Special Section 2
S2.A. Cooling and Liquefactions of Gases
S2.B. Entropy of Mixing and the Gibbs Paradox
S2.C. Osmotic Pressure in Dilute Solutions
S2.D. The Thermodynamics of Chemical Reactions
S2.E. The Thermodynamics of Electrolytes
S2.A. Cooling and Liquefactions of Gases
S2.A.1. The Joule Effect: Free Expansion
S2.A.2. The Joule-Kelvin Effect: Throttling
S2.A.1. The Joule Effect: Free Expansion
Consider the adiabatic free expansion of a gas for which
Q W n U 0
During the expansion, the gas is not in equilibrium so that thermodynamical analysis
is applicable only to the initial and final equilibrium states. On the reversible path
linking these 2 states, we have
U U
dU n 0 dT dV (2.188)
T Vn V Vn
U
V Tn T
dT dV dV (2.189)
U V Un
T Vn
The quantity
T 1 U
V Un CVn V Tn
is called the Joule coefficient. From
dU TdS PdV dn
P T
and using the diagram , we have
S V
U S P
T P T P (2.190)
V Tn V Tn T Vn
For an ideal gas, PV nRT , we have
U nR T
T P PP0 0
V Tn V V Un
and there is no temperature change during a free expansion.
For a van der Waals gas
an 2 an 2 nRT
P 2 V nb nRT P 2 (2.190a)
V V V nb
so that
U nR an 2
T P (2.191)
V Tn V nb V2
T an 2
(2.191a)
V Un CVnV 2
To find CVn , we start with
CVn S S 2P
T T T 2 (2.192)
V Tn V T Vn Tn T V Tn Vn T Vn
Now, from (2.190a), we have
P nR 2P
and 2 0
T Vn V nb T Vn
so that by (2.192), we have
CVn CVn T , n (2.193)
Since CVn contains no volume correction, it is thermodynamically consistent for it to
have the same value as the ideal gas. Neglecting any contributions from internal
3
degrees of freedom, we have CVn nR . Eq(2.191a) thus becomes
2
T 2an
(2.194)
V Un 3RV 2
Putting this into (2.189) gives
2an
dT dV
3RV 2
2an 1 1
Vf
2an dV
3R Vi V 2
T f Ti (2.196)
3R V f Vi
2an
or T V
3RV f Vi
so that there is cooling for free expansion. However, the effect is quite small.
S2.A.2. The Joule-Kelvin Effect: Throttling
S2.B. Entropy of Mixing and the Gibbs Paradox
Since entropy is a measurement of disorder, it should increase in the mixing of
distinguishable substances. Consider an ideal gas of volume V, temperature T and
consisting of nj moles of type j molecules with j 1, , m . The equation of state is
m m
nRT RT
P nj Pj (2.210)
V j 1 V j 1
where
m
RT
n nj Pj n j (2.211)
j 1 V
The molar fraction of type j molecules is defined as
nj Pj
xj (2.212)
n P
Consider now the Gibbs free energy G G T , P, n j with
m
dG SdT VdP j dn j (2.213)
j 1
From eq(2) of Exercise 2.3 and eq(2.107) we see that for a pure gas
T 5/ 2
G T , P, n nRT ln G
0
(2.214)
P
where G is some constant.
0
For the case of m types of molecules, we begin by putting each type of molecules in
its own compartment with the same overall temperature T and pressure P. The total
Gibbs free energy is therefore just the sum of the energies of m pure phase
compartments, i.e.,
T 5/ 2
GI T , P, n j n j RT ln
m
0
GI (2.215)
j 1 P
T 5/ 2 0
nRT ln GI
P
where GI
0
is a constant. If the walls of the compartments are now removed, the
gases will mix while keeping the overall temperature T and pressure P unchanged.
The total Gibbs free energy is again the sum of the energies of m pure phase gases but
of partial pressures given by (2.211), i.e.,
T 5/ 2
GF T , P, n j n j RT ln
m
0
GF (2.216a)
j 1 P
j
where GF is a constant. Now, (2.216a) (2.215) gives
0
m
P
G GF GI n j RT ln j GF 0 GI 0
j 1 P
m
n j RT ln x j GF 0 GI 0 (2.217)
j 1
The entropy of mixing is therefore
m
Smix n j R ln x j (2.218)
j 1
For m 1 , we have x1 1 and Smix 0 as it should be.
1
For m 2 and n1 n2 1 , we have x1 x2 so that Smix 2 R ln 2 .
2
Eq(2.218) seems reasonable as long as the gases to be mixed are distinct. However,
it also gives a finite entropy increase for the mixing of identical gases, which is absurd.
This is called the Gibbs paradox and can be resolved only by introducing the
indistinguishability of identical particles in quantum mechanics.
S2.C. Osmotic Pressure in Dilute Solutions
S2.D. The Thermodynamics of Chemical Reactions
S2.D.1. The Affinity
S2.D.2. Stability
S2.D.1. The Affinity
S2.D.1.1. Degree of Reaction
S2.D.1.2. Gibbs Free Energy
S2.D.1.1. Degree of Reaction
Consider a chemical reaction of the form
bA A bB B 1
bC C bD D
k
(2.227a)
k 2
where ki is the rate constant equal to the probability per unit time that the indicated
reaction can take place. Each b j 0 is called a stoichiometric coefficient and is
proportional to the number N j of the jth kind of molecules required in the reaction.
The reaction rate of (2.227) is defined as
dN A
k1 N AbA N BbB k2 N CbC N DbD (2.228a)
dt
One can write (2.227a) as
bA A bB B bC C bD D 0
or more generally,
X
i
i i 0 (2.227b)
where X i stand for the chemical substances and i bi so that the changes in the
numbers of molecules are given by
N i i i (2.227c)
where 0 is some constant.
