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2.10 CALCULATING q, w, ¢U , AND ¢H FOR PROCESSES INVOLVING IDEAL GASES 37

EXAMPLE PROBLEM 2.5 1 2
-1
A system containing 2.50 mol of an ideal gas for which C V,m = 20.79 J mol K -1 is 16.6 bar, 16.6 bar,
25L
1.00L
taken through the cycle in the following diagram in the direction indicated by the
arrows. The curved path corresponds to PV = nRT , where T = T = T 3 . P/bar
1
a. Calculate q, w, ¢U , and ¢H for each segment and for the cycle assuming that
the heat capacity is independent of temperature.
3
b. Calculate q, w, ¢U , and ¢H for each segment and for the cycle in which the
direction of each process is reversed. V/L
Solution
We begin by asking whether we can evaluate q, w, ¢U , or ¢H for any of the
segments without any calculations. Because the path between states 1 and 3 is
isothermal, ¢U and ¢H are zero for this segment. Therefore, from the first law,
q 3:1 =-w 3:1 . For this reason, we only need to calculate one of these two quan-
tities. Because ¢V = 0 along the path between states 2 and 3, w 2:3 = . 0
Therefore, ¢U 2:3 = q 2:3 . Again, we only need to calculate one of these two
quantities. Because the total process is cyclic, the change in any state function is
zero. Therefore, ¢U =¢H = 0 for the cycle, no matter which direction is cho-
sen. We now deal with each segment individually.

Segment 1 : 2
The values of n, P and V , and P and V are known. Therefore, T and T can be cal-
1
2
2
2
1
1
culated using the ideal gas law. We use these temperatures to calculate ¢U as follows:
nC V,m
¢U 1:2 = nC V,m (T - T ) = (P V - P V )
2
1 1
1
2 2
nR
-1
20.79 J mol K -1
=
-1
0.08314 L bar K mol -1
* (16.6 bar * 25.0 L - 16.6bar * 1.00 L)
= 99.6kJ
The process takes place at constant pressure, so
5
10 Nm -2
w =-P external (V - V ) =-16.6bar *
2
1
bar
-3
3
3
-3
* (25.0 * 10 m - 1.00 * 10 m )
=-39.8kJ
Using the first law,
q =¢U - w = 99.6 kJ + 39.8 kJ = 139.4 kJ
We next calculate T :
2
P V 16.6bar * 25.0L
2 2
3
T = = = 2.00 * 10 K
2
-1
nR 2.50mol * 0.08314LbarK mol -1
We next calculate T = T 1 and then ¢H 1:2 :
3
P V 16.6bar * 1.00L
1 1
T = = = 79.9 K
1
-1
nR 2.50mol * 0.08314LbarK mol -1
¢H 1:2 =¢U 1:2 +¢(PV) =¢U 1:2 + nR(T - T )
2
1
3
= 99.6 * 10 J + 2.5mol * 8.314Jmol -1 K -1
* (2000K - 79.9K) = 139.4kJ