TITLE.PM5

SECOND LAW OF THERMODYNAMICS AND ENTROPY 271

dharm

/M-therm/th5-3.pm5

∆U = 0, Q = W

i.e., Qpdv
v

v

=z.

1

2

=

RT

v

dv

v

v

.

1

2

z Q pv RT p

RT

v

L ==

NM

O

QP

and

= RT loge

v

v

2

1

∴ Q

T

= R loge v

v

2
1
But this is the expression for change in entropy of the system. Entropy being the property of
the system, its change is same whether it is reversible or irreversible process.
∴ For the given process,

∆s = R loge

v

v

2

1

F

HG

I
KJ
= 287 loge (2) [Q v 2 = 2v 1 (given)]
= 198.9 kJ/kg K
Hence change in entropy = 198.9 kJ/kg K. (Ans.)
Example 5.26. 0.04 kg of carbon dioxide (molecular weight = 44) is compressed from 1
bar, 20°C, until the pressure is 9 bar, and the volume is then 0.003 m^3. Calculate the change of
entropy.
Take cp for carbon dioxide as 0.88 kJ/kg K, and assume carbon dioxide to be a perfect gas.
Solution. Mass of carbon dioxide, m = 0.04 kg
Molecular weight, M =44
Initial pressure, p 1 = 1 bar = 1 × 10^5 N/m^2
Initial temperature, T 1 = 20 + 273 = 293 K
Final pressure, p 2 = 9 bar
Final volume, V 2 = 0.003 m^3
cp for carbon dioxide = 0.88 kJ/kg K
Change of entropy :

T(K)

T 2

T

= 293 K

1

s 2 s 1 sA s (kJ/kg K)

1 bar

A

0.003 m^3 9 bar

2

1

Fig. 5.34