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
)
TdV+(
∂S
∂T
)
VdT]
−PdV.Similarly applying (5.45) todU, we find
dU=(
∂U
∂V
)
TdV+(
∂U
∂T
)
VdT.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 thatcan 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−PdVwhen 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.