Page 141 - Physical Chemistry

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122
Chapter 4 Exercise
Material Equilibrium
1
1
For water at 95.0°C and 1 atm: a 7.232 10 4 K , k 4.81 10 5 bar ,
3
c 4.210 J/(g K), and r 0.96189 g/cm . Find c for water at 95.0°C and
P V
1 atm. [Answer: 3.794 J/(g K).]
The use of (4.53) and experimental C values to find C for solids and liquids
P,m V,m
gives the following results at 25°C and 1 atm:

Substance Cu(s) NaCl(s) I (s) C H (l) CS (l) CCl (l)
2 6 6 2 4
C /[J/(mol K)] 23.8 47.7 48 95 47 91
V,m
C /[J/(mol K)] 24.4 50.5 54 136 76 132
P,m
C and C usually do not differ by much for solids but differ greatly for liquids.
P,m V,m
Ideal-Gas (
U/
V)
T
An ideal gas obeys the equation of state PV nRT, whereas a perfect gas obeys both
PV nRT and (
U/
V) 0. For an ideal gas, (
P/
T) nR/V, and Eq. (4.47) gives
T V
(
U/
V) nRT/V P P P 0.
T
10U>0V2 0 ideal gas (4.55)
T
We have proved that all ideal gases are perfect, so there is no distinction between an
ideal gas and a perfect gas. From now on, we shall drop the term “perfect gas.”

(
U/
V) of Solids, Liquids, and Nonideal Gases
T
The internal pressure (
U/
V) is, as noted in Sec. 2.6, a measure of intermolecular in-
T
teractions in a substance. The relation (
U/
V) aT/k P [Eq. (4.47)] enables one
T
to find (
U/
V) from experimental data. For solids, the typical values a 10 4.5 K 1
T
and k 10 5.5 atm 1 (Sec. 1.7) give at 25°C and 1 atm
1
5.5
10U>0V2 110 4.5 K 21300 K2110 atm2 1 atm 3000 atm 300 J>cm 3
T
For liquids, the typical a and k values give at 25°C and 1 atm
4
1
3
10U>0V2 110 K 21300 K2110 atm2 3000 atm 300 J>cm 3
T
The large (
U/
V) values indicate strong intermolecular forces in solids and liquids.
T
EXAMPLE 4.3 (
U/
V) for a nonideal gas
T
Estimate (
U/
V) for N gas at 25°C and 1 atm using the van der Waals equa-
T 2
tion and the van der Waals constants of Sec. 8.4.
The van der Waals equation (1.39) is
2
2
1P an >V 21V nb2 nRT (4.56)
We have (
U/
V) T(
P/
T) P [Eq. (4.47)]. Solving the van der Waals
T V
equation for P and taking (
/
T ) , we have
V
nRT an 2 0P nR
P and a b
V nb V 2 0T V V nb
0U 0P nRT nRT an 2 an 2
a b Ta b P a b (4.57)
0V 0T V nb V nb V 2 V 2
T V