TITLE.PM5

SECOND LAW OF THERMODYNAMICS AND ENTROPY 287

dharm

/M-therm/th5-4.pm5

Solution. Given : VA = 0.25 m^3 ; pAi = 1.4 bar ; TAi = 290 K ; VB = 0.25 m^3 ; pBi = 4.2 bar ;

TBi = 440 K ; cv = 0.715 kJ/kg K.

(i)Final equilibrium temperature, Tf :

Mass of air, mA =

pV

RT

Ai A

Ai

= ××

××

(. ).

(. )

14 10 025

0 287 1000 290

5

= 0.4205 kg

mB = pV

RT

Bi B

Bi

= ××

××

(. ).

(. )

42 10 025

0 287 1000 440

5

= 0.8315 kg

Let Tf be the final equilibrium temperature (K). Since the diaphragm is conducting,
Heat gained by air in chamber A = Heat lost by air in chamber B
mA cv (Tf – 290) = mB cv(440 – Tf)
or, 0.4205 × (Tf – 290) = 0.8315(440 – Tf)
or, 0.4205 Tf – 121.94 = 365.86 – 0.8315 Tf
∴ Tf = 389.6 K. (Ans.)
(ii)Final pressure on each side of the diaphragm : pAf ; pBf

pAf = 1 4 389 6

290

..× = 1.88 bar. (Ans.)

pBf = 4 2 389 6

440

..× = 3.72 bar. (Ans.)

(iii)Entropy change of the system :

Entropy change for chamber A = mA cv loge

T

T

f

Ai

F

HG

I

KJ

= 0.4205 × 0.715 × loge 389 6

290

F.

HG

I

KJ = 0.0888 kJ/K

Entropy change for chamber B = mB cv loge

T

T

f

Bi

F

HG

I

KJ

= 0.8315 × 0.715 × loge 389 6

440

F.

HG

I

KJ

= – 0.0723 kJ/K

∴ Net change of entropy = 0.0888 + (– 0.0723) = 0.0165 kJ/K. (Ans.)
Example 5.42. A certain gas has a specific heat at constant volume of 1.25 kJ/kg K. When
it is expanded reversibly and adiabatically from a specific volume of 0.0624 m^3 /kg and a tempera-
ture of 530 K to a specific volume of 0.186 m^3 /kg its temperature falls by 165 K. When it is
expanded into an evacuated space from the same initial condition to the same final specific
volume its temperature falls only by 25 K.
Find the change in entropy in each of the adiabatic processes.
Solution. Refer Fig. 5.44.
Specific heat of gas at constant volume, cv = 1.25 kJ/kg K
Initial specific volume, v 1 = 0.0624 m^3 /kg
Initial temperature, T 1 = 530 K
Final specific volume, v 2 = 0.186 m^3 /kg