Let us take relation

CP = CV + R or

CP − CV = R .

(6.25)

From example (6.25) we see that heat capacity at constant pressure is higher than the heat capacity at constant volume to the value equal to gas universal constant R . If this is so, in the event of heating of one mole of gas up to one kelvin (1К) at constant pressure, its heat capacity is higher than the heat capacity at constant volume to the value of work, that is equal to p dV  .  dT  p Thus, when one mole of ideal gas increases with isobaric process being heated to one kelvin, numeric value of performed work is equal to universal gas constant. 183

In view of formula (6.20), we write formula (6.25) in the following way: 3 5 (6.26) CP = R + R = R . 2 2 Value of the universal gas constant R = 8,314 J / mole, then the ratio CP (designated by letter γ) is the constant value for ideal CV monoatomic gas and is equal to: γ =

CP = 1,66 . CV

(6.27)

Work in isobaric process under the influence of external forces or forces that influence from gas side to external bodies is determined by equation (6.8): 2

A = ∫ pdV = p(V2 − V1 ) .

(6.28)

1

Fig. 6.7

If V2  V1 , then A  0 ; V2  V1 , A  0 . In fig. 6.7 p − V – in diagram gas work in the process of transformation from the state 1 into the state 2 is equal to area of rectangular 12V2V11 . 184

6.5.4. Enthalpy. Heat function If heat quantity is transferred to gas without gas pressure change, in other words, if there is opportunity for its increase, then the first law of thermodynamics can be written as follows:

(dQ )P

= d (U + pV ) .

(6.29)

Differentiating the right side of the equation (6.29), we receive:

d (U + pV ) = dU + pdV + Vdp . In this event, if p = const , then dp = 0 , consequently,

d (U + pV ) = dU + pdV .

(6.30)

Thus, upon the application of heat or cold on isobaric gas, the generated or evolved heat is equal to value change:

I = U + pV .

(6.31)

This value I is called enthalpy or heat function. If gas state change occurs isobarically, then enthalpy is a parameter that indicates the gas state. Therefore, when the pressure is constant, gas heat capacity may be determined with the use of enthalpy:

d (U + pV ) dI  dQ  . =  = dT dT  dT  P

CP = 

(6.32)

Gas enthalpy for monoatomic ideal gas with certain mass is proportional to temperature and does not depend on the volume:

I = U + pV =

3 5 RT + RT = RT . 2 2

(6.33)

Then according to (6.32), CP = 5 R , correspond to value (6,26), 2 determined above. Real gas enthalpy depends on the volume. 185

6.6. Relations between gas heat capacity and their designation through the number of degree of molecules freedom 6.6.1. Relation of heat capacity with the number of degree of molecules freedom and monoatomic gas heat capacity Molecules of ideal gas are considered as material points. Mean energy of such particles is determined by the mean kinetic energy of 2 translational movement m0υ . This energy may be considered as 2 sum of three components of kinetic energy of molecule traveling in three mutually perpendicular directions:

m0υ 2 2

=

m0υ X2 2

+

m0υY2 2

+

m0υ Z2 2

,

where υ X , υY , υ Z - velocity components of molecule along the three axes of coordinates. Due to chaotic motion of molecules, the values of mean kinetic energy in three directions will be equal to each other:

m0υ X2 2

=

m0υY2 2

=

m0υ Z2 2

=

1 m0υ 2 . 3 2

(6.34)

In accordance with the main equation of kinetic theory mυ 2 3 . = kT 2 2 2 Consequently, each component of energy m0υ x in equation

2

(6.34) is equal to kT / 2 . Division of kinetic energy of particle into three free components is related to the consideration of it as material point that has three degrees of freedom. Number of independent coordinates determining its location in space and configuration is called the number of degrees of freedom. 186

Therefore the number of degrees of freedom of monoatomic molecule is equal to three, then the energy that is equal to kT / 2 correspond to each degree of freedom. If so, the polyatomic gas heat capacity will not be equal to monoatomic heat capacity. In table 6.1. heat capacities of polyatomic and monoatomic gases are presented, that were obtained experimentally. Table 6.1 C p − CV R

Cp CV

Polyatomic gases

CV

R

C p − CV R

Cp CV

1,519

1,001

1,659

3,02 2,45

1,090 1,005

1,360 1,404

1,500

1,008

1,64 1,67

4,04

1,030

1,250

-

-

1,68

3,25

1,010

1,310

-

-

1,66

Chlorine (Сl2) Nitrogen (N2) Ethylene (C2H4) Methane (CH4) Ammonia (NH3)

3,42

1,060

1,310

Monoatomic gases

CV

Helium (Не) Neon (Ne) Argon (Ar) Krypton (Kr) Xenon (Xe)

R

Having compared the values of CV / R and CP / CV for gases based on the Table 6.1, molecules of which are monoatomic and gases molecules of which consist of two or more atoms, we will see the difference. Ratio of CP − CV / R for all gases is approximately equal. This difference between heat capacities C P and CV is independent from number of atoms in molecule, we see that R is equal to constant value. Therefore in the event of any ideal gas mole increase, if the temperature rises under constant pressure for 1K, the work is carried out that is equal to R . Gases in this table can be divided into two groups: diatomic gases, ratio CV / R of which is approximately 2,5, and CP / CV − 1,4 ; three or more atomic gases, value CV / R of which is about 3-4, and CP / CV − 1,3 . Heat capacity of gases that are composed of diatomic molecules of the first group are close to each other and are equal to: 187

CV = CP =

5 R ≈ 20,8 J/moleК, 2

5 R + R ≈ 29,1 J/mole·К, 2

therefore CP / CV = 7 / 5 ≈ 1,4 . Values of molar heat capacities of gases that are composed of three atomic and polyatomic molecules (table 6.1) are as follows: CV = 3R ≈ 24,9 ,

CP = 3R + R ≈ 3,33 J/mole·К, CP / CV = 8 / 6 ≈ 1,33 J/mole·К. Then we will see, that CP / CV = 8 / 6 ≈ 1,33 . Experimental values of gases heat capacities presented in table 6.1, were determined under atmospheric pressure at room temperature. In these cases, gases properties will be close to ideal. And in order explain deviations, that are observed in polyatomic gases, it is necessary to use the law of equal energy distribution.

6.6.2. Equipartition Law Mean kinetic energy kT / 2 is discharged for each translational degree of freedom of monoatomic gas molecule. If molecule has degrees of freedom beside translational movement, is this energy ratio conserved? The answer to this question was given by Boltzmann. Boltzmann suggested and proved the theorem of equal energy partition. The main point of the theorem: if molecule system is in thermal balance at the T temperature, then the mean kinetic energy is uniformly partitioned between all degrees of freedom and for each degree of freedom of molecule it is equal to kT/2. This theorem is called the law of equal kinetic energy partition by degrees of freedom, or the equipartition law. It means that in order to calculate the internal energy of gas and, consequently, heat 188

capacity it is necessary to be able to determine the number of degrees of freedom of gas molecules. Let us look at the diatomic molecule. It can be represented in the form of the system consisting of two atoms that are located at some distance from each other (fig. 6.8).

Fig. 6.8

If the distance between these atoms does not change, such molecules are called rigid. Such rigid diatomic molecule has six degrees of freedom. Because position and configuration of such molecule are determined: by the three coordinates of its mass center by which translational movement of molecule is determined, and by three coordinates, determining possible rotation of molecule near mutually perpendicular axes Оx, Оy, Оz . Nevertheless, it is known from practice that molecule rotation along the line connecting the centers of two atoms (axis Оx , fig.6.8), can be stirred only under very high temperatures. Thus, molecule rotation along Оx axis is not observed under normal temperature, therefore it is that molecule rotation has two degrees of freedom. In view of this, the number of degrees of freedom of rigid diatomic molecule s equal to 5, among them three are translational and two are rotational degrees of freedom. But atoms in molecule may oscillate in respect to each other, in other words, they are not always rigidly connected with each other. The one more coordinate is needed for determination of molecule configuration. This is the distance between atoms. Consequently, 189

generally diatomic molecule has six (6) degrees of freedom: three are translational, two are rotational and one is oscillating.

Fig. 6.9

If molecule is composed of n atoms, not rigidly connected, the it has 3n degrees of freedom. Each atom has three degrees of freedom. Therefrom three degrees of freedom are rotational, if atoms are not located on one right line. In fig. 6.9 the model of three atomic molecule is presented. In a general way, nonlinear n -atomic molecule has translational degrees of freedom 3 n − 6 , and linear molecule (3 n − 5) . In many cases there is no excitation of oscillative motions of atoms. But if oscillative motions of atoms do exist, and if their amplitudes are small comparing to the distance between atoms, then such oscillations can be considered as harmonic, and atoms are harmonic oscillators (lat. oscillare – oscillate). Such oscillator has kinetic and potential energy. Potential energy is connected with forces that bring oscillating atom to balanced state. For harmonic oscillator, as it is known from mechanics, average values of kinetic and potential energy are equal between each other. Consequently, in the event of excitation of atoms harmonic oscillation, then according to the equipartition law for each oscillating degree of freedom of molecule there is form of kinetic energy kT / 2 and the form potential energy kT / 2 . For non-harmonic oscillators the above 190

mentioned is not applicable. Therefore energy, that falls on one oscillating degree of freedom of molecule is equal not to 1 kT , but to 2 1 2 ⋅ kT = kT . 2 Taking this into account, we may estimate polyatomic gas heat capacity. Let us designate the number of degrees of freedom of gas molecules i , its mean energy i kT , and internal energy of one mole of gas, 2

U =

i RT . 2

(6.35)

Molar heat capacity of gas by degrees of freedom is determined in the following way: dU i (6.36) CV = = R dT 2 and i (6.37) C P = CV + R = ( + 1) R . 2 Estimating the number of degrees of freedom of molecule i , it is necessary to double the oscillating degree of freedom. 5 Heat capacity of diatomic gases in table 6.1 is equal to R . The 2 quantity of degrees of freedom of molecules of these gases is equal to 5, which means that molecules are rigid, oscillating degrees of freedom are not observed. Heat capacity values of some three atomic gases, experimentally discovered, are in compliance with values, estimated from theory. Nevertheless, there are significant deviations of experiment results from theory. According to equation (6.36), gases heat capacities that are composed of three atomic molecules 6 must be equal to CV = R = 3R under constant volume of gas. 2 However, heat capacity of three atomic gases in table 6.1. is a little larger than this value. 191

For example, heat capacity of Chlorine CV = 3,02 R corresponds to six degrees of freedom of molecules. Nevertheless, Chlorine mole5 cule is diatomic. If atoms are in rigid contact, then CV = R , and if 2 molecules are in oscillation, then the number of degrees of freedom 7 is equal to seven, heat capacity CV = R . 2 The result of this experiment does not correspond to heat capacity theory. According to theory (equation (6,36)), heat capacity does not depend on the temperature. The result of this experiment, however, shows the dependence from the temperature. Practice shows that when the temperature decreases, the heat capacity decreases as well. Dependence of heat capacity from the temperature proves the dependence that equal partition of energy by degrees of freedom of molecule occurs in limited conditions. In heat capacity theory we assumed that molecule is similar to solid globe and its motions follow the law of classical mechanics. Molecules consist of interacting atoms, structure of atoms is very complex. Motion of its particles does not follow the law of classical mechanics. Their motion is regulated by the laws of quantum mechanics. Therefore in polyatomic molecule the internal processes take important place, such as oscillative motions, oscillating degrees of freedom are connected with them. Therefore, classical theory does not take into account quantum properties of atomic systems, as result of which there are deviations between the data of theory and practice. Quantum theory fully explains the results of experiments on heat capacity.

6.6.3. Insufficiency of classical theory of heat capacity. The concept of heat capacity quantum theory For example, let us look at the dependence of hydrogen heat capacity from the temperature (6.10). Hydrogen heat capacity under constant volume at room temperature 5 cal /mole К (1 cal / mole К= 4,1868 J / mole), and at 192

the temperature of 50 К (-223 °С) around 3 cal / mole К. Thus, the heat capacity of diatomic hydrogen gas will be the same as of the two atomic gas that has three degrees of freedom.

