6.5 Third-Law Entropy and Residual Entropy
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S(T ) = S(0) +
T
0
dq rev
T
.
The second term can be calculated from a knowledge of c p (T ) for an ideal gas,
since
dq rev = dU + P dV
and, since for an ideal gas U = U(T ) only,
C V =
∂U
∂T
V
=
dU
dT
,
and
dq rev = C V dT + P dV .
Now, as all equilibrium states of an ideal gas are represented by
P V = Nk B T ,
we have
P dV + V dP = Nk B dT
so that
dq rev = (C V + Nk B ) dT − V dP
= C P dT − V dP .
Hence, we see that S(T ) can be obtained from
S(T ) = S(0) +
T
0
C P (T )
T
dT ,
in which C P (T ) is obtained from calorimetric data.
By comparing values of S(T ) calculated from the statistical mechanical formula
and spectroscopic data with those calculated calorimetrically, S(0) is found to be
zero for most gases, a result that agrees with Planck’s version of the third Law
of Thermodynamics: ‘S(0) for any pure material is zero’. However, some notable
exceptions do occur. We might well ask why nonzero values are found for S(0).
The answer is almost invariably that it is a consequence of unattainability of the
lowest state of the system. Let us consider H 2 O, CO, N 2 O, and CH 3 D, all having