Schaum’s Outline of Physical Chemistry [2nd Edition] 0070417156

Table of contents :
Contents......Page 5
Chapter 1......Page 10
Chapter 2......Page 30
Chapter 3......Page 50
Chapter 4......Page 76
Chapter 5......Page 101
Chapter 6......Page 126
Chapter 7......Page 150
Chapter 8......Page 171
Chapter 9......Page 187
Chapter 10......Page 206
Chapter 11......Page 222
Chapter 12......Page 245
Chapter 13......Page 272
Chapter 14......Page 288
Chapter 15......Page 308
Chapter 16......Page 328
Chapter 17......Page 344
Chapter 18......Page 357
Chapter 19......Page 387
Chapter 20......Page 400
Chapter 21......Page 429
Chapter 22......Page 441
Chapter 23......Page 465
Chapter 24......Page 479
Appendix......Page 495
Index......Page 499

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SCHAUM’S

ouT

!

PHYSICAL CHEMISTRY

Second Edition CLYDE R. METZ

The perfect aid for better grades! Covers ail course fundamentals— supplements any class text Teaches effective problem-solving 704 fully worked problems Ideal for self-study!

USB

With tllBSB GOllPS6S:

[^Biochem istry [^M olecular Biology [^Therm odynam ics □ Health Science □ Analytical Chemistry ^ Electro ch em istry □ Medicine Organic Chemistry □ Intensive General Chemistry

SCH AU M 'S OUTLINE OF

THEORY AND PROBLEMS of

PHYSICAL CHEMISTRY SECOND EDITION

by

CLYDE R. METZ, Ph.D. Professor of Chemistry College of Charleston

SCHAUM’S OUTLINE SERIES McGRAW-HILL New York San Francisco Washington, D.C. Auckland Bogota Caracas London Madrid Mexico City Milan Montreal New Dehli San Juan Singapore Sydney Tokyo Toronto

Lisbon

Clyde Metz, currently Professor of Chemistry at The College of Charleston, received his Ph.D. from Indiana University. He is a coauthor of a general chemistry textbook and related laboratory materials, solutions manuals, and study guides and has published several student research papers. Dr. Metz is active in the Division of Chemical Education of the American Chemical Society; is a member of the Electrochemical Society, Alpha Chi Sigma, Sigma Xi, Tau Beta Pi, Phi Lambda Upsilon, and the South Carolina Academy of Science; and is a Fellow of the Indiana Academy of Science.

Schaum's Outline of Theory and Problems of PHYSICAL CHEMISTRY Copyright © 1989, 1976 by The McGraw-Hill Companies, Inc. All rights reserved. Printed in the United States of America. Except as permitted under the Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a data base or retrieval system, without the prior written permission of the publisher. 14 15 16 17 18 19 20 RHR 12 11 10

ISBN

4 1 7 1 5 -b

Sponsoring Editor, John Aliano Production Editor, Leroy Young Editing Supervisor, Marthe Grice Project Supervision, The Total Book

Library of Congress Cataloging-in-Publication Data Metz, Clyde R. Schaum's outline of theory and problems of physical chemistry. (Schaum's outline series) Includes index. 1. Chemistry, Physical and theoretical—Problems, exercises, etc. ( Title. II. Title: Theory and problems of physical chemistry. QD45* M-S 1988 541.3 076 87-29839 :5 B \ (MU-041715-6

McGraw-Hill A Division of The McGraw-Hill Companies

Preface This supplementary book has been written for students at all levels in physical chemistry and contains the solved problems and additional exercises that most texts omit because of space limitations. Each chapter is divided into three sections: The important concepts for each subject are summarized in word and equation form, usually followed by an example problem; a series of completely solved problems is presented to illustrate the concepts both singly and in combination; and a set of supplementary problems (with answers) is provided for additional drill. Since the publication of the first edition of this book, the changeover to the SI system of units from the modified cgs system that had been used in physical chemistry for many years has been nearly completed. In fact, even a new thermodynamic standard state pressure (1 bar) has been chosen. Nearly all the data in this book are given in the newer units—the older systems are discussed briefly and conversion factors given both in the text and in the appendix. Data for the problems have been taken from the Handbook o f Chemistry and Physics with permission of the Chemical Rubber Company, from various original papers, and from publications of the National Bureau of Standards. Appreciation is extended to Chemical Education Resources, Inc. for permission to reprint several figures and tables found in Chapters 18 and 22; to W. H. Freeman and Co., Inc. for permission to reprint several figures in Chapter 22; to Academic Press, Inc. for permission to use tables in the Appendix; to Mr. Thomas J. Dembofsky and Mr. David Beckwith and their staff for their cooperation in making many improvements in the manuscript; to Miss Linda S. Hill for reworking all the problems in the first edition; to an anonymous reviewer who made a number of valuable suggestions, to two outstanding teachers of physical chemistry—Dr. Oran M. Knudsen and Dr. Ralph L. Seifert; and to my family—Jennie, Curtis, and Michele. C

lyde

R.

M

etz

Contents Chapter /

GASES AND THE KINETIC-MOLECULAR T H E O R Y ..................................

1

TEMPERATURE AND PRESSURE 1.1 Temperature 1.2 Pressure LAWS FOR IDEAL GASES 1.3 Boyle’s or Mariotte’s Law 1.4 Charles’s or Gay-Lussac’s Law 1.5 Ideal Gas Law 1.6 Molar Mass of an Ideal Gas MIXTURES OF IDEAL GASES 1.7 Dalton’s Law of Partial Pressures 1.8 Amagat’s Law of Partial Volumes 1.9 Molar Mass of a Gaseous Mixture REAL GASES 1.10 Critical Point 1.11 Compressibility Factor 1.12 Virial Equations 1.13 Van der Waals Equation 1.14 Molar Mass of a Real Gas KINETIC-MOLECULAR THEORY (KMT) 1.15 KMT Postulates for Gases 1.16 P-V-KE Relationships

Chapter 2

TRANSLATION AND TRANSPORT PHENOMENA

..................................

21

VELOCITY AND ENERGY DISTRIBUTIONS OF GASES 2.1 Velocity Distribution 2.2 Energy Distribution COLLISION PARAMETERS 2.3 Collision Numbers of Gases 2.4 Mean Free Path TRANSPORT PROPERTIES 2.5 General Transport Law 2.6 Viscosity 2.7 Fluid Flow 2.8 Diffusion

Chapter 3

FIRST LAW OF T H E R M O D Y N A M IC S...............................................................

41

INTERNAL ENERGY AND ENTHALPY 3.1 Internal Energy, U 3.2 Enthalpy, H 3.3 Thermal Energy of an Ideal Gas 3.4 Thermal Enthalpy of an Ideal Gas HEAT CAPACITY 3.5 Definitions 3.6 Relationship Between CP and Cv 3.7 Empirical Heat-Capacity Expressions 3.8 Cv for Ideal Gases 3.9 Cv for Condensed States INTERNAL ENERGY, WORK, AND HEAT 3.10 Work, w 3.11 Heat, q 3.12 Statement 3.13 General A U and AH Calculations SPECIFIC APPLICATIONS OF THE FIRST LAW 3.14 Reversible Isothermal Expansion of an Ideal Gas 3.15 Isothermal Isobaric Expansion of an Ideal Gas 3.16 Isothermal Isobaric Phase Change 3.17 Reversible Adiabatic Expansion of an Ideal Gas 3.18 Isobaric Adiabatic Expansion of an Ideal Gas 3.19 Joule-Thomson Effect

Chapter 4

THERMOCHEMISTRY

.............................................................................................

67

HEAT OF REACTION 4.1 Introduction 4.2 Temperature Dependence of the Heat of Reaction 4.3 Calorimetry CALCULATIONS INVOLVING THERMOCHEMICAL EQUATIONS 4.4 Law of Hess 4.5 Enthalpy of Formation 4.6 Enthalpy of Combustion 4.7 Bond Enthalpy 4.8 Thermal Energies and Enthalpies APPLICATIONS TO SELECTED CHEMICAL REACTIONS 4.9 Ionization Energy and Electron Affinity 4.10 Lattice Energy 4.11 Enthalpy of Neutralization 4.12 Enthalpy of Solution 4.13 Enthalpy of Dilution APPLICATIONS TO PHYSICAL CHANGES 4.14 States of Matter 4.15 Approximate Values of Enthalpies of Transition 4.16 Enthalpy of Heating 4.17 Clapeyron Equation 4.18 Enthalpy of Formation Diagram

Chapter 5

ENTROPY

......................................................................................................................

