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Table of contents :
Chapter 11
11.1
11.2
11.3
11.4
11.5
11.6
11.7
11.8
11.9
Chapter 11 Problems
Chapter 11 Answers
Chapter 12
12.1
12.2
12.3
12.4
12.5
12.6
Chapter 12 Problems
Chapter 12 Answers
Chapter 13
13.1
13.2
13.3
13.4
13.5
13.6
13.7
13.8
13.9
Chapter 13 Problems
Chapter 13 Answers
Chapter 14
14.1
14.2
14.3
14.4
14.5
14.6
14.7
14.8
Chapter 14 Problems
Chapter 14 Answers
Chapter 15
15.1
15.2
15.3
15.4
15.5
15.6
Chapter 15 Problems
Chapter 15 Answers
Chapter 16
16.1
16.2
16.3
Chapter 16 Problems
Chapter 16 Answers
Preface
I.
xiv
The Nature ofThermodynamics 1.1
What is ther1nodyn
1.2
Definitions
I
3
ics?
5
1.3
T he kilomole
1.4
Limits of the continuum
1.5
More definitions
1.6
Units
I.7
Temperature and the zeroth law of thermodynamics
1.8
Temperature scales
•
6 6
7
9
Problems
I0
12
16
v
Contents
•
Yl
2.
19
Equations of State 2.1
Introduction
2.1
Equation of state of an ideal gas
21 22
.
2.3
23
s' equation for a real gas
2.4
PvT surfaces for real substances
2.5
Expansivity and compressibility
2.6
An application Problems
25 27
29
31 •
3.
The First Law ofThermodynamics Configuration work
3.2
Dissipative work
3.3
Adiabatic work and internal energy
3.4
Heat
3.5
Units of heat
3.6
The mechanical equivalent of heat
3.7
Summary of the first law
3.8
Some calculations of work
40 41
43
Problems
4.
37
3.1
45 46
46 47
48
Applications of the First Law 4.1
Heat capacity
4.2
Mayer's equation
4.3
Enthalpy and heats of transformation
4.4
Relationships involving enthalpy
4.5
Comparison of
4.6
Work done in an adiabatic process Problems
64
SI
53
u
54
and h
59
61 61
57
35
Contents
5.
••
VII
5.1
The GayLussacJoule experiment
5.2
The joule"Thomson experiment
5.3
Heat engines and the Camot cycle Problems
6.
69 72 74
80
The Second Law ofThermodynamics 6.1
Introduction
6.2
The mathematical concept of entropy
6.3
lrreversible processes
6.4
Carnot's theorem
6.5
The Clausius inequality and the second law
6.6
Entropy and available energy
6.7
Absolute temperature
6.8
Combined first and second laws· Problems
7.
67
Consequences of the First Law
85
87 88
89
91 94
97
98 I03
I 04
I 07
Applications of the Second Law I 09
7.I
Entropy changes in reversible processes
7.2
Temperatureentropy diagrams
7.3
Entropy change of the surroundings fc>r a �eversible process
7.4
Entropy change for an ideal gas
7.S
The Tds equations
7.6
Entropy change in irreversible processes
7. 7
Free expansion of an ideal gas
7.8
Entropy change for a liquid or solid Problems
Ill
I I0
Ill
113
I 21
1 22
118
Ill
Contents
VIII •••
8.
Thermodynamic Potentials 81
Introduction
8.1
The Legendre transformation
8.3
Definition of the thersrnodynamic potentials
8.4
The Maxwell relations
8.5
The Helmholtz function
8.6
The Gibbs function
8.7
Application of the Gibbs function to phase
8.8
An application of the Maxwell relations
8.9
Conditions of
.·
Problems
.
9.
. .
.
1 29 1 30 1 31
1 34 1 34
1 16
e equilibrium
s
1 37
142
1 43
1 46
.
..
.
.
.
.
.
.
.
. .
.
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.
.
.
.
.
:
. .
.
:
.
.
.
.
.
.
. .
.
.
.
·.
;
'
...
· :: ..
:
. . :.
. .·
..
.
.
. : · . .
:
. .. .
The Chemical Potential and Open
149
Systems 9. I
The chemical potential
9.2
Phase equilibrium
9.3
The Gibbs phase rule
9.4
Chemical reactions
9.5
Mixing processes Problems
.
I 0.
127
·.
.
•,
.
.
.
15I
•
ISS 1 57
160
1 62
166
.
.
.•
.
. .: .
· .. .
.
.
.
. .
. .
·.
.
·:
. . .:
.
.
.
.
. ...
•
.
.
.
.<
erature of water is 0.01 :)C, that is� 0.01 degrees above the i cc point at atrilospheri
This equation can be easily integrated to give ( 4.34)
where K is a constant of in t e g ra t ion . This is the relationship between the pres
sure and volume for an adiabatic process involving an ideal gas. Since y > 1
�
it follows that P falls off more rapidly with v for an adiabatic process than it
does for an isothermal process (for which J:lv
==
constant).
The work done in the adiabatic process is V2
1
•
Ptlv
1C =
=
vYdv
K
==
•
1 (Kv Y)
1)2
.
. .
1'VI

