Question

A gaseous mixture of $He$ & $C{H}_4$ gases with mole ratio 2 : 1 respectively is placed in a container and allowed to effuse through a small orifice. $Xml$ of the $He$ gas effuses in 5 seconds. The time required to effuse same volume of $C{H}_4$ gas is:

• A. 5 $sec$
• B. 20 $sec$
• C. 40 $sec$
• D. 10 $sec$

Answer 1

Answer: (C)

40 $sec$

The rate of effusion is directly proportional to the mole fraction and inversely proportional to the square root of molar mass.

$\frac{r}{r^{\prime }}=\frac{V}{V^{\prime }}\times \frac{t^{\prime }}{t}=\frac{X}{X^{\prime }}\times \sqrt{\frac{M^{\prime }}{M}}$

Here, $r,V,t,X$ and $M$ are the rate of diffusion, volume diffused, time for diffusion, mole fraction and the molar mass for He respectively
and $r^{\prime },V^{\prime },t^{\prime },X^{\prime }$ and $M^{\prime }$ are the rate of diffusion, volume diffused, time for diffusion, mole fraction and the molar mass for methane respectively

$\frac{X}{X}\times \frac{t^{\prime }}{5}=\frac{2}{1}\times \sqrt{\frac{16}{4}}$

$t^{\prime }=40sec$
Hence, the time required to effuse same volume of methane gas is 40 $sec$.