GASES AND VAPOUR MIXTURES 435

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\M-therm\Th9-3.pm5

Now pN 2 =

nRT

V

N (^20) ...(i)
and pO 2 =
nRT
V
O (^20) ...(ii)
Dividing (ii) by (i), we get
p
p
O
N
2
2

n
n
O
N
2
2
∴ nO 2 =
pn
p
ON
N
22
2
×

5 10 0 0893
15 10
5
5
××
×
.
= 0.0297
∴ Mass of O 2 added,
mO 2 = nO 2 × MO 2
= 0.0297 × 32 = 0.9504 kg. (Ans.)
Example 9.15. Given that air consists of 21% oxygen and 79% nitrogen by volume. De-
termine :
(i)The moles of nitrogen per mole of oxygen ;
(ii)The partial pressure of oxygen and nitrogen if the total pressure is atmosphere ;
(iii)The kg of nitrogen per kg of mixture.
Solution. Let nO 2 = 1.0 and V = volume of air
so that VO 2 = 0.21V ; VN 2 = 0.79V
Let V contain n = nO 2 + nN 2 moles of air at p and T.
(i)Moles of N 2 per mole of O 2 :
Now pVO 2 = nO 2 R 0 T ...(i)
and pVN 2 = nN 2 R 0 T ...(ii)
Dividing (i) by (ii), we get
V
V
O
N
2
2

n
n
O
N
2
2
∴ nN 2 =
nV
V
ON
O
2 × 2
2

1079
021
×.
.
V
V
= 3.76 moles. (Ans.)
(ii)pO 2 and pN 2 :
Now xO 2 =
nO 2
4

p
p
O (^2) (n = nn
ON 22 + )
(where p = total pressure = 1 atm.)
∴ pO 2 =
n
n
O (^2) × p =^1
(.) 1376 + × 1 = 0.21 atm. (Ans.)