TITLE.PM5

THERMODYNAMIC RELATIONS 363

dharm
\M-therm\Th7-2.pm5

∂

∂

F

H

I

K

v

T p = –

∂

∂

F

H

I

K

∂

∂

F

H

I

K

p

T

v

vT. p

= – FH∂∂psIK FH∂∂TsIK FH∂∂psIK FH∂∂vsKI

vvTT

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

∂

∂

F

H

I

K

v

T p =

∂

∂

F

HG

I

KJ

∂

∂

F

HG

I

KJ

∂

∂

F

HG

I

KJ

∂

∂

F

HG

I

KJ

T

v s

s

T

s

p

v

v T sT

= HF∂∂ IK HF∂∂IK FH∂∂ IK

R

S

T

U

V

W

∂

∂

F

H

I

K

T

v s

v

s

s

T

s

..TvTp = –

∂

∂

F

H

I

K

s

pT

∴ FH∂∂TvIK

p

= – FH∂∂spIK

T

This is Maxwell fourth relation.

Example 7.9. Derive the following relations :

(i)u = a – T (^) HF∂∂TaIK
v
(ii)h = g – T (^) HF∂∂TgIK
p
(iii)cv = – T ∂
∂
F
HG
I
KJ
2
2
v
a
T
(iv)cp = – T ∂
∂
F
H
G
I
K
J
2
2
p
g
T
where a = Helmholtz function (per unit mass), and
g = Gibbs function (per unit mass).
Solution. (i) Let a = f(v, T)
Then da = HF∂∂avIK
T
dv + HF∂∂TaIK
v
dT
Also da = – pdv – sdT
Comparing the co-efficients of dT, we get
∂
∂
F
H
I
K
a
T v = – s
Also a = u – Ts
or u = a + Ts = a – T FH∂∂TaIK
v
Hence u = a – T FH∂∂TaIK
v

. (Ans.)
(ii) Let g = f(p, T)

Then dg = HF∂∂gpIK

T

dp + HF∂∂TgIK

p

dT

Also dg = vdp – sdT

Comparing the co-efficients of dT, we get

∂

∂

F

H

I

K

g

T p = – s