Question

A vessel contains equal no of moles of helium and methane . Due to a hole in the vessel , half of the gaseous mixture effused out. What is the ratio of the number of moles of helium and methane remaining in vess

Answer 1

A vessel contains equal no of moles of helium and methane. due to a hole in the vessel, half of the gaseous mixture effused out.

To find : The ratio of the number of moles of helium and methane remaining in the vessel.

solution : from Graham's law of diffusion/effusion,

$\frac{r_{He}}{r_{CH_4}}=\sqrt{\frac{M_{CH_4}}{M_{He}}}$

here, $M_{CH_4}$ = 16 g/mol

and $M_{He}$ = 4 g/mol

so, $\frac{r_{He}}{r_{CH_4}}=\sqrt{\frac{16}{4}}=\frac{2}{1}$

i.e., rate of effusion of He is 2 times faster than that of CH₄.

at specific time, evolved volume of He = 2 × evolved volume of CH₄

i.e., $\frac{V_{He}}{V_{CH_4}}=\frac{2}{1}$

at constant temperature and pressure, V ∝ n

so, $\frac{n_{He}}{n_{CH_4}}=\frac{2}{1}$

given, A vessel contains equal no of moles of helium and methane.

let no of moles of helium/methane = n

so no of moles of evolved helium gas = n × 2/(2 + 1) = 2n/3

so remaining helium gas = n - 2n/3 = n/3

and no of moles of evolved methane gas = 2n × 1/(2 + 1) = n/3

so remaining methane gas = n - n/3 = 2n/3

Therefore the ratio of moles of Helium and methane remaining in the vessel is (n/3)/(2n/3) = 1/2