Fig. 6.10

According to experiment, hydrogen heat capacity slowly decreases in the process of the temperature decreasing (fig. 6.10). It is difficult to explain this change based on the classical theory. Therefore the quantum theory is used in order to explain these deviations. Because it explains affections of bodies based on the interacting motions of microparticles in its composition. If values chariacterizing the system are comparable with Plank’s constant h, then the method of description of physical systems and law of motion are determined by the quantum mechanics (wave mechanics). This condition satisfy the moton of microparticles. In some cases microscopic systems also have quantum properties. Quantum mechanics laid a foundation for sciences of substance structure. Many macroevents are explained based on the quantum theory, for example, temperature dependence of solid bodies and gases heat capacities; structure of solid substances, etc. Foundations of the quantum theory is explained at special course. Quantum theory explains the dependence of hydrogen heat capacity from the temperature in the following way: hydrogen molecules may be in two different states – in parahydrogen state and in orthohydrogen state. It is follows from the quantum theory that atomic nucleuses have some moment of impulse. When molecule develops out of two hydrogen atoms, these nuclear moments (vector 193

values) may be located in parallel (↑↑) or in antiparallel to each other (↑↓). Hydrogen which molecules consist of atoms with paralleloriented nuclear moments is called orthohydrogen, contrary to antiparallel nuclear moments of atoms in molecule, which is called parahydrogen. In hydrogen the molecules content ratio depends on the temperature. At room temperature in normal hydrogen there is about 25% of parahydrogen, and with the decrease of temperature the content of parahydrogen increases, therefore at 20 К hydrogen is almost fully consists of parahydrogen (99,8%). Different values of rotational motion energy correspond to ortho- and parastates of hydrogen. Therefore, in these two states of hydrogen the heat capacity value is different. Under low temperatures (about 50 К, fig. 6.10) heat capacity that depends on rotational motion of molecules, in both states becomes equal to zero. This explains why the hydrogen heat capacity becomes the same as of monoatomic gas. Heat capacity measurement allow making important conclusions on molecule structures, therefore such measurements, especially under low temperatures, are very important. Besides that, knowledge of heat capacity value and its dependence from temperature is necessary in the process of solving of many technical problems

6.7. Use of the first law of thermodynamics and the of the equation of ideal gas state for description of isoparametric processes 6.7.1. Isometric process Operation under isometric change of ideal gas volume. In order to determine the work under isometric change of gas volume, it is necessary, according to equation (6.8), to calculate integral along the isotherm.

∫ dA = ∫ pdV . 194

(6.38)

Pressure p during expansion continuously change with the change of volume under the Boyle-Mariotte’s law, in other words the equation of state pV = RT = const , pV = const , hence

p=

RT . V

Placing this expression for (6.38) under the sign of integral, we will receive for work А under the change of volume from V1 to V2 : 2

V2

V2

1

V1

V1

A = ∫ dA =

∫ pdV = ∫ RT

V dV = RT ln 2 . V1 V

(6.39)

(6.39) is the very expression for work of isothermal expansion (or compression) of 1 mole of ideal gas. If gas mass is equal to m , then the formula (6.39) becomes as follows:

A=

m V RT ln 2 , M V1

(6.40)

where m / M = ν – number of moles, M – molar mass.

Fig. 6.11

From formulas (6.39) and (6.40) 29.1) it is seen that the work of isothermal expansion (or compression) depends not on the difference 195

of volumes, between which expansion occurs, but from logarithmetic volumetric ratio (V2 /V1 ) in the process of transition from the state

(V1) to the state (V2) Work under isothermic process p = f (V ) may be determined with the help of dependence graphic (fig. 6.11). Shaded area of trapzoid (12V1V21) in fig. 6.11 determines numeric value of the work А. This area is calculated according to formula (6.40). According to Boyle-Marriotte’s law the formula (2.58):

p1V1 = p2V2 , then

V2 p1 . = V1 p2

Therefore in formulas (6.39) and (6.40) it is possible to place inverse ratio of pressures. Then

A=

m p RT ln 1 . p2 M

(6.41)

The first law of thermodynamics for isotherm process. Isotherm process does not lead to the change of gas internal energy, and consequently dU = 0 , or ∆U = U 2 − U1 = 0 , because T = const . The first law of thermodynamics for isotherm process may be written as follows: Q=A. (6.42) In other words, generated (or evolution) by gas of the heat quantity is spent for work performance. Heat capacity under isotherm process. In the process of isotherm expansion the heat that is being braught to gas is spent only for the performance of external work. On the contrary, under isotherm compression the work of external forces is directed on the increase of internal energy (warming) of surrouning bodies. Formally it is corresponding to the idea that gas heat capacity is equal to infinity, in other words 196

 dQ  CT =   = ∞,  dT T

(6.42)

where dT = 0 , T = const .

6.7.2. Adiabatic process The first law of thermodynamics for adiabatic process. In adiabatic process gas does not give up the heat to bodies, as well as does not receive from outside. Therefore adiabatic process goes without heat exchange with the environment, dQ = 0 . Law of conservation of energy in this case takes the following form: (6.43) − pdV = CV dT dQ = pdV + CV dT = 0 ; or dA= -dU. It means that the work, related to the gas volume change, must be accompanied by internal energy change and therefore gas temperature as well. Minus sign in the expression (6.43) means that the gas volume increase is accompanied by the decrease of its temperature, and compression – by the increase of the temperature.

Fig. 6.12

In the first case the work is carried out by gas at the expense of its own internal energy, therefore its temperature decreases. Then the 197

change of internal energy ∆U < 0 , and the performed work A > 0 . In the second case, the work is carried out by external force and due to this work the internal energy increases and therefore does the gas temperature. Then ∆U > 0 , and the work is negative, A < 0 . This change of temperature can be explained with the use of the following experiment. Let us imagine the vessel with gas in the form of cylinder with displaceable piston (fig. 6.12). Let the piston be  lifted up with the speed u , then gas expands. Let us review certain  molecule moving with speed υ in the same direction with piston. If the speed of this molecule in relation to the walls of the vessel is equal to υ , then in relation to piston its speed is equal to υ − u . After elastic collision of molecule and piston, it will fly back. The speed of molecule in relation to piston does not change, but its speed in respect to the walls of vessel must be υ − 2 u . Thus, all molecules that collide with displaceable piston, bounce back off from it with the lesser speed than before collision. This is what exactly leads to decrease of the average speed of molecules, and therefore to the decrease of the temperature. In the same way during the reverse motion of the piston the increase of the temperature must occur. Poisson equation. Gas state equation under adiabatic process During adiabatic process the change of gas volume by pressure will not be determined by the Boyle-Marriotte’s law. During this process the temperature changes. According to (6.43)

CV dT + pdV = 0 .

(6.44)

In order to solve this equation, it is necessary to determine the change in dT . Using the ideal gas state equation pV = RT , differentiation of the equation gives: pdV + Vdp = RdT , change of temperature determines dT :

dT =

pdV + Vdp . R 198

(6.45)

Placing in (6.44) this value of dT , we will receive: CV

pdV + Vdp + pdV = 0 . R

Instead of R we place its value C P − CV = R and will receive the following expression:

CVVdp + CP pdV = 0 .

(6.46)

Using the heat capacity ratio C P / CV = γ , it is possible to write formula (6.46) in the form:

dp dV +γ = 0. p V

(6.47)

If in example (6.47) γ is constant, the this equation can be written in the following way:

∫

dp dV +γ ∫ = 0. p V

After integrating this equation, we receive:

ln p + γ ln V = const .

(6.48)

Using potential, the equation (6.48) can be written in such way:

pV γ = const .

(6.49)

Expression (6.49) is the very ratio between the pressure and ideal gas volume under adiabatic process of volume change. This expression is called Poisson equation or the equation of ideal gas under adiabatic process, γ = CP – value of adiabatic. CV 199

When integrating we accepted value from the constant. In the strict sense, this is not exactly for real gases. Heat capacities C P , CV in real gases may vary with the change of volume, pressure and temperature. Therefore, Poisson equation (6.49) is strictly correct for limited interval of values of pressures and volumes. But differential equation (6.47) is exact in all cases.

Fig. 6.13

In the process of adiabatic change of gas volume its pressure changes in inverse proportion to the volume in degree of γ , which means that p ~ 1 γ . And in the isotherm process, when the pressure V changes in inverse proportion to volume, the volume of degree is equal to one, p ~ 1 . V Adiabatic value γ is more than one, in other words γ  1 , It is clear that the dependency graph of pressure from the volume under adiabatic process will not be hyperbola like under isotherm process (fig. 6.13). Criss-cross curve p = f (V ) in adiabatic process called adiabatic is steeper than isotherm. Therefore upon gas adiabatic increase the criss-cross curve of dependency p from V – adiabatic (6.13 fig) is lying lower than isotherm. The steeper pressure drop with the volume increase under adiabatic process can be explained by the fact that in case of ideal gas adiabatic expansion its pressure decreases not only due to volume increase, but also due to gas temperature decrease that takes place at the same time. 200

Let us find ratio between the temperature and gas volume in during adiabatic change. Let us find pressure value from Mendeleyev-Clapeyron equation p =

RT , by placing value in equation V

(6.49) we receive:

pV γ =

RT γ V = const V

or

TV γ −1 = const .

(6.50)

In the same way, by placing in equation (6.49) the value of V, discovered according to Mendeleyev-Clapeyron law V = RT , we p will receive ration between pressure and temperature during adia-batic process: γ

γ 1− γ  RT   = const or T p = const . pV γ = p  p 

(6.51)

Taking (6.51) in degree of 1 we receive: γ 1− γ

Tp

γ

= const .

(6.52)

So, formulas (6.49), (6.50) and (6.52) are the equations of gas state during adiabatic process. Equation (6.49) determines ratio between pressure and ideal gas volume during adiabatic process, equation (6.50) – ratio between temperature and volume, and (6.52) – ratio between temperature and gas volume. Work during adiabatic change of ideal gas volume Using Poisson equation, let us calculate the work, performed by gas during its adiabatic expansion or the work that is carried out by external forces during gas compression. 201

We will calculate the expansion work of 1 mole of gas from the volume V1 to V2 (fig. 6.14). Elementary work during volume change to dV is equal to dA = pdV :

Fig. 6.14

Relation between gas presssure and its volume is determined by the adiabatic equation, in other words, by Poisson equation:

pV γ = const , which we may write in the following way: pV γ = p1V1γ where

p1 – initial gas pressure, V1 – its initial volume. Hence p =

p1V1γ . Vγ

By placing this pressure value into formula for work, we will receive: p Vγ (6.53) dA = pdV = 1 γ1 dV . V In order to receive expansion work from V1 to V2 it is ecessary to integrate expression (6.53) within these volumes: γ

A = p1V1

V2

1

202

dV

∫ γ V V

.

For integral in the right part of the equality: V2

dV

∫ Vγ

V1

=

1  1 1   γ −1 − γ −1  .  γ − 1  V1 V2 

Therefore the expression for work А takes the following form: γ −1

1  p Vγ 1  V  p Vγ  1 A = 1 1  γ −1 − γ −1  = 1 1 ⋅ γ −1 1 −  1  γ − 1  V1 V2  γ − 1 V1   V2  

 . 

After finding p1 from Mendeleyev-Clapeyron equation it follows that p1 = RT1 from where we get the final exprerssion: V1 A=

γ −1 RT1   V1   . 1 −    γ − 1   V2    

(6.54)

Equation (6.54) determines the work during adiabatic increas of ideal gas volume from V1 to V2 . If gas mass is equal to m , then formula (6.54) takes the following form: A=

γ −1 m RT1   V1  1 −   M γ − 1   V2  

 . 

