Mathematical Methods for Physics and Engineering : A Comprehensive Guide

PARTIAL DIFFERENTIATION

5.11 Thermodynamic relations

Thermodynamic relations provide a useful set of physical examples of partial

differentiation. The relations we will derive are calledMaxwell’s thermodynamic

relations. They express relationships between four thermodynamic quantities de-

scribing a unit mass of a substance. The quantities are the pressureP, the volume

V, the thermodynamic temperatureTand the entropySof the substance. These

four quantities are not independent; any two of them can be varied independently,

but the other two are then determined.

The first law of thermodynamics may be expressed as

dU=TdS−PdV, (5.44)

whereUis the internal energy of the substance. Essentially this is a conservation

of energy equation, but we shall concern ourselves, not with the physics, but rather

with the use of partial differentials to relate the four basic quantities discussed

above. The method involves writing a total differential,dUsay, in terms of the

differentials of two variables, sayXandY, thus

dU=

(

∂U

∂X

)

Y

dX+

(

∂U

∂Y

)

X

dY , (5.45)

and then using the relationship

∂^2 U

∂X∂Y

=

∂^2 U

∂Y ∂X

to obtain the required Maxwell relation. The variablesXandYaretobechosen

fromP,V,TandS.

Show that(∂T /∂V)S=−(∂P /∂S)V.

Here the two variables that have to be held constant, in turn, happen to be those whose
differentials appear on the RHS of (5.44). And so, takingXasSandYasVin (5.45), we
have

TdS−PdV=dU=

(

∂U

∂S

)

V

dS+

(

∂U

∂V

)

S

dV ,

and find directly that
(
∂U
∂S

)

V

=T and

(

∂U

∂V

)

S

=−P.

Differentiating the first expression with respect toVand the second with respect toS,and
using

∂^2 U

∂V ∂S

=

∂^2 U

∂S∂V

,

we find the Maxwell relation
(
∂T
∂V

)

S

=−

(

∂P

∂S

)

V

.