TITLE.PM5

THERMODYNAMIC RELATIONS 365

dharm
\M-therm\Th7-2.pm5

Now substituting this in eqn. (i), we get

ds = cp

dT

T

s

T p

−F

HG

I

KJ

∂

∂

. dp ...(ii)

But β =^1 v Tv

p

∂

∂

F

H

I

K

Substituting this in eqn. (ii), we get

ds = cp dTT – βvdp (Ans.)

Example 7.11. Derive the following relations :

(i)HF∂∂TpIK

s

= Tvc

p

β (ii) ∂

∂

F

H

I

K

T

v s = –

T

cKv

β.

where β = Co-efficient of cubical expansion, and

K = Isothermal compressibility.

Solution. (i) Using the Maxwell relation (7.19), we have

∂

∂

F

H

I

K

T

p s =

∂

∂

F

H

I

K

v

s p =

∂

∂

F

H

I

K

∂

∂

F

H

I

K

v

T

T

p s p

Also cp = T (^) HF∂∂TsIK
p
From eqn. (7.34), β =^1
v
v
T p
∂
∂
F
H
I
K
∂
∂
F
H
I
K =
T
p
vT
s cp
β
i.e., ∂
∂
F
H
I
K =
T
p
Tv
s cp
β. (Ans.)
(ii) Using the Maxwell relation (7.18)
∂
∂
F
H
I
K
T
v s = –
∂
∂
F
H
I
K
p
sv = –
∂
∂
F
H
I
K
∂
∂
F
H
I
K
p
T
T
vvs
Also cv = T (^) HF∂∂TsIK
v
(Eqn. 7.23)
K = –^1 v vp
T
∂
∂
F
H
I
K
(Eqn. 7.36)
Then FH∂∂TvIK
s
= – Tc Tp
v v
∂
∂
F
H
I
K
Also FH∂∂pvIK FH∂∂TvIKHF∂∂TpIK
T p v
= – 1
i.e., FH∂∂TpIK
v
= – ∂
∂
F
H
I
K
∂
∂
F
H
I
K
p
v
v
T T p
= – HF−^1 vKKIβv = Kβ