214 ENGINEERING THERMODYNAMICS

dharm

/M-therm/Th4-7.pm5

and – 12°C on the other side. The cylinder walls and piston may be regarded as perfect thermal
insulators but the oxygen may be heated electrically. Heating takes place so that the volume of
oxygen doubles. Find :
(i)Final state condition ; (ii)Work done by the piston ;
(iii)Heat transferred to oxygen.
Treat both gases as perfect and take :
For oxygen cp = 0.88 kJ/kg K, R = 0.24 kJ/kg K
For methane cp = 1.92 kJ/kg K, R = 0.496 kJ/kg K.
Solution. For oxygen :
Initial volume, V 1 = 0.035 m^3
Initial pressure, p 1 = 4.5 bar
Initial temperature, T 1 = 60 + 273 = 333 K
For methane :
Initial volume, V 1 = 0.07 m^3
Final volume, V 2 = 0.035 m^3
Initial pressure, p 1 = 4.5 bar
Initial temperature of methane,
T 1 = – 12 + 273 = 261 K.
For Methane :

cp = R ×

γ

γ− 1 or 1.92 = 0.496

γ

γ−

F

H

I

(^1) K
or

1. 
   

   .496
   92
   01

   
   

   −
   γ
   γ or 1.92 (γ – 1) = 0.496 γ
   ∴γ =

   
   


2. 
   


3. 
   

   .496

   
   

92

()92 0−

= 1.348 say 1.35

For Oxygen :

cv = cp – R = 0.88 – 0.24 = 0.64 kJ/kg K.

(i) According to problem ; for methane

pVγ = constant holds good

∴ p 1 V 1 γ = p 2 V 2 γ

p 2 = p 1.

V

V

1

2

F

HG

I

KJ

γ

= 4.5 (2)1.35 = 11.47 bar. (Ans.)

Also,

pV

T

pV

T

11

1

22

2

=

or T 2 =

pVT

pV

221

11

11.47 035 261

40

=

××

×

0.

.5 .07

= 332.6 K. (Ans.)

∴ Work done = −

−

= ×× − ××

−

pV 11 pV 22 55

1

4 10 0 07 11.47 10 0

γ 35 1

.5 .035

1.

.

()

J

= ×− ×

×

10 4 0 11.47 035

1000

(^5) ().5 .07 0.
0.35
kJ
= – 24.7 kJ (done on the methane)