VI
v,
Pv1 at bot h limits: if we use K Now, K P1v{Y at the lower limit �'e ohtain K =
=
P2vl at the upper limit and
=
lV =
For an expansion� v2
> v1
..
lV >
l
1 y
t.
0. and the work is done bj� the gas: for
pression the work is done by the surroundings
reve rs i ble adiabatic process.
1�' = u!
ful expression for an ideal gas.
( 4.35)
 r P2L2 ),  PI v IJ·1

lt2
==
cr(T1
C. Assume that all the processes are isobaric. Calculate the c hange in enthalpy in kilocalo ries and in
joules.
(The specific heats at constant pressu;e of ice, liq uid water,
and steam are 0.55, 1.00, and 0.48 kcal kg1, respectively.)
414 An auto mob ile tire is inflated to a pressure of 270 kPa at After three hours of high speed 
driving the
the beginning of a trip.
pressure is 300 kPa. What is the
internal energy change of the air in the tire between pressure measurements? Assume that air is an ideal gas with a constant specific hea t capacity Cv
and that the internal volume of the tire remains constant at 0.057 m3.
=
5R/2
Chap. 4
66 415 Shortly after detonation the fireball of
a
Applications of the First law
uranium fission bomb consists of a
sphere of gas of radius 15 m and temperature 3 x lDc; K. Assuming that the
e xpa n sion is adiabatic and that the fireball remains sphe ri c a L estimate the radius of the ball when the temperature is 3000 K. (Take y =
1.4.)
5.1
The GayLussac oule Experiment
69
5.2
The oule Thomson Experiment
72
5.3
Heat Engines and the Carnot Cycle
74
67
5.1 THE GAY·LUSSACJOULE EXPERIMENT
We have postulated that in general u
=
u(T, v ) . The first efforts undertaken
to measure the dependence of the internal energy of a gas on its volume were made by GayLussac and later by Joule in the middle of the nineteenth cen tury. The results of the experiments showed that the internal energy is a func tion of T only and does not depend on the volume v. This, like many other properties of gases, is on ly approximately true for real gases and is assumed to hold exactly for ideal gases. Before describing the experiments, we consider the following arguments. Using the cyclical and reciprocal relations for partial derivatives given in Appendix A, we can write •
au iJ·v
aT 

av

T •
dlt
II
aT
For a reversible process, we have (Equation
(5.1)
,,
(4.9))
au
Ct
,
=
aT
·
Thus Equation (5.1) gives aT
...
dU
av
= c 1.
v
av
.
(5.2)
ll
Equation (5.2) implies that if we can measure the change in temperature of a gas as the volume increases while the internal energy is held constant, we can
69
Consequences of the First Law
Chap. S
70
diaphragm I
�as sa ...
mpl�
,
I .·
vl  vo
vo. To 
thermal insula t1on" •
Figure 5.1
�


vacuum
Setup for the Joule experin1cnt prior to rupture of the diaphragm.
determine the dependence of the internal energy on the volume. The question is: how can we keep
11
constant duri ng the expansion? Since
ev i dently we want an adiabatic process in \\'hich no work is done. The pr inci ple used by Joule is shown �chematically in Figure 5.1. A sam ple of the gas unde r investigation is enclosed in one insul�ted vesseL separated by
a
d i a phragm from an initially evacuated cham
ber . The t empe r atu re of the gas is measured w i th
diaphragm is rup t u red
�
the
gas under goes
tial specific volume of the gas.
c1
portion of a thermal1y
a
a
thern1ometer. When t he
.free epl·pansiv11. He re
V0
i s the ini
the final specific volume:�� and T1 are the ini
tial and final temp e r a t u re s respectively. In the .
e xpa n sion
no work is done and
the internal energy is unchanged since no �nergy is lost or gain ed . * The fina] temperature� which can he measured. is u, oT Vo
av
dv. u
from basic mathematics. The subscript on the d e ri va t i ve is appro priat e since
the internal
e ne rgy
coefficient. Joule
\\'as
is held c onsta n t. The· integrand is kno\\'n as TJ, the
unable
to
measure
any
temperature
change
Joule
for
air.
Subsequent experitnents using various gases ind ica t e that the temperature d e cr e as es bv at most a verv sn1all amount. Summarizing these resul ts .
.
�
a v.
0.001 K kilomole m3.
u
is finite. However. the rupture of the diaphragm results in a rapid expansion during which the gas is so highly nonuniform that we cannot assign a distinct value to the pressure. See Sect ton 7.7 for a further discussion of free expansion. )jl
Although
Chap. 5
Consequences of the First Law
aQ ttw.
(5.14)
78
We can write the first law dV
=
On the isotherms. for an ideal gas d U alone. Thus aQ
Now
b and
=
l!W. Here Q2
=
=
0 since U depends on the temperature
W2 and Q1
are on the same adiabat pyr
c
W1. Hence
=
(>
0 since vb >
(
.· : '· ...:'!:: .. 0 . . 0. . 0 o• 0 0 0o o' ooO o o
0 .
. :/ · ;::·:· :. :0::::: 0 0 0 0
0
� :0�.� �.
�
•
.
•
?
X
Q
. . ..
0
���;�:
:S ��::;:
o:' :00 0:0 .
. :;:::: : ; . ·•· ·.. . ....•, .. ..... . .. ·.·.... ..... . . . ..� . . .. ·.. ... · .. . , .·.� ·.. . .. ·. ·. . ·. :· ���� �� ·:� ::·:· . . .... , ........: �.. . ..· ··· ··. . ... · ·. . . ·. . . .. . . · . .. . .. .�.. .. .· . .· •..· · . . .•,·· ·. . . ··.� . . . ....,..·.·: .· ;,;. · . .. . . . · · : .· { · . . . . : :: ::; ·=} ..:�: �::: �;: .:: :: :� ' .::.: .:::;:: : .;. , :;.::, : : ;. : : :. '·::�: ::::�:·:·.':'· ;: ::. � ·.: ::· · :: :. :::.. .::;: · � .· · : : ::::: :::::::;::} .; . ::::� :.: � ·:: ·::::< ·:· . . . . . . . . · . : · · · · . · · · · · . , . .. . . . . ·.·. . . : .;. . . : ... •.; ,. .... •• ,. : . . . . ......· .. ...... .··. .:· � : .·. . . ,.;..; .....,. .. . ··•··:. . •. ·. .·.· ...·..•.·.••.· ...· . ·· . .. . , .··.· ·...·..·:...:....... . .;. ·�··.; .. ....·.·;·.....·· . ... ·,. ...•,.,..:.. : .... X..:;:.• ·.". . ·: ·· . · . ·. ··,.7• ·.· •··......·,.. . . • . · · · ·..· ' .· . · ::: ' · .. : ;.t,; •. i : : .; : ;:  � ; ; .: : · . : : : ::.· � :·: · � ·:·: ·�;  :· ; . .
11· That is,
(6.5) The engines are depicted schematically in Figure
6.7.
Since a Carnot engine is reversible, it may be driven backward (as a
refrigerator) by the work from M'. We can arrange the position of the adiabats T,
...
M
M'
r, (a) Fi811re 6.7
(b)
Two engines: (a) a Camot engine M and (b) a
hypothetical engine M' with efficiency assumed to exceed that of the Carnot engine.
92
Chap. 6
The Second Law ofThermodynamics •
Q,

