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) $\dfrac{{3\sqrt 2 }}{4}$
(B) $\dfrac{3}{{4\sqrt 2 }}$
(C) $\sqrt 2 $
(D) $\dfrac{3}{2}$

Answer 1

Hint: In this question, we have to identify the rate of effusion of given compounds. The rate of effusion was defined with the help of graham’s law of effusion. Rate of effusion depends on the molar mass of the compound. Graham’s law tells that the rate of diffusion of a gas is inversely proportional to the square root of the molecular mass of its particle.
$\text{Rate of effusion} \propto \dfrac{1}{{\sqrt M }}$$\dfrac{{{\text{Rat}}{{\text{e}}_{\text{1}}}}}{{{\text{Rat}}{{\text{e}}_{\text{2}}}}}{\text{ = }}\sqrt {\dfrac{{{{\text{M}}_{\text{1}}}}}{{{{\text{M}}_{\text{2}}}}}} $$\dfrac{{{\text{Rat}}{{\text{e}}_{\text{1}}}}}{{{\text{Rat}}{{\text{e}}_{\text{2}}}}}{\text{ = }}\sqrt {\dfrac{{{{\text{M}}_{\text{1}}}}}{{{{\text{M}}_{\text{2}}}}}} $

Complete answer:
The rate of effusion gas with molar mass M at pressure P, and the whole area A at temperature T can be written as:
$r = \dfrac{{PA{N_A}}}{{\sqrt {2\pi MRT} }} \propto \dfrac{1}{{\sqrt M }}$
Here, R indicates the gas constant, ${N_A}$ represents Avogadro’s number and r indicates the rate of effusion of gas.
Thus for a mixture of gases with a mass ratio ${m_1}:{m_2}$
$\dfrac{{{r_1}}}{{{r_2}}} = \dfrac{{{P_1}}}{{{P_2}}} \times \dfrac{{\sqrt {{M_2}} }}{{\sqrt {{M_1}} }}$
 $P \propto n$(number of moles)
$n = $mass/Molar mass
Thus for a mixture of ${O_2}$ and $C{H_4}$in $3:2$ mass ratio:
$\dfrac{{{P_{{O_2}}}}}{{{P_{C{H_4}}}}} = \dfrac{{{m_{{O_2}}}}}{{{M_{{O_2}}}}} \times \dfrac{{{M_{C{H_4}}}}}{{{m_{C{H_4}}}}}$
So, here pressure is defined in terms of given mass and molar mass of oxygen and methane.
Given $\dfrac{{{m_{{O_2}}}}}{{{m_{C{H_4}}}}} = \dfrac{3}{4}$and ${M_{C{H_4}}} = 16$ and ${M_{{O_2}}} = 32$
Here, the molar mass of methane is 16 and oxygen is 32.
Upon substitution we get:
$\dfrac{{{P_{{O_2}}}}}{{{P_{C{H_4}}}}} = \dfrac{3}{{32}} \times \dfrac{{16}}{2} = \dfrac{3}{4}$
\[\dfrac{{{r_{{O_2}}}}}{{{r_{C{H_4}}}}} = \dfrac{{{P_{{O_2}}}\sqrt {{M_{C{H_4}}}} }}{{{P_{C{H_4}}}\sqrt {{M_{{O_2}}}} }} = \dfrac{3}{4}\dfrac{{\sqrt {16} }}{{\sqrt {32} }} = \dfrac{3}{{4\sqrt 2 }}\]
The relative rate of effusion of ${O_2}$ to $C{H_4}$ through a container containing ${O_2}$ and $C{H_4}$ through a container containing ${O_2}$ and $C{H_4}$ in $3:2$ mass ratio will be $\dfrac{3}{{4\sqrt 2 }}$

Thus the answer is (B).

Note:
Graham’s law was identified by Thomas Graham. This law tells about the effusion of lighter molecules being faster than the higher molecular mass particles at constant temperature and pressure.
The ratio of the square roots of the masses of their particles:
$\dfrac{{\text{rate of effusion of A}}}{{\text{rate of effusion of B}}} = \dfrac{{\sqrt {{M_B}} }}{{\sqrt {{M_A}} }}$
The most important is Graham’s law is used to determine the rate of effusion for two different gases at equal temperature and pressure.