A gas mixture consisting of He and `CH_(4)` gases in mole ratio 4:1 is present in a vessel at a pressure of 20 bar. Due to a fine hole in the vessel, the gas mixture undergoes effusion. What is the composition (or ratio) of the initial gas mixture that is effused out?

Explanation

If the total number of moles in the mixture be n, then number of moles of He and `CH_(4)` are `(4)/(5)n and (1)/(5)n` or , 0.8 n and 0.2 n respectively.
In the mixture, the partial pressure of `He(p_(He))`
`=(0.8n)/(n)xx20=16` bar and that of `CH_(4)(p_(CH_(4)))`
`=(0.2)/(n)xx20=4` bar
Rate of effusion of a gas at constant temperature `r prop(P)/(sqrt(M))`,
where P=pressure of the gas and M=molar mass of the gas.
Therefore, the rate of effusion of He gas, `r_(1)prop(p_(He))/(sqrt(M_(He)))`
or, `r_(1)prop(p_(He))/(sqrt(4))` and that of `CH_(4)` gas, `r_(2)prop(p_(CH_(4)))/(sqrt(M_(CH_(4)))) or, r_(2)prop(p_(CH_(4)))/(sqrt(16))`
`therefore (r_(1))/(r_(2))=(p_(He))/(p_(CH_(4)))xxsqrt((16)/(4))=(2p_(He))/(p_(CH_(4)))=(2xx16)/(4)=8`
`therefore` The rato of number of moles of He and `CH_(4)` (composition) in the initial gas mixture effused out is 8:1.