THE SECOND LAW OF THERMODYNAMICS 5.1 Statements 5.2 The Carnot Cycle 5.3 Efficiency of a Heat Engine 5.4 Refrigerators ENTROPY CALCULATIONS 5.5 Definition of Entropy 5.6 AA(system) for Heat Transfer 5.7 AA(system) for Volume-Pressure-Temperature Changes 5.8 AA(system) for

92

CONTENTS

Isothermal Mixing 5.9 AS(surroundings) THE THIRD LAW OF THERMODYNAMICS 5.10 Statement 5.11 Values of S°r AS FOR CHEMICAL REACTIONS 5.12 ArS°T from Third Law Entropies 5.13 Temperature Dependence of ArS°T 5.14 Pressure Dependence of ArS°T

Chapter 6

FREE E N E R G Y ............................................................................................................. 117 FREE ENERGY 6.1 Definition and Significance 6.2 Pressure Dependence of G 6.3 Temperature Dependence of G 6.4 Free Energy Calculations 6.5 Temperature Dependence of A,G° 6.6 Chemical Potential ACTIVITIES 6.7 Introduction 6.8 Activities for Ideal Gases 6.9 Activities for Real Gases 6.10 Activities for Liquids and Solids 6.11 Activities for Electrolytic Solutions 6.12 Reaction Quotient THERMODYNAMIC RELATIONS 6.13 Maxwell Relations 6.14 Transformations

Chapter 7

CHEMICAL EQUILIBRIUM

.................................................................................... 141

EQUILIBRIUM CONSTANTS 7.1 Equilibrium Constant Expressions 7.2 Temperature Dependence of K 7.3 Free Energy Curves EQUILIBRIUM AND GASES 7.4 Gaseous Equilibrium Constants 7.5 Calculations for Heterogeneous Systems 7.6 Le Chatelier’s Principle EQUILIBRIUM IN AQUEOUS SOLUTIONS 7.7 Monoprotic Acids and Conjugate Bases 7.8 Aqueous Solutions of Weak Acids 7.9 Aqueous Solutions of Weak Bases 7.10 Buffer Solutions 7.11 Solutions of Ampholytes 7.12 Hydrolysis of Ions 7.13 Consecutive Equilibria 7.14 Slightly Soluble Salts

Chapter 8

STATISTICAL T H E R M O D Y N A M IC S ................................................................... 162 ENSEMBLES 8.1 Introduction 8.2 Ensemble Energy States and Probabilities 8.3 Most Probable Distribution IDEAL-GAS PARTITION FUNCTIONS 8.4 Introduction 8.5 Molecular Translational Partition Function 8.6 Molecular Rotational Partition Function 8.7 Molecular Vibrational Partition Function 8.8 Molecular Electronic Partition Function 8.9 Molecular Nuclear Partition Function APPLICATION TO THERMODYNAMICS INVOLVING IDEAL GASES 8.10 General Thermodynamic Functions 8.11 Molar Thermal Energy 8.12 Molar Entropy 8.13 Free Energy Function MONATOMIC CRYSTALS 8.14 Partition Functions and Molar Heat Capacities 8.15 Other Thermodynamic Properties

Chapter 9

ELECTROCHEMISTRY

.............................................................................................178

OXIDATION-REDUCTION 9.1 Stoichiometry 9.2 Galvanic and Electrolytic Cells CONDUCTIVITY 9.3 Molar Conductivity 9.4 Transport Numbers 9.5 Ionic Mobilities 9.6 Ionic Molar Conductivity ELECTROCHEMICAL CELLS 9.7 Sign Convention and Diagrams 9.8 Standard State Potentials 9.9 Nonstandard State Potentials 9.10 Concentration Cells and Thermocells

Chapter 10

HETEROGENEOUS EQUILIBRIA

........................................................................197

PHASE RULE 10.1 Phases 10.2 Components 10.3 Degrees of Freedom (Variance) 10.4 Gibbs Phase Rule PHASE DIAGRAMS FOR ONE-COMPONENT SYSTEMS 10.5 Introduction PHASE DIAGRAMS FOR TWO-COMPONENT SYSTEMS 10.6 Introduction 10.7 Liquid-Liquid and Liquid-Vapor Diagrams 10.8 Solid-Liquid Diagrams PHASE DIAGRAMS FOR THREE-COMPONENT SYSTEMS 10.9 Introduction

CONTENTS

Chapter 11

SOLUTIONS

..................................................................................................................213

CONCENTRATIONS 11.1 Introduction 11.2 Concentration Units 11.3 Dilutions 11.4 Henry’s Law 11.5 Distribution Coefficients THERMODYNAMIC PROPERTIES OF SOLUTIONS 11.6 Ideal Solutions 11.7 Vapor Pressure 11.8 AS, AH, and AG of Mixing 11.9 Activities and Activity Coefficients COLLIGATIVE PROPERTIES OF SOLUTIONS CONTAINING NONELECTROLYTIC SOLUTES 11.10 Vapor Pressure Lowering 11.11 Boiling Point Elevation 11.12 Freezing Point Depression 11.13 Osmotic Pressure SOLUTIONS OF ELECTROLYTES 11.14 Conductivity 11.15 Colligative Properties of Strong Electrolytes 11.16 Colligative Properties of Weak Electrolytes PARTIAL MOLAR QUANTITIES 11.17 Concept

Chapter 12

RATES OF CHEMICAL REACTIONS

............................................................... 236

RATE EQUATIONS FOR SIMPLE REACTIONS 12.1 Reaction Rate 12.2 Concentration Dependence 12.3 Zero-Order Reactions 12.4 First-Order Reactions 12.5 Second-Order Reactions 12.6 Third-Order Reactions 12.7 PseudoOrder Reactions DETERMINATION OF REACTION ORDER AND RATE CONSTANTS 12.8 Differential Method 12.9 Integral Methods 12.10 Half-Life Method 12.11 Powell-Plot Method 12.12 Relaxation Methods 12.13 Experimental Parameters RATE EQUATIONS FOR COMPLEX REACTIONS 12.14 Differential Rate Equations 12.15 Steady-State Approximations 12.16 Opposing Reactions and Equilibrium 12.17 Consecutive First-Order Reactions 12.18 Competing (Parallel) Reactions RADIOACTIVE DECAY 12.19 Decay Constant and Half-Life 12.20 Successive Decays 12.21 Radioactive Dating

Chapter 13

REACTION KINETICS

.............................................................................................263

INFLUENCE OF TEMPERATURE 13.1 Arrhenius Equation REACTION RATE THEORY 13.2 Collision Theory of Bimolecular Reactions 13.3 Transition-State Theory 13.4 Thermodynamic Considerations CATALYSIS 13.5 Homogeneous Catalysis PHOTOCHEMISTRY 13.6 Introduction

Chapter 14

INTRODUCTION TO QUANTUM MECHANICS

.......................................... 279

PRELIMINARIES 14.1 Electromagnetic Radiation 14.2 De Broglie Wavelength 14.3 Heisenberg Uncertainty (Indeterminacy) Principle 14.4 Rydberg Equation 14.5 Bohr Theory for Hydrogenlike Atoms POSTULATES OF QUANTUM MECHANICS 14.6 Wave Functions 14.7 Operators 14.8 Eigenfunctions and Eigenvalues 14.9 Expectation Values 14.10 Time Dependence 14.11 The Correspondence Principle APPROXIMATION METHODS 14.12 The Variation Method 14.13 Nondegenerate Perturbation Theory

Chapter IS

ATOMIC STRUCTURE AND S P E C T R O S C O P Y .............................................. 299 HYDROGENLIKE ATOMS 15.1 System Description 15.2 The Angular Function 15.3 The Radial Function 15.4 Electron Position 15.5 Energy Values QUANTUM THEORY OF POLYELECTRONIC ATOMS 15.6 Electron Spin Wave Functions 15.7 Hamiltonian Operator and Wave Function 15.8 Energy Levels ATOMIC TERM SYMBOLS 15.9 Russell-Saunders Coupling 15.10 Polyelectronic Atom Term Symbols SPECTRA OF POLYELECTRONIC ATOMS 15.11 Selection Rules 15.12 The Normal Zeeman Effect

CONTENTS

Chapter 16

ELECTRONIC STRUCTURE OF DIATOMIC M O L E C U L E S ..................... 319 QUANTUM THEORY OF DIATOMIC MOLECULES 16.1 Hamiltonian Operator 16.2 Wave Functions APPLICATION OF THE VARIATION METHOD 16.3 Energy 16.4 Molecular Orbitals BOND DESCRIPTION 16.5 Electronegativity 16.6 Dipole Moment 16.7 Ionic Character MOLECULAR TERM SYMBOLS 16.8 Classification of Electronic States 16.9 Term Symbols for Electronic Configurations

Chapter 17

SPECTROSCOPY OF DIATOMIC MOLECULES

.......................................... 335

ROTATIONAL AND VIBRATIONAL SPECTRA 17.1 Rotational Spectra 17.2 Vibrational Spectra 17.3 Anharmonic Oscillator 17.4 Vibrational-Rotational Spectra 17.5 The Raman Effect ELECTRONIC SPECTRA 17.6 Selection Rules 17.7 Deslandres Table

Chapter 18

ELECTRONIC STRUCTURE OF POLYATOMIC MOLECULES

. . . .

