438 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th9-3.pm5

Solution. Mass of oxygen, mO 2 = 4 kg

Mass of nitrogen, mN 2 = 6 kg

Pressure, p = 4 bar

Temperature, T = 27 + 273 = 300 K.

(i)The mole fraction of each component :

n = m

M

∴ nO 2 =

4

32

= 0.125 and nN 2 =

6

28

= 0.214

Now xO 2 =

0.125

0.125 + 0.214 = 0.3687. (Ans.)

and xN 2 =
0.214
0.125 + 0.214
= 0.6313. (Ans.)
(ii)The average molecular weight, M :

M =

nM n M

nn

OO N N

ON

22 2 2

22

+×

+

= 0125 32 0 214^28

0125 0 214

..

..

×+ ×

+

= 29.475

i.e., Average molecular weight = 29.475. (Ans.)
(iii)The specific gas constant, R :

R = R

M

(^0) = 8 314
29 475
.
.
= 0.282 kJ/kg K. (Ans.)
(iv)The volume and density :
pV = mRT for mixture
V =
mRT
p

().4 6 0 282 10 300
410
3
5
+× × ×
×
= 2.115 m^3. (Ans.)
Density, ρ = ρρON 22 +
ρO 2 =
m
V
O (^2) = 4
2115.
= 1.891 kg/m^3
ρN 2 =
m
V
N (^2) =^6
2115.
= 2.837 kg/m^3
∴ρ = 1.891 + 2.837 = 4.728 kg/m^3. (Ans.)
(v)The partial pressures and partial volumes :
pVO 2 = nRTO 2 0
∴ pO^2 =
nRT
V
O0 (^2) = 0125 8 314 10^300
2115 10
3
5
..
.
×××
×
= 1.474 bar. (Ans.)
and pN 2 = 4 – 1.474 = 2.526 bar. (Ans.)
VO 2 = xO 2 V = 0.3687 × 2.115 = 0.779 m^3. (Ans.)
VN 2 = xN 2 V = 0.6313 × 2.115 = 1.335 m^3. (Ans.)