Question

A gas is a mixture of H 2 and HD in the proportion 7000:1. As the gas effuses through a small hole from a vessel at constant temperature into a vacuum, the composition of the remaining mixture changes. By what factor will the pressure in the vessel have fallen when the remaining mixture consists of H 2 and HD in the proportion 700:1. (H=hydrogen, D=deuterium)

hint: use the differential equation for number density dn/dt = -0.25*A**n/V

note: I'm aware of another post with the same question but the answer is clearly wrong; please don't copy and paste from that question

Answer 1

Given,

The molar ratio of H2 and HD is 2:3

Let the initial pressure be P₁and the final pressure be P₂

The partial pressure ratio of H2 and HD is 2:3 ​

Total number of molecule = 7001K

Using,         PV = nRT      

We get 7001K = (P₁×V) / (RT) --------------(1)

Since the time of diffusion for both is same

The partial pressure ratio of H2 and HD is 2:3 ​​

Total number of molecule = 701K

Using,         PV = nRT      

We get 701K = (P₂×V) / (RT) --------------(2)

from (1) and (2),

P₂/P₁ = 701/7001  

=> P₂= P₁ (701/7001) = 0.1 P₁

Thus the pressure in the vessel have fallen to a factor of 0.1