M'
"'
M '
Figure 6.8 The two engines of Figure 6.7 harnessed together, with the hypothetical engine driving the Carnot engine configured as a refrigerator.
so that in one cycle of each engine� M uses exactly as much work as M' pro duces. Thus I WI I W' 1. The Hcomposite" engine is shown in Figure 6.8. Paying careful attention to signs� we note that =
W'l1'I, I  ;Q'IIQ I 2
and
or
The equality
f WI
=
j W' I therefore gives
or
Taking account of the signs in Equation (6.5) and noting that
W
=
; W' i, we
see that jQ2 > IQ2'1 and therefore Q11 > Q1'j. We conclude that the compos' I Q 1 I from the cold reservoir ite engi1;1e does no work� but extracts heat I Q1 and delivers an amount Q2  �Q2': to the hot reservoir. This conclusion vio, lates the Clausius statement, so our hypothetical engine cannot exist. It fol
lows
77'. We can show that if the Clausius statem�nt is violated, the KelvinPlanck statement is also violated. In this sense the two statements are equivalent.
that TJ
>
Sec. 6.4
Carnot's Theorem
93
•
M'
M
w
M
w
I
(a)
(b)
Figure 6.9
(a) A composite engine in violation of the Clausius statement; and (b) the equivalent engine, in violation of the KelvinPlanck statement.
Referring to Figure
6.9, we consider
a hypothetical machine M' that, in viola
tion of the Clausius statement, extracts heat from a cold reservoir and delivers the same amount of heat to the hot reservoir (no work is done on the machine). The Carnot engine M absorbs heat Q2 from work
W, and rejects heat Q1• The work done is lV
ite engine,
no
=
Q2
the hot reservoir, does +
Q1• For the compos
heat is exchanged at the cold reservoir, but heat Q2 + Q1 is
extracted from the hot reservoir and an equal amount of work is done. This violates the PlanckKelvin statement.
Continuing the proof of Carnot's theorem, we have shown that 11 > 11'· Suppose that the hypothetical engine were reversible. Then TJc
>
(6.6)
1J,,
where the subscript C denotes the Carnot engine and
r
the reversible hypo
thetical engine. Since both engines are reversible, the Carnot engine could have been used to drive the other engine backward (as a refrigerator), attd we would have the result 1Jc If both Equations
0.
The Helmholtz Function
Sec. 8.5
TABLE 8. I
135
Relationships involving the thermodynamic potentials.
Thermodynamic
Independent
Reciprocity
Maxwell
Potential
Variables
Relations
Relations
T=
S, V
lntemal energy
as
v
av
aP 
=

·

as
v
a2u =
·
av
'

5
au
P
dU = T dS  P d'l
u
ar
au\
s

av as
•
H
=
F= U TS
=

=
P
dG = SdT + V dP
U TS + PV
as
s =
s

riP as aP
as
aF

ar
v
av

ar
r
V=

av
T
ar
aG
p
aP


r
T
av ar av
as
aG
liP
v
a2F
aF =
p
a2H
aP
S= 
T,P
Gibbs function =
S dT  P dV
aP
p

aH
S=
T,V
dF
as
V=
dH = T dS + V dP
U + PV
Helmholtz function
G
T=
S.P
Enthalpy
av
ar
riH
ar
p
a2G aP ar
H  TS
where �S is the entropy change of the system and �So is the entropy change of the reservoir. If the reservoir transfers heat to the system, then �So Q/T and therefore, =

Q
0, implies that
{8.40)
I(> 0.
We conclude that for thermal stability, the addition of heat to a body causes the temperature to increase:
Cv
=
For mechanical stability, a decrease of pressure causes an increase in vol.ume:
K

1
v
iJV aP
> 0. r
That is, for a closed system, stablethermodytzamic equilibrilJffi consists of ther mal and tnechanical equilibrium. There are a few exceptions to this otherwise general result. At critical points the heat capacities and the compressibility can diverge and the stability conditions are then violated. It is also interesting to note that longrange forces can give rise to negative heat capacities. Such modifications have by no means been fully investigated as yet.
Thermodynamic Potentials
Chap.8
146
PROBLEMS
81 A van der Waals gas and an ideal gas are originally at the same pressure, tem perature .. and volume. If each gas undergoes a reversible isothermal compres sion.. which gas will experience the greater change in entropy?
82 Show that for the an ideal gas
(a) f
=
C0(T

To)
c]

In
T
To .
(b) g = c,.(T

T0)
c,.T In

 RT In
T
T
�
v
Vo .

