0

Guest

Courses New

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 )

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 )

Grade:12th pass

1 Answers

Vikas TU
14149 Points
7 years ago
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.

Think You Can Provide A Better Answer ?

Other Related Questions on Physical Chemistry

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More