Question

A gaseous mixture of $He$ and $C{H_4}$ gases with a mole fraction $2:1$ respectively is placed in a container and allowed to effuse through a small orifice. ${\text{X}}$ ml of the $He$ gas effuses in 5 seconds. The time to the effuse same volume of $C{H_4}$ gas is;
A.5 sec
B.20 sec
C.40 sec
D.10sec

Answer 1

We know that the rate of effusion is inversely proportional to the square root of its molar mass and the rate of effusion is also directly proportional to its mole fraction. And it is given by Graham's law of effusion.

Formula Used:
${\text{r = }}\dfrac{{\text{X}}}{{\sqrt {\text{M}} }}$ (equation 1)
Here, r = rate of effusion
X = molar fraction of gas
M = molar mass of gas

Complete step by step answer:
We know that the effusion is the special case of diffusion when a gas is allowed to escape through a small orifice or hole.
So the rate of effusion and diffusion is given based on Graham's law.
We know that the rate is directly proportional to the volume of gas and inversely proportional to the time of effusion.
${\text{r = }}\dfrac{{\text{V}}}{{\text{t}}}$
Here, r = rate of effusion
V = volume of gas
t = time of effusion
So we can write the ratio of the rate of effusion for helium and methane gas as:
$\dfrac{{{{\text{r}}_{{\text{He}}}}}}{{{{\text{r}}_{{\text{C}}{{\text{H}}_{\text{4}}}}}}}{\text{ = }}\dfrac{{{{\text{V}}_{{\text{He}}}}}}{{{{\text{t}}_{{\text{He}}}}}} \times \dfrac{{{{\text{t}}_{{\text{C}}{{\text{H}}_{\text{4}}}}}}}{{{{\text{V}}_{{\text{C}}{{\text{H}}_{\text{4}}}}}}}$ (equation 2)
Now from equation 1, we can write the ratio of the rate of effusion:
$\dfrac{{{{\text{r}}_{{\text{He}}}}}}{{{{\text{r}}_{{\text{C}}{{\text{H}}_{\text{4}}}}}}}{\text{ = }}\dfrac{{{{\text{X}}_{{\text{He}}}}}}{{\sqrt {{{\text{M}}_{{\text{He}}}}} }} \times \dfrac{{\sqrt {{{\text{M}}_{{\text{C}}{{\text{H}}_{\text{4}}}}}} }}{{{{\text{X}}_{{\text{C}}{{\text{H}}_{\text{4}}}}}}}$ (equation 3)
Now by comparing equation 2 and equation 3 we get:
$\dfrac{{{{\text{V}}_{{\text{He}}}}}}{{{{\text{t}}_{{\text{He}}}}}} \times \dfrac{{{{\text{t}}_{{\text{C}}{{\text{H}}_{\text{4}}}}}}}{{{{\text{V}}_{{\text{C}}{{\text{H}}_{\text{4}}}}}}}{\text{ = }}\dfrac{{{{\text{X}}_{{\text{He}}}}}}{{\sqrt {{{\text{M}}_{{\text{He}}}}} }} \times \dfrac{{\sqrt {{{\text{M}}_{{\text{C}}{{\text{H}}_{\text{4}}}}}} }}{{{{\text{X}}_{{\text{C}}{{\text{H}}_{\text{4}}}}}}}$
Now by substituting the value we get:
$\dfrac{{{\text{X ml}}}}{{{\text{5 sec}}}} \times \dfrac{{{{\text{t}}_{{\text{C}}{{\text{H}}_{\text{4}}}}}}}{{{\text{X ml}}}} = \dfrac{2}{{\sqrt 4 }} \times \dfrac{{\sqrt {16} }}{1}$
Now by solving we get:
${{\text{t}}_{{\text{C}}{{\text{H}}_{\text{4}}}}} = 20\sec$

Therefore, we can conclude that the correct answer to this question is option B.

Note: Here, we have to equate the two formulas for the rate of effusion because we are given with volume, time, and mole fraction and there is no relation between the volume and mole fraction and also no relation between time and mole fraction.