p + RT ln p, '

soT. s0T.
·
0
83 A cylinder contains a piston on each side of which is one kilomole of an ideal gas. The walls of the cylinder are diathermal and the system is in contact with a heat reservoir at a temperature of 0°C. The initial volumes of the gaseous sub systems on either side of the piston are 12 liters and 2 liters. respectively. The piston is now moved reversibly so that the final volumes are each 7 liters. What is the change in the Hemhohz potential? (Note that this is the work delivered to the system by the reservoir.)
84 Derive the following equations: (a) F
(b) Cr
=
U + T
=
aF
:
ar
\
T
(reversible process)� . •
(c) H
=
T
G 
(d) C,
=

·
rJG aT
a 2G T 2 ar
p:
(reversible process):

p
(The third relation is the GibbsHelmholtz equation. alluded to in Chapter
10.)
8S The Helmholtz function of a certain gas is F
n2a =

V

nRT ln(V nb) + J(T).
where J is a function of T only. Derive an expression for the pressure of the gas.
86 The Gibbs function of a certain gas is G where A� 8,
C.
=
nRT lnP
+
A + BP
+
CP2 2
DP3 + 3 .
and D are constants. Find the equation of state of the gas.
Problems
87
147
The specific Gibbs function of a gas is given by ' g
p
RT ln
=
 AP.
Po
where A is a function of T. Find expressions for:
88
the equation of state;
(a) (b) (c)
the specific Helmholtz function.
(a)
A van der Waals gas undergoes an isothetrnal expansion from specific vol
the sp ecific entropy ;
ume v1 to specific volume
v2•
Calculate the cha nge in the specific Helmholtz
function.
(b)
Calculate the change in the specific internal energy in terrns of
89 Start with the first Maxwell relation li�ted in Table
t't
and v2•
8.1 and derive the second by
using the cyclical and recipirocal relations (Appendix A) and the identity
az
ay
ax
ay I oz
I
iJx
=
f
1.
(The remaining Maxwell relations can be derived in a similar manner.)
810
(a)
Prove that Cp
=
T
and use the result to show that
(b)
Prove that Cp for an ideal gas is a function ofT o nly
.
811 ln Figure 2.2 depicting the isotherms for a van der Waals gas, the shaded areas were shown by Maxwell to be equal. The path
bed is the segment of an isotherrn
along which liquid and vapor are in equilibrium. Prove Maxwell's result by not ing that Ag
=
0 for a phase change (P and T constant) and calculating �f
812 The Pv diagram in Figure 8.4 shows two neighboring isotherrns in the region of a liquidgas phase transition. By considering a Carnot cycle .between tempera tures T and T + d1. in the region shown, derive the ClausiusClapeyron equa tion
dP/dT
=
f23/T(�v).
Here
�vis the
specific volume change between gas
and liquid.
813 The equations of the sublimation and the vaporization curves of a particular material are given by In
P
In
P
=
=
0.04 0.03


6fT
(sublimation),
4/T
(vaporization),
148
Chap. 8
Thermodynamic Potentials
p
+
T ·.
.
. .
.;· �. . · . � .· . ....
T Figure 8.4
P v diagram for
dT
·
. .
.
·. · .:... ·
· �
·
···. . . .. .
.
.
.
= const.
.. .
.
.
· . . ·. .·.
.
·. : ...:� . . .. . .. . . . .
·
.
.
.
.
. ·
·.#
.
..
.
..
.. . . · .· · .
. .
.
I
I
= const.
'
an infinitesimal Carnot cycle.
�· ·
t'
where P is in atmosp heres
.
(a) Find the temperature of the tri ple point. (b) Show that the specific latent heats of vaporization and sublimation are 4R and 6R, resp ective ly (You may assume that the specific volume in the vapor .
phase is much larger than the specific volume in the liquid and solid
phases . )
(c) Find the latent heat of fusion. 814 (a) Calculate the slope of the fusion curve of ice in PaK _, at the normal melting point . At this temperature, the heat of fusion is 3.34 X 1 a� Jkg. i and the change in specific volume on melting is  9.05 x 1 o � m3kg 1• (b) Ice at 2°C and atmospheri c pressure is compressed i so th e rmally Find the 
pressure at which the ice starts to melt (in atmospheres).
.
e en
•
9 .I
The Chemical Potential
lSI
9.2
Phase Equilibrium
ISS
9.3
The Gibbs Phase Rule
157
9.4
Chemical Reactions
160
9.5
Mixing Processes
162
149
9.1 THE CHEMICAL POTENTIAL
Until now we have confmed our discussion to closed physical systems, which cannot exchange matter with their sutToundings. We turn our attention in this chapter to open systems, in which the quantity of matter is not fixed. Suppose that
dn kilomoles of matter are introduced into a system. Each
kilomole of added matter has its own internal energy that is released to the rest of the system, possibly in a chemical reactio11. The added energy is propor tional to
dn and may be written as p,dn. The quantity #L is called the chemical
potential. The chemical potential is associated with intertttolecular forces. An elec trically polarized molecule experiences a Coulomb attraction when it is brought into the vicinity of another such molecule.* 'Ibis force is expressed as a negative potential energy, a sort of "potential well." As the new particle approaches its neighbor, it gains kinetic energy while losing potential energy. The kinetic energy is imparted to other particles through collisions, so the sys tem gains internal energy in the process. Consider a motionless molecule infinitely distant from other molecules. Its kinetic energy and potential energy are both zero. The molecule is moved into the force field of a second molecule. This can, in principle, be done slowly so that the kinetic energy is negligibly small. Left by itself, however, the mole cule picks up kinetic energy equal in magnitude to the depth of the potential well (Figure 9.1). Quantitatively, E=K+V (r), where E is the total energy, K is the kinetic energy, and V(r) is the potential energy; r is the distance between the molecule and its neighbor. At r = oo, K=
0, and V
•
=
0, so E
=
0 everywhere. At r = r0,
See Section 11.1 for a discussion of the molecular interaction potential. lSI
The Chemical Potential and Open Systems
Chap. 9
152
V(r)
r
Figure 9.1 Schematic diag ram of a potential well due to intermolecular forces. A more descriptive graph is given in Section 11.1.
I I· K I I
Energy is conserved� but a conversion from potential energy to kinetic energy takes place. The kinetic energy is added to the internal energy of the system. It is reasonable to ask what the magnitude of J.L is. In a standard labora tory experiment� sulfuric acid is added to water, producing an increase in tem perature. Imagine that 105 kilomoles of acid at room temperature are added to a liter of water
(5.56
x Jo·:! kilomoles). also at room temperature. The tem
perature is observed to rise
0.1 °C. We wish to determine the chemical poten
tial of acid in water. We shall assume that the interaction among the acid molecules is small compared with their interaction with tl1e water molecules (both acid and water molecules are eJectrically polarized). 'llle specific heat capacity of water is
7.52
x
104 J kilomole 1
K 1• Thus the heat gained by the water through the
addition of the acid is
Q
=
lnc,�T
=
(5.56
�)(7.52
X 10
X
10')(0.1)
=
4181.
Since the mass of the water is more than 103 times the mass of the acid .. we can ignore the heat capacity of the latter. The chemical potentiaL then� is
" ,
=
41X
Q

