Question

$S{O}_2$ and $C{H}_4$ are introduced in a vessel in the molar ratio $1:2$ . The ratio of molecules of two gases present in the container when their rate of effusion becomes equal is :

• A. $1:2$
• B. $2:1$
• C. $4:1$
• D. $1:4$

Answer 1

The correct option is B $2:1$

Rate of effusion for a gas $\left(r\right)\propto \frac{P}{\sqrt{M}}$, where $P$ is pressure and $M$ is molar mass of gas.
Since the rate of effusion for both gases is equal so,
$\frac{P_1}{\sqrt{M_1}}=\frac{P_2}{\sqrt{M_2}}$
$\Rightarrow \frac{P_1}{P_2}=\frac{\sqrt{M_1}}{\sqrt{M_2}}$
During effusion all conditions are identical so $Pressureofgas\left(P\right)\propto Molesofgas\left(n\right)\propto Moleculesofgas\left(N\right)$
$=\frac{N_1}{N_2}\Rightarrow \frac{N_{S{O}_2}}{N_{C{H}_4}}=\sqrt{\frac{M_{S{O}_2}}{M_{C{H}_4}}}=\sqrt{\frac{64}{16}}=\frac{2}{1}$