348

HYBRIDIZATION 18.1 Angular Wave Functions 18.2 Relative Bond Strength LOCALIZED MULTIPLE BONDS 18.3 Molecular Orbital Theory CONJUGATED BONDS 18.4 Chain Molecules 18.5 Cyclic Molecules 18.6 Bond Order and Length COORDINATION COMPOUNDS 18.7 Valence Bond Theory 18.8 Crystal Field Theory 18.9 Molecular Orbital Theory for Complexes SPATIAL RELATIONSHIPS 18.10 Introduction 18.11 Lewis Structures 18.12 Structure Number and Shape

Chapter 19

SPECTROSCOPY OF POLYATOMIC MOLECULES

..................................377

ROTATIONAL SPECTRA 19.1 Moments of Inertia for a Rigid Molecule 19.2 Spherical Top Molecules 19.3 Symmetrical Top Molecules 19.4 Asymmetrical Top Molecules VIBRATIONAL SPECTRA 19.5 Degrees of Freedom 19.6 Infrared Spectra ELECTRON MAGNETIC PROPERTIES 19.7 Magnetic Susceptibility 19.8 Electron Spin (Magnetic or Paramagnetic) Resonance NUCLEAR MAGNETIC RESONANCE 19.9 Introduction 19.10 Chemical Shifts 19.11 Spin-Spin Splittings

Chapter 20

SYMMETRY AND GROUP THEORY

............................................................... 390

SYMMETRY OPERATIONS AND ELEMENTS 20.1 Introduction 20.2 Identity 20.3 Axis of Proper Rotation 20.4 Center of Symmetry and Inversion 20.5 Mirror Plane 20.6 Rotoreflection 20.7 Rotoinversion 20.8 Translation 20.9 Screw Axis 20.10 Glide Planes POINT GROUPS 20.11 Concept 20.12 Mathematical Properties of a Point Group 20.13 Determination of a Point Group REPRESENTATION OF GROUPS 20.14 Matrix Expressions for Operations 20.15 Representations 20.16 Character 20.17 Character Tables APPLICATIONS OF GROUP THEORY TO MOLECULAR PROPERTIES 20.18 Optical Activity 20.19 Dipole Moment 20.20 Molecular Translational Motion 20.21 Molecular Rotational Motion 20.22 Vibrational Motion for Polyatomic Molecules

Chapter 21

INTERMOLECULAR BONDING

............................................................................419

EXTENDED COVALENT BONDING 21.1 Covalent Bonding 21.2 Hydrogen Bonding METALLIC BONDING 21.3 The Free-Electron Model 21.4 The Band Theory IONIC BONDING 21.5 Born-Haber Cycle 21.6 Potential-Energy Functions VAN DER WAALS FORCES 21.7 Dipole Moments 21.8 Potential-Energy Functions

CONTENTS

Chapter 22

CRYSTALS

......................................................................................................................431

UNIT CELL 22.1 Introduction 22.2 Unit Cell Content 22.3 Unit Cell Coordinates 22.4 Crystallographic Projections 22.5 Coordination Number 22.6 Theoretical Density 22.7 Crystal Radii 22.8 Separation of Atoms CRYSTAL LORMS 22.9 Metallic Crystals 22.10 Covalently Bonded Crystals 22.11 Ionic Crystals 22.12 Molecular Crystals CRYSTALLOGRAPHY 22.13 Miller Indices 22.14 d-Spacings 22.15 Point Group Symmetry X-RAY SPECTRA 22.16 Bragg Equation 22.17 Extinctions 22.18 Method of Ito 22.19 Intensities

Chapter 23

PHENOMENA AT IN T E R F A C E S ............................................................................455 SURFACE TENSION OF LIQUIDS 23.1 Measurement of Surface Tension 23.2 Temperature Dependence 23.3 Vapor Pressure of Droplets 23.4 Parachor SURFACE TENSION IN BINARY SYSTEMS 23.5 Interfacial Tension 23.6 Surface Excess Concentration ADSORPTION 23.7 Adsorption Isotherms 23.8 Heterogeneous Catalysis

Chapter 24

M A C R O M O L E C U L E S .................................................................................................469 MOLAR MASS 24.1 Average Molar Mass 24.2 Distribution Functions SOLUTIONS OF MACROMOLECULES 24.3 Thermodynamic and Colligative Properties 24.4 Osmotic Pressure 24.5 Viscosity 24.6 Ultracentrifugation 24.7 Light Scattering THERMODYNAMIC PROPERTIES 24.8 General Properties 24.9 Fusion of Polymers

A P P E N D IX E S ..................................................................................................................485 Base Units

INDEX

Derived Units

Fundamental Constants

Periodic Table of the Elements

489

Chapter 1 Gases and the Kinetic-Molecular Theory T e m p e r a tu r e a n d P r e s su re

1.1 TEMPERATURE A thermal equilibrium exists between two systems provided no change in any observable property occurs when the systems are in thermal contact. The “zeroth law of thermodynamics” states that “two systems that are separately in thermal equilibrium with a third system are in thermal equilibrium with each other.” This law implies that there must be a property of a system that signifies the existence of a condition of thermal equilibrium that is independent of the composition and size of the system. This property is called temperature. The thermodynamic temperature ( T ) is defined by assigning the exact value 273.16 K to the triple point of water, and the unit of thermodynamic temperature, the kelvin (K), is the fraction 1/273.16 of the thermodynamic temperature of the triple point of water. The Celsius temperature (t ) is defined as f/(°C) = 7 / ( K ) - 273.15

(1.1)

EXAMPLE 1.1. The freezing point of a saturated NaCl-water solution is - 17.78 °C = 0.00 °F, and the freezing point of pure water is 0.0 °C = 32.00 °F. Derive an equation relating the Celsius and Fahrenheit temperature scales. The relative sizes of the respective degrees are found by comparing the numbers of degrees for the same span, giving 32.00 °F - 0.00 °F 1.800 °F °C“l 0.00 ° C - ( - 17.78 °C) The general form of the desired relation between the scales is

t / (°F) = 1.800t/(°C) + k where k corrects for the difference between the zero points of the scales. This constant can be evaluated by substituting the freezing point data for the pure water, giving 32.00 = (1.800)(0) + fc fc =32.00 The desired equation is f/(°F) = 1.800//(°C) + 32.00

( 1. 2 )

1.2 PRESSURE Pressure (P ) is defined as a force distributed over an area. The SI unit for expressing pressure is the pascal (Pa), which is equal to 1 N m-2 or 1 kg n T 1s-2. Because a pressure of 1 Pa is very small, the more convenient bar ( lb a r = 1 0 5Pa) is commonly used for expressing pressures near normal atmospheric pressure, and 1 bar has been chosen as the standard pressure for reporting various thermodynamic data. Other pressure units with appropriate conversion factors are given in Table 1-1. The absolute pressure of a system is defined as the gauge pressure plus the ambient atmospheric pressure.

GASES AND THE KINETIC-MOLECULAR THEORY

2

[CHAP. 1

Table 1-1 Pressure Unit

Symbol

Conversion Factor

Bar Standard atmosphere Conventional millimeter of mercury Torr Pound per square inch

bar atm mmHg torr psi

1 bar= 105 Pa* 1 atm = 1.013 2 5 x l0 5 Pa* 1 mmHg = (13.595 1)(9.806 65) Pa* 1 torr= (1.013 25 x l0 5/760) Pa* 1 psi = 6 894.757 2 Pa

* Exact conversion factor.

EXAMPLE 1.2.

Calculate the height of a column of water equivalent to 1.00 bar.

The height of a mercury column equal to 1.00 bar is 1 mmHg

/1 0 5 P a \ / (1

\/

1m \

^

'°0 b a r ) \ 1 bar / \ (13.595 1)(9.80665) P a / \ 1 0 3 m m / "

750 m

The heights of various liquids that are equivalent to the same pressure are inversely proportional to the densities (p) of the liquids, so fc(H20 ) = fc(Hg)

13.6 x 103 kg m p(Hg) = (0.750 m) 1.00 x 103 kg m~ p(H 2o )

= 10.2 m = 33.5 ft

L a w s f o r Id e a l G ases

1.3 BOYLE’S OR MARIOTTE’S LAW For a fixed amount of gas under isothermal (constant-temperature) conditions, P V = constant

(1.3)

where P is the absolute pressure and V is the volume occupied by the gas. EXAMPLE 1.3. to 35 m3.

Calculate the pressure necessary to compress isothermally a 105-m3 sample of air at 1.05 bar

For the fixed amount of gas, (1.3) gives Pt V, = P2 V2, or 1*2 =

V, , , 105 m3 ^ 1 1 7 - 0 .0 5 b a r)— - T = 3.2bar V2 35 m

EXAMPLE 1.4. Assuming the constant in (1.3) to be 22.7 dm 3 bar for 1.00 mol of gas at 273 K and 227 dm 3 bar at 2 730K, prepare isothermal plots of V against P for values of pressure between 0 and 100 bar. Note: 1 cubic decimeter equals 1 liter (1 dm 3 = 10-3 m3 = 1 L). Substituting P = 0.01, 0.05, 0.1, 1, 5, 10, 50, and 100 bar into V = (22.7 dm 3 b ar)/P gives V = 2 270, 454, 227, 22.7, 4.5, 2.3, 0.5, and 0.2 dm3, respectively, at 273 K. These results and the results of calculations for 2 730 K are plotted in Fig. 1-1.

1.4

CHARLES’S OR GAY-LUSSAC’S LAW For a fixed amount of gas under isobaric (constant-pressure) conditions, V

— = constant where V is the volume and T is the absolute temperature.

L

(1-4)

CHAP. 1]

GASES AND THE KINETIC-MOLECULAR THEORY

3

Fig. 1-1

EXAMPLE 1.5. Calculate the resulting temperature change as a 52-dm3 sample of gas at 25 °C is isobarically expanded to 275 dm3. For a fixed amount of gas, (1.4) gives V J T, = V2/ T2, or V2 , , 275 dm 3 J 2= T . — = (25 + 273) K -------- r =1 600K 2 ' V, 52 dm 3 The temperature change is A T = T 2 - T 1 = 1600 K -2 9 8 K = 1 300K EXAMPLE 1.6. Equation (1.4) can be stated as V = V0[l + ar/(°C)], where V0 is the volume of the gas at 0°C, a is a constant, and t is the temperature in °C. Find the value of a. Equations (1.4) and (1.1) give V=

(r + 273)K (0 + 273)K

Upon comparison of terms, a = 1/273 = 3.66 x 10 3.