· �n
=

1()) _
=

7 4.1 8 x 10 J kilomole
1• _
The sign i s negat i ve because heat is transferred jro111 the acid to the \Vater. We are interested in the chemical energy per particle, which is essentially the depth of the potential well. Since a kil()ffiole has 6.02 x 1026 n1olecules. this value is 6.94 x 10· 20 J or 0.43 e\t'. (One electron volt is equal to
1.6
x
1019 joules.) Most chemical potentials are of this order of magnitude.
To account for the effect of adding ma ss to
a
system. we need to add a
term to our fundamental equation of thermodynamics: dU
=
Ttl�'

P dV
+
J.Ldn.
(9.1)
Sec. 9.1 Here
153
The Chemical Potential
dn is
the increment of mass added (in kilomoles) and J.L is the chemical
potential in joules per kilomole. If, in an open system, U
iJS
dS
au
+
OV
V,n
dV
+
S.n
=
U (S, V,
au dn
n)
and
(9.2)
dn, S,V
then iJU
J.L=
an.
(9.3)
•
s.v
That is, the chemical potential is defined as the internal energy per kilomole added under conditions of constant entropy and volume. If there is more than one type of particle added to the system (say types), then Equation
(9.1) becomes
m
m
dU
=
TdS  PdV +
j=l
J.Li
(9.4)
d nj,
with
(9.5) The subscript nk means that all other n's except ni are held constant. Another way of seeing the relationship between U and J.Li is to integrate Equation
(9.4). This
can be done by using Euler's theorem for homogenous
functions. Euler's theorem states that if
Af(x, y, z)
=
f(Ax, Ay, Az),
(9.6)
at yay
(9.7)
then
at f =X ax
+
+
at . Z az
The theorem can be easily proved by differentiating Equation (9.6) with respect to
A
and then setting
A
equal to unity. Now U
=
U(S, V,
n1
•
•
•
nm)·
Suppose the amounts of all the types of substance, called constituents, in the system were doubled or halved or, more generally, changed by the factor
A
without changing any of the fundamental state variables. Then the extensive variable U would be changed by
A
and all the independent, extensive state •
The Chemical Potential and Open Systems
Chap. 9
154
variables would also be changed by the factor A. Thus U is a homogeneous function and Euler's theorem can be applied to it:
u
=
au
s
au
+ v
as
\"
111
+
rlv
,/l 1
,·= 1
s·"
1
iJU
lli
•
rin,·
(9.8)
, v.fl J.. �)' .
From the differential of U we know that au as
=
V.nJ
iJU
T ..
av
= I

P
au ..
s.v.n ..
S.n,
= J.lj .
(9.9)
Substituting the relations of Equation (9.9) in Equation (9.8)� we have
U
=
ST PV +
Jl;n;.
..
(9.10)
r= I
Recall that the Gibbs function is defined as
G
==
U

ST + PV It is
therefore immediatelv clear that �
111
(9.11) If only one constituent is present .. then G
==
p.11 or JL ==
Gjn: so 1L in this case is
simply the Gibbs function per kilomole of the substance. Finally.. if we take the differential of Equation (9.1 0). we obtain .
I
J
Equating Equations
(9.4) an d (9.12) we have ..
Ill
SdT VdP
+ (=· I
11;dJ.L,
=
•
0..
(9.13)
a relation known as the GibbsDuhem equation. Taking the differential of Equation
(9.11) gives
1
.
Note that
G
=
G(T P n1 ..
..
•
•
•
(9.14) . .
I
11,,). Thus if we have
a
process that takes place
at constant temperature and pressure, the first two terms of Equation
(9.13)
Sec. 9.2
Phase Equilibrium
155
vanish so the sum also vanishes. Equation
(9.14)
then yields the important
result
(9.15) •
1
We now have the mathematical tools to look at some applications.
.
.
.
·
.
.
. . .
.
.
. .
.
·. . .
·..
.
.
. .
'
.
:
.
.
.
.
9.2 PHASE EQUILIBRIUM We Ylish to find the condition of equilibrium for two subsystems under particle exchange where the subsystems are two phases of the saane substance. An example would be ice melting in liq.uid water; an exchange of H20 molecules occurs between the two subsystetns_ We assume that the total system is enclosed in a rigid adiabatic wall (Figure 9.2). Since there can be heat and particle flow across the boundary between the two phases, the boundary will move as the process evolves. The various interactions are subject to the foUowing conditions:
nA
+
n8
VA
+
V8
UA
+
U8
=n = constant
= V =constant =
U
=
constant
(conservation of mass),
(9.16)
(conservation of volume),
(9.17)
(conservation of energy).
(9.18)
The total internal energy is constant, according to the first law, since no heat flows into or out of the combined system and no work is done. At equilibrium, the entropy of the combined system will be a maximttm:
SA
+
S8
Let all the qu�ntities n, V, U, and
dS
=
S
(maximun1).
(9.19)
S change by infinitesimal amounts.. Then
=
dS.4 + dSB
=
0.
phase boundary
Figure 9.2 1\vo phases of a substance in thermal equilibrium.
(9 20) .
156
Chap. 9
The Chemical Potential and Open Systems •
But
(9.2 1 ) and
(9.22) We can eliminate three infinitesimals by invoking Equations
(9.16), ( 9.17), and
(9.18):
.
Combining these ex pressions with Equations (9.21) and (9.22), and substitut
ing in Equation (9.20) gives the result 1 ·
1
(9.23)