(6.55)

In the event of using T1V1γ −1 = T2V2γ −1 (example (6.50)), then γ −1

 V1  T   = 2 . Therefore, the formula for work during adiabatic T1  V2  change of volume can be presented in the following form:

A=

where

R

γ −1

RT1  T2  R 1 −  = (T1 − T2 ) = CV (T1 − T2 ) , γ − 1  T1  γ − 1

= CV . 203

(6.56)

Comparison of expressions (6.55) and (6.40) shows that during adiabatic expansion the lesser work is performed than during isotherm one under the same change of volume. The work during adiabatic expansion depends on adiabatic value γ = CP . If adiaCV

batic value tends to γ → 1 , then the work during adiabatic expansion is equal to work during isotherm process. This, the above mentioned can be seen from the description of gas volume changing process (fig 6.13). Initial conditions of gas during adiabatic and isotherm process are the same, change of volume – also ∆V = V2 − V1 . But the work during isotherm process is equal to area 12′V2V11 АВ, shown as изображенная isotherm curve 12′ , the work during adiabatic process is equal to area 12V2V11 АВ, shown as adiabatic curve 12, which means lesser, because S12′V2V1 1  S12V2V1 1 . Heat capacity during adiabatic process. Gas heat capacity during adiabatic process is equal to zero, because dQ = 0 :

0  dQ  C ад =  = 0.  =  dT  ад dT

(6.57)

6.7.3. Polytropic process Revision of the above isotherm and adiabatic processes of gas state change are particular cases of more general process which is called polytropic process. Each state change process is called like dQ that, during which heat capacity С remains constant and equal to dT

C=

dQ or dQ = CdT . dT

(6.58)

Let is find common equation of polytropic process. According to the first law of thermodynamic 204

dQ = CdT = CV dT + pdV or

(C − CV )dT =

pdV .

(6.59)

Instead of temperature change dT let us put the value discovered from (6.45) pdV + Vdp pdV + Vdp dT = = R CP − CV let us write in the form of: C − CР ( pdV + Vdp ) = pdV CP − CV

or

dP C − CP dV = 0. + P C − CV V

After performing simple transformations, we will discover:  C − CV  C − CV  Vdp − 1 pdV = − CP − CV  CP − CV 

let us write in the form of:

C − CV dp C − CV dV =− CP − CV p CP − CV V integrating this equation, we will write it in the following form: ln P +

where n=

C − CP ln V = const , C − CV

C − CP  or C = CV  n − γ n − 1  C − CV

is called polytropic exponent.

205

(6.60)

(6.61)

Let us write the equation (6.61) with the help of potential in the following form: pV n = const . (6.62) This formula (6.62), constitutive relation between pressure and volume of gas is called gas equation during polytropic process or polytropic equation. Here polytropic value n may possess the value from (-∞) to (+∞).

Fig 6.15

In table 6.2 the values n of polytropic process are presented during its equality with the isoprocesses that are under our consideration, as well as isoprocesses corresponding to it. It is seen from the table that polytropic value n is subjected to formula (6.61). In isobaric process polytropic value n = 0 , p = const , C = C p . During isotherm process C = ±∞ (because dT = 0 ), then n = 1 . During adiabatic process C = 0 (because dQ =0), then according to formula (6.61) polytropic value is equal to n = CP =γ , CV during isochoric process C = CV , then n = ±∞ . During adiabatic process the dependence of heat capacity from polytropic value in different isoprocesses is show in fig. 6.15 206

Table 6.2 Polytropic value, n 0 1

γ

±∞

Heat capacity, Сх

Cp ±∞ 0

CV

State equation,

pV n = const p = const pV = const pV γ = const V = const

Isoprocess Isobaric Isotherm Adiabatic Isochoric

Adiabatic compressibility of ideal gas. Adiabatic compressibility of ideal gas is determined based on the formula (2.59) so: χ = − 1p.

Gas compressibility in adiabatic conditions will be different. It is related to the fact that during adiabatic change of pressure the change occurs not only in respect to volume but to gas temperature as well. If gas is compressed, the temperature rises. Gas compressibility during adiabatic change of pressure is lesser than during isotherm one. Coefficient of ideal gas adiabatic compressibility χ = 1  dV   

V  dp T

can be found from equation (6.48):

χ = − 1γp .

(6.63)

Thus, adiabatic compressibility of gas γ times lesser than isotherm one.

207

7 SECOND LAW OF THERMODYNAMICS 7.1. Introduction Irreversibility and Probability. On the basis of the thermal molecular motion, processes are irreversible. Concluding basic equations of an ideal gas based on the kinetic theory, we assumed that the motion of individual molecules depends on the laws of mechanics. But only motion of a huge number of molecules forms irreversible changes, while motion of an individual molecule follows the law of mechanics and is reversible. So why is the motion of numerous particles not like the motion of an individual molecule? The reason is in a large number of particles and in a chaotic motion. Imagine that a vessel with a volume of V is full of air. Mentally, we divide the volume into five v parts of the volume and we launch a molecule from one side. The molecule, which has penetrated with the particles, randomly moves chaos and gets into the initial volume of the vessel, which remains in the initial v volume. The probability of finding a molecule in a v volume is 1/5. This is a high probability. If instead of one molecule we launch two molecules into the initial v volume, then their probability will be equal to (1/5)2 based on the probable numbers addition theorem. This is a fairly low figure, unlike the previous one, and therefore irreversibility of the transition process of two molecules from V to decreases v . If we put 3,4,5, ... molecules in volume v , the probability of finding in the initial volume v will be as follows: (1/5)3; (1/5)4 ; (1/5)5; … .

208

If from volume V we take volume v , it will be v =

V , and m

then the probability of finding particle N in this volume will be N

1 equal to Ρ =   . If N is greater, then P will be very small.  m Gas that is in the volume v or gas in the v state completely fills volume V . Thus, it goes into V state. The gas probability in the first state is N

N

1 v Ρ =   or Ρ =   ,  m V  hence m =

(7.1)

V . v

The probability of finding gas in the volume is one, because volume v is a part of V; therefore, gas cannot be in a different volume other than this vessel. If (7.1) v = V , then Р = 1 . If the number of particles N is very large, then only in one part of the volume V the probability of gas location will be low, even if volume v is approximately equal to the volume. For example, gas simultaneously fills the atmospheric pressure and room temperature (T = 300K), but not the vessel of 1 ⋅ 10−9 cm3 of the part, with a volume of V = 1 cm3.

V−v v = 1 − = α = 10− 9 . V V p Number of molecules N = ≈ 2.4 ⋅ 1019 , based on the kT Then

Formula (7.1) probability Р is equal to: N

v N Ρ =   = (1 − α ) . V  Using the conversion method for low values, 209

(7.2)

(1 − α ) lim α →0

−1

2

= e = 10 lg e = 10 0, 43 ,

we reduce the following formula (7.2) as follows: N

1 1 1 v N ≈ 0, 43⋅2, 4⋅1019 ⋅10 − 9 ≈ 10 . Ρ =   = (1 − α ) = − 1 Nα α V 10 10   (1 − α )

The sum of dividing the equation is very large. Therefore, there is a low probability that molecules will accumulate in a small part of the vessel. This is the cause of irreversibility among natural processes. If gas will fill the vessel by itself, then perhaps it can gather in one place. Molecular motion, following the laws of mechanics, confirms that these two processes are possible, but because of the large number of particles the probability of an independent gas arrangement in the vessel is very low, so it is impossible to observe such a process. This explains the probability of heat transfer from the cold body to hot one or independent decomposition of a gas mixture into components. In such similar situations, processes go into a high state of equality. But the reverse process of inequality state is impossible, because probability of such a state is low. Therefore, the motion of each molecule in the molecular system is reversible, despite following the law of mechanics, a large number of these particles and their chaotic motion leads to an irreversible process. If their number was smaller, then there were no irreversible processes in the system.

7.2. Change of heat into mechanical work Such concepts as equality, reversibility and irreversibility apply equally to every process in nature. In nature, one of the frequent processes of transition to a state of equilibrium is the conversion of mechanical energy into heat. For example, the release of heat in the friction of solids. Mechanical energy is a macroscopic energy, i.e., potential energy, which is 210

regarded as a whole system interacting with external and internal fields through the kinetic motion of bodies. And kinetic energy of thermal motion of the molecules in a body and potential energy of interaction is called internal energy. In this case, the heat released by mechanical energy can be called the process of conversion of the thermal motion of macroscopic energy into microscopic energy. In XVII century and XVIII century the century of thermal levers began. The use of thermal levers required the solution of the problems of heat and mechanical work interaction and laid the theoretical foundation. The methods of obtaining mechanical work with the help of heat led to the emergence of the term thermodynamics in the history of the industry. For example, the use of nuclear energy in the work is an important event in our time. But, nuclear energy does not immediately go into mechanical work, this only happens with the help of heat. Thus, the study of the laws of mutual conversion of mechanical work and heat energy are important problems. Thermodynamics has developed in history as a branch of physics, studying the laws of mechanics and heat. Now thermodynamics considers the general laws of thermal energy and the relationship with other energies (chemical, electrical, nuclear, etc.). Mechanical and thermal energy are connected by the first law of thermodynamics (the law of conservation of energy, 6.3). If heat energy dQ is released into a system, its state will change and it will work dA . Then the law of conservation of energy according to formula (6.4) will be equal to the sum of the work performed and the change of internal energy:

dQ = dU + dA , where dA = pdV . This formula can be transformed as follows:

dU = dQ − dA . If the state of the system undergoes macroscopic changes, then it is necessary to subtract the sum dQ and dA ; therefore, when the system passes from the first state to the second, the internal energy is subtracted as follows: 211

2

2

2

∫ dU = ∫ dQ − ∫ dA . 1

1

(7.3)

1

The released volume of heat and the work completed (or excessively completed) by the system will depend on transition from the first state to the second state. A change in internal energy dU does not depend on the transition, but is only calculated with the help of the initial and final states. Therefore, equation (7.3) can be written as follows: 2

2

2

1

1

1

∫ dU = U 2 − U1 = ∫ dQ − ∫ dA , but this equation 2

2

2

∫ dU = U 2 − U1 = ∫ dQ − ∫ dA = (Q 1

1

2

− Q1 ) − ( A2 − A1 )

1

cannot be written in this way. In this regard, the system in any state has a certain value of internal energy, but this does not mean that it has a volume of heat and work. Therefore, heat Q and work A are the functions of the process of changing the system of work. Mathematical dQ and dA are not a total differential, but dU is considered a total differential. Now consider the conversion process of heat transfer to mechanical energy. The energy obtained when passing from a body with a high temperature to a body with a low temperature, for example, in friction or glowing, is called the volume of heat. With this exchange of energy, no work is performed, because there is no movement of bodies. In this case, the internal heat energy increases and when temperatures are compared, the heat exchange process stops. But if in heat exchange a body expands, then work takes place. According to the law of energy conservation, the work will be equal to: 212

dA = dQ − dU , where dU – is the change of internal energy. A high value of work occurs in an isothermal process, and only then dA = dQ . No greater work than that can be performed. Therefore, in order to perform the maximum work, there should be no difference between the expanded body and the heat source. If there is no difference in temperature between the heat source and the body, then there will be no heat exchange. But, for the transfer of heat, the infinitely low temperature difference suffices, and only then isothermal conditions are maintained. Such conditions require an infinitely long process of heat transfer, hence the process is reversible.

7.3. Cyclic Processes. Cyclic Work Returning to the initial state after numerous changes is a very important process. Such processes are called cyclic or circular, where the change in internal energy is zero:

∫ dU = 0 . But this does not prove that Q and A will be zero. In cyclic processes, a body consumes or releases enough heat, but internal energy does not change. In this case, the first law of thermodynamics is as follows:

∫ dQ = ∫ dA , where

∫

is a circular integral sign.

