TITLE.PM5

SECOND LAW OF THERMODYNAMICS AND ENTROPY 277

dharm

/M-therm/th5-4.pm5

(v)Work transfer, W1–2 :

Q1–2 = (U 2 – U 1 ) + W1–2

∴ W1–2= Q1–2 – (U 2 – U 1 )

= 0 – 187.84 = – 187.84 kJ

Hence, work done on the fluid = 187.84 kJ. (Ans.)

+Example 5.30. An insulated cylinder of volume capacity 4 m^3 contains 20 kg of nitro-

gen. Paddle work is done on the gas by stirring it till the pressure in the vessel gets increased

from 4 bar to 8 bar. Determine :

(i)Change in internal energy,

(ii)Work done,

(iii)Heat transferred, and

(iv)Change in entropy.

Take for nitrogen : cp = 1.04 kJ/kg K, and cv = 0.7432 kJ/kg K.

Solution. Pressure, p 1 = 4 bar = 4 × 10^5 N/m^2

Pressure, p 2 = 8 bar = 8 × 10^5 N/m^2

Volume, V 1 = V 2 = 4 m^3

and it is constant for both end states.

Now,

p

T

p

T

1

1

2

2

=

or TT^2 pp

1

2

1

5

5

810

410

==×

×

= 2

Also, R = cp – cv = 1.04 – 0.7432 = 0.2968 kJ/kg K.

The mass of the gas in the cylinder is given by

m=RTpV or mT=pVR

∴ mT 1 =pVR^11 = 0 2968 1000410 4

××^5

. × = 5390.8 kg K

and mT 2 =pV^22 R = 810 4
2968 1000

××^5

0. ×

= 10781.6 kg K.
(i)Change in internal energy,
∆U = (U 2 – U 1 )
= mcv (T 2 – T 1 ) = cv (mT 2 – mT 1 )
= 0.7432 (10781.6 – 5390.8) = 4006.4 kJ. (Ans.)
(ii)Work done, W :
Energy in the form of paddle work crosses into the system, but there is no change in system
boundary or pdv work is absent. No heat is transferred to the system. We have
Q1–2 = (U 2 – U 1 ) + W1–2
But Q1–2 = 0
∴ W1–2 = – (U 2 – U 1 ) = – 4006.4 kJ or kN-m. (Ans.)
(iii)Heat transferred, Q1–2 = 0. (Ans.)
(iv)Change in entropy,
S 2 – S 1 = mcv loge T
T

2

1