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Grade: 12th pass

                        

In a vessel an equilibrium mixture at 300k contained N2O2 and NO2 gases at partial pressures 100 atm and 10 atm respectively . the volume of the container is doubled at the same temperature. then the partial pressures of N2O2 and NO2 gases are (in atm )

4 years ago

Answers : (1)

Vikas TU
12280 Points
							
Total Pressure = sum of individual partial pressures of gases.
Pt =  100  + 10 = > 110 atm.
From the ideal gas eqn.
PtV = nRT
110V = (n1+n2)RT.........(1)
Pt(2V) = (n1+n2)RT..............(2)
Dividing the eqns. we get,
Pt = 55 atm.
 
Now for both gases individually,
100V = n1RT  and   10V = n2RT
from these two relations,
n1 = 10n2.................(3)
and
after doubling the volume,
P1(2V) = n1RT
and
P2(2V) = n2RT
from these two relations,
P1/P2 = n1/n2  = 10
or
P1 = 10P2
Now total pressure calculated after doubling the volume,
Pt => 55 = P1 + P2 
      or
10P2 + P2 = 55
11P2 = 55
P2 = 5 atm
and 
P1 = 55 - 5 => 50 atm.
4 years ago
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