The first law of thermodynamics is applied in processes of both equilibrium and inequality. But the first law of thermodynamics is not applied in calculating the direction of the process of changing the state of the system. In accordance with the first law, heat can pass from a hot body to a cold one and vice versa (without the influence of the external environment). Whatever it was, it is enough for the 213

heat volume received from one body and transferred to another body to be equal. Therefore, the first law of thermodynamics does not determine the irreversibility of processes in nature. From the above, the conversion of heat into mechanical work is carried out with the repeated return of heat to the workinf medium. But, with the technical use, such a method of converting heat into mechanical work is not suitable. Devices that convert heat into work must operate cyclically, because the periods of heat transfer and conversion to work are repeated. For this, the body, performing the work, after receiving heat returns to its original state and starts the process anew. Hence, the circular process begins. As a result, the state of the system, subject to numerous changes, and with a repeated initial state is called a cycle. A circular process (cycle) is represented by a closed 1a 2b1 line in the diagram p − V , (Fig. 7.1). If you take work on the la2 part of the cycle on the right side, then at the bottom of this line 1a 2V2V11 it will be equal to the area (right-hashed).

Fig. 7.1

The work of 1b2 part of cycle is not correct and quantitatively is equal to the area under this line 2b1V1V2 2 (left-hashed). Consequently, the work performed in the full cycle from the quantitative side will be equal to 1a 2b1 line surrounding the area. 214

Then at the bottom of this line will be equal to the area (shaded with a slope to the right side). The work of part of cycle 1b2 is not correct and quantitatively equal to the area under this line (shaded with a slope to the left). Consequently, the work done in the full cycle from the quantitative side will be equal to the line surrounding the area. A very important question: is the work equal to the amount of heat obtained from a heat source in a cyclic process? On the one hand, this question can be answered positively, because as a result of a cycle the workinf medium returns to its initial state, the internal energy remains unchanged and the work must be equal to the heat consumed. But in fact, the results of experiments show the opposite. The answer to this question was given by V. Thomson (Calvin) in 1854. "It is impossible to carry out a cyclic process, the only result of which would be conversion into mechanical work of heat taken from some body, without any changes taking place in another body or bodies". This answer is called the Kelvin principle. Thus, the heat borrowed from the heat source can be converted into work in a cyclic process, with the indispensable condition that the state of some other bodies must change Without this, heat cannot convert into mechanical work. Hence, some third body must also participate in the process of conversion of heat into work, in addition to the source of heat and the body that performs the work. In order to convert heat Q0 into work, it has to be obtained from a source with temperature T0 and transfer it to a body with a temperature lower (Т T1 in heat transfer upon contact will not be accompanied by performance of effective work. Therefore, the workinf medium must first be cooled to the refrigerator temperature T1 and only after that they can be brought into contact (Fig. 7.2). To cool the workinf medium, it is necessary to isolate the workinf medium from the heater, and to enable it to expand adiabatically. When the body is adiabatically expanded, the temperature will drop and it will take the temperature of the refrigerator T1 (ВС curve in Fig. 7.3). We know that adiabatically expanded body has cooling properties. At this time, the workinf medium expands and moves the piston, while performing additional mechanical work (Figure 7.3). Thus, the cooled workinf medium is brought into contact with the condenser. This completes the first stage of the cycle and, due to the heat obtained from the heater, the workinf medium performs efficient work. Now it is necessary to return the workinf medium to the initial state, i.e. to restore the initial pressure and temperature. Only then can the workinf medium be compressed and come into contact with the heater. This contact does not occur immediately, because the body temperature is below the temperature of the heater. Therefore, the return to the original state is carried out in two stages. First, the workinf medium is compressed, without interrupting its contact with refrigerator, i.e., compression process is isothermal (CD curve in 218

Figure 7.3). Then, by isolating the workinf medium from refrigerator, it is additionally compressed adiabatically before heating to the heater temperature (DA curve in Fig. 7.3). Under adiabatic compression, the body heats up due to external work (6.7.2). After in the process of adiabatic compression the temperature of the workinf medium becomes equal to the temperature of the heater, they are brought into contact, and the cycle ends. The workinf medium is in its initial state and the process can be started anew. The described circular process consists, therefore, of two isothermal and two adiabatic extensions and contractions. In the extensions, the workinf medium performs efficient work, while the work of external forces results in compression. In the examined circular process, the different bodies are not in contact, i.e. there is no irreversible process of heat conductivity. The entire cycle is reversible, these processes can be called quasi-static if the changes, expansion and contraction are infinitely slow. French scientist Saadi Carnot had used the described cycle in 1824 when examining the operation of an ideal heat engine. Therefore, this cycle is called the Carnot cycle. Iideal gas was used as the working medium. In the work of this machine it was believed that heat spreads into the environment, and there is absolutely no annihilation. Let's calculate the proportion of heat received from the heater. As a workinf medium, we take 1 mole of ideal gas and at its point A (Fig. 7.3) its initial state of pressure P0 and volume V0 are described. By condition, gas temperature is equal to that of the heater

T0 = p0V0 R . Let us denote the temperature of the refrigerator as T1 . Hence, T0  T1 . In the initial state, the working medium contacts the heater. At the first stage of the circular process, the ideal gas expands to an isothermal volume V1 . Then the pressure drops to p1 value (Fig. 7.3 point B), the positive value A1 works as follows:

A1 = RTо ln

V1 = Q0 . V0

219

(7.4)

Where Q0 – is the amount of heat produced by gas from heater. Work A1 is performed using Q0 heat. In the second cycle, gas is isolated from the heater, and further expansion occurs adiabatically, at which point the gas cools. When gas temperature equals the temperature of the refrigerator, i.e. before T1 , adiabatic expansion stops. We find gas volume in the adiabatic expansion using equation (6.50).

T0V1γ −1 = T1V2γ −1 .

(7.5)

From this the following formula follows: γ −1

 V2     V1 

=

T0 . T1

(7.6)

Pressure p2 in this case varies adiabatically (point C, Fig. 7.3). The work performed in the second stage is calculated as follows: (6.56) γ −1 RT   V1   R(T0 − T1 ) . (7.7) 1 −    = A2 = γ − 1   V2   γ −1   At the third stage of the cyclic process, gas is isothermally compressed by external forces at the refrigerator temperature T1 from volume V2 to volume process is equal to:

V3 . The work performed over gas in this

A3 = RT1 ln

V3 V = − RT2 ln 2 = Q1 . V2 V3

(7.8)

At the expense of work A3 , heat Q1 is released and transferred to the refrigerator with which the gas contacts. 220

In order to return to the initial state, it is necessary to go through the last stage of adiabatic compression, where gas passes through three states: volume V0 , pressure P0 , and temperature T0 . In the third stage, the gas is compressed to volume V3 , with this condition being fulfilled: γ −1

 V3     V0 

=

T0 , T1

(7.9)

because T1V3γ −1 = T0V0γ −1 . The compression work at the last stage of the cycle equals:

A4 =

R(T1 − T0 ) R(T0 − T1 ) . =− γ −1 γ −1

(7.10)

Now gas is in its original state, the process of the Carnot cycle is completed, and the process is ready to start again. What is the result of the cycle? Has the task been accomplished to achieve the conversion of heat into mechanical work? The work performed independently by gas and over gas is equal to: A = A1 + A2 + A3 + A4 . We will put their values to problems (7.4), (7.7), (7.8) and rearrange: A = RT0 ln

V V1 R(T0 − T1 ) V R(T0 − T1 ) V = RT0 ln 1 − RT2 ln 2 − RT2 ln 2 − + γ −1 γ −1 V3 V0 V3 V0

from problems (7.6) and (7.9)

V V V2 V3 or 1 = 2 it follows. = V1 V0 V0 V3 221

Hence

ln

V1 V = ln 2 = ln r , V0 V3

(7.11)

where V1  V0 and V2  V3 , i.e. ln r  0 . The worked performed as part of the cycle shall be equal to:

A = R(T0 − T1 )ln r ,

(7.12)

where T0  T1 , i.е. A  0 . Hence, the work performed by gas during expansion is greater than the work of external forces spent on its compression. Due to the heat received by the working medium from the heater, the working medium (ideal gas) performs efficient useful work. But, this work is not equal to the amount of heat Q0 that the working body has received from the heater. From the amount of heat Q0 released by the heater Q0 = RT0 ln

V1 . V0

Q1 = RT1 ln

V3 V2

Part equal to Q1

was transferred to the refrigerator under isothermal compression of gas from volume V2 to volume V3 . Therefore, only part of the received heat was possible to convert into effective work, equal to:

Q0 − Q1 = R(T0 − T1 )ln r = A .

(7.13)

The area of the ABCDA curve (Figure 7.3), the digital value of work A is determined by the graph. 222

There is a difference between the conversion of heat into work and the conversion process. Due to certain situations, mechanical work can completely convert into heat. Bit only some of the heat will convert into work. Equations (7.4) and (7.8) can be written as follows:

Q0 V Q V = R ln 1 , − 1 = R ln 2 . T0 V0 T1 V3 Combining them and taking into account (7.11), we obtain the following formula:

Q0 Q − 1 = 0, T0 T1 from where

Q0 Q1 , = T0 T1

(7.14)

i.e., in thermodynamics this comes to a very important equation. 7.4.2. Efficiency Factor in the Carnot Cycle When analyzing the circular Carnot process, it follows that through it, it is impossible to completely convert heat, borrowed from a heater, into mechanical energy. Part of this heat must necessarily be transferred to the refrigerator, i.e. to a body with a lower temperature than the heater. If the amount of heat Q0 received by the working medium from the heater, and Q0 − Q1 will convert to work, then:

η=

Q0 − Q1 Q0

(7.15)

the ratio is the efficiency factor of the circular process of the machine operation. The efficiency factor of the Carnot cycle is determined by the formula (7.14): 223

η=

Q0 − Q1 Q T = 1− 1 = 1− 0 . Q0 Q0 T1

(7.16)

The efficiency factor (EF), therefore, is always less than one ( η < 1 ) and depends on the ratio between the temperatures of the heater and the refrigerator. It is impossible to get a value greater than the EF by the equation (7.16). Because in the Carnot cycle, bodies with different temperatures never touch each other, which eliminates the possibility of irreversible heat conductivity processes. The change in the volume of the working medium is reversible. In this case, the maximum work is performed; therefore, the most useful conditions for thermal energy are fulfilled. 7.4.3. Carnot Theorems A thermal machine operating at certain temperatures of a heater and a refrigerator cannot have efficiency that is greater than a machine operating under the reverse Carnot cycle at the same temperatures of the heater and the refrigerator. This statement is called the first Carnot theorem. Formula (7.16) shows that the efficiency factor of the Carnot cycle does not depend on the nature of the working medium, but only on the temperatures of the heater and the refrigerator. This statement is the content of the second Carnot theorem. In calculating, we have chosen the ideal gas as the working medium only because the equation of state is precisely known for it, which made it possible to easily calculate the value of the efficiency factor. The fact that the efficiency factor of the machine operating in the Carnot cycle is maximal, is due, as we see, to the circumstance that this circular process is completely reversible. 7.4.4. Refrigerating Machine If the reversible process is carried out in the opposite direction, the body participating in the process will pass through the same 224

states, but in the reverse order. As applied to the Carnot cycle, this means that heat will not be transferred from the heater to the refrigerator, but vice versa – from the refrigerator to the heater.

