TITLE.PM5

356 ENGINEERING THERMODYNAMICS

dharm
\M-therm\Th7-1.pm5

Since the above equation is true for any process, therefore, it will also be true for the case

when dp = 0 and hence

(cp – cv) (dT)p = p

u

+ vT

∂

∂

F

H

I

K

L

N

M

O

Q

P (dv)p

or (cp – cp) = p uv

T

L +FH∂∂ IK

N

M

O

Q

P^

∂

∂

F

H

I

K

v

T p

By definition, β =

1

v

v

T p

∂

∂

F

H

I

K

∴ The above equation becomes,

cp – cv = p uv

T

L +FH∂∂ IK

N

M

O

Q

P^ vβ

or = pvβ + vβ

∂

∂

F

H

I

K

u

vT Proved.

+Example 7.2. Find the value of co-efficient of volume expansion β and isothermal

compressibility K for a Van der Waals’ gas obeying

p a

v

FH + 2 IK()vb− = RT.

Solution. Van der Waals equation is

p a

v

FH + (^2) KI()vb− = RT
Rearranging this equation, we can write
p = vbRT a
− v
− 2
Now for β we require F∂∂
H
I
K
u
T p. This can be found by writing the cyclic relation,
∂
∂
F
H
I
K
∂
∂
F
H
I
K
∂
∂
F
H
I
K
v
T
T
p
p
p vTv = – 1
Hence
∂
∂
F
H
I
K
v
T p = –
∂
∂
F
H
I
K
∂
∂
F
H
I
K
p
T
p
v
v
T
From the Van der Waals equation,
∂
∂
F
H
I
K
p
T v =
R
vb−
Also FH∂∂pvIK
T
= – RT
()vb−^2

• 2
  3
  a
  v
  Hence β =^1 v Tv
  p
  ∂
  ∂
  F
  H
  I
  K =
  1
  v
  p
  T
  p
  v
  v
  T

∂

∂

F

HG

I

KJ

∂

∂

F

HG

I

KJ

L

N

M

M

M

M

O

Q

P

P

P

P