Since Equation
(9.23) must be true for
ar b i t ra ry
increments
dUA, d\lA· dnA all ..
of the coefficients must be zero: TA
PA
=
=
J.L 11
PH
=
(thermal equilibrium) ..
T8
(9.24)
(mechanical equilibr ium)
J.L B
( diffusive tquilibrium)
(9.25)
.
(9.26)
.
We conclude that if we have two phases or s u bsys t ems in thermal and mechanical equil i bri um then they will also be in .
diffitsive equilibri u m ( equi
librium against particle exchange) if their chemical potentials are equal. Suppose that the two interacting systems are not yet in equilibriu m
.
Then the entropy of the combined system \l:ill not have reached its max i mum value bu t will still be increasing:
·
(9.27) Consider a state very close to an equilibriurr1 state and differing from it only in that there is a small excess of mass �'zA(>O) in s ubsystem A over the equ i l ib
rium value
nA,
and a correspondi n g defici� (>f n1ass

d11A in subsystem B. As
Sec. 9.3
The Gibbs Phase Rule
157
the combined system moves toward equilibrium, subsys tem A will give up its excess mass with an accompanying entropy change
(9.28) Simultaneously, the entropy change of subsystem B as a result of acquiring the additional mass will be
(9.29)
The temperature is the same in both expressions. Substituting Equations (9.28) and (9.29) in Equation (9.27) gives
or (9.30)
J.LA > J.LB·
Thus, in the approach to equilibrium, heat energy must flow from the hotter to the cooler body, volume will be gained by the body under lower pres sure at the expense of the other, and mass will tend to flow from the body of higher chemical potential toward the one of lower chemical potential. All of these tendencies follow from the second Jaw of thertnodynamics. •
9.3 THE GIBBS PHASE RULE Consider a system in equilibrium with k constituents in 7r phases
.
*
We note the
following: 1. Only one gaseous phase can exist because of diffusion . .
2. Several liquids can coexist in equilibrium if they are immiscible. 3. Several solids can coexist. 4. Only rarely do more than three phases of a given constituent coexist.
*The symbol here
with 3.14. .
.
.
1r,
traditionally used to denote the number of phases, is not to be confused
158
Chap. 9
The Chemical Potential and Open Systems
Let P.l be the chemical potential of the ith constituent in the yth phase. According to Equation (9. 1 1) the G i bbs functi on is
(9.31) where n? is the number of kilomoles of the ith constituent in the yth phase. For equilibrium�
(dG)r.P
=
0. so Equation (9.15) gives
1L? dn? ..,..
=
0� 1· and P fixed.
i=] y=l
(9.32)
It also follows from the GibbsDuhem equation (Equation (9.13)) that for constant temperature and pressure.
(9.33)
t=J y= I
If we consider the special case of a closed �ystem in which mass
dn; of the ith
constituent is transferred from one phase t(> another with the total 1nass of the constituent unchanged.. then 
"
dll? 
:=:
0, i
=
(9.34)
1, 2. . . . k.
'Y =� 1
•
This equation is simply a statement of the conse rvati on of mass: No con stituent is created or destr oy ed. Suppose that there are two constituents ( i
(y
=
cr,
=
1. 2 ) and two phases
J3). Then Equation (9.32) becomes
and Equation ( 9 34) giv es .
(9.36) .
Combining these e q ua tion s gives •
(9.37) Since the differentials dn1u and £in2o are arbitrary, the coefficients must vanish. Hence (9.38)
Sec. 9.3
The Gibbs Phase Rule
159
In other words, the equality of the chemical potentials for the two phases of each of the constituents is the condition for equiljbriutn with T and P fixed. From the conservation of mass it follows that the number of kilomoles of each constituent is fixed provided no chemical reaction occurs. Thus
·
(9.39) where
n1
and
n2
are the given nu·mber of kilomoles of constituents
and 2,
1
respectively. It. is convenient to introduce for each constituent the concept of the kilomole fraction, the ratio of the number of kilomoles of a given phase to the total number of kilomoles of all constituents in that phase: .
. .
(9.40)

From Equations (9.39) and (9.40), it is evident that
(9.41) Therefore, in our example of k
=
2, 1r
=
2, the state of the system is detertnined
by four independent variables: T, P, x1a, and x2a. tlowever, since there are two equilibrium conditions ( Equation (9.38)) , the ttumber of variables is reduced to two. The generalization of this specia l case leads to the socalled Gibbs phase rule. For a multiconstituent, multiphase system with T and P fixed, the condition for equilibrium is II. Q
ri

