Page 140 - Physical Chemistry

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pressure, a fact we shall use frequently. For gases, V is large and G increases rapidly Section 4.4
m
m
as P increases (due mainly to the decrease in S as P increases). Thermodynamic Relations for a
System in Equilibrium
We also have (
G/
T ) S. However, thermodynamics does not define absolute
P
entropies, only entropy differences. The entropy S has an arbitrary additive constant.
Thus (
G/
T) has no physical meaning in thermodynamics, and it is impossible to
P
measure (
G/
T) of a system. However, from (
G/
T) S, we can derive
P
P
(
G/
T ) S. This equation has physical meaning.
P
In summary: For solids and liquids, temperature changes usually have significant
effects on thermodynamic properties, but pressure effects are small unless very large
pressure changes are involved. For gases not at high pressure, temperature changes usu-
ally have significant effects on thermodynamic properties and pressure changes have
significant effects on properties that involve the entropy (for example, S, A, G) but usu-
ally have only slight effects on properties not involving S (for example, U, H, C ).
P
Joule–Thomson Coefficient
We now express some more thermodynamic properties in terms of easily measured
quantities. We begin with the Joule–Thomson coefficient m JT (
T/
P) . Equa-
H
tion (2.65) gives m (
H/
P) /C . Substitution of (4.48) for (
H/
P) gives
JT
T
P
T
m 11>C 23T 10V>0T2 V4 1V>C 21aT 12 (4.52)
P
P
P
JT
which relates m to a and C .
JT P
Heat-Capacity Difference
Equation (2.61) gives C C [(
U/
V) P](
V/
T) . Substitution of
T
V
P
P
(
U/
V) aT/k P [Eq. (4.47)] gives C C (aT/k)(
V/
T) . Use of a
T P V P
V 1 (
V/
T) gives
P
2
C C TVa >k (4.53)
P V
For a condensed phase (liquid or solid), C is readily measured, but C is hard to mea-
P V
sure. Equation (4.53) gives a way to calculate C from the measured C .
V P
Note the following: (1) As T → 0, C → C . (2) The compressibility k can be
P V
proved to be always positive (Zemansky and Dittman, sec. 14-9). Hence C C . (3) If
P V
a 0, then C C . For liquid water at 1 atm, the molar volume reaches a minimum
P V
at 3.98°C (Fig. 1.5). Hence (
V/
T) 0 and a 0 for water at this temperature. Thus
P
C C for water at 1 atm and 3.98°C.
P V
EXAMPLE 4.2 C C V
P
1
For water at 30°C and 1 atm: a 3.04 10 4 K , k 4.52 10 5 atm 1
2
3
4.46 10 10 m /N, C 75.3 J/(mol K), V 18.1 cm /mol. Find C of
P,m m V,m
water at 30°C and 1 atm.
Division of (4.53) by the number of moles of water gives C C
P,m V,m
2
TV a /k. We find
m
1 2
3
6
1
4
TV a 2 1303 K2118.1 10 m mol 213.04 10 K 2
m

2
k 4.46 10 10 m >N
1
1
2
TV a >k 1.14 J mol K
m
C V,m 74.2 J>1mol K2 (4.54)
For liquid water at 1 atm and 30°C, there is little difference between C and
P,m
C . This is due to the rather small a value of 30°C water; a is zero at 4°C and
V,m
is still small at 30°C.