PHYSICAL CHEMISTRY IN BRIEF

CHAP. 3: FUNDAMENTALS OF THERMODYNAMICS [CONTENTS] 78

Solution

By comparing with (3.23) we obtain

M=

(

∂z

∂x

)

y

= 10xy^3 + 7 and N=

(

∂z

∂y

)

x

= 15x^2 y^2.

It holds that

(

∂M

∂y

)

=

∂^2 z

∂x∂y

= 30xy^2 and

(

∂N

∂x

)

=

∂^2 z

∂y∂x

= 30xy^2.

The mixed second derivatives are identical and consequently functionzhas a total differential.

Sometimes we need to know the derivative ofxwith respect toyat a fixedz. Equation
(3.25) rearranges to

0 =

(

∂z

∂x

)

y

dx+

(

∂z

∂y

)

x

dy , [z] (3.27)

and from this relation we obtain

(

∂x

∂y

)

z

=−

(

∂z

∂y

)

( x

∂z

∂x

)

y

. (3.28)

This formula is identical with that for the differentiation of an implicitly defined function
z(x, y) = 0 (see a course of differential calculus).

Example

Using the van der Waals equation of state

p=

RT

Vm−b

−

a

Vm^2

calculate the derivative of molar volume with respect to temperature at constant pressure.