Fig. 7.4

The reversed Carnot cycle begins with the fact that the working medium A, for example, in a state corresponding to point D (Figure 7.4), expands adiabatically to the state corresponding to the point (Fig. 7.3). On the straight cycle the process starts with isothermal expansion). Then isothermal expansion to the state C follows. In these first two stages of the cycle, the working medium expands and performs work. In the second half of the cycle, along CB adiabat and BA isotherm, the working medium compresses. The work of compression is performed over the body by an external source of energy. As has just been shown, this work surpasses the work performed by the body itself in the first half of the cycle. Therefore, the result of the reverse Carnot cycle will not be the external useful work, but heat transfer from refrigerator to the heater, i.e., from a less heated body to a more heated one. If a plant operating in a direct Carnot cycle serves to convert heat into mechanical work, that is, it is a heat engine, then the machine operating on the reverse Carnot cycle is used to transfer heat from a less heated body to a more heated one, i.e., is a refrigerator. With its help due to external mechanical work, heat is taken away from the cold body and transferred to the body with a higher temperature. 225

7.4.5. Free Energy In the Carnot cycle, the working medium performs work in the first half of the cycle, first in the processes of isothermal and then adiabatic expansion. In the adiabatic process, the work, as is known, is performed using internal energy, and this work is equal to the loss of internal energy: dA = −dU . In isothermal process, the temperature of the working medium does not change, so that part of the internal energy, which is related to the kinetic energy of molecular motions, cannot be used to be converted into mechanical energy. This circumstance motivates us to distinguish between the total energy that the body or the body system possesses, and that part of it that under the given conditions can be used to obtain work. Let us imagine gas in which isothermal reversible processes of expansion and contraction can occur. For this purpose, gas must be placed in a thermostat, i.e., it must be brought into contact with a body of a large heat capacity, the temperature of which is constant. When expanding, gas can produce mechanical work, therefore, our system, consisting of a thermostat and gas, has some energy. That part of the energy of the system, which under given conditions can be used to convert into mechanical work, is called free energy. The system cannot perform work exceeding the value of its free energy. As is known, in mechanics, the energy of a body or a system of bodies is equal to the sum of potential and kinetic energies. Both these types of energy of macroscopic bodies can be wholly converted into mechanical work. The internal energy of the molecular system in the case of interest to us cannot be wholly converted into work. Therefore, the work performed in the isothermal process does not describe the internal energy. If you need to find out the working capacity of the system, you need to calculate the internal energy. Because, the work performed in the adiabatic process is equal to the change in internal energy. Free energy characterizes the system during change in the isothermal state. 226

Thus, free energy of the system is measured by the work performed by the system, changing its state isothermally and reversibly from the state it is in, to the initial state chosen by us, at which the free energy is assumed to be zero (the reference point). If we denote free energy of the system as F , then the work performed by the system in the reversible isothermal process dA , is

dA = −dF .

(7.17)

If, for example, a change in the state of a system reduces to an isothermal expansion of a body (an increase in its volume) at which the work is positive, then the minus sign means that free energy decreases. Conversely, when body is compressed (work is negative), free energy increases due to external forces that compress the body (gas). In particular, for an ideal gas, in its isothermal expansion from volume V1 to volume V2 , the work is known to be expressed by the following equation (for one mole):

A = RT ln

V2 . V1

(7.18)

The right-hand side of (7.17) during isothermal expansion 1 of the mole of an ideal gas, we can calculate the drop in the free energy. At a given temperature, free energy of a given mass of gas is the greater, than the smaller is the volume occupied by it, i.e., the more it is compressed. The internal energy of an ideal gas does not depend on the volume occupied by it; one mole of gas compressed in a cylinder has the same internal energy as uncompressed gas at the same temperature. But compressed gas has a large free energy because it can perform a greater work in isothermal expansion. If the process of isothermal change in volume is irreversible, then, since the work performed here is less than in the case of a reversible process, the change in free energy will be greater than the work performed, so formula (7.18) should be written in the following form: 227

dA ≤ dF , where inequality sign refers to the irreversible, and the equal sign – to the reversible process. There are also cases when a change in free energy is not accompanied at all by the performance of work. For example, if ideal gas expands into a vacuum, no work is performed. Gas temperature and internal energy remain unchanged. Meanwhile, free energy of gas has decreased, since the work that gas can perform has decreased. This is due to the fact that the process of expanding gas into vacuum, although isothermal, is completely irreversible. Earlier it was emphasized that free energy characterizes the state of the body. When a body passes from one state to another, it is isothermal and reversible, the work performed, equal to the difference in the free energies of the body in these states, does not depend on the path of the transition. This follows immediately from the fact that under an isothermal reversible circular process, the work is zero. In fact, the body can go from state 1 to state 2 by two different isothermal paths, performing work A1 on the first path and A2 in the second. But, in that case, we can transfer our body from state 1 to state 2 along a single path and return it back, having made a circular process along another path. The general work performed in this process, is A1 − A2 = 0 , therefore, A1 = A 2 . This means that the work performed by the body only depends on the initial and final states of the body. Consequently, free energy is a state function.

7.5. Entropy. Second Law of Thermodynamics 7.5.1. Entropy. Entropy State Function Let us consider the change in the state of the Carnot circular process. From the initial state A , characterized by pressure P0 and 228

temperature T0 , the working medium undergoes successive isothermal and adiabatic extensions of states C, when it takes the temperature of the refrigerator. The return of the body from state C back to state A took place by two successive isothermal and adiabatic compressions of the body. When returning to the initial state, the released amount of heat Q1 is less than that obtained from the heater

Q0 , i.e. Q1  Q0 . Therefore, the amount of heat obtained during transition from state A to state C and its return from state C to state A will not be equal. This is because two processes pass by two different methods (Figure 7.3) - in one case from the expansion process (from A to C), in the second stage due to compression at a higher pressure (from C to A). An important conclusion follows from this. The amount of heat that must be delivered to the body or taken from it during transition from one state to another is not determined by the initial and final states, but essentially depends on the mode of implementation of this transition. In other words, the amount of heat Q is not a body state function, just as free energy is not a body state function. If the amount of heat Q0 delivered to the body from the heater at the temperature T0 transferred by it to refrigerator at a temperature

T1 are not equal ((7.13) formula), the ratios at which they were absorbed or given are numerically equal to each other ((7.14) formula), but have opposite signs:

Q0 Q = 1 , T0 T1 where

(7.19)

Q – the ratio is called reduced heat. T

In a circular process, the heat received and given by the working medium will be equal to ((7.19) formula). This feature of heat allows us to introduce a special thermodynamic variable – entropy, which is of fundamental importance in physics. The importance of this variable is determined by the fact 229

that it is a state function and the role it plays in all processes in nature, in particular, in the process of converting heat into work. Rudolf Clausius had first introduced the term of entropy (en+tropø from the Greek is transformation, change) in 1865. We agree to consider dQ positive when the system absorbs heat, and negative when it releases it. If the system, as a result of any changes in state, reversibly passes from state A to state B, then the B

sum of the reduced amounts of heat, i.e., the variable

dQ

∫T

Does not

A

depend on the path by which the transition from A to B occurs. To do this, it suffices to show that for a circular process, when the initial and final states coincide, this integral is equal to zero:

φ

dQ = 0. T

(7.20)

In any circular process the integral

φ

dQ T

(7.21)

cannot be a positive variable. Suppose that when the state changes, the body G produced a circular process. During the process, the body gave away and absorbed heat. Let us imagine that the heat emitted by the body G is transferred to a certain heat reservoir, with a large heat capacity, the temperature of which is equal to T0 . This transfer can be carried out in a reversible way, for example, with the aid of an intermediate body performing the Carnot circular process, so that the body G will serve as a heater for the intermediate body and the reservoir will serve as a refrigerator. When considering the Carnot cycle (Figure 7.3), the amount of heat dQ0 removed from the body G at the temperature T0 and the amount of heat transferred to the reservoir at the temperature T1 are not equal. 230

Q0 Q1 . = T0 T1 Therefore,

dQ1 =

T1 dQ0 . T0

(7.22)

If T1  T0 , the reservoir serves as a refrigerator, and the body G serves as a heater. If, T1  T0 , then the reservoir and the body G switch roles. After body G completes the process, the total amount of heat lost by the body must be equal, as is clear from (7.22), since heat capacity of the reservoir is large and its temperature remains constant, this quantity is equal to:

φ T1

dQ0 . T0

The heat capacity of the reservoir is large, so its temperature does not change, i.e. T1 can be deduced from the integral:

Tφ 1

dQ0 . T0

The completed process is circular, so there will be no changes at the end. The intermediate body also performs a circular process. If the integral (7.21) was positive, this would mean that the amount of heat lost by the body Tφ 1

dQ0 , has turned entirely into T0

work. Then neither body G nor the intermediate body changed their state. But this is impossible based on the Thomson principle. Hence, the assumption that φ

dQ0  0 is impossible. T0

The integral (7.21) can not be negative. In fact, if you make all the changes the opposite, then the heat coefficient will change. If in 231

dQ  0 , in the reverse process T the same integral becomes a plus sign, but as we noted, it is not possible. Therefore this integral cannot be negative. If the integral cannot be positive or negative, then for reversible cycles it equals zero

the direct process it looks like this φ

φ

dQ = 0. T

Therefore, in any reversible cycle

∫

(7.23)

dQ the integral value will T

not depend on the process. There is a co-factor of the state function that describes the process, we mark it with an S sign. In case if a system’s transitions from state A to state B is reversible, then the function of changes of this state is determined as follows: B

SB − S A =

dQ . T A

∫

(7.24)

According to (7.24) equation, the change of this function can be observed only when passing from one state to another. The absolute total is not determined. But, you can note one state by making the S function value equal to zero, thus comparing other states of the system. Therefore, taking into account all that was said for formula (7.21), function S is equal to the integral being determined:

S=∫

dQ . T

(7.25)

The S coefficient determined in this way is called entropy. In practice, the it is not the S coefficient that is determined, when the state system is changed, only entropy is calculated. 232

Thus, zero entropy value can be determined by the state of any system. The temperature in the absolute zero value is equal to the zero entropy value. When the system is given infinitely little heat dQ , the change in its entropy is determined as follows:

dS =

dQ . T

(7.26)

The amount of heat dQ that cannot be a total differential relative to the temperature, which is a function in this equation, turns into a total differential. Therefore, the 1/T coefficient is an integral product. The formally negative value of temperature is equal to the integral product of the volume of heat dQ .

7.5.2. The second beginning of thermodynamics. Clausius Inequality Using equation (7.26), we can write the first combined form of the beginning of thermodynamics ((6.4) formula):

TdS = dU + dA .

(7.27)

This equation is called thermodynamic equality. This equation is often called the second beginning of thermodynamics for recurrent processes. The system in which the recurrent processes take place, described by the state of entropy functions, is called the second beginning of thermodynamics. If the cycle which the system passes is irretrievable, then:

φ

dQ  0. T 233

(7.28)

This inequality is called the Clausius inequality. For the irreversible Carnot cycle, the heat connection looks like this:

Q1 Q2  0. + T1 T2 Concerning any system in the Carnot cycle, inequality is the analysis of the process result (7.28).

7.6. Differential equation of thermodynamics The properties of objects can be studied using thermodynamic laws. This thermodynamic method helps, to firstly find the connection between the properties of the system in the state of thermodynamic equality. Secondly, it helps to establish the conditions that determine the state of equality. To solve this problem, differential equations of thermodynamics are used. These equations are solved by means of the (7.27) equation, based on the combination of the first and second law. This (7.27) sign is written as follows for the equality process:

dU = TdS − pdV .

(7.29)

Here, dA = pdV . (7.29) the front coefficients on the right-hand side of the differentials are generalized forces, and coefficients at the bottom of the differential sign are called generalized coordinates. T , S , P and V parameters describing the system state, taking into account that their differentials are total, are the differential equations of thermodynamics. Therefore, T , S , P and V , any of these coefficients are called variables, the rest can be found through them. 234

If x and y are independent variables, U ( x, y ) , then total diffe-

 ∂U   ∂U   dx +   dy and  ∂x  y  ∂y  x

rential is determined as follows: dU = 

∂ 2U ∂ 2U . = ∂x∂y ∂ydx If we differentiate (7.29), we obtain the following:  ∂p   ∂V  ∂T   ∂S   ∂T   ∂S     =        −   ∂x  y  ∂y  x  ∂y  x  ∂x  y  ∂x  y  ∂y

  ∂p   ∂V   −     . (7.30) ∂ y ∂ x  y x  x

In this equation, any two coefficients can be put in instead of coefficients T , S , P and V . For example, the change in volume to dV and of temperature to dT is due to a change in the state of the system. This process leads to the entropy change by dS . Then, in equation (7.30), we need to put x = V and y = T instead of x, y .  ∂p   ∂p   ∂p   ∂V  . (7.31)  ∂T   ∂S   ∂T   ∂S      −    =    −     ∂V T  ∂T V  ∂T V  ∂V T  ∂V T  ∂T V  ∂T V  ∂V T

(

∂V

Here V and T are independent inconstant coefficient, hence, = ∂T = 0 . Therefore, ∂T V ∂V T

) (

)

 ∂p   ∂S   .  =   ∂V T  ∂T V

(7.32)

Total entropy differential S (V , T ) will be equal.

