Question

The relative rates of effusion of $O_2$ to $C{H}_4$ through a container containing $O_2$ and $C{H}_4$ in $3:2$ mass ratio will be :

• A. $\frac{3\sqrt{2}}{4}$
• B. $\frac{3}{4\sqrt{2}}$
• C. $\sqrt{2}$
• D. $\frac{3}{2}$

Answer 1

Answer: (B)

$\frac{3}{4\sqrt{2}}$

The rate of effusion of a gas with molar mass M at pressure P, and the hole area A at a temperature T is :

$r=\frac{PA}{\sqrt{2\pi MRT}}$

Thus for a mixture of gases with mass ratio $m_1:m_2$

$\frac{r_1}{r_2}=\frac{P_1}{P_2}\times \sqrt{\frac{M_2}{M_1}}$

$P\propto n$ (number of moles)

n=mass/Molar mass

Thus for a mixture of $O_2$ and $C{H}_4$ in 3:2 mass ratio :

$\frac{P_{O_2}}{P_{{CH}_4}}=\frac{m_{O_2}}{M_{O_2}}\times \frac{M_{{CH}_4}}{m_{{CH}_4}}$

Given $\frac{m_{O_2}}{m_{{CH}_4}}=\frac{3}{2}$

and $M_{C{H}_4}=16$ and $M_{O_2}=32$

upon substitution we get :

$\frac{P_{O_2}}{P_{{CH}_4}}=\frac{3}{32}\times \frac{16}{2}=\frac{3}{4}\frac{r_{O_2}}{r_{{CH}_4}}=\frac{P_{O_2}}{P_{{CH}_4}}\sqrt{\frac{M_{{CH}_4}}{M_{O_2}}}=\frac{3}{4}\sqrt{\frac{16}{32}}=\frac{3}{4\sqrt{2}}$

The relative rates of effusion of O2 to CH4 through a container containing $O_2$ and $C{H}_4$ in 3:2 mass ratio will be $\frac{3}{4\sqrt{2}}$

The correct option is B.