TITLE.PM5

270 ENGINEERING THERMODYNAMICS

dharm

/M-therm/th5-3.pm5

Now, using characteristic gas equation (to find mass ‘m’ of nitrogen), we have :

p 1 V 1 = mRT 1

m pV

RT

==××

×

11

1

105 10^5004

297 288

..

= 0.0491 kg

(i)The change of entropy,

S 2 – S 1 = mR loge

p

p

1

2

= 0.0491 ×

297

103

loge

105

48

.

.

F

HG

I
KJ
i.e., S 2 – S 1 = – 0.02216 kJ/K.
∴ Decrease in entropy, S 1 – S 2 = 0.02216 kJ/K. (Ans.)
(ii)Heat rejected = Shaded area on Fig. 5.32 (b)
= T(S 1 – S 2 ) = 288 × 0.02216 = 6.382 kJ. (Ans.)
(iii) For an isothermal process for a perfect gas,
W = Q = 6.382 kJ
Hence, the work done on air = 6.382 kJ. (Ans.)
Example 5.25. 1 kg of gas enclosed in an isolated box of volume v 1 , temperature T 1 and
pressure p 1 is allowed to expand freely till volume increases to v 2 = 2v 1.
Determine the change in entropy.
Take R for gas as 287 kJ/kg K.
Solution. During the process of free expansion in an isolated box,
∆ U = 0, W = 0 and Q = ∆ U + W = 0
The process is represented by dotted line on p-v diagram as shown in Fig. 5.33 (a) where
v 2 = 2v 1.

(a)(b)
Fig. 5.33
To calculate the entropy change, assume that the irreversible free expansion process is
replaced by a reversible isothermal process as temperature in free expansion remains constant, in
such a way that the volume increases to double of its original as shown in Fig. 5.33 (b). As the work
is developed by the system and heat is given to the system at constant temperature, during iso-
thermal reversible system then as per first law of thermodynamics :