Question

A 4:1 molar mixture of He and CH4 is contained in a vessel at 20 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. What is the composition of the mixture effusing out initially?

Answer 1

M(CH4)= 16M(He)= 4
Let total pressure=P
Then, P(he) = (4/5)P & P(CH4) = (1/5)P
Now r(He)/r(CH4) = [P(He)/P(CH4)]× √[M(CH4)/M(He)]
ANS : r(he)/r(CH4) = 8:1

Answer 2

Hey dear,

● Answer -
He : CH4 = 8 : 1.

● Explaination -
# Given -
P = 20 bar
n1/n2 = 4

# Explanation -
For He,
M(He) = 4
P(He) = P.X(He)
P(He) = 20 × 4 / (4+1) P(He) = 16 bar

For CH4,
M(CH4) = 16
P(CH4) = P.X(CH4)
P(CH4) = 20 × 1 / (4+1)
P(CH4) = 4 bar

Rate of diffusion is given by -
r1/r2 = √[M(CH4)/M(He)] × P(He)/P(CH4)
r1/r2 = √(16/4) × 16/4
r1/r2 = 8

Therefore, composition of initial mixture is He : CH4 = 8 : 1.