2.4 Calculation of Amounts of Heat and Energy Changes
69
E X A M P L E 2.18
Show that for a reversible adiabatic process the van der Waals gas obeys
C V dT −
nRT
V − nb
dV
(2.4-23)
Solution
Consider a system containing 1.000 mol of gas. Using Eq. (2.4-22) we can write for a closed
system
dU
∂U
∂T
V ,n
dT +
∂U
∂V
T ,n
dV C V dT +
a
V 2
m
dV
For a reversible process
dw −PdV −
RT
V m − b
−
a
V 2
m
dV
For an adiabatic process
dU C V dT +
a
V 2
m
dV dw −
RT
V m − b
−
a
V 2
m
dV
so that when terms are canceled
C V dT −
RT
V m − b
dV
Exercise 2.15
Show that for a reversible adiabatic process in a van der Waals gas with constant C V,m ,
T 2
T 1
V 1 − nb
V 2 − nb
nR/C V
(2.4-24)
E X A M P L E 2.19
Find the final temperature for the process of Example 2.17, using Eq. (2.4-24) instead of
Eq. (2.4-21), but still assuming that C V 3nR/2.
Solution
From Table A.3, the van der Waals parameter b is equal to 3.219 × 10 −5 m 3 mol −1 for
argon.
T 2 (373.15 K)
5.000 × 10 −3 m 3 mol −1 − 3.22 × 10 −5 m 3 mol −1
20.00 × 10 −3 m 3 mol −1 − 3.22 × 10 −5 m 3 mol −1
2/3
147.6 K
This value differs from the ideal value by 0.5 K.