 ∂S   ∂S  dS =   dV .  dT +   ∂V T  ∂T V 235

(7.33)

In the latter equation, total change in entropy is equal to the combination of its growth in temperature and volume. Taking (7.32) into account, we find (7.33) as follows:

 ∂p   ∂S   dV .  dT +   ∂T V  ∂T V

dS = 

(7.34)

The first part of (7.34) ( dV = 0 ), when the volume is constant, determines the change in entropy only because of the change in temperature (dS )V . By definition (7.26), (dS )V =

(dQ )V T

, there the

heat volume (dQ )V is needed to change the body state. Then (dQ )V = CV dT .

Here CV is the capacity of heat at constant volume.

C  ∂S   dT = V dT , (7.34) value is written as T  ∂T V

Therefore,  follows:

dS =

CV  ∂p  dT +   dV . T  ∂T V

(7.35)

In this equation, by determining p , V , T and CV coefficients, you can find the change in entropy dS . If we take p and T as independent variables, then changes in entropy can be found through the change of dT and dp . To do this, we must put p and obtain

T

in (7.30), in place of x and y. Then we

 ∂T   ∂S   ∂T   ∂S   ∂p   ∂V       −   =    ∂T  P  ∂p T  ∂p T  ∂T  P  ∂T  P  ∂p 236

  ∂p   ∂V   −     T  ∂p T  ∂T  P

 ∂T  ∂p  But,   = 0 , because T and p are independent  =   ∂T  P  ∂p T changed variables. As a result of transformation comes something similar to (7.32). I.e.  ∂S   ∂V  . (7.36)   = −   ∂T  P  ∂p T And (7.35) change of dS of entropy is found using dT and

dp : dS =

CP  ∂V  . dT −  dp T  ∂T 

(7.37)

By integrating (7.35) and (7.37), we can find out a certain body mass, for example, S (V , T ) or S ( p, T ) of the modules or entropy for certain units V and T or T and P . In order to make calculations, according to the entropy parameters, p0 , T0 and V0 ,T0 , there should be S ( p0 , T0 ) and S (V0 ,T0 ) present. In this case

CV dT T ,V  ∂p  S (V , T ) − S (V0 , T0 ) = ∫ + ∫   dV T ∂T  V T0 ,V0 To ,V0  T ,V

and

S (P, T ) − S (P0 , T0 ) =

C P dT P ,V  ∂V  + ∫   dV . ∫ T ∂T  V T0 , P0 P0 ,Vo  T ,P

For example, per one mole of ideal gas: T

∫ CV

T0

V

 ∂p 

V

dV

∫   dV = V∫ R V V  ∂V V 0

T , dT = CV ln T0 T

= R ln

0

237

V , because V0

R  ∂p    = .  ∂T V T

Thus, the change in entropy equals to:

∆S = S (V , T ) − S (V0 , T0 ) = CV ln

T V + R ln . T0 V0

(7.38)

In the last equation it is clear that when temperature and volume of gas increase, the entropy grows. If the isotherm of an ideal gas increases, then T = To, in this case the change in entropy is determined as follows:

∆S = R ln

V . V0

(7.39)

7.7. Examples of application of differentiated thermodynamic equations The differentiated thermodynamic equations establish a connection between the properties of substances, for example, establish a relationship between p, V, T (thermal properties) and S, U, F, I, Cv, Cp (calorific properties). Let us consider examples of the application of differentiated thermodynamic equations

7.7.1. Dependence of internal energy on volume. Internal energy of Van der Waals gas Internal energy of the exact gases U varies depending on the gas temperature T and the gas volume V, meaning

U = U (T ,V ) .

(7.40)

Let us determine the dependence of the internal energy on the volume using differentiated thermodynamic equations. The total differential of internal energy looks like this: 238

 ∂U   ∂U   dT +   dV .  ∂V T  ∂T V

dU = 

(7.41)

This (7.41) consists of the equation of total internal energy and exact gases: first:  ∂U  dT is a temperature change,  ∂T V second,  ∂U  dV is change in the volume.  ∂V T Thermodynamic equation ((7.27) formula) is calculated like this:

dS =

1 dU + pdV 1 = dU + pdV . T T T

(7.42)

If this formula (7.42) is set as the determining element dU (7.41), then the entropy change will be equal to:

dS =

 1  ∂U  1  ∂U   + p  dV +   dT .  T  ∂V T T  ∂T V 

Now let us compare this equation with the above-mentioned sample (7.34):

 ∂S   ∂p   dT .  dV +   ∂T V  ∂T V

dS = 

In both equations, the coefficients are dV , therefore

 1  ∂U   ∂p   + p .  =    ∂T V T  ∂V T 

(7.43)

From this  ∂U  we determine the value, which will be equal to:  ∂V T

239

 ∂U   ∂P    = T  − P . ∂ V  T  ∂T V

(7.44)

For an ideal gas, we find the first part on the right-hand side of equation (7.44), differentiating the Mendeleev’s-Clapeyron equation.

pV = RT , pdV + Vdp = RdT . From this then V=const

R  ∂p   ∂p   = , in this case T   = p.   ∂T V V  ∂T V Therefore,

 ∂U   = 0.   ∂V T

(7.45)

Hence the internal energy of the ideal gas does not depend on the volume. The foregoing does not apply to exact gases. Condition:

P=

RT a − 2. V −b V

We calculate the internal energy of the exact gas, which is determined by the van der Waals equation (formula 5.23), we in use formula (7.44). By differentiating the Van der Waals equation, we take the following combination:

RT  ∂U  a  ∂P  T ,  = 2.  =  ∂T V V − b  ∂V T V Setting these coefficients in (7.41), we determine the internal energy differential: 240

а  ∂U   ∂U  dU =   dT = CV dT + 2 dV .  dT +  V ∂ V T ∂ T  V  By integrating this last equation, we find the internal energy of the Van der Waals gas:

U = ∫ CV dT + ∫

a a dV = CV T − + B . 2 V V

Here B is the constant integrating variable. Its value can be found through the formula V → ∞ , in this case the gas is infinitely diluted, therefore its qualities are identical to the ideal gas, hence,

i RT , here i is the degree of gas molecule 2 independence. Then, B = 0 . U = CV T =

Thus, the internal energy of van der Waals gas is:

U = CV T −

a . V

And so, the internal energy of the exact gases

(7.46)

(CV T )

and

potential energy of the molecular particles  − a  is equal to this  V equation. When the distance between molecules increases, the quantitative number of energy increases, too, then consequently intermolecular forces decrease when the volume of gases increases.

7.7.2. Heat Capacity of non-ideal gasses For ideal gases, the molar heat capacity is independent of either temperature or volume occupied by gas. This is due to the fact that the internal energy U of an ideal gas does not depend on the volume occupied by mole (or mass unit) of this gas, i.e. of density, and is only determined by temperature. But this is only true for an ideal gas. 241

For nonideal gas, as in general for any body, the internal energy U can depend not only on the temperature, but also on the volume occupied by a given gas mass. This is due to the fact that in nonideal gases, the internal energy is composed of the kinetic energy of molecules, which depends on the temperature. Consequently, for nonideal gases, the internal energy U of one mole is a function of temperature T and the volume V occupied by it, that is, U = U (T ,V ) . In this case molar heat capacity is C=

dQ . dT

Can no longer be expressed by simple formulas (6.25), (6.26) or (6.36), (6.37). Let us try to calculate heat capacity of a non-ideal gas. In (7.47), we substitute it in the expression in accordance with the first law of thermodynamics, then dU + pdV . C= dT But now the change in internal energy is made up of two parts: 1) a part that only depends on the change in temperature with an unchanged volume, which we denote as (dU )V , and 2) the part that only depends on the volume change at a constant temperature (dU )T .

 ∂U   per unit of temperature  ∂T V

The change in internal energy 

change (dT = 1); at a constant volume of (dV = 1). Similarly,

 ∂U   .   ∂V T Therefore,

 ∂U   ∂U  dU =   dT +   dV .  ∂T   ∂V  V

T

242

Correspondingly, heat capacity С is

 dV  ∂U   ∂U  .  + p  +   ∂T V  ∂V T  dT

C =

(7.48)

The equation (7.42) for the heat capacity is general, suitable for all isotropic bodies. It differs from the expression obtained earlier for heat capacity of ideal gases in that it includes the added variable  ∂U  dV . For ideal gas it equals zero  ∂U  = 0 . For heat      ∂V T  ∂V T dT capacity at a constant volume (7.49), we obtain an expression dV = 0 that is already known, hence:  ∂U  . CV =    ∂T V

(7.49)

The specific heat at constant pressure, equation (7.48), is determined as follows:   ∂U   ∂V  . (7.50)  ∂U   ∂V   ∂U   CP =      = CV +  p +     + p +  ∂V T  ∂T  p  ∂V T  ∂T  P  ∂T V  

All variables included the right-hand side of (7.50) can be measured experimentally, except for the variable  ∂U  that is not  ∂V T measured experimentally.  ∂U  is determined using equation (7.44):    ∂V T  ∂U   ∂P    = T  − P .  ∂V T  ∂T V

By substituting this formula in (7.50), we obtain C P : 243

 ∂P   ∂V  CP = CV + T     .  ∂T V  ∂T  P Therefore:

 ∂P   ∂V  CP − CV = T     .  ∂T V  ∂T  P

(7.51)

For ideal gas

 ∂P   ∂V  T    = R.  ∂T V  ∂T  P This is based on pV = RT equation of state. For nonideal gases, the difference in heat capacities C P − CV is not equal to the constancy of universal gas. According to differential equations of thermodynamics, heat capacity of exact gases is determined as follows:

 ∂2 p   ∂Cv    = T  2  ;  ∂V T  ∂T V  ∂V   ∂p  .     ∂T  p  ∂V T 2

c p − cV = −T 

(7.52)

7.8. Entropy Properties 7.8.1. Entropy in reversible processes in a closed system If an adiabatic process occurs in any system (6.7.2), then this means that our system is closed. Closed in the sense that it is isolated from external sources of heat, both giving it heat, and absorbing heat. Obviously, if the process of change in the state in such a system is 244

reversible, then the change in entropy equals zero ( δQ = 0 ), since in equation dQ , is equal to zero, value dQ . When the state of a dS = T closed system changes adiabatically, its entropy remains unchanged. External bodies can perform work on such a system, and the system itself can perform work on external bodies. In this sense, the system is not closed. With any reversible change in the state of a closed system, the entropy does not change. In fact, let some body able to expand or contract, exchange heat with sources of heat - receive heat from some sources or transfer heat to others. Let also work per performed on the body or the body perform the work itself. Let us call the bodies that perform the work or on which the work is performed, the sources of work. Consider a closed system, including both the body, and sources of heat, and sources of work. Let the state of the body reversibly change due to the fact that it exchanges heat with sources of heat, and because it performs work or work is performed on it. Performance of work (of any sign), as we have just seen, does not result in any change in entropy. The entropy only changes when heat is exchanged between the body and heat sources. If the body, for example, has received heat from the source, which we denote as dQbody , then its entropy has changed by the dQbody

T , where T is the body temperature. In this case, the source lost the same amount of heat. If we denote the amount of heat lost through dQsource then it is obvious that

value

dQbody = −dQsource . In this case, the entropy of the heat source will change by the amount of ( −

dQsource

T ) where T is the temperature of the source.

Since the heat exchange process is reversible, the body temperature must be equal to the temperature of the source. Otherwise, an irreversible thermal conductivity process will occur. Therefore 245

dQbody T

= −

dQsource

T or dSbody = −dS source .

The total change in the entropy of the entire closed system is zero:

dS = dSbody + dS source = 0 .

(7.53)

Consequently, the entropy of adiabatic closed system remains unchanged in any reversible process.

