TITLE.PM5

GASES AND VAPOUR MIXTURES 431

dharm

\M-therm\Th9-2.pm5

Solution. Using the relation, n =

pV

RT 0

∴ nO 2 =

810 18

8 314 10 323

5

3

××

××

.

(. )

= 0.536 (where TO 2 = 50 + 273 = 323 K)

and nCO =
110 36
8 314 10 293

5

3

××

××

.

(. )

= 0.1478 (where TCO = 20 + 273 = 293 K).

(i)Final temperature (T) and pressure (p) of the mixture :

Before mixing :

U 1 = Σ niCviTi = 0.536 × 21.07 × 323 + 0.1478 × 20.86 × 293

i.e., U 1 = 4551.15 kJ

After mixing :

U 2 = T Σ niCvi = T (0.536 × 21.07 + 0.1478 × 20.86)

i.e., U 2 = 14.37 T

For adiabatic mixing,

U 1 = U 2

∴ 4551.15 = 14.37 T

∴ T =

455115

14 37

.

.

= 316.7 K

∴ Temperature of the mixture = 316.7 – 273 = 43.7°C. (Ans.)

Now p =

nR T

V

0

∴ p =

(.. )..

(.. )

0 536 01478 8 314 10 316 7

18 36 10

3

5

+×××

+×

= 3.33 bar

i.e., Pressure after mixing = 3.33 bar. (Ans.)

(ii)Change of entropy of the system :

Change of entropy of the system = change of entropy of the O 2 + change of entropy of CO

...... Gibbs-Dalton law

Referring to Fig. 9.7, the change of entropy of O 2 can be calculated by replacing the process

undergone by the oxygen by the two processes 1 to A and A to 2.

V 2

V 1

1

T^2

2

T 1 B

R loge

C logve

T

S

V 2

V (^1) V 1
V 2
1 Isothermal
2
A
T
S
C logve
R loge
V 2
V 1 Constant volume T
T
1
2
T
T
2
1
Fig. 9.7 Fig. 9.8