436 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th9-3.pm5

Again, xN 2 =

n

n

N (^2) = p
p
N 2
∴ pN 2 =
n
n
N (^2) × p =^376
1376
.
+. × 1 = 0.79 atm. (Ans.)
(iii)The kg of nitrogen per kg of mixture :
m
mm
N
NO
2
22 +

nM
nM nM
NN
NN OO
22
22 22+

376 28
376 28 1 32
.
.
×
×+×
= 0.77 kg N 2 /kg mix. (Ans.)
Example 9.16. Air (N 2 = 77%, O 2 = 23% by weight) at 25°C and 12 bar is contained in a
vessel of capacity 0.6 m^3. Some quantity of CO 2 is forced into the vessel so that the temperature
remains at 25°C but the pressure rises to 18 bar.
Find the masses of O 2 , N 2 and CO 2 in the cylinder.
Solution. Volume of the vessel, V = 0.6 m^3
Temperature (constant), T = 25 + 273 = 298 K
Initial pressure = 12 bar
Final pressure = 18 bar
Now pV = nR 0 T
or n =
pV
RT 0 =
12 10 0 6
8 314 10 298
5
3
××
××
.
.
= 0.29
Also Rair =
m
M
m
M
R
m
O
O
N
N
2
2
2
2
F + 0
H
G
I
K
J (m = n × M)
Considering 100 kg of air
mO 2 = 23 kg, mN 2 = 77 kg
∴ Rair =
23
32
77
28
F +
HG
I
KJ ×
8 314
100
.
= 0.288 kJ/kg K
Mair =
R
R
0
air
= 8 314
0 288
.
.
= 28.87
Now pV = mRT
∴ m =
pV
RT

12 10 0 6
0 288 298 10
5
3
××
××
.
.
= 8.39 kg of air
Mass of O 2 , mO 2 = 0.23 × 8.39 = 1.93 kg. (Ans.)
Mass of N 2 , mN 2 = 0.77 × 8.39 = 6.46 kg. (Ans.)
After adding CO 2 in the vessel :
ppNO 22 + = 12 bar ... before adding CO 2
pCO 2 + ()ppNO 22 + = 18 bar ... after adding CO 2