7.8.2. Entropy in irreversible processes in a closed system, entropy increase law An important entropy feature is its behavior in irreversible processes. In (7.29) we saw that for an irreversible circular process the following ratio is correct

φ

dQ  0. T

It is a generalization of partial equation

Q1 Q2 0 + T1 T2 which, in turn, is a consequence of Carnot's first theorem. Let us consider a process where the system is irreversibly transferred from the equilibrium state 1 to the equilibrium state 2 (in Fig. 7.5 it is shown by a solid line). Irreversibility of the transition means that intermediate states are unequal. How does the entropy of the system change with this transition? To find this out, we will return the system to its original state in some reversible way, for example, as shown in Fig. 7,5 with the dashed line. 246

Fig. 7.5

The resulting circular process is irreversible, because one part of it is irreversible. Therefore, the following equation is correct:

φ

dQ 0 T

but

φ

2

1

dQ dQ dQ . =∫ +∫ T T T 1 2

(7.54)

The second of the two integrals, since it refers to a reversible process, is equal to 1

dQ = S1 − S 2 . T 2

∫

(7.55)

Thus, taking into account the equation (7.55) 2

2

dQ dQ ∫ T + S1 − S2  0 or S2 − S1  ∫ T . 1 1

(7.56)

If adiabatic system is closed, that is, isolated from heat sources, then 247

dQ = 0 and S 2 − S1  0 , or S 2  S1 .

(7.57)

It follows that entropy of a closed (i.e., adiabatically isolated) system increases with an irreversible process. The entropy of a closed system either remains constant or increases. The entropy increase law obtained by us under irreversible processes is one of the most important features of the entropy value. It is all the more important that, as was already indicated, the notion of a reversible process is an idealization. After all, with a reversible process, the system at any stage must be in a state of thermodynamic equilibrium. Establishment of equilibrium takes time, and therefore the process, in order to be completely reversible, must proceed infinitely slowly, which, of course, never happens. For irreversible processes in closed systems, entropy, as experience and theory show, always increases, and this property is also inherent in entropy, as it is typical of energy to conserve in all processes in closed systems. Precisely because energy has the property of being conserved in a closed system, it (energy) cannot serve as a function that shows the direction in which processes take place in such a system; in fact, for any changes in the energy state at the beginning and at the end of the process is the same, and it therefore does not make it possible to distinguish between the initial and final states. Entropy, however, which always increases in naturally occurring processes, allows us to judge which process direction is possible and which one is not, which state is the initial state and which is the terminal state. If, for example, we mix two masses of water with different temperatures, then it is not difficult to see that the sum of entropies of both masses before mixing is less than the entropy of the mixture having an intermediate temperature. It is clear that the process of mixing can go on by itself, but the reverse process of mixed masses separation can in no way occur, since it would be accompanied by a decrease in entropy. Entropy increase in any process does not continue without limit, but only up to a certain maximum value characteristic of the given 248

system. This maximum value of entropy corresponds to the equilibrium state, and after it is reached, any changes in the state without external influence cease. Thus, the entropy as a state function differs substantially from the energy. While energy cannot be created or destroyed, entropy can be created and it is constantly created in every process of transition to equilibrium. But, once created, it can no longer be destroyed: the reverse process cannot take place with a decrease in entropy. The entropy increase law in irreversible processes is also often called the second law of thermodynamics. Here are some examples that illustrate this law.

7.9. Examples of the proof of entropy growth in an irreversible process. 7.9.1. Entropy increase in heat transmission. If we bring into contact two bodies of dQ temperature, which are equal to A and B, then the heat will pass from the more heated body to the less heated, resulting in equalization of the temperature of both bodies. Assume, TA  TB . Let’s calculate the entropy change that accompanies this irreversible process. The state of the body A changes due to the loss of a certain amount of heat ( − dQ ); respectively, body B changes its state by obtaining the same amount of heat dQ . In order to determine the change in the entropy of a system consisting of both bodies, it is necessary to calculate the dS =

dQ T

values for some reversible process, which results in the same change in the state of the system. Such a process can be, for example, the process of heat transmission from body A to body B by means of a third working medium, as was done in the consideration of the Carnot process, which, as is known, is reversible at all stages. Then for the body A and, respectively, for the body B 249

dS A = −

dQ , TA

dS B =

dQ . TB

The overall change in the entropy of both bodies is equal to:

 1 1  dS = dS A + dS B =  − dQ .  TB TA 

(7.58)

Since TA  TB , i.e. dS  0 , the system’s entropy increases. The above reasoning does not depend on the process by which the heat is transferred from body A to body B – by thermal conductivity or radiation. It is only significant that the temperatures of both bodies are different.

7.9.2. Entropy increase during adiabatic expansion of an ideal gas in a vacuum Gas expansion in a vacuum is an irreversible process. We now show that this process is accompanied by an increase in the entropy.

Fig. 7.6

Imagine a vessel with heat-insulating walls, divided into two parts by a partition with an aperture V1 and V2 , closed with a flap 250

(Figure 7.6). Let one of the vessel parts, with V volume be filled with 1 mole of ideal gas, while the other one is free of gas. If you open the flap, gas will expand adiabatically and fill the entire volume V (V = V1 + V2 ) of the vessel. It is known that the gas temperature does not change (the Joule experiment). At first glance it seems that gas entropy under such an expansion should not change, since heat is not diverted from it and is not supplied to it. However, it is not true. The expansion process in the described experiment is irreversible, and the equation dS = cannot be applied to it. In an irreversible process, the dQ

dQ T

variable

T dQ integral is not a differential of any state function. In particular, ∫ T

is not equal to the change in entropy. In fact, gas entropy changes during adiabatic expansion into a vacuum. To find this change, it is necessary to calculate the entropy change in some reversible process, leading to the same state change. Such a process can be, for example, a reversible isothermal expansion of gas at the same temperature. We have already made such a calculation, which has shown that under isothermal reversible expansion of 1 mole of gas into the void, the change in entropy equals [see Formula (7.40)]

∆S = R ln

V . here V  V1 , then ∆S  0 , hence, V1

Entropy increases with the expansion of gas.

7.9.3. Entropy increase in mutual diffusion of gases If two different gases are brought into contact, they become mixed due to mutual diffusion on their own, without any external influence. The reverse process, that is, decomposition of a gas 251

mixture into its components, does not happen by itself and is only possible with a certain external influence. Mixing of gases is an irreversible process, and it must therefore be accompanied by an increase in entropy. Indeed, let us imagine that there is 1 mole of a certain ideal gas in the vessel with a V1 volume. Another vessel, with the V2 volume contains 1 mole of a different gas. We combine the two vessels. The gases then mix and the resulting mixture will take up the volume V = V1 + V2 . This process can be regarded as an expansion of each of the gases: the first has expanded from volume V1 to volume V2 ; the second - from volume V2 to volume V1 . The entropy of the first gas in this process, as we have just seen, changes by the amount of

V V , the second by the amount R ln . The total change in V1 V2 entropy ∆S is expressed by the following equation: R ln

 V V  ∆S = R ln + ln  . V2   V1

(7.59)

Thus, both in V1  V and V2  V , hence, ∆S  0 i.e. entropy of the system had increased.

7.10. Statistical description of the second law of thermodynamics. The physical meaning of entropy The second law of thermodynamics establishes that irreversible processes (and such are practically all thermal processes and in any case all naturally occurring processes) flow in such a way that entropy of the system of bodies participating in the process increases, tending to the maximum value. The maximum value of entropy is reached when the system comes to a state of equilibrium. At the same time, we have seen that the very irreversibility of thermal processes is due to the fact that the transition to the 252

equilibrium state is overwhelmingly more probable in comparison with all other transitions. Therefore, only those changes in the state under which the system passes from the less probable state to the more probable state are observed. The similarity in the behavior of both quantities, entropy and probability, is striking: both increase when passing to equilibrium. It is therefore natural to relate entropy systems in this or that state to the probability of this state. This relation was discovered by Boltzmann. First of all, we need to more accurately define the concept of the state probability. This is easiest to do, if again, we refer to the example of the distribution of gas particles in the volume of the vessel. Imagine a vessel divided into two parts, and let this vessel contain, for example, eight molecules like billiard balls, each of which we mentally assign a certain number to, according to which they can be "distinguished" from each other. It is easy to see by a simple calculation that these eight molecules can be placed in both halves of the vessel in various ways.

Each of them corresponds to a certain state of the system. For example, the state in which there is 1 molecule in the left part of the vessel, and 5 in the right part, differs from the state where there are 2 molecules in the left part, and 4 in the right part. Let us see how many of the total number of molecular placement methods will implement each state of our system. The greatest number of arrangements is achieved by a state in which three molecules are in the left part and three are in the right part; that is, a uniform 253

molecules distribution between the two halves of the vessel. According to probability theory:

CN =

N! . n!(N − n )!

(7.60)

Here N is the number of molecules, N = 8; n is the number of molecules on one side. For example, 1) all N = 8 molecules are in the left part (Fig. 7.7), and n = 0 are in the right part, then based on (7.60) this state is only implemented one way; 2) there are 7 on the left side and 1 on the right side, then 8 methods are implemented; In the course of such a calculation, with a decrease in the number of molecules in the right and left parts, the combination of methods increases, with the equation of the molecules in both parts reaching a maximum. In our example, the number of molecules of the right and left sides were leveled, and thus, 70 ways of sets appeared. According to Boltzmann, the smaller the number of sets is, the higher is the order. In our example, where eight parts were arranged in one direction, the state of the system is equal to the highest order, and the parts arranged equally in fours are determined by the maximum disorder. Boltzmann's concept of order is thermodynamics, since molecules are constantly on the move, and our experiment is just a hypothesis. The point of view of thermodynamics somewhat differs from the point of view of a mechanical order. Let's give an example from life. Fill a bag with the sand of two colors (white and black). The lower half is black sand, and the upper part is white. The first state of our system has a high level of order, since only one set is possible. If you shake the bag, white and black sand will mix. This mixing process will stop when a homogeneous gray mixture appears. Thus, there is a state of equilibrium, meaning that this is the maximum disorder. As a result of these reflections, Boltzmann suggested that a free closed system from a state with a low probability passes to a high 254

probability. Thus, the system transition from order to chaos means a transition from a low probability to a high probability. The larger the number of molecules is, the greater is the probability of a transition from order to chaos, and the system that has fallen into the most probable state will remain in this state for an infinitely long time. Hence, probability in practice determines the validity of a certain quality. Thus, the entropy of a closed system is equal to its probability: they can increase or remain unchanged, it depends on the processes. Boltzmann related entropy and thermodynamic probability using the following equation:

S = k ln W ,

(7.61)

macroscopic state by the statistical weight of the state; W ≥ 1 K; k is the Boltzmann's constant. The physical significance of entropy is determined by the logarithm of probability of thermodynamics. Since entropy and probability of thermodynamics are closely related: both reach the maximum value in the equality, the transition of any exact system to the state of equality leads to an increase in entropy and probability of thermodynamics, which means that it is equal to the direction of the process. The probability of thermodynamics W is related to P, but P is always less than one, and W is determined by large figures. In our imaginary experiment (Fig. 7.7), the probability (7.1) of the arrangement of gas consisting of N particles in the v part of the vessel equals: N

v Ρ=  . V  The probability of thermodynamics W is determined in the opposite way: N

W=

1 V  =  . Ρ v 255

(7.62)

Thus, irrevocability of gas distribution process throughout the V volume, because the reverse process probability is lower. In this connection, the second law of thermodynamics determines that the system passes into a state with a high probability in all natural processes. Summing up, let's take into account the qualities of entropy:

dS = d e S + di S ,

(7.63)

di S ≥ 0 , here the equal sign is related to the state of equality; i represents internal, and e represents external parameters of the sign; diS = 0 (reversible processes), diS > 0 (irreversible processes), If (7.63),

deS = 0, then dS ≥ 0

one type of the second law of thermodynamics. For closed systems exchanging only energy with the environment according to the Carnot-Clausius theorem:

de S =

dQ , here dQ is the heat transmitted to the system from T

the external environment; Т is temperature. For closed systems dS ≥ dQ . T A certain type of the second law of thermodynamics. The maximum value of entropy is equal to the state of equality of the system, hence, S = S max entropy measures the chaos of the system. In an irreversible process, unfavorable sides of energy are established in which the entropy growth and its transformation into mechanical work are systematized. The energy of the system in a state of equality, when the maximum value is reached, does not even turn into work. 256

