TITLE.PM5

362 ENGINEERING THERMODYNAMICS

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\M-therm\Th7-2.pm5

Example 7.8. Using the first Maxwell equation, derive the remaining three.

Solution. The first Maxwell relation is as follows :

∂

∂

F

H

I

K =−

∂

∂

F

H

I

K

T

v

p

svs ...(i) (Eqn. 7.18)

(1) Using the cyclic relation

∂

∂

F

H

I

K

∂

∂

F

H

I

K

∂

∂

F

H

I

K

T

v

v

s

s

sT v..T = – 1

∴

∂

∂

F

H

I

K

s

vT = –

∂

∂

F

H

I

K

T

v s.

∂

∂

F

H

I

K

s

T v ...(ii)

Substituting the value from eqn. (i) in eqn. (ii), we get

∂

∂

F

H

I

K

s

vT =

∂

∂

F

H

I

K

∂

∂

F

H

I

K

p

s

s

vv. T ...(iii)

Using the chain rule,

∂

∂

F

H

I

K

∂

∂

F

H

I

K

∂

∂

F

H

I

K

p

s

s

T

T

vvv..p = 1 ...(iv)

Substituting the value of eqn. (iv) in eqn. (iii), we get

∂

∂

F

H

I

K

s

vT =

∂

∂

F

H

I

K

p

T v

This is Maxwell Third relation.

(2) Again using the cyclic relation

∂

∂

F

H

I

K

∂

∂

F

H

I

K

∂

∂

F

H

I

K

s

p

p

v

v

vs..s p = – 1

∴

∂

∂

F

H

I

K

v

s p = –

∂

∂

F

H

I

K

∂

∂

F

H

I

K

p

s

v

vs. p ...(v)

Substituting the value from eqn. (i) into eqn. (v)

∂

∂

F

H

I

K

v

s p =

∂

∂

F

H

I

K

∂

∂

F

H

I

K

T

v

v

ss. p ...(vi)

Again using the chain rule,

∂

∂

F

H

I

K

∂

∂

F

H

I

K

∂

∂

F

H

I

K

T

v

v

p

p

ss s..T = 1

Substituting the value of (vi) into (v), we get

∂

∂

F

H

I

K =

∂

∂

F

H

I

K

v

s

T

p p s

This is Maxwell second relation.

(3)

∂

∂

F

H

I

K

∂

∂

F

H

I

K

∂

∂

F

H

I

K

v

T

T

p

p

p..vTv = – 1