1.5 IDEAL GAS LAW Combining (1.3) with (1.4) gives PV — = constant

(1-5)

for a fixed amount of gas. Under Avogadro’s hypothesis that equal volumes of all ideal gases under identical pressure and temperature conditions contain equal numbers of molecules, it can be shown that (1.5) becomes (

where n is the amount o f substance and R is the gcs c o n sta n t ( R = 8.314 41 J K 'm o l ').

1- 6 )

GASES AND THE KINETIC-MOLECULAR THEORY

4

[CHAP. 1

EXAMPLE 1.7. Consider a 42.5-dm3 sample of gas at 25 °C and 748 torr. If the volume is expanded to 52.5 dm 3 and the pressure changed to 760.0 torr, what is the final temperature of the gas? For a fixed amount of gas, (1.5) gives P1 V J T X= P2V2/ T2 (the combined gas law) or P V 760.0 torr 52.5 dm 3 = 374 K T, — — = (298 K) P ,V t 748 torr 42.5 dm 3 EXAMPLE 1.8. The value of the gas constant is usually given as R =8.314 41 J K -1 mol-1, where 1 J = 1 kg m2 s-2. Although this value is convenient for calculating energy, it is rather inconvenient to use in (1.6) because of the need to repeatedly make the same unit conversions for volume and pressure. For this reason, evaluate R in units of m3 Pa K -1 mol-1, dm 3 Pa K 1 mol-1, and dm 1 bar K -1 mol-1 for use in ( 1.6). The values of R in the desired units are / I kgm 2s 2\ t( l pa \ V 1J / ’U kg m 1 s 2j , . . / l 0 3 dm3\ , , ■ R = (8.314 41 m3 Pa K-1 mol- ‘)l —— — 1 = 8 314.41 dm 3 Pa K -1 mol-1 R = (8 314.41 dm 3 Pa K-1 mol-1)( \

1.6

1^ )= 10 5 Pa /

0.083 144 1 dm 3 bar K-1 mol-1

MOLAR MASS OF AN IDEAL GAS Substituting n = m /M , where M is molar mass, into (1.6) gives M =

m RT PV

(1.7)

M =

pR T P

(1.8)

or, in terms of density,

M ix tu r e s o f I d e a l G a se s

1.7

DALTON’S LAW OF PARTIAL PRESSURES The total pressure (P,) of a mixture of r gases is given by

P ,= Z Pi

(1-9)

1=1

where Pi is the partial pressure of component i in the mixture. If P, and the molar composition of the mixture are known, P, can be calculated from Pi = x,P,

(1.10)

where xit the mole fraction for component i, is defined as

to ta l

EXAMPLE 1.9. The vapor pressure (Pv) of a material can be measured by passing an inert gas over a sample of the material and analyzing the composition of the gaseous mixture. For these conditions, (1.10) and (1.11) become Py P,

n n + nmc„

CHAP. 1]

GASES AND THE KINETIC-MOLECULAR THEORY

5

Calculate Pv for Hg at 23 °C if a 50.40-g sample of nitrogen-mercury vapor at 745 torr contains 0.702 mg of Hg. The amount of each gas in the mixture is , , 7.02 x 10-4 g n Hg = — —-------= 3.50 x 10 ~6 mol 200.59 g mol 1 , , [50.40 - (7.02 x l 0 “4)]g "(N 2) = ---- _gn , -------------- = L799 mol 28.013 4 g mol giving P- ‘ 5|0 -» )V l.799- , J S ,0"3' “ r ’“ 0193 Pa

1.8

AMAGATS LAW OF PARTIAL VOLUMES The total volume ( V,) o f a m ixture o f r gases is given by

V, = I Vi

(1.12)

1= 1

where V; is the partial volume o f com ponent i in the m ixture. If V, and the m olar com position o f the m ixture are know n, Vt can be calculated from Vi = x tV,

(1.13)

[F o r a m ixture o f real gases, (1.12) is usually m ore accurate th an ( 1.9).]

1.9

MOLAR MASS OF A GASEOUS MIXTURE The average molar mass (M ), o f a m ixture o f r gases is given by M = £ XjMj i= 1

(1.14)

EXAMPLE 1.10. The density of dry air at 0.986 6 bar and 27.0 °C is 1.146 g dm-3. Calculate the composition of air assuming only N 2 and 0 2 to be present. The average molar mass of air can be calculated from (1.8) as A-

(1.146 g dm~3)(0.083 14 dm 3 bar K~‘ m or'IP O O ^ K) 0.986 6 bar = 28.99 g mol-1

Letting the mole fraction of N 2 be x, we obtain x(28.013 4 )+ ( l —x)(31.998 8 ) = 28.99

or

x = 0.755

Thus x(N 2) = 0.755, and x ( 0 2) = 0.245.

R e a l G ases

1.10 CRITICAL POINT As real gases are cooled, the nicely shaped isotherm s show n in Fig. 1-1 becom e distorted. T he isotherm th a t exhibits an inflection point w here the tangent is vertical, i.e., dP/d V - d2P/$ V2 = 0, corresponds to the critical temperature (Tc). T he values o f P and V at the p o int of inflection are the critical pressure (Pc) and critical volum e (Vc), respectively.

6

GASES AND THE KINETIC-MOLECULAR THEORY

[CHAP. 1

One way to determine the critical temperature and density of a material that does not associate in the liquid phase is to use the rectilinear diameter law o f Cailletet and Mathias, which can be stated as L 2(Pv,q+ PgJ = A + B T + C T 2+ ---

{1.15)

where A, B, and C are constants. This law implies that plots of the orthobaric densities of the gas and of the liquid will intersect at the critical point, giving pc = A + B T C+ C T f+ • • •

{1.16)

EXAMPLE 1.11. Estimate the critical temperature fora substance having gaseous densities of 1.03 x 103, 1.07 x 103, 1.14X103 and 1.25 x 103 kg m ~3 and liquid densities of 1.46 xlO3, 1.43 xlO 3, 1.39 xlO3, and 1.33 x 103 kg m ~3 at 100, 290, 450, and 540 °C, respectively. The plot of these densities (see Fig. 1-2) shows that the curves converge at about 560 °C= tc. The point of intersection can be identified more easily by extrapolation of the (puq + pgas)/2 plot.

Fig. 1-2

1.11

COMPRESSIBILITY FACTOR

The compressibility factor (Z) is an empirical correction for the nonideal behavior of real gases that allows the simple form of the combined gas law to be retained. Thus, we write for a real gas PV = Z nR T

{1.17)

The factor Z is determined by first calculating the reduced pressure {Pr) and the reduced temperature ( T,), defined as

and

Pr=P /Pc

{1.18)

Tr = T /T c

{1.19)

Then the value of Z is read from a graph of Z plotted against Pr for various reduced temperatures (Fig. 1-3). EXAMPLE 1.12. Calculate the volume that 1.50 mol of (C 2H5)2S would occupy at 275 °C and 12.5 bar. Pc = 39.6 bar and Tc = 283.8 °C for (C 2H 5)2S.

GASES AND THE KINETIC-MOLECULAR THEORY

7

Fig. 1-3

CHAP. 1]

8

GASES AND THE KINETIC-MOLECULAR THEORY

[CHAP. 1

Using (1.18) and (1.19) gives 12.5 bar P,= — : =0.320 39.6 bar

(275 + 273)K Tr = — ------------------ = 0.983 (283.8 + 273.15)K

From Fig. 1-3, Z = 0.87, so (1.17) gives (0.87)(1.50 mol)(0.083 14 dm 3 bar K _1 mol“')(548 K) V= 12.5 bar = 4.8 dm 3

1.12 VIRIAL EQUATIONS There are tw o virial equations, one describing P V as a function o f 1 / V and the second using the variable P. F or 1 mol o f gas, these are PVm = A v + B v( 1 / V J + Cv( l / Vm)2+ ■• •

(1.20)

PVm = A p + BpP + CpP 2 + ■• •

(1.21)

H ere Vm is the m olar volum e, an d A v, B v, etc., are constants for a p articular gas. The values o f these virial coefficients can be determ ined from the van der W aals constants, statistical m echanics, or experim ental data. EXAMPLE 1.13. Evaluate A v and Ap in (1.20) and (1.21), assuming that a real gas will approach ideality as (1/ Vm)-»0 and as P-* 0. As the limits of zero are approached, the polynomial terms in (1.20) and (1.21) become insignificant, and both equations approach ( 1.6), giving A v = Ap = RT.