8 TRANSFER PROCESSES 8.1. Transfer phenomenon Thermal conductivity, diffusion and viscosity phenomenon are similar to each other because the transfer of physical quantity of gas or liquid is observed. Diffusion is the system of transfer the weight from one part to another by the reason of concentration gradient availability; viscosity is displacement of gas or liquid momentum by means of gradient flow rate; thermal conductivity is transfer of thermal energy from one system to another with help of gradient’s temperature. Altogether these three processes are called transfer processes. In the processes of substance transfer the discharge of thermal energy and momentum is observed and time separation is also visibly. Perhaps these segmentations connected with “clean” mechanical (hydrodynamic) motion system (solid) as homogeneous substance, and also thermal motion of molecules in substance composition. Transfer in «pure form» is determined by motion of thermal molecules. That is why diffusion is allocation or transfer of weight (of substance) by the time or in space as a result of thermal motion of molecules in the substance composition. Thus, momentum transfer during thermal motion of molecules is called viscosity, this process leads to deformation of material (gas, liquid), as a result the voltage which depends on deformation speed appears. Thermal conductivity is transfer of thermal energy as the result of molecules motion (entities with high energy divided with entities with low energy). It is known from practice that these processes are noninvertible. Their noninvertibility related to the feature of thermal motion of molecules. Empirical bracings, describing the ratio between permanence rules and gradients of laboratory-size magnitudes and energy transfer, weight and momentum, are based on experiments. For example, it is law on internal friction of Newton, Fou257

rier’s law on thermal conductivity, Fick’s first diffusion law. Transfer rations included in these laws, for example, viscosity ratios, thermal conductivities and diffusion are tied with interaction forces between molecules. These laws are very important upon studying of thermal phenomenon. Gas molecules are in random motion, they are in erratic manner colliding with each others and moving through the space by themselves. While changing of space from on place to another, molecules take different physical phenomenon (molecular-based condition) with itself, for example weight, momentum or energy. If the propagation of some physical phenomenon is not of the same kind, the molecules intimately divide it from bigger to smaller part. This transfer process does not stop untill its random distribution in the space. Transfer phenomenon in science – is transfer of some sort of physical parameters from one place to another during thermal motion of molecules in the space. It is worth to remark that transfer of physical paramentres (weight, momentum, energy) does not have any external factors, it get over only as a result of thermal motion. Mechanism of physical transfer for this appearance is the same, it based on chaotic motion of molecules. In this regad, these three appearance have a lot of general properties.

8.2. Molecular collisions 8.2.1. Molecular collision number per unit time Chaotic walking gas molecules collide with each other and with vascular wall. We clarified distribution of quantity of collided molecules with vascular wall in artice 4.6, applying velocity profiles upon Maxwell. Analyzed the geometrical encounter of two molecules in the center of weight, we find frequency of encounters of molecules with each others through cumulative distribution function. It is detailed scrutinized in the section of special courses. And now we define the quantity of collided molecules in dispersed gas according to elementary kinetic theory. Real gas molecules are continuing move to all directions and their velocity refer to wide range. During approximation of two 258

molecules to each other the complex interaction is observed, despite of it, the simple model of transfer phenomenon characteristics can be defined. According to this model: a) gas molecules are hard spheres with diameter σ (in the form of billiard ball), they are not attracted to one another; b) all molecules move to all directions with the same speed, υ = (8kT πm0 )

12

is arithmetical average speed; c) 1 6 part

of molecules move to (+x ) direction, the other 1 6 move to (-x) direction , the other one part moves to (-y) direction, etc. Molecular collisions occur in choatic way. As a result of such collisions trajectory of gas molecules take place in broken line, as it shown on Figure 8.1. Continuance of molucules in the same period of time is shown on the Figure by dots. Molecules at collision exchange with momentums and energy. Let’s compute average number of collisions of one molecule with others per unit time. Compute is made by two stages. On the first stage we will consider that molecules do not move, apart from one molecule, only one molecule is in motion. On second stage all molecules are in motion. Let us suppose that A molecule moves in the line of direct line and the other molecules do not move. In the motion A molecule meets molecules which located not far from diameter center. On the Figure 8.2 these molecules are marked by 1 and 2 numerical characters. Collisions with molecule number three do not take place because the distance between A molecule more then σ -. By that reason radious of molecular diameter equal to σ - , and length is equal to average molecular velocity υ (∆t = 1) A collides with molecule located inside of cylinder or in lateral surface (Figure 8.2). Collisions number of z ′ molecular identities in it will be equal to the number of molecules in the volume identity n = N V multiply by cylinder volume therefore, z ′ = πσ 2 υn , (8.1) herein πσ 2υ -

volume of cylinder. But if we say that all molecues are in motion and each molecule moves with different motion speed, then in formula (8.1) needs to multiply on coefficient 2 (for granted). So average number of molecules colliding with A molecule per unit of time is: 259

z = 2πσ 2 υn .

(8.2)

Average number of molecular collisions per unit of time is collisions frequency. Now we determine Average number of molecular collisions per unit of volume z v . For that we multiply z v on molecular quantity in unit of volume. In view of the fact that in collision take part two molecules, the result needs to be devided on two (two molecules one collision). Probability of collision of three or more molecules is very little, that is why it does not taken into consideration. Number of collision per unit of time per unit of gas volume is:

n

zv = z = 2

2 πσ 2υn 2 . 2

(8.3)

If there will be collision of molecules with weight m01 and m02 , so unit of quantity, so the number of collision per unit of time in unit of gas volume may be defined by the cumulative distribution function (for granted):

 2πkT (m01 + m02 )   . z12 = 2n1n2σ   m01m0 2   12

2 12

Herein σ 12 =

σ1 + σ 2 2

(8.4)

, σ 1 ,σ 2 – diameter of molecules; n1 , n2 –

number density. The number of collision per unit of time in unit of gas volume between molecules I and J, so I frequency of molecular collisions ν i identified by the following way:

νi =

∑j zij ni 260

.

(8.5)

That is why average time τ i of free path and therefore the time of collision of two traces will be equal to:

τ i = ν i−1 .

(8.6)

8.2.2. Average length and mean free time It follows that molecule takes the path of two collision by different length. Average length λ (lambda) of free path is equal to way per unit of time. Quantitively it can be defined by division of average collision velocity: υ 1 υ . (8.7)

λ=

z

=

2πσ 2nυ

=

2πσ 2n

But question appears then, does molecule can easy transfer after binary collision. It is known there is internaction forces between molecules of real gas. Molecules line of march a hundred times more then its diameter after binary collision. Consequently, with a rise of distance strength of inter-molecular efforts decrease very quick. That is why it may be said that molecules intercommunicate only at a distance equal to 2-3 deametres. Molecules get over further way without mutual effect. This way thousand times bigger then way of molecules getting through coordinate action (it indicates that time and length of free path connected with it). By definition from mechanics, length ( λ ) divided by velocity, determinate the time of free motion τ (tau):

τ=

λ 1 . = v 2πσ 2 nυ

(8.8)

Temporal value of free motion is insignificantly For example, in usual situation (p = 1 атм, Т = 273,2 K, σ = 3,8·10-10 м) for molecules is equal to: τ ≈ 1,33 ⋅ 10−10 c. 261

8.2.3. Connection of free path of travel from pressure and temperature In order to reply on question of what dependence of length of free path of travel from pressure, it is nesessary to conduct analysis of connection of quantity occuring in formula (8.7) with pressure. By example, diameter of molecule does not depend on pressure, neither temperature. And only intensity value of molecule n number density p depend on pressure. Relation p of pressure with n we show in this formula: (8.9) p = nkT . At a constant temperature, (8.9) number density of moleculas will be in direct proportion depending on pressure. In view of this, average length of free path of moleculas is in inverse ration to pressure, therefore:

λ~

1

p

.

(8.10)

Interaction is only fulfilled on determined pressure intervals (8.10). In the low-pressure vessel molecules do not come into contact, they only collide with vessel wall. In this case, average length of free path as average interval of collide in vessel determined by vessel size (altitude, width) and shape, but does not depend on number of molecules in vessel, (quantity density of molecules). Such a special form of gas is named extra rarified . It is necessary to mark that extra low-density gas is not a gas under low pressure. For example, gas in very small fises of the stones, (sizes ~ 10-6 centimeters) under atmosphere pressure will be rarified. In the time of determination of extra rarified gas condition, the most important thing is if vessel dimension ratio, calculated by formula (8.7) and average size of length of free path are more than unit, so in this case it is possible to receive extra rarified gas. In gas where molecules are in the forms of solid globe, collision of three or more molecules in time of high-pressure is possible. In this case, it is impossible to determine molecules collision number by the formula (8.2) that is why average length of free path formula (8.7) will not be accurate. 262

Average clear track of fixed density gas flow does not depend on density , and under fixed pressure does not observed in practice, that it is directly proportional to temperature. It can be explained as follows. Length of average path of free movement formula πσ 2 called section of effective collision or gas-kinetic horizontal section of rigid spheres. Formula includes these ratio values which determine all transfers. This example does not include other molecules center. By collision of two spherical molecules, distance between full-sphere will be equal to their double radius (8.2). These forces, in such a way, change motion direction of molecules in the distance. Because of that molecules transverse section depends on ther velocity (energy). During inreasing of molecular velocity and therefore ingresing of gas temperature, molecules transverse section decrease. Average length of molecules free path increase during temperature increase. Thus it is very important to measure average length of molecules free path, because λ depends on effective size of molecules, corresponding to their force interaction. Example 1. If nitrogen diameter equal to 3,22 ⋅ 10−8 см , so in temperature 27 °С and pressure 0,1 Мпа inside 1с please find molecules collision number. By using of quantitive link and molecules density we find n -,: n = N A ⋅ p RT . By plugging average molecules velocity into equation 8 RT (8.2)-move away υ=

πM

z=

4 π N Aσ 2 ⋅ p

.

RTM

If we put numerical symbols to this equation, so that:

4 3,14 ⋅ 6,02 ⋅ 1023 ⋅ (3,22 ) ⋅ 10−20 ⋅ 1 ⋅ 105 = 5,3 ⋅ 109 c −1. 8,31 ⋅ 300 ⋅ 0,028 2

z=

263

As a result of this calculation we will see how 1 nitrogen molecule will collide 5 ⋅ 103 times in one second with other nitrogen molecules. Example 2. Taking previous example data into account calculate the number of collide molecules in one cubical second. Total number of molecular collisions in a unit volume (8.3) equals half the product of the number of molecules in this unit by the number of unit times. Number of unit capacity molecules (in cubical meters) equal to:

n = p kT . So zv = n z = pz = 6,36 ⋅ 1028 см − 3 ⋅ с −1 . 2

2kT

In atmospheric pressure and in temperature of 27 °С in one cubical centimeter huge ~1028 amount of nitrogen collision gets through. Diffusion, viscosity and thermal conductivity in kinetic theory of elemetary particles are called the phenomena related to a free path length. Average length of molecules’ free path in decreasing gas pressure (sufficient rarefaction, low pressure) can be higher than in-line vessel size. For example d, with an approximate p ≈ 10−6 mm of mercury column, average length of molecules free path can reach ten meters ( d ~ 25 cм , P = 10 −7 аtм, λ ~ 25 см ). If the vessel size is about ten centimeters, with such pressure gas molecules will not come in contact with each other. Collision will be only be with vessel walls. Situation when average length of molecules free path is more than vessel size in which it situated calls vacuum (in the translation from the latin vacuum means– empty space). Pressure of Vacuum environment is much lower than atmosphere pressure. Average length of gas molecules free λ path in the capacity of in-line vessel designation d uses as description λ d of vacuum. For example, the distance between two walls of a vessel containing gas, and the distance between electrovacuum electrodes are linear dimensions. λ d is devided in relation to the value of vacuum modes: λ d