TITLE.PM5

SECOND LAW OF THERMODYNAMICS AND ENTROPY 267

dharm
/M-therm/th5-3.pm5

Solution. Change in entropy during constant volume process

= m cv ln

T

T

2

1

F

HG

I

KJ

...(i)

Change in entropy during polytropic process (pvn = constant)

= m cv γ−

−

F

HG

I

KJ

n

n 1

ln T

T

2

1

F

HG

I

KJ

...(ii)

For the same entropy, equating (i) and (ii), we have

γ−

−

n

n 1 = 1, or (γ – n) = (n – 1) or^2 n = γ + 1

∴ n = γ+ 21. (Ans).
Example 5.23. Air at 20°C and 1.05 bar occupies 0.025 m^3. The air is heated at constant
volume until the pressure is 4.5 bar, and then cooled at constant pressure back to original tem-
perature. Calculate :
(i)The net heat flow from the air.
(ii)The net entropy change.
Sketch the process on T-s diagram.
Solution. The processes are shown on a T-s diagram in Fig. 5.31.

Fig. 5.31

For air :

Temperature, T 1 = 20 + 273 = 293 K

Volume, V 1 = V 3 = 0.025 m^3

Pressure, p 1 = 1.05 bar = 1.05 × 10^5 N/m^2

Pressure, p 2 = 4.5 bar = 4.5 × 10^5 N/m^2.

(i)Net heat flow :

For a perfect gas (corresponding to point 1 of air),

m =

pV

RT

11

1

=

1 05 10 0 025

0 287 10 293

5

3

..

.

××

××

= 0.0312 kg