59. Now
dQ = nC
p
dT = n(C
V
+ R)dT =
3
2
nR + nR
dT =
5
2
nR dT ,
so we need to replace the factor 3/2 in the last problem by 5/2. The rest is the same. Thus the answer
now is
∆S =
5
2
nR ln
T
f
T
i
=
5
2
(1.00 mol)
8.31
J
mol
· K
ln
400 K
300 K
= 5.98 J/K .