TITLE.PM5

THERMODYNAMIC RELATIONS 369

dharm
\M-therm\Th7-2.pm5

With the aid of eqn. (ii) show that

∂

∂

F

H

I

K

u

pT = – T^

∂

∂

F

H

I

K

v

T p – p

∂

∂

F

H

I

K

v

pT

The quantity cp (^) HF∂∂TpIK
h
is known as Joule-Thomson cooling effect. Show that this cooling
effect for a gas obeying the equation of state (v – b) =
RT
p^ –
C
T^2 is equal to^
3C
T
FHG 2 IKJ−b.
Solution. We know that
∂
∂
F
H
I
K
h
pT = – μcp ...[Eqn. (7.44)]
Also μ = c^1 T Tv v
p p
∂
∂
F
H
I
K −
L
N
M
M
O
Q
P
P
...[Eqn. (7.46)]
∴
∂
∂
F
H
I
K
h
pT = – T
v
T p v
∂
∂
F
H
I
K −
L
N
M
M
O
Q
P
P
= v – T (^) HF∂∂TvIK
p
... Proved.
Also μ =
∂
∂
F
H
I
K
T
p h
∴
∂
∂
F
H
I
K
h
pT = – cp^
∂
∂
F
H
I
K
T
p h.
(ii) Let u = f(T, v)
du =
∂
∂
F
H
I
K
u
T v^ dT +
∂
∂
F
H
I
K
u
vT^ dv
= cv dT +
∂
∂
F
H
I
K
u
vT^ dv ...(i)
Also du = Tds – pdv
Substituting the value of Tds [from eqn. 7.24], we get
du = cv dT + T
∂
∂
F
HG
I
KJ
p
T v^ dv – pdv
= cv dT + T
p
T v p
∂
∂
F
HG
I
KJ −
L
N
M
O
Q
P dv ...(ii)
From (i) and (ii), we get
∂
∂
F
H
I
K
u
vT = T^
∂
∂
F
HG
I
KJ
p
T v – p ...Proved.
Also
∂
∂
F
HG
I
KJ
u
pT =
∂
∂
F
H
I
K
∂
∂
F
HG
I
KJ
u
v
v
T p T
or
∂
∂
F
HG
I
KJ
u
pT =
∂
∂
F
HG
I
KJ
∂
∂
F
HG
I
KJ −
L
N
M
O
Q
P
v
p T
p
TvT p
or
∂
∂
F
HG
I
KJ
u
pT = T
p
T
v
vTp
∂
∂
F
HG
I
KJ
∂
∂
F
HG
I
KJ – p^
∂
∂
F
HG
I
KJ
v
pT
...Proved.