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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 69
Assuming that CV = 3
2R, the ideal gas limit, for the temperature range of
interest, the molar entropy at 250 K is given by
Sm(250 K) = Sm(298 K) + ∫
250 K
298 K
3
2R
dT
T
= Sm(298 K) + 3
2R × ln(
250 K
298 K
)
= (154.84 JK−1mol−1)
+ ( 32 × 8.3145 JK
−1mol−1) × ln(250 K
298 K
)
= 153 JK−1mol−1 .
E3B.5(b) Two identical blocks must come to their average temperature. �erefore the
�nal temperature is
Tf = 1
2 (T1 + T2) =
1
2 × (100 ○C + 25 ○C)
= 62.5... ○C = 3.35... × 102 K = 336 K .
Although the above result may seem self-evident, the more detailed explaina-
tion is as follows. �e heat capacity at constant volume is de�ned in [2A.14–
43], CV = (∂U/∂T)V . As shown in Section 2A.4(b) on page 43, if the heat
capacity is constant, the internal energy changes linearly with the change in
temperature. �at is ∆U = CV∆T = CV(Tf − Ti). For the two blocks at
the initial temperatures of T1 and T2, the change in internal energy to reach
the �nal temperature Tf is ∆U1 = CV ,1(Tf − T1) and ∆U2 = CV ,2(Tf − T2),
respectively. �e blocks of metal are made of the same substance and are of
the same size, therefore CV ,1 = CV ,2 = CV . Because the system is isolated
the total change in internal energy is ∆U = ∆U1 + ∆U2 = 0. �is means that
∆U = CV ((Tf − T1) − (Tf − T2)) = CV × (2Tf − (T1 +T2)) = 0, which implies
that the �nal temperature is Tf = 1
2 (T1 + T2), as stated above.
�e temperature variation of the entropy at constant volume is given by [3B.7–
90], ∆S = CV ln (Tf/Ti), with Cp replaced by CV . Expressed with the speci�c
heat CV ,s = CV/m it becomes
∆S = mCV ,s ln(
Tf
Ti
)
Note that for a solid the internal energy does not change signi�cantly with the
volume or pressure, thus it can be assumed that CV = Cp = C. �e entropy