·
ll.f3  ll. y ri
 ,.,

.
•
=
.
J.L/r'
i
1' 2,
=
.
.
.
k.
( 9 42 )
7T.
(9.43)
.
The kilomole fractions are:
xY l
=
n;·y
k
i= 1
'
i
=
1' 2 ..k; .
1'
=
1' 2
...
,,Ir •
Thus, there would b� k1r k ilomole fractions were it not for the identity k
x? i
=
1
=
(9.44)
1,
which subtracts 1r kilomole fractions from the total. Counti ng the two prescribed auantities. T an d P. there are therefore 2 + k1r

1T
inde oenden t variables
The Chemical Potential and Open Systems
Chap. 9
160
and k ( 1T
l) equilibrium conditions (Equation (9.42)) . The number of

remaining .. degrees of freedom," called the variance f, is
f
==
[2
+
k7i 1T] [k(1T
J)l
or
1·
=
k

1r +
(9.45)
(no chtmical reaction).
2
•
This is the usual form of the Gibbs phase rule. If f
f
==
=
0. the system has zero variance and is completely determined. If
1, the system is monovariant and is not determined until one additional
variable is specified. Consider the following examples: 1. A homogeneous fluid with one constituent and one phase. Here k 71'
=
1
�
f
=
=
1,
2. Thus we can choose T and P arbitrarily. This is the case of.
an ideal gas� for instance. 2. A homogeneous system consisting ot two chemically different gases in
the same phase. Here k
=
2,
r.
=
I, f
=
3. Here we can choose
T, P. and
one kilomole fraction. 3. Water in equilibrium with its saturated vapor. In this case k
f
=
1. Thus we can choose only the temperature
trarily. Equilibrium is defined by
a
=
1,
7T
==
2,
(or the pressure) arbi
p1 and c/J2 are functions of the same temperature T, the functions themselves are different because
Cp
will in general not have the same value for
the two gases. The final value of the Gibbs function is
with
Since T doesn �t change in the mixing process, c/>1 and c/>2 will be the same for the final state as for the initial state. Using P1 g11
=
RT( ln P + 1 +Inx1)�
g21
=
=
x1P and P2
=
x2P, we obtain
RT(ln P + c/J2 +Inx2),
so G1
=
n1RT(In P
+
1 +In x1) + 1z2RT(ln
P
+
4>2 +In x2).
(9.55)
Suppose that we define J.l
=
RT(ln P + 4> + lnx) g
==
+
RTin x.
Then
Comparing this with Eq u ati on (9.11) we see that J.L is the chemical potential! Since g1 and g2 are characteristics of the constituents, they are the same for the initial and final states and we can drop the subscripts on gin Equation (9.53), which becomes
(9.56) Our final result is therefore �G
=
G,.  G,
�
nt (p.1 •

g,) + n2(J.l2

g2)
•
or
(9.57)
Sec. 9.5
Mixing Processes
165
�s
==

a(aG) aT
•
p
Hence (9.58) Clearly
as > 0, as must be the case.
Entropy increase is to be expected when two different gases are mixed. But what if the two gases are the same? Does the removal of a diaphragm sep . aratmg the two halves of a volume V filled with one kind of gas change its entropy or not? Then
with
so that
and LlS
=
nR In 2 > 0.
(9.59)
Sttppose that the properties of gas 1 approach those of gas 2. The pre\'l ous result gives a finite entropy increase. However, if the two gases are identi cal there can be no change in entropy when the diaphragm is removed ( LlS
=
0). This is the Gibbs paradox.
TI1� resolution will be fully discussed when we treat statistical thermody namics. Suffice it to say now that there is a positive change in entropy when different, distinguishable gases are mixed. The transition from differettt to identical gases is not a continuous change, so Equation (9.59) does not apply to the case where the gases are the same.
The Chemical Potential and Open Systems
Chap. 9
166
. .
:
• '
•
.c
.
:
• .
.
•
•
.
.
.
.
•
.
'
:
.
:
:
•
.
.
.
. .
.
.
.
.
•
•
.
.
.
.
.
•
.
•
•
.
•
.
PROBLEMS
91 For a onecomponent open system tlU
==
.
TdS
PcJV

dn.
+ 11
(a) Use Eule r s theorem for homogeneous functions to show that ·
U
(b) Prove that (dG) r.r
==
:::
TS

PV
+ JLn.
J.Ldn or� equivalently, aG
J.l
fin
=
.
:·
r.P
(c) Prove the alternative definition of the ch e nt ical potent i al iiF
:
li=
dn
f.\·
.
92 (a) E xpress the chemical pote n t ia l of an ideal gas in terms of the te 1n pe rature T and the volume V.· •
J.L
==
crT

