TITLE.PM5

GASES AND VAPOUR MIXTURES 421

dharm

\M-therm\Th9-1.pm5

Also, the characteristic gas equation is given by

pV = mRT ...(ii)

∴ pV = mR T

M

(^0) ...(iii)
i.e., p =
mR T
MV
(^0) ...(iv)
Hence, for a constituent,
pi = mRT
MV
i
i
(^0) ...(v)
Substituting the values, we get the partial pressures as follows :
(i)Partial pressures :
For O 2 , pO 2 =
0 233 8 314 10 293
32 0 35 10
3
5
.(. )
.
×××
××
= 0.5068 bar. (Ans.)
For N 2 , pN 2 =
0 767 8 314 10 293
28 0 35 10
3
5
.(. )
.
×××
××
= 1.9065 bar. (Ans.)
For CO, pCO=
0 40 8 314 10 293
28 0 35 10
3
5
.(. )
.
×××
××
= 0.9943 bar. (Ans.)
(ii)Total pressure in the vessel, p :
p = Σ pi = pO 2 + pN 2 + pCO
= 0.5068 + 1.9065 + 0.9943 = 3.4076 bar. (Ans.)
Example 9.2. The gravimetric analysis of air and other data are as follows :
Constituent Percentage Molecular weight
Oxygen 23.14 32
Nitrogen 75.53 28
Argon 1.28 40
Carbon dioxide 0.05 44
Calculate : (i) Gas constant for air ;
(ii) Apparent molecular weight.
Solution. Using the relation, R = R
M
(^0) ...(i)
RO 2 = 8 314
32
.
= 0.2598 kJ/kg K
RN 2 = 8 314
28
.
= 0.2969 kJ/kg K
RAr =^8314
40
.
= 0.2078 kJ/kg K
RCO 2 =^8314
44

. = 0.1889 kJ/kg K