1.13 VAN DER WAALS EQUATION The van d er W aals eq uation o f state ( p + ^ ) ( V - n b ) = nRT

(1.22)

corrects the ideal gas law, (1.6), fo rth e excluded volum e o f the m olecules (b) and the force o f interaction betw een the m olecules (a ). EXAMPLE 1.14. Using (1.6) and (1.22), calculate the volume that 1.50 mol of (C 2H5)2S would occupy at 105 °C and 0.750 bar. Assume that a = 19.00 dm 6 bar m oF 2 and b = 0.121 4 dm 3 mol- '. If we assume the gas to be ideal, (1.6) gives (1.50 mol)(0.083 14 dm 3 bar K~' m or')(378 K) = 62.9 dm 3 (0.750 bar) On the other hand, (1.22) gives ^0.750 +

(19.00)(1.50)2 [V-(1.50)(0.121 4)] = (1.50)(0.083 14)(378) V2

or 0.750 V3 - 47.2 V2 + 42.8 V - 7.79 = 0. Of the several ways to solve a polynomial equation for the various roots, one of the most useful is known as the Newton-Raphson iterative process. For the equation f ( x ) = 0, the (n + 1)th estimate of the root is given by

f'(x„)

(1.23)

CHAP. 1]

GASES AND THE KINETIC-MOLECULAR THEORY

9

where f(x„) is the value of the function evaluated at x„ and f ' ( x n) is the first derivative of the function evaluated at xn. Denoting the left-hand side of the above equation for volume as f ( V ) , we have / ' ( V) = 2.250 V2 - 94.4 Y + 42.8 and (1.23) becomes 0.750 V„3 -47.2 V„2 + 42.8 Vn - 7.79 Vn+,= Vn — 2.250 V„2- 94.4 Vn+ 42.8 If we assume for V, the value predicted by (1.6), the second estimate is (0.750)(62.9)3- (47.2)(62.9)2+ (42.8)(62.9) - 7.79 V2 = 6 2 .9 -(2.250)(62.9)2- (94.4)(62.9) + 42.8 2 585 = 62.9-------- = 62.0 dm 3 3 007 The third estimate, and the answer to three significant figures, is /(62.0) -4 5 V3 = 62.0 - -~4;------ = 62.0--------- = 62.0 dm 3 / (62.0) 2 839 EXAMPLE 1.15. Critical-point data can be used to determine approximate values of the van der Waals constants. Derive these values. Solving (1.22) for P and taking the necessary derivatives gives, for 1 mol, P=

RT V r m - bu

a V J m2

-R T

2a

(\d—Vm)) t - (Vm- b ) 2+ V j ~ ° \d V m2) T

2 RT (Vm~ b ?

6a ;=

Vm4

0

Assigning the values of P, Vm, and T as Pc, Vc, and Tc and solving the three simultaneous equations gives a = 3PCV 2

b -Z 3

3 Tr

Because R is usually determined from other sources, the value of Vc can be eliminated, giving 21R 2 tTc2 64 Pc

1.14

RTC 8 Pc

M O L A R M A SS O F A REAL GAS F or n m oles o f a real gas at rather low pressures, ( 1.21) becom es

PV = n (R T + BpP) U pon substituting n = m / M an d p = m /V , th e above equation becom es p

M /R T

P ~ \ + (BpP /R T ) At low pressures, [l + (BpP /R T )]~ ' * = l-(B pP /R T ) and p P

M -+ RT

(1.24)

Thus an isotherm al plot o f p / P against P will have a slope o f ( M / R T ) ( - Bp/ R T ) and an intercept of M /R T .

[CHAP. 1

GASES AND THE KINETIC-MOLECULAR THEORY

10

EXAMPLE 1.16.

E. Moles reports the following density-pressure data for S 0 2 at 0°C:

P / ( Pa) ( p /P ) /( g d m -3 bar-1)

101 325

50 662.5

10132.5

1 013.25

101.325

10.132 5

2.888 411

2.854 584

2.827 510

2.821416

2.820 807

2.820 746

Determine the molar mass for S 0 2 from these data. A plot of p / P against P gives the intercept as 2.820 741 g dm -3 bar-1; thus M = (intercept)(RT) = (2.820 741 g dm -3 bar-1)(0.083 144 1 dm 3 bar K -1 mol-1)(273.15 K) = 64.061 3 g mol-1

K in e tic -M o le c u la r T h e o ry ( K M T )

1.15

KMT POSTULATES FOR GASES

The m odel for an ideal gas is based on the assum ptions th at (1) the gas consists o f sufficiently m any particles to allow statistical averaging to be perform ed; ( 2 ) the intrinsic volum e o f the particles is sm all com pared to the distances betw een the particles and negligible relative to the volum e o f the container, w hich im plies th at the particles may move freely throughout the entire volum e; (3) the particles are in random m otion and have no m utual attractions; (4) the collisions betw een the particles and betw een the particles and the walls o f the container are elastic, which im plies conservation of energy and m om entum ; an d (5) the average translational kinetic energy o f the particles is proportional to the absolute tem perature. EXAMPLE 1.17.

The average translational kinetic energy for a molecule (etrans) is given by e tr a n s = W v 2

where m is the mass of the molecule and v2 is the average of the square of the velocity. Given t>2 = 3 kT / m, where k is Boltzmann's constant, calculate the ratio of the kinetic energies at 200 °C and 100 °C. e-ran5(200 °C) J m Q k T / m ) 200 473 K ®trans(100 °C) \m C ik T /m ) 100 373 K

1.16 P-V-KE RELATIONSHIPS The K M T im plies the following series o f relationships for 1 mol o f an ideal gas;

PVm= W

= f £ trans= R 7

where £ , rans is the average translational kinetic energy for defined previously. EXAMPLE 1.18.

1

(1.25)

mol o f gas and the other term s have been

For oxygen molecules at 25 °C, find the root mean square speed (t>rms), which is defined by

Substituting (1.25) into (1.26) gives /3 R T \

1/2

T3(8.314 J K -1 mol- ‘>(298 K )1 1/2 = ------------------x----------- :----= 482 1 L 32.0 x i o -3 kg mol-1 J

CHAP. 1]

GASES AND THE KINETIC-MOLECULAR THEORY

11

Solved Problems TEMPERATURE AND PRESSURE 1.1.

C om bine (1.1) w ith (1.2) to define the R ankine scale (°R), which is the absolute F ahrenheit scale. The general form of the desired equation is T/(°R) = t/(°F) + k. Substituting (1.2) gives T/(°R) = [1.800t/(°C) + 32.00] + /c and substituting (1.1) gives r/(°R ) = (1.800)[r/(K ) —273.15]+ 32.00-Ht The value of k can be determined by recognizing that 0°R = 0 K, giving 0 = (1.800)(0 —273.15) + 32.00+ k k = 459.67 Hence T/(°R) = t/(°F) + 459.67.

1.2.

The triple poin t o f w ater is 273.16 K. Express this tem perature as a C elsius tem perature. Using (1.1) gives t/(°C) = 273.16-273.15 = 0.01 / = 0.01 °c

1.3.

The gauge pressure o f the air in an autom obile tire is 32 psi, and the am bient atm ospheric pressure is 14.8 psi. Express the absolute pressure in the tire in bars. The absolute pressure is found by adding the gauge pressure and the atmospheric pressure, giving ,

.

/6

895 P a\ / 1 bar \

(14.8p., + 3 2 p . , ) r r^ r X Tj W - 3 . 2 b a ,

LAWS FOR IDEAL GASES 1.4.

A vacuum m anifold was calibrated using Boyle’s law. A 0.503-dm 3 flask containing dry nitrogen at 7 4 6 to rr was attach ed to the m anifold, which was at 13 m torr. A fter the stopcock was opened an d the system allow ed to reach equilibrium , the pressure o f the com bined system was 273 torr. A ssum ing isotherm al conditions, w hat is the volum e o f the m anifold? Before opening the stopcock, the original condition of the system was given by p, V, = (746 torr)(0.503 dm3) + (13 x 10“ 3 torr)( V) and after opening the stopcock, the condition of the system was given by P2V2 = (273 torr)(0.503 + V) dm 3 Equating these PV terms via (1.3) gives (746)(0.503) + (13 x 10-3) V = (273)(0.503+ V) V = 0.872 dm 3

1.5.

Show for a fixed am ount o f an ideal gas u n d er isotherm al conditions th at ( d V / d P ) ngPT~0 n0dRT0

Va ~

and

nlaRTl

Vd

nldRT0

va

vd

n0a + nod = nla + nld

1.34.

Equation (1.6) can be used to calculate the value of R from experimental data. Assuming that the molar volume of most gases is 22.4 dm 3 at STP (0°C and 1.00 atm), calculate R in units of dm 3 atm K -1 mol-1. Ans. 0.082 1 dm 3 atm K -1 mol” 1

1.35.

An ideal gas at 175 K contains 5x 1020 molecules m-3. What is the pressure of this gas?

1.36.

The density of steam at 100 °C and 760.0 torr is 0.597 4 kg m-3. Calculate the molar mass for water from these data. Explain any discrepancy from the value of 18.015 2 g mol-1. Ans.

Ans.

1.2 Pa

18.3 g mol-1; the gas shows deviation from ideal behavior because it is very near the liquid state.

MIXTURES OF IDEAL GASES 1.37.

Repeat Problem 1.8 for a 0.21-mol sample of 0 2 under the same conditions.Doesair at an altitude of 1 500 m contain a larger or smaller fraction of 0 2 than air at sea level? Ans.

1.38.

P0(O2) = 0.21 bar and x(O2) = 0.21 1 500 m; smaller fraction of 0 2

1.185 kgm ~3 for the dry air; difference

level;

P ( 0 2) = 0.17 bar

and

x (0 2) = 0.20

V(N2) = 44 m3,

P(H 20 ) = 11.8 torr, 1.178 kg m -3 for the wet air;

0.007 kg m

3

V (02) = l l m 3,

V(C02) = 2m 3

The average molar mass of the vapor above NH 4C1(s) is 26.5 g mol-1. Give an interpretation for this value. Ans.

The vapor consists of NH, and HC1, with M = 26.7 g mol-1.

REAL GASES 1.41.

at

The volume of air in a room is 56 m3. Assuming the composition of the air to be x(N 2) = 0.78, x (0 2) = 0.19, and x (C 0 2) = 0.03, find the partial volume for each of the gases. Ans.

1.40.

sea

What is the difference in the density of dry air at 1 atm and 25 °C and moist air with 50% relative humidity under the same conditions? The vapor pressure of water at 25 °C is 23.7 torr. See Example 1.10 for additional data. Ans.