�
well"' dis
cussed at the beginning of the chapter There we observed tha t � E has a mag .
nitude ranging from about 0.0002
eV
to ().05
e V.
I 1.4 EQUIPARTITION OF ENERGY Let the mol ec u lar velocity have cornponents ( vx� v2
==
+
v2,
vY, vz ) .
Then
v2Y + v2,.
since taking an average is a linear op e r at i on By assumption, there arc no pre .
ferred directions, so •
2 V
X
2
== V
y
::_:
2 V
l
=
I ,  V". 3
Thus the mean kinetic energy per molecule aSS(lCiatcd with any one component the xcomponent, for example I
is
 nzv.;.X
2
l
'\
=
1  n1v
6
..
l =
2
·
kT.
Sec. I 1.4
Equipartition of Energy
191
stnce •
1
mv2 2
=
3 , kr 2 ,
(11.15) .
It follows that the average translational kinetic energy associated with each
component of velocity is onethird th� totai and has a magnitude Each component represents a
of kT /2.
degree of freedom f of a mechanical system. An
atom or a monatomic molecule, for example, has three translational degrees of freedom so that f
=
3.
A diatomic molecule is modeled as tv.'o masses connected by a s pri ng
.
As such, it has seven degrees of freedom: three translational, two rotational, and two vibrational. There are two rotational degrees of freedom, because t\\·o angles are needed to specify the orientation of an axially symmetric mol
ecule. The vibrational motion is represented by· a onedimensional harmonic oscillator. For simple harmonic motion, the energy is partly kinetic and partly potential; the average values of the two are equal, and both are expressible as quadratic forms. Thus these are two vibrational degrees of freedom. The equipartition theorem (proved in Chapter 14) states that (1) there is one degree of freedom associated with every form of energy quadratic fortn, and
(2)
of the energy of
the mean value
sponding to each degree of freedom is
expressible in
a molecule corre
kT/2. Thus the total average energy of a
molecule 8 is f kT/2. It follows that the internal energy of a gas must be the total average energy of its N molecules. That is,
(11.16). ..
or
u
u
=
n
=
f
1_RT.
(11.17)
11.5 SPECIFIC HEAT CAPACITY OF AN IDEAL GAS For a reversible process� the specific heat capacity at constant volume is given by
v
c
From Equation
au =
(11.17), this gives Cv
=
d f__RT T 2 d
The Kinetic Theory of Gases
Chap. II
192
TABLE
11.1
Ratio of specific heat capacities of various
values the velocity of so u nd in the gas.
gases at temperatures close to room temperature. The a re obtained hy measuring Gas
Gas
y
Gas
co,
He
1.66
H"'
1.40
Ne
l.M
0"
1.40
NH�
Ar Kr
1.67 1.69
f\., co
1.40 1.42
CH_. Air
.


Mayer's equation for an ideal gas is
Cp
=
and the ratio of specifi� heats
fR 2
a
+
R. Hence,
c,. ·+
=
j' + 2\
R=

2
R
,
is
y= For
cp
1.29 1.33 1.30 1.40
Cp C I'
=
f.
2
+
.
,
•
.
•
mo n ato mi c gas,
f
•
=
3�
c,.
3
= iR.
Cp
=
5
:;R. I
5
'Y = 1 =
1.67 .
...
These results are in good agreement with mfasured values at tem per atures in the vicinity of room temperature ( Ta ble 11.1 ). For a diatomic eas.. '
7
9
'Y = 7 =
1.28.
Here the agreement is poor: the number of degrees of freedom is too large. We try 
.f
=
5.
c,.
)
= 2_R.
c I'
==
7 
·
2
R
..
y =
7 
5
==
1.40.
Good agreement is obtained wi th this adjustment. Evidently� near room tem perature the rotational degrees of freedom
(Jr
dom are excited, but not blJth. Maxwell
C(lnsidered
the vibrational degrees of free
the resolution of this
q ue st io n the most pressing challenge confronting the kinetic theory. It took statistical thermodyn ami cs to provide the final answer (see Ch ap ter 15).
Sec. 11.6
Distribution of Molecular Speeds
193
Two other observations are noteworthy. For large molecules� f will be large and y � 1. In general,
1
n
.
Three
speeds that characterize
the MaxwellBoltzmann distribution arc of 5pcciaJ importance: '
1.
Mean speed:
,
I

·v
=
..
vN(·'')tlv
N
o
m
.

•

2r.kT
8kT 1Tn1
( 1 I .29)

sqttare speed: The mean square speed is given
41T 
.
by
1'11
21TkT
3kT_ 
m
•
nz
2. Root mean
=
)
=
47T 
=
.
•
� .
1 
. .
v4e··nl'.tlv
.
Sec. I I .6
Distribution of Molecular Speeds
197
The root mean square ( rtns) speed is, then, 3kT m
m
(11.30)
•
3. Most probable speed: This is the value
Vm
for which N (v) j N is a maximum, or for which
d (v2eav�) dv
a=
"
The derivative is zero for
vm
v
=
0
'
m 
2kT.
12 , which yields 1 ) a 1/ (
=
112 2kT =
=
m
112 kT 1.414 m
(11.31)
•
It is evident that (Figure 11.7) Vm
: V:
V,ms =
For a nitrogen molecule
=
1.414 : 1.596:
1 : 1.128
(m
=
4.65
:
1.732
1.225. X
(11.32)
1026 kg) at T = 273
K,
vrms
=
493 ms t, which is slightly greater than the speed of a sound wave in nitrogen gas. Numerical values of the mean speed and root rnean square speed for vari ous gases are given in Table 11.2.
N(v) N
0.6 0.4
 ... 6"' .... �+
�'_....__._ ��+· 6+�.+4+_.... .....____.___�
o� 0
Fi
� 2.0
��������.
0.4
0.8
1.0
1.2
1.6
11.7
The MaxwellBoltzmann speed distribution function at a particular temperature showing the most probable speed vm, the mean speed v, and the root mean square speed Vrms·
The Kinetic Theory of Gases
Chap. 11
198 TABLE 11.2 Molecular Speeds at 273 K. in ms 1 Gas
v
v,ms
1690 1210 570 530 450 420 400
1840 1310 620 580 490 460 430
'
•
There are other speed distributions of less interest to us� one is the num ber of particles whose �tcomponent of velocity lies in th e range
vr f()
v.\
+
dv,
regardless of the values of the other components:
(11.33) ( Fi g ure 11.8). Clearly�
vx
=
vY
vz since the velocity directions are assumed to
=
be un i formly distributed over all a ng 1 e s in s p ace
.
I I. 7
MEAN FREE PATH AND COLLISION FREQUENCY
We are interested in ho\\' far the molecules of a gas travel bet\\'cen collisions and how often they undergo collisi