TITLE.PM5

286 ENGINEERING THERMODYNAMICS

dharm

/M-therm/th5-4.pm5

Entropy variation for a constant volume process is given by :

dS = mcv dT

T

, or, dS = 4 × (0.48 + 0.0096T) dT

T

Integrating both sides, we get,

S 2 – S 1 = 4 × 0.48

T

T

T

T

dT dT

1

2

1

2

zz+×4 0 0096.

= 192 2

1

.loge

T

T

F

HG

I

KJ

+ 0.0384 (T 2 – T 1 )

= 1.92 loge

500

400

F

HG

I
KJ + 0.0384(500 – 400) = 4.268 kJ/K
i.e., S 2 – S 1 = 4.268 kJ. (Ans.)
Example 5.40. The specific heats of a gas vary linearly with absolute temperature accord-
ing to the following relations :
cp = (0.85 + 0.00025 T) kg/kg K, and
cv = (0.56 + 0.00025 T) kJ/kg K
If the entropy of the gas at 1 bar pressure and 273 K is zero, find the entropy of the gas at
25 bar and 750 K temperature.
Solution. Given : cp = (0.85 + 0.00025 T) kJ/kg K ; cv = (0.56 + 0.00025 T) kJ/kg K ;
p 1 = 1 bar ; T 1 = 273 K ; p 2 = 25 bar ; T 2 = 750 K.
We know that, ds = c dT
T

p

T

dv c dT

T

Rdv

vvv

+= +

Integrating both sides, we get,

s 2 – s 1 = c dT

T

R v

vev

+

F

HG

I

z KJ

log^2

1

= c

dT

T

cc p

p

T

v p ve T

+− ×

F

HG

I

z KJ

()log^1

2

2

1

=^056 0 00025 0 29^1

25

750

273

. ..log
T

F + dT e

HG

I

KJ

+×F

HG

I

z KJ

= 0 56^2 0 00025 0 6404

1

.loge .( ). 21

T

T

TT

F

HG

I

KJ

+−

L

N

M

M

O

Q

P

P

−

= 056 750

273

.logeFHG IKJ + 0.00025(750 – 273) – 0.6404 = 0.0448 kJ/kg K

i.e., s 2 – s 1 = 0.0448 kJ/kg K. (Ans.)
Example 5.41. An insulated vessel of 0.5 m^3 capacity is divided by a rigid conducting
diaphragm into two chambers A and B, each having a capacity of 0.25 m^3. Chamber A contains
air at 1.4 bar pressure and 290 K temperature and the corresponding parameters for air in cham-
ber B are 4.2 bar and 440 K. Calculate :
(i)Final equilibrium temperature,
(ii)Final pressure on each side of the diaphragm, and
(iii)Entropy change of system.
For air take cv = 0.715 kJ/kg K and R = 0.287 kJ/kg K.