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54 CHAPTER 3 The Importance of State Functions: Internal Energy and Enthalpy

V f V f
2
0U n a 1 1
b. ¢U = a b dV = dV = n aa - b
2
T
0V T 3 V 2 V i V f
3
V i V i
Note that ¢U T is zero if the attractive part of the intermolecular potential is zero.
Example Problem 3.5 demonstrates that in general (0U>0V) Z 0 , and that ¢U T
T
can be calculated if the equation of state of the real gas is known. This allows the rela-
tive importance of ¢U = 1 V i V f (0U>0V) dV and ¢U = 1 T i T f C dT to be determined
V
V
T
T
in a process in which both T and V change, as shown in Example Problem 3.6.
EXAMPLE PROBLEM 3.6

One mole of N 2 gas undergoes a change from an initial state described by T = 200. K
and P = 5.00 bar to a final state described by T = 400. K and P = 20.0 bar . Treat
f
i
6
N 2 as a van der Waals gas with the parameters a = 0.137 Pa m mol -2 and
3
-5
b = 3.87 * 10 m mol -1 . We use the path N (g, T = 200. K, P = 5.00 bar) :
2
N (g, T = 200. K, P = 20.0 bar) : N (g, T = 400. K, P = 20.0 bar) , keeping in
2
2
mind that all paths will give the same answer for ¢U of the overall process.
= V f
a. Calculate ¢U T (0U>0V) dV using the result of Example Problem 3.5.
T
1 V i -3 3 -4 3
Note that V = 3.28 * 10 m and V f = 7.88 * 10 m at 200. K, as calcu-
i
lated using the van der Waals equation of state.
b. Calculate ¢U = n 1 T i T f C V, m dT using the following relationship for C V, m in this
V
temperature range:
C V,m T T 2 T 3
-8
-5
= 22.50 - 1.187 * 10 -2 + 2.3968 * 10 - 1.0176 * 10
-1 -1 2 3
J K mol K K K
n
The ratios T >K n ensure that C V, m has the correct units.
c. Compare the two contributions to ¢U . Can ¢U T be neglected relative to ¢U V ?
Solution
a. Using the result of Example Problem 3.5,
1 1
¢U = n aa - b = 0.137 Pa m 6
2
T
V m, i V m, f
1 1
* a - b =-132 J
-3
3.28 * 10 m 3 7.88 * 10 -4 m 3
T f
b. ¢U = n C V, m dT
V
3
T i
T T 2
-5
400. 22.50 - 1.187 * 10 -2 + 2.3968 * 10 2
= § K K ¥ da T b J
T 3 K
-8
200. -1.0176 * 10 3
3
K
= (4.50 - 0.712 + 0.447 - 0.0610)kJ = 4.17 kJ
c. ¢U T is 3.2% of ¢U V for this case. In this example, and for most processes, ¢U T
for real gases.
can be neglected relative to ¢U V
The calculations in Example Problems 3.5 and 3.6 show that to a good approxima-
tion ¢U = 1 V i V f (0U>0V) dV L 0 for real gases under most conditions. Therefore, it
T
T