1.39.

at

The densities of the gas and liquid in units of kg m -3 for a substance are given by pgas = 20.0 + (0.175 0) T + (1.500 x 10-4) T2 p,iq = 1 000.0 - (0.500 0) T - (2.000 x 10-4) T2 Find Tc, pc, A , B, and C for this substance. Ans.

967 K,

329.5 kg m-3, 510.0 kg m-3,

-0.162 5 kg m -3 K-1,

-2 .5 0 x 10-5 kg m -3 K -2

CHAP. 1]

1.42.

Bp is the vertical intercept, and Cp is the initial slope of a plot of (PVm —R T ) /P against P.

0.059 3 dm 3 mol-1, about 84% larger

Transform (1.22) into a general polynomial as a function of volume. Arts.

1.47.

Vm(Ar) = 0.880 dm 3 mol-1; the values of Z are very difficult to read

Two molecules of a gas will collide when their centers are within a volume of jm r3. The excluded volume per molecule is fircr3; for a mole it is b = ^Liro-3. Using the diameter of argon as 0.361 nm calculate b, and compare the answer to the van der Waals value of 0.032 19 dm 3 mol-1. Arts.

1.46.

Vm(N2) = 0.892 dm 3 mol-1, from Fig. 1-3.

Discuss how Bp and Cp in (1.21) could be evaluated from experimental PVT data. Arts.

1.45.

Tr = 1.1, giving ^Boyle 0-1)7;

At low pressures, only the first two terms on the right side of (1.21) are important. Find the molar volume of N 2 at 100 °C and 35 bar given that Bp = 6 cm3 mol-1. Likewise, find the molar volume of Ar under the same conditions given that Bp = - 6 cm 3 mol-1. Given that the critical temperatures of those gases are near - 150 °C and the critical pressures are near 45 bar, why is (1.17) difficult to use to calculate these volumes? Arts.

1.44.

19

The Boyle temperature for a real gas is defined as the temperature at which d Z /d P = 0. Show that this definition can also be written as (1 / Pc)(dZ/ dPr). Using Fig. 1-3, find the Boyle temperature for Pr = 2.0. Arts.

1.43.

GASES AND THE KINETIC-MOLECULAR THEORY

PV3- n ( b P + R T ) V 2+ n2a V - n iab = 0

Substitute 1/ V m = P / R T into the expression for PVm derived in Problem 1.19, and rearrange the result into the form given by (1.21). Determine Ap, Bp, and Cp. Arts.

Ap = RT,

Bp = b - a / R T ,

Cp = a b /(R T )2

1.48.

Assuming that Z = PVm/ R T = (AP+ BpP)/ RT, show that Bp = 0 at the Boyle temperature (see Problem 1.42). Using the results of Problem 1.47 and Example 1.15, derive an equation for 7Boyle in terms of Tc. Arts. TBoyle = 27Tc/8

1.49.

Moles reported the following values of p / P as a function of P for C 0 2 at 0 °C:

P /(Pa) (p /T O A g d m ^b ar-1)

101 325

50 662.5

10 132.5

1 013.25

101.325

10.132 5

1.970 916

1.944 469

1.939 082

1.937 867

1.937 746

1.937 734

From these data calculate the molar mass of C 0 2, and, using the accepted value of 12.011 15 u for the relative atomic mass of C, find the relative atomic mass for O. Arts. 1.50.

Determine the value of Bp for S 0 2 from the data in Example 1.16. Arts.

1.51.

Intercept = 1.937 733, 44.007 5 g mol-1; 15.998 2 u

Slope = 6.678 9 x 10-7 g dm -3 bar-1 Pa-1, -0.538 dm 3 mol-1

The Dieterici equation,

20

GASES AND THE KINETIC-MOLECULAR THEORY

[CHAP. 1

may be expressed in a form similar to (1.21) by multiplying both sides of (1.28) by Vm -b; solving for PVm; expanding the exponential as x2 ex = \ + x + — +■ ■■ 2

substituting the expression derived for PVm the Pb term; collecting terms; determining A v, Bv, and Cv by comparison with (1.22); and using the results of Problem 1.15 to find A p, Bp, and Cp. a a2 Ans. Ap = RT, Bp = b - — , Cp = j ^ r 3 1.52.

Evaluate the constants a and b in the Dieterici equation, ( 1.28), in terms of the critical point data. Determine the reduced form of this equation of state. Ans.

1.53.

a=

4 R 2T 2 e2Pc ’

p

Tr = r 2 V ,-l

e 2 - ( 2 / T rVr)

Calculate the pressure for 1.00 mol of argon at 0°C and 5.00 dm 3 using (a) the ideal gas law, (b) the van der Waals equation (a = 1.363 dm 6 bar mol~2 and b = 0.032 19 dm 3 m o f 1), and (c) the Dieterici equation (a = 1.75 dm 6 bar mol” 2 and b = 0.035 dm 3 mol-1)Ans.

(a) 4.54 bar,

(b) 4.52 bar,

(c) 4.50 bar

KINETIC-MOLECULAR THEORY 1.54.

Calculate the ratio of vrms of

1.55.

Show that for a fixed amount of gas at constant temperature (1.25) becomes Boyle’s law, and for a fixed amount of gas at constant pressure (1.25) becomes Charles’s law.

1.56.

What is the average translational kinetic energy of a molecule of an ideal gas at 25 °C? Ans.

6.17 x 10-21 J

238UF6

to vrmi of

235UF 6

at room temperature.

Ans.

0.996

Chapter 2 Translation and Transport Phenomena V e lo c ity a n d E n e r g y D is tr ib u tio n s o f G a s e s

2.1

VELOCITY DISTRIBUTION

The Maxwell relation gives the fraction of molecules in 1 mol ( d N / L ) having a velocity in the x direction between vx and vx + dvx as d N /L

= A e ~mvx2/ 2kT

( 2 . 1)

dvr

where A is a constant and L is Avogadro’s number. In three dimensions, the M axwell-Boltzm ann relation for the fraction of molecules in 1 mol having a velocity between v and v + dv is

2 )

[( h - D / 2 ]!

(for odd n > 1 )

2 ( J (n - H ) /2

In particular, for h(u)= v2 we have

' 2“ 1

v‘ ‘/ N / L - 4lr( ^ f )

m /2kT) /2 k T ) ',2v, The integral can be transformed byi letting z = ((m ' v, which gives

The integral is equal to

37 t

'

/8

(see Table 2-1), giving

r —S/2,-1/2

\2iTkT/

\ m )

8

3 kT m

So for v„

which, because k = R / L and m = M / L, can also be written as

EXAMPLE 2.3.

The fraction of molecules having a velocity r a c is given by ^ - e = -?73 ( — )

e ~ (c/v^ )1+

1 - e rf(c / ump)

(2.8)

where the definition of the error function erf(z) is given in Table 2-1. Find the fraction of neon molecules having a velocity equal to or greater than 775 m s -1 at 298 K. The most probable velocity for neon at 298 K is given by (2.6) as Ump

/ 2(8.314 J K~‘ m or')(298 K)[(l kgm 2 s~2)/(l J); \ (20.18 g mol_l)[( 10-3 kg ) / (1 g)]

r-

496 m s

The argument of the error function is c/ vmp = (775 m s“‘)/(491 m s”1) = 1.58 which corresponds to erf(1.58) = 0.97 (see Fig. 2-1). Substituting into (2.8) gives the fraction as -^2i£ = —775 (1.58) e“(l 58)2 + 1 -0 .9 7 = 0.18 L 7Tl/2

CHAP. 2]

TRANSLATION AND TRANSPORT PHENOMENA

23

c Fig. 2-1

2.2

ENERGY DISTRIBUTION The fraction of molecules in 1 mol ( d N / L) having a translational kinetic energy between etrans and + detl is 2e t1/2 ra n s

d N /L

g - e t r a n s /' k T

(2.P)

b r a n s ' 7T1/ 2(fcT ) 3/2

and the fraction of molecules having a translational kinetic energy e.t r a n s >— N

7

^ - 2( i r r V ' 7" + ' -

4

( C .

° '

e

is ( 2 / 0)

In the limiting case where e ’> k T , (2.10) becomes a special case of the Boltzmann distribution law, which can be used to compare the numbers of molecules having two different energies:

Nj

( 2 . 11 )

gj

where g represents the statistical weight or degeneracy of the energy level (the number of states with the same energy). For these calculations, g = 1. EXAMPLE 2.4.

Determine e.,

Using (2.9) with (2.7) gives

-r

2 e ,/2 trans ,rans n l/2(kT)3/2

2

n ,/2(k T )3/2

r

Ae

3/2

JkT

de„

The integral can be transformed by letting z 2 = eIrans/kT, which gives tra n s

r

2 (2)(kT)s/2 | TTl,2(kT) ,3/2 1/2

z4

dz

The integral is equal to 3 tt /2/ 8 (see Table 2-1), giving ^trans tra n s

2

3 tt1/2 ( 2 )(kT)5/2:— - = |JtT

1/ 2 , 3 /2 ' 7T1 / 2 ( L T )

( 2 / 2 a)

TRANSLATION AND TRANSPORT PHENOMENA

24

[CHAP. 2

which can also be written in molar terms as Elrans = l R T EXAMPLE 2.5.

(2.12b)

Compare the number of molecules with elrans = 3.0fcT to that with etrans = 2.5fcT.

