TITLE.PM5

SECOND LAW OF THERMODYNAMICS AND ENTROPY 273

dharm
/M-therm/th5-4.pm5

Fig. 5.35

or T 2 = 873 ×^105
7

025

. 125

.

F.

HG

I

KJ

= 873 × (0.15)0.2 = 597.3 K.

Now replace the process 1 to 2 by processes, 1 to A and A to 2.

Then at constant temperature from 1 to A,

sA – s 1 = R loge

v

v

2

1 = R loge^

p

p

1

2 = 0.287 loge^

7

051.

F

HG

I

KJ = 0.544 kJ/kg K.

Fig. 5.36

At constant pressure from A to 2

sA – s 2 = cp loge

T

T

1

2 = 1.005 loge^

873

597.3

= 0.3814 kJ/kg K
Then s 2 – s 1 = 0.544 – 0.3814 = 0.1626 kJ/kg K
i.e., Increase in entropy = 0.1626 kJ/kg K. (Ans.)