TITLE.PM5

THERMODYNAMIC RELATIONS 361

dharm
\M-therm\Th7-2.pm5

Equating the co-efficients of dT in the two equations of ds, we have

c

T

s

T

v

v

= ∂

∂

F

HG

I

KJ

cv = T

∂

∂

F

H

I

K

s

T v

∂

∂

F

HG

I

KJ

= ∂

∂∂

c

v

T s

Tv

v

T

2

From eqn. (7.20),

∂

∂

F

H

I

K =

∂

∂

F

H

I

K

s

v

p

TvT

∂

∂∂

= ∂

∂

F

HG

I

KJ

22

2

s

vT

p

T v

∂

∂

F

HG

I

KJ =

∂

∂

F

HG

I

KJ

c

v

T p

T

v

T v

2

2

Also p = RTv ...(Given)

∂

∂

F

H

I

K =

p

T

R

v v

∂

∂

F

HG

I

KJ

=

2

2 0

p

T v

or ∂

∂

F

HG

I

KJ

cvv =

T

0

This shows that cv is a function of T alone, or cv is independent of pressure.

Also, cp = T (^) HF∂∂TsIK
p
∂
∂
F
HG
I
KJ
= ∂∂∂
c
p T
s
Tp
p
T
2
From eqn. (7.21),
∂
∂
F
H
I
K =−
∂
∂
F
H
I
K
s
p
v
T T p
∂
∂∂ =−
∂
∂
F
HG
I
KJ
22
2
s
pT
v
T p
∂
∂
F
HG
I
KJ
=− ∂
∂
F
HG
I
KJ
c
p T
v
T
p
T p
2
2
Again, v = R
p
...(Given)
∂
∂
F
HG
I
KJ
v =
T
R
p p
and
∂
∂
F
HG
I
KJ
2
2
v
T p = 0 ;
∂
∂
F
HG
I
KJ
c
p
p
T
= 0
This shows that cp is a function of T alone or cp is independent of pressure.