5.11 THERMODYNAMIC RELATIONS
Show that(∂S/∂V)T=(∂P /∂T)V.
Applying (5.45) todS, with independent variablesVandT, we find
dU=TdS−PdV=T
[(
∂S
∂V
)
T
dV+
(
∂S
∂T
)
V
dT
]
−PdV.
Similarly applying (5.45) todU, we find
dU=
(
∂U
∂V
)
T
dV+
(
∂U
∂T
)
V
dT.
Thus, equating partial derivatives,
(
∂U
∂V
)
T
=T
(
∂S
∂V
)
T
−P and
(
∂U
∂T
)
V
=T
(
∂S
∂T
)
V
.
But, since
∂^2 U
∂T ∂V
=
∂^2 U
∂V ∂T
, i.e.
∂
∂T
(
∂U
∂V
)
T
=
∂
∂V
(
∂U
∂T
)
V
,
it follows that
(
∂S
∂V
)
T
+T
∂^2 S
∂T ∂V
−
(
∂P
∂T
)
V
=
∂
∂V
[
T
(
∂S
∂T
)
V
]
T
=T
∂^2 S
∂V ∂T
.
Thus finally we get the Maxwell relation
(
∂S
∂V
)
T
=
(
∂P
∂T
)
V
.
The above derivation is rather cumbersome, however, and a useful trick that
can simplify the working is to define a new function, called apotential.The
internal energyUdiscussed above is one example of a potential but three others
are commonly defined and they are described below.
Show that(∂S/∂V)T=(∂P /∂T)Vby considering the potentialU−ST.
We first consider the differentiald(U−ST). From (5.5), we obtain
d(U−ST)=dU−SdT−TdS=−SdT−PdV
when use is made of (5.44). We rewriteU−STasFfor convenience of notation;Fis
called theHelmholtz potential. Thus
dF=−SdT−PdV,
and it follows that (
∂F
∂T
)
V
=−S and
(
∂F
∂V
)
T
=−P.
Using these results together with
∂^2 F
∂T ∂V
=
∂^2 F
∂V ∂T
,
we can see immediately that
(
∂S
∂V
)
T
=
(
∂P
∂T
)
V
,
which is the same Maxwell relation as before.