As an example, the reaction (2.227a) is written as
A A B B CC D D 0
or, to preserve the structure of (2.227a),
A A B B C C D D (2.227)
For the reaction (k1) going to the right, we have
A , B 0 and C , D 0
n A n B nC n D
(2.235)
A B C D
where is called the degree of reaction. Thus, to consume n A bAn0 An0
moles of A, the resultant changes in the numbers of moles of the substances are
B
n A n A An0 n B n B n0 bB n0
A A
C D
nC n n b n n D n D n0 bD n0
A A C 0 C 0 A A
If, initially, the numbers of moles of the substances are
n A An0 nB B n0 nB
nC C n0 nD D n0 nD
the reaction will be completed when the numbers of moles of the substances become
nA 0 nB nB
nC C n0 n0 nD D n0 n0 nD
S2.D.1.2. Gibbs Free Energy
With the help of (2.235), the differential of the Gibbs energy G G T , P, n j
becomes
m
dG SdT VdP j dn j
j 1
m
SdT VdP j j d (2.236)
j 1
SdT VdP Ad
where m is the number of chemical species in the reaction and A is the affinity defined
by
m
G
A j j (2.238,7)
j 1 TP
At chemical equilibrium where T and P are constant, G is a minimum. Hence
0
G
A
0
0 (2.239)
TP
where the superscript 0 denotes an equilibrium value. In other words, the condition
for chemical equilibrium is
m
A j j 0 (2.239a)
j 1
Now, the fact that G is a minimum at equilibrium means that in the approach to
equilibrium, dG 0 . To be more precisely, a spontaneous reaction must obey
G
dG TP d Ad 0 (2.240)
TP
Using the convention that d < 0 if the reaction (2.227) goes to the right, (2.240) then
implies A > 0. Conversely, if the reaction goes to the left, d > 0 and A < 0.
If there are r reactions involving m chemical species, there will be r degree of
reactions k defined by
dn j
dk k with k 1, , r and j 1, , m
jk
where dn j is the change of n j in the reaction k and jk is the (signed)
k
stoichiometric coefficient of species j in the reaction k. Since
r
dn j dn j for j 1, , m
k
k 1
we have
r
dn j jk d k for j 1, , m (2.241)
k 1
According to (2.214) [see section S2.B], the chemical potential of the jth species in a
mixture of ideal gases can be written as
T 5 / 2 P
j T , Pj T0 , P0 RT ln 0
0
(2.242)
T0 Pj
j
where Pj is its partial pressure and the subscript 0 denotes some convenient
reference state. Some values of 0 at atmospheric pressure and room temperature
are listed in table 2.2. In terms of (2.242), the Gibbs free energy can be written as
m
G T , P, n j j
j 1
m T 5 / 2 P
m
n j T0 , P0 RT n j ln 0
0
T0 Pj
j
j 1 j 1
m T 5 / 2 P
m
n j T0 , P0 RT n j ln 0
0
T0 Px j
j
j 1 j 1
m T 5 / 2 P
m m
n j T0 , P0 RT n j ln 0 RT n j ln x j
0
j
j 1 j 1 T0 P j 1
m T 5 / 2 P m
n j T0 , P0 RTn ln RTn x j ln x j
0
j
0
(2.243a)
j 1 T0 P j 1
nj Pj
where x j with P being the total pressure and n the total number of moles
n P
of gases [see 2.212]. Note that we can write
m m m nj
n ln x j ln x ln x j
nj
j j
j 1 j 1 j 1
so that, for example, the Gibbs free energy for the reaction (2.227) is
m T 5 / 2 P
G T , P, n j T0 , P0 RTn ln 0
0
j
j 1 T0 P
RT ln x An A xBnB xCnC xDnD (2.243)
Similarly, the affinity can be written as
m
A j j
j 1
m T 5 / 2 P
m m
j T0 , P0 RT j ln 0 RT j ln x j
0
j (2.244a)
j 1 j 1 T0 P j 1
with
m m m
j ln x j ln x j ln x j j
j
j 1 j 1 j 1
which, for the reaction (2.227), becomes
x bC x bD
j ln x j ln xA xB xC xD
m
A B C D
ln CbA DbB (2.244b)
j 1 x A xB
At equilibrium, A 0 so that (2.244a) becomes
x bC x bD 1 m T 5 / 2 P
m
ln CbA DbB j T0 , P0 j ln 0
0
j
T0 P
(2.245)
x A xB RT j 1 j 1
which is one form of the law of mass reaction. [see Exercise 2.11]
S2.D.2. Stability
Since, at equilibrium, G G T , P, is a minimum for fixed T and P, we have
0
G
A 0
0
(2.246)
TP
0 0
2G A
2 0 (2.247)
TP TP
From the fundamental equation
G H TS U PV TS
we have, from (2.246),
0 0 0
G H S
0 T
TP TP TP
0 0
H S
T (2.248)
TP TP
0 0
G G
T T
T P
TP T PT P
A
0
T (2.249)
T P
0
H
The term is called the heat of reaction. It is positive (negative) for
TP
endothermic (exothermic) reactions.
For an ideal gas reaction, we can use (2.244a) in (2.249) to get
H
0
m T 5/ 2
P0 5 m
m j
RT j ln T RT j RT ln x j (2.250)
TP j 1 0 P 2 j 1 j 1
m 5 T 5 / 2 P m
RT j ln 0 RT ln x j j (2.250a)
j 1 2 T0 P j 1
The last term is due the entropy of mixing [cf (2.218) of section S2.B] and, for the
case where the total number of particles remains unchanged, i.e.,
m m
dn j d 0 j 0
j 1 j 1
it is the only contribution.
S2.E. The Thermodynamics of Electrolytes
