2.3 New Thermodynamic State Functions
43
P=P f
P=P i
dG T =
P f
P i
V dP .
(2.3.7a)
For an ideal gas we thus obtain
((G) T ≡ G(T , P f ) − G(T , P i ) = Nk B T ln
P f
P i
.
(2.3.7b)
If we choose the initial pressure to be P ◦ ≡ 1 bar, we obtain
G(T , P f ) − G(T , P
◦ ) = Nk B T ln
P
P ◦
and, if we define the standard Gibbs energy G ◦ (T ) via G ◦ (T ) ≡ G(T , P ◦ ), we may
then express the Gibbs energy in the form
G(T , P ) = G
◦ (T ) + Nk B T ln
P
P ◦
.
(2.3.7c)
Example 2.2 Illustration of the usefulness of a Maxwell relation.
We begin with the differential form for the combined first and second laws of
thermodynamics as given in Eq. (2.2.12), namely, dU = T dS − P dV . We shall
consider the entropy to be a function of T and V , thereby allowing us to replace the
total differential dS by
dS =
∂S
∂T
V
dT +
∂S
∂V
T
dV ,
to obtain a modified version of the combined first and second laws, which we may
express as
dU = T
∂S
∂T
V
dT +
T
∂S
∂V
T
− P
dV
≡
∂U
∂T
V
dT +
∂U
∂V
T
dV .
(2.3.8)
Upon comparing the coefficients of dV in each of these two equations for dU , we
see that
∂U
∂V
T
= T
∂S
∂V
T
− P .
(2.3.9)
To eliminate the partial derivative involving the entropy S from our expression, we
may employ one of the four best-known Maxwell relations, namely the reciprocal
of Eq. (2.3.4b),