TITLE.PM5

THERMODYNAMIC RELATIONS 367

dharm
\M-therm\Th7-2.pm5

Example 7.14. Derive the following relations

∂

∂

F

H

I

K

T

v u =

T p

T

p

c

v

v

∂

∂

F

HG

I

KJ −

.

Solution.

∂

∂

F

H

I

K

T

v u can be expressed as follows :

∂

∂

F

H

I

K

T

v u =

−FH∂∂ IK

∂

∂

F

H

I

K

T

u

v

u

v

T

= –

∂

∂

F

H

I

K

∂

∂

F

H

I

K

u

v

u

T

T

v

Also Tds = du + pdv

or du = Tds – pdv

or

∂

∂

F

H

I

K

u

vT = T^

∂

∂

F

H

I

K

s

vT – p^

∂

∂

F

H

I

K

u

vT

or

∂

∂

F

H

I

K

u

vT = T^

∂

∂

F

H

I

K

s

vT – p ...(i)

or

∂

∂

F

H

I

K

u

T v = T^

∂

∂

F

H

I

K

s

T v ...(ii)

Dividing eqn. (i) by eqn. (ii), we get

∂

∂

F

H

I

K

T

v u =

T sv p

T Ts

T

v

∂

∂

F

H

I

K −

∂

∂

F

H

I

K

...(iii)

Also cv = T

∂

∂

F

H

I

K

s

T v

and

∂

∂

F

HG

I

KJ

s

vT =

∂

∂

F

H

I

K

p

T v ... Maxwell relation

Substituting these value in eqn. (iii), we get

∂

∂

F

H

I

K

T

v u =

T p

T

p

c

v

v

∂

∂

F

HG

I

KJ −

...Proved.

+Example 7.15. Prove that for any fluid

(i)

∂

∂

F

H

I

K

h

vT = v^

∂

∂

F

H

I

K

p

vT + T^

∂

∂

F

H

I

K

p

T v (ii)

∂

∂

F

H

I

K

h

pT = v – T^

∂

∂

F

H

I

K

v

T p

Show that for a fluid obeying van der Waal’s equation

p = vbRT− – a

v^2

where R, a and b are constants

h (enthalpy) =

RTb

vb

a

− − v

2

+ f(T)

where f(T) is arbitrary.