The ratio is given by (2.11) as N 3 .0 k T _ e - O . O k T - 2 . 5 k T ) / k T _

q

^ 2 .5 kT

C o llis io n P a r a m e te r s

2.3

COLLISION NUMBERS OF GASES

F or a b inary gaseous m ixture consisting o f m olecules having collision diam eters cr, an d o), the total num ber of collisions per unit time per unit volume between unlike molecules (Z,y) js given by 1/2

Z ^ N fN fn a ll— ) \1TH /

(2.13)

w here the molecular density (N *), the num ber o f m olecules p er unit volum e, can be determ ined from

'

V

(2.14)

RT

The average collision diam eter (o-y) for the m ixture is given by o',7 = (o-j + /kT j

(3.3d) (3.8a) (3.8b)

The degeneracy of each electronic energy level of energy sj is given by g,. If nuclear contributions to the internal energy are included, these contributions are given by equations similar to (3.5d) and (3.#). EXAMPLE 3.3. Derive general expressions for E(thermal) for ideal monatomic, diatomic, linear polyatomic, and nonlinear polyatomic gases. If a molecule of the ideal gas contains A atoms, then the total number of degrees of freedom is given by 3A. Three of these are assigned to the translational motion of the molecule, leaving 3 A -3 degrees of freedom for rotational and vibrational motion. For an ideal monatomic gas, 3 (1 )-3 = 0; thus the entire contribution to E (thermal) is from translational motion, as given by (3.5a): E(thermal) = 3((RT) = \ R T

(3.9a)

For a linear molecule there are two degrees of rotational motion, leaving 3 A -5 degrees of freedom for vibrational motion. For an ideal diatomic gas, 3(2) —5 = 1; thus E(thermal) is given by (3.5) as E(thermal) = 3 ^ R T ^ + 2 Q RT^j 5

2

RT+

RTx ex —1

(3.9b)

and for an ideal linear polyatomic gas, 5 3A75 R T x , E(thermal) = - R T + Y ------2 , =i e ‘ - 1

(3.9c)

For a nonlinear molecule there are three degrees of rotational motion, leaving 3 A -6 degrees of freedom for vibrational motion; thus, RTx Y (-1R r j\+ 3 (/-lR T )\ + 3A~6 3a 7 6

= 3RT + I

i=i

R T

x

,

—— L e*‘- l

(3.9d)

44

FIRST LAW OF THERMODYNAMICS

[CHAP. 3

EXAMPLE 3.4. Compare the values of AU for heating an ideal monatomic gas and a nonlinear triatomic gas from 25 °C to 50. °C. Assume that the triatomic gas has three vibrational frequencies near 2 000 cm-1. For the ideal monatomic gas, (3.9a) gives £(therm al)298 = |(8.314 J K "1 m0r ‘)(298 K) = 3 716 J mol-1 £(therm al)323 = §(8.314)(323) = 4 028 J m oP 1 and (3.1) gives A U = A£(thermal) = 4 028 - 3 716 = 312 J mol” 1 For the ideal nonlinear triatomic gas, (3.7a) gives ""

(1.438 8)(2 000) ^ =8-910

(1.438 8)(2 000) = 9.658 298

which upon substitution into (3.9d) gives

3(8.314,(29 8 H 3(83» : g 8_)f - 658) = 7 433 +

- ■ °-9 = 7 438 J mol-1 1.57 xlO4

£(therm al)323 = 8 056+10 = 8 066 J m oP 1 and (3.1) gives A t/ = 8 0 6 6 -7 438 = 628 J m o P 1. The difference in the values is 316 J m oP 1, a factor of 2.

3.4 THERMAL ENTHALPY OF AN IDEAL GAS The thermal enthalpy [//(therm al)] is defined as

where

//(therm al) = HT —E0

(3.10)

//(therm al) = £(therm al)+ R T

(3.11)

H e a t C a p a c ity

3.5 DEFINITIONS The heat capacity (C) is defined as C = lim - y AT- 0 A T

(3.12)

where q is the heat transferred corresponding to a temperature change AT. The specific heat capacity (c) is related to C by c= C /m

(3.13)

For constant-volume processes, q = AU, and (3.12) gives (3.14) and for constant-pressure processes, q = AH, which gives

FIRST LAW OF THERMODYNAMICS

CHAP. 3]

45

EXAMPLE 3.5. For the heating or cooling of an ideal gas, A t/ is a function of temperature only. Calculate A t/ for heating 2.50 mol of an ideal monatomic gas from 25 °C to 125 °C using Cv =lR. Rearranging (3.14) and integrating gives dU = Cv dT A t/= [ 2 Cv d T = C v (T2- T l) J t, = (2.50mol)(|)(8.314 J K“‘ mol-1)(398 K -2 9 8 K) = 3 120 J

EXAMPLE 3.6. The pressure dependence of CP is given by (3.16) Newer tabulations of CP values are at P= 1.00 bar, and older tabulations of CP were at P= 1.00 atm. What is the difference between these CP values for an ideal gas? For 1 mol of an ideal gas,

giving no difference between the values of CP.

3 .6

R E L A T IO N S H IP B E T W E E N

CP

AND

Cy

The difference between the heat capacities can be shown to be

(3.17) For an ideal gas, V = R T /P and (dU /dV)T = 0, so (3.17) becomes

(3.18)

Cp-Cy=R For real gases, solids, and liquids, (3.17) yields „

^

« 2v t

(3.19)

CP —Cy —-----where a (the cubic expansion coefficient) and 1

and

k

(the isothermal compressibility) are given by (d V \

/d ln V \

a ~ V \d T )P~ \

dT ) P

_ -i/a v \ K- V W / r -

( a in v \ V dP I

EXAMPLE 3.7. Assuming a = 2.21 x 10“5 K-1, difference between CP and Cv for A1 at 25 °C.

k

(3.21) t

= 1.32 x 10 6 bar-1, and density of 2.702 x 103 kg m-3, find the

Substituting these values into (3.19) gives (2.21 x 10-s k -'H(26.98 Cp “ Cv '

= 0.11 J K -1 mol-1

IQ-3 kg mol- i)/(2.702 x 1Q3 kg m~3)](298 KX(1 J)/(lP a m^)] (1.32 x 10-6 bar-1)[(l bar)/(105 Pa)]

x

(3.20)

46

3 .7

for

FIRST LAW OF THERMODYNAMICS

[CHAP. 3

E M P IR IC A L H E A T -C A P A C IT Y E X P R E S S I O N S

Although theoretical values of C v can be determined for ideal gases (see Sec. 3.8), the equations for the condensed states of matter are much more complex, and empirical relations of the types

C y

C P =

a + bT+ cT2+ dT3

(3.22a)

a + bT+ c'T~2

(3.22b)

C P =

are often used for these phases and also to represent actual data for gases. EXAMPLE 3.8. The values of a and b in (3.22a) for aluminum are 20.7 J K~‘ mol- 1 and 0.012 4 J K~ 2 mol-1, respectively. Calculate A// for heating aluminum from 25 °C to 100. °C. Using (3.15) gives A //= f J =

2 t

(a + bT) dT = a(T2~ T,) + jb (X 22- T,2)

,

(20.7 J K~‘ mol- ')(373 K - 298 K) +|(0.012 4 J KT2 moP')[(373 K) 2 --(298 K)2]

= 1 860Jm or‘

3 .8

F O R ID E A L G A S E S

C y

For

1

mol of an ideal gas,

Cv =

(d£(therm al)/dT) v. The contributions are (3.23a)

CV(trans) = §R , ^ fR Cv(rot) = t . „ ( §R

for a diatomic or linear polyatomic molecule , . for a nonlinear polyatomic molecule

3A-5or3A-6 c * (v ib )‘

(3.23b)

P y 2pxf

(3.23c)

« V -,)=

(3.23d) where xh Q, and Q' have been previously defined by (3.7) and (3.8) and (3.8c) If nuclear contributions are included, (3.23d) is used to determine these contributions. EXAMPLE 3.9. The heat capacity ratio CP/C V for a gas was experimentally measured as 1.38. If the empirical formula is ABA, what conclusions can be made concerning the structure? Assuming the gas to be ideal, (3.18) gives

CP= CV+ R which upon substitution into the desired ratio gives CV

R

— =1.38

or

Cv = 2.63 R

C y

If vibrational contributions are neglected, C V = 2 . 5 R for a linear triatomic gas and CV = 3.0R for a nonlinear triatomic gas. Assuming the difference of 0.13R to be from vibrational contributions, the gas is linear.

