TITLE.PM5

432 ENGINEERING THERMODYNAMICS

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\M-therm\Th9-2.pm5

For an isothermal process 1-A :

SA – S 1 = mR loge

V

V

A

1

or SA – S 1 = nR 0 loge

V

V

A

1

= 0.536 × 8.314 × loge

54

18

.

.

= 4.896 kJ/K

For constant volume process A-2 :

SA – S 2 = mcv dT

T

A

z 2

= mcv loge

T

T

1

2

i.e., SA – S 2 = nCv loge

T

T

1

2

= 0.536 × 21.07 × loge
323
316 7.
= 0.222 kJ/K
∴ S 2 – S 1 = (SA – S 1 ) – (SA – S 2 )
= 4.896 – 0.222 = 4.674 kJ/K
Referring to Fig. 9.8, the change of entropy of CO can be found in a similar way to the above,
i.e., S 2 – S 1 = (SB – S 1 ) + (S 2 – SB)

∴ S 2 – S 1 = nR 0 loge

V

V

B

1

+ nCv loge

T

T

2

1

= 0.1478 × 8.314 × loge

54

36

.

. + 0.1478 × 20.86 loge^

316 7

293

.

= 0.498 + 0.239 = 0.737 kJ/K

Hence the change of entropy of the whole system is given by

(S 2 – S 1 )system = ()SS 21 − O 2 + ()SS 21 − CO

i.e., Change of entropy of system = 4.674 + 0.737 = 5.411 kJ/K. (Ans.)

+Example 9.12. Two vessels, A and B, both containing nitrogen, are connected by a

valve which is opened to allow the contents to mix and achieve an equilibrium temperature of

30 °C. Before mixing the following information is known about the gases in the two vessels.

Vessel Pressure (p) Tem. (t) Contents

A 16 bar 55°C 0.6 kg mole

B 6.4 bar 25°C 3.6 kg

(a) Calculate : (i) The final equilibrium pressure ;

(ii) The amount of heat transferred to the surroundings ;

(b) If the vessel had been perfectly insulated, calculate :

(i) The final temperature ;

(ii) The final pressure.

Assume γ = 1.4.

Solution. Refer Fig. 9.9.

Vessel A :

pAVA = nAR 0 TA (where VA = volume of vessel A)