Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(Darren Dugan) #1

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.

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