3 .9

C y

FO R C O N D E N S E D STATES

Many metals at room temperature have an average value of C P = 25.9 J K - ' m ol"'. Using this in combination with (3.13) generates the law. o f Du'ong and Petit, which can be used to determine

CHAP. 3]

FIRST LAW OF THERMODYNAMICS

47

a p p r o x im a te a t o m ic m a s s e s fo r m e ta ls fr o m v a lu e s o f s p e c if ic h e a t c a p a c it y . T h is v a lu e , a p p r o x im a te ly e q u a l to 3 R , is th e h ig h -te m p e r a tu r e lim it p r e d ic t e d b y s e v e r a l th e o r ie s . A t lo w e r te m p e r a tu r e s , th e v a lu e s o f C P d r o p o ff c o n s id e r a b ly in a n o n lin e a r r e la t io n s h ip . A p lo t o f th e E i n s t e i n h e a t c a p a c i t y r e l a ti o n s h ip fo r m e t a ls ,

(3 .2 4 )

fits th e e x p e r im e n ta l c u r v e s r e a s o n a b ly w e ll. In th is e q u a t io n th e E i n s t e i n c h a r a c t e r i s t i c t e m p e r a t u r e ( 0 E ) d e s c r ib e s th e v ib r a t io n o f th e a to m s in th e m e t a llic c r y sta l. S o m e w h a t b e tte r o v e r a ll a g r e e m e n t b e tw e e n e x p e r im e n t a n d th e o r y is o b t a in e d u s in g th e D e b y e h e a t c a p a c i t y r e l a t i o n s h i p g iv e n b y 9R

' 0 ° /T

x V

(© o /T )3 J,

Cv =-

(ex ~ l ):

dx

(3 .2 5 )

w h e r e 0 D is th e D e b y e c h a r a c t e r is ti c te m p e r a t u r e . E v a lu a tio n o f ( 3 . 2 5 ) is u s u a lly d o n e u s in g C v = 3 R S (©

d/T

)

(3 .2 6 )

w h e r e ® ( 0 D / T ) r e p r e s e n ts th e D e b y e f u n c t i o n ( s e e T a b le 3 -2 ) . C o m b in in g th e D e b y e r e su lt w ith a te r m fo r t h e e le c t r o n ic c o n tr ib u tio n a s p r e d ic t e d b y th e f r e e - e le c tr o n m o d e l fo r m e ta ls ( s e e S e c . 2 1 .3 ) g iv e s

r e D/ r

9R Cv —

(® n /T )

x V

(ex -l y (

Jo

d x + 17 T

(3 .2 7 )

w h e r e 17 is a c o n s t a n t th a t c a n b e d e te r m in e d a s Tr2R k / 2 E f , w h e r e Ef is th e F e r m i e n e rg y . F o r liq u id s , e x a c t v a lu e s o f h e a t c a p a c it y c a n n o t b e p r e d ic t e d . H o w e v e r , a p p r o x im a t io n s in c lu d e 3R

fo r th e v ib r a t io n a l m o d e s o f th e c e n te r s o f m a s s o f th e m o l e c u le s , R x 2e x/ ( e x - 1 )2 fo r e a c h

in t r a m o le c u la r v ib r a t io n a l m o d e , a n d a c o n tr ib u tio n o f \ R to R fo r e a c h m o d e o f r o ta t io n a l m o t io n .

Table 3-2. T. 9 (0 n /T ) = ,_ . (O d/ I ) Jo

v , dx A (e - 1 1'

® d/ T

2 ( & d/ T )

©d/T

9 ( O d/ T )

Q o /T

9 (O JT )

Q o /T

m o „ /T )

0.5

0.987 6 0.951 7 0.896 0 0.825 4 0.745 9 0.662 8 0.580 7 0.503 1

4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0

0.432 0 0.368 6 0.313 3 0.265 6 0.225 1 0.190 9 0.162 2 0.138 2

8.5 9.0 9.5 10.0 10.5 11.0 11.5 12.0

0.118 2 0.101 5 0.087 5 0.075 8 0.066 0 0.057 7 0.050 7 0.044 8

12.5 13.0 13.5 14.0 14.5 15.0 15.5 16.0

0.039 7 0.035 4 0.031 6 0.028 4 0.025 5 0.023 1 0.020 9 0.019 0

1.0

1.5 2.0 2.5 3.0 3.5 4.0

SO(0d /T ) = 77.927(T /0d )3

0 d/ T > 16 for

© VI r

( S n/T ) 2 (0 n/T )4 (0„ /T)61 1------------ + —— 20 560 18 144 J

i

\

for

EXAMPLE 3.10. Calculate Cv for silver at 75 K using (3.24) and (3.25). The value of 0 D is 215 K for silver. Compare the answers to the observed value of 16.9 J KT1 mol-1.

FIRST LAW OF THERMODYNAMICS

48

[CHAP. 3

The value of 0 £ can be calculated from 0 D as 0 £ = ( |) 1/20 d = ( |) i/2(215 K.) = 167 K giving

Cv = 3(8.314 J K

.

mol

)

(167 K \ 2

e167/75

= 16.8 J K-1 m o F 1 This value is 0.6% lower than the experimental value. The value of &D/ T is (215 K)/(75 K) = 2.87, and the value of 2d(&D/ T ) interpolated from Table 3-2 is 0.684, giving Cv = (3)(8.314 J K-1 mol“ l)(0.684) = 17.1 J K“' mol-1 This value is 1% higher than the experimental value.

I n te r n a l E n e r g y , W o r k , a n d H e a t

3.10 WORK, w In thermodynamics, work is defined as the energy that is transferred between a system and its su rro u n d in gs during a change in the state of the system and is completely convertible into some form

of mechanical work in the surroundings. Work is given by the product of an intensity factor X (such as force) and a capacity factor y (such as distance): dw = X dy

(3.28)

where X is related to y and the potential energy Ep by (3.29) The symbol d implies that work is a path function : Its value depends on the actual process used during a change of the system. The types of work commonly encountered in physical chemistry are summarized in Table 3-3. The sign convention used in this book for work is that a negative value means that the system under consideration has performed work on the surroundings and a positive value means that the surroundings have done work on the system. EXAMPLE 3.11. The symbol for the differential work, dw, used in (3.28) indicates that work is, in general, an inexact differential and the work involved in a process is path-dependent. To illustrate this, consider the expansion of 1 mol of an ideal gas from 0.010 0 m3 to 0.100 0 m3 at 25 °C by the following processes: (1) against a constant external pressure of 0.100 bar, (2) from 0.010 0 m3 to 0.025 0 m3 against a constant external pressure of 0.333 bar, followed by a second expansion from 0.025 0 m3 to 0.050 0 m3 against a constant pressure of 0.200 bar, followed by a third expansion from 0.050 0 m3 to 0.100 0 m3 against a constant pressure of 0.100 bar; (3) a reversible expansion. For the first process (3.28) gives (see Table 3-3)

j:

dV = —P,x, A V

, , /TO5 P a\ / 1 J \ = -(0.100 bar)(0.100 0 m3 -0.010 0 m3) -------- -------- r = -900 J ' \ 1 bar / \ 1 Pa m3/ For the second process, repeating the above calculation for each step gives w = -[(0.333)(0.025 0-0.010 0) + (0.200)(0.050 0-0.025 0) + (0.100)(0.100 0-0.050 0)] = -0.015 0 m3 b ar= -1 500 J

CHAP. 3]

FIRST LAW OF THERMODYNAMICS

49

Under reversible conditions, the external pressure and the internal pressure differ only by dP, so substituting (1.6) into (3.28) gives

- j ;

* nRT d V = - n R T In — V V,

, , / 0.100 0 m3\ = -(1.00 mol)(8.314 J K”1 mol_1)(298 K) In ------------r = - 5 705 J \0.010 0 m V

From the above calculations it can be seen that w is dependent on the process chosen and is greatest for a reversible process and smallest for the most irreversible process.

Table 3-3 Process

3.11

dw = X dy

Mechanical work

dw = Fexl dl

Stretching work

dw = kldl

Gravitational work

dw = mg dl

Expansion work

dw = —Pcxl dV

Surface work

dw = y d A

Electrochemical cell work

dw = AVdQ = A VI dt

Comments Fext = external force / = displacement kl = tension / = displacement m = mass g = gravitational constant / = displacement Pext = external pressure V = volume y = surface tension A = area A V = electric potential difference Q = quantity of electricity I = electric current t = time

H EA T, q

In therm odynam ics, heat is defined as the energy th at is transferred betw een a system and its surroundings during a change in the state o f the system and is transferred as a result o f a difference in tem p eratu re betw een the system and its surroundings. Unless work is done, heat transfer will be directed from the poin t o f higher tem perature to the p o int o f low er tem perature. The sign convention for heat in this b ook is th at a negative value represents heat flow fro m a system to the surroundings (an exotherm ic process) and a positive value represents heat flow fro m the surroundings to the system (an en dotherm ic process). EXAMPLE 3.12. The heat transferred between a system and its surroundings under constant external pressure conditions is given by q = nCP A T = mcP A T

(3.30)

What is q for heating 125 g of water from 15 °C to 95 °C? For liquid water, cP =4.2 J K.” ' g~'. Equation (3.30) gives q —(125 g)(4.2 J K 1 g~‘)(368 K.-2 8 8 K) = 42 000 J = 42 kJ [Note that the positive value means that heat is absorbed by the system (the water) from the surroundings (a laboratory burner, etc.).]

FIRST LAW OF THERMODYNAMICS

50

3.12

[CHAP. 3

STATEMENT The first law o f thermodynamics can be written as AU = q + w

(3.31a)

d U ^ d q + dw

(3.31b)

or, in differential form, as For a system capable of performing only expansion work, if dV = 0 then w = 0, which upon substitution into (3.31a) gives A U = qv

(3.32)

Thus the heat transferred under constant-volume (isochoric) conditions is a direct measurement of A U. Heat of combustion measurements for substances are made under these conditions using a “bomb” calorimeter and hence are listed as A (/(combustion). For a system operating under isobaric conditions, (3.3) becomes AH = A t/ + PA V Since under these conditions P A V = —w and A U = qP + w, AH = qP

(3.33)

Thus the heat transferred under constant-pressure conditions (such as the heat of reaction for a chemical reaction performed in a “solution” calorimeter, beaker, flask, etc.) is AH. EXAMPLE 3.13. Consider a system consisting of 1 mol of a monatomic gas contained in a piston. What is the temperature change of the gas if