TITLE.PM5

THERMODYNAMIC RELATIONS 359

dharm
\M-therm\Th7-1.pm5

Substituting this value in eqn. (3), we get

(dh)T = v

p

v T

p

TvT

∂

∂

F

H

I

K +

∂

∂

F

H

I

K

L

N

M

O

Q

P dv ...(4)

8.7. Van der Waal’s Equation

∂

∂

F

H

I

K

p

vT =

∂

∂−

F

H

I

K−

L

NM

O

v QP

RT

vb

a

v^2 T

= – ()vbRT− 23 +^2 va ...(5)

∂

∂

F

H

I

K

p

T v =

∂

∂−

F −

HG

I

KJ

L

N

M

M

O

Q

P

T P

RT

vb

a

v^2 v

= vbR− ...(6)

Substituting the values of eqns. (5) and (6) in equation (1), we get

(dh)T = v RT

vb

a

v

T R

vb

−

−

+

R

S

T

U

V

W

+

−

F

HG

I

KJ

L

N

M

M

O

Q

P

()^23 P

(^2) dv
∴ ()dhT
1
2
z = – RT^
v
z (^1) ()vb− 2
2
dv + 2a
dv
(^1) v^2
2
z + RT^
dv
z 1 ()vb−
2
∴ (h 2 – h 1 )T = – RT loge vbvb^2 b vbvb
121
− 11
−
F
HG
I
KJ
− RS − − −
T
U
V
W
L
N
M
O
Q
P

• 2a 11
  21

2

vv 1

− RT e vbvb

F

HG

I

KJ

+ −−

F

HG

I

KJ

log

= bRT (^) (v^1 b) (v^1 b)
21 −
L − −
NM
O
QP

• 2a 1
  v

1

21 v

−

L

N

M

O

Q

P. (Ans.)

(iii)Change in entropy :

The change in entropy is given by

ds = cp dTT Tp

v

+FH∂∂ IK. dv

For Van der Waals equation,

∂

∂

F

H

I

K = −

p

T

R

v vb ...as per eqn. (6)

∴ ds = cv dTT +vbR− dv

∴ ds

1

2

z = cv^

dT

T

R dv

vb

L

NM

O

QP

+

zz 1 ()−

2

1

2

∴ s 2 – s 1 = cv loge T

T

2

1

L

N

M

O

Q

P + R loge^

vb

vb

2

1

−

−

L

N

M

O

Q

P. (Ans.)

Example 7.6. The equation of state in the given range of pressure and temperature is

given by

v =

RT

p

C

T

− 3

where C is constant.

Derive an expression for change of enthalpy and entropy for this substance during an

isothermal process.