TITLE.PM5

THERMODYNAMIC RELATIONS 373

dharm
\M-therm\Th7-2.pm5

Differentiating both sides, we get

1

2.302p.

dp

dT =

3276 6.

T^2

• 0.652
  2.302T

or dp

dT

= 2.302 × 3276.6 ×

p

T^2 – 0.652

p

T ...(ii)

From (i) and (ii), we have

or

h

vT

fg

g = 2.302 × 3276.6 ×

p

T^2 – 0.652

p

T ...(iii)

We know that log 10 p = 7.0323 –

3276.6

T – 0.652 log^10 T ... (given)

At p = 0.1 bar,

log 10 (0.1) = 7.0332 –

3276.6

T – 0.652 log^10 T

• 1 = 7.0323 –

3276.6

T – 0.652 log^10 T

or 0.652 log 10 T = 8.0323 –

3276.6

T

or log 10 T = 12.319 –

5025.4

T

Solving by hit and trial method, we get

T = 523 K

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

294.54 10

523

×^3

vg× = 2.302 × 3276.6 ×

0.1 10

523

5

2

×

()

• 0.652 ×
  0.1 10
  523

×^5

563.17

vg = 275.75 – 12.46

i.e., vg = 2.139 m^3 /kg. (Ans.)

Highlights

1. Maxwell relations are given by
   ∂
   ∂

F

HG

I

KJ

T

v s = –

∂

∂

F

HG

I

KJ

p

sv ;

∂

∂

F

HG

I

KJ

T

p s =

∂

∂

F

HG

I

KJ

v

sp

∂

∂

F

HG

I

KJ =

∂

∂

F

HG

I

KJ

p

T

s

vTv ;

∂

∂

F

HG

I

KJ =−

∂

∂

F

HG

I

KJ

v

T

s

p pT.

2. The specific heat relations are
   cp – cv = vTKβ

2

; cv = T GHF∂∂TsIKJ

v

; cp = T

∂

∂

F

HG

I

KJ

s

T p.

3. Joule-Thomson co-efficient is expressed as
   μ = GHF∂∂TpIKJ
   h

.