What is the mole fraction of dihydrogen in a mixture, given the relative rates of effusion of its components?

Question

A mixture of $\ce{H2}$ and $\ce{He}$ at $\pu{300 K}$ effuses from a small hole in a gas chamber. What is the mole fraction of $\ce{H_2}$ in the original mixture, if $3.0$ times as many $\ce{He}$ atoms as $\ce{H2}$ molecules escape from the hole in one minute?

The answer apparently relies on the fact that $\ce{H2}$ collides with the walls $\displaystyle\frac{1}{\sqrt2}$ as often as $\ce{He}$, and we therefore need $\sqrt{2}$ as many $\ce{He}$ atoms as $\ce{H2}$ molecules to yield an equal amount of substance escaping. Considering that the ratio is in fact 3, the final answer is then that the mole fraction of $\ce{H2}$ is

$$x(\ce{H2}) = \frac{1}{3\sqrt{2} + 1}.$$

Can anybody please explain how they got the $\displaystyle\frac{1}{\sqrt2}$ part?

Answer 1

If you are familiar with Graham's Law, you will know that the ratio of the rates of effusion of two gases is inversely proportional to the square root of the ratio of their molar masses. That is:

$$\frac{R_1}{R_2}=\sqrt{\frac{M_2}{M_1}}$$

Here, $M_1 = M(\ce{H2}) = \pu{2 g//mol}$ and $M_2 = M(\ce{He}) = \pu{4 g//mol}$, so the lighter of the two, hydrogen, effuses at $\sqrt{2}$ the rate of helium, or conversely, the heavier, helium, effuses at $\frac{1}{\sqrt{2}}$ the rate of hydrogen. That's where the fraction came from.

Since we know that helium is escaping three times faster than hydrogen, there must be three times more helium than would product an equal rate of flow for the gases. An equal flow would occur when there is $\sqrt{2}$ times as much helium, and triple flow would occur with $3\sqrt{2}$ times as much helium.

The total amount of gas present is then: $$\ce{H2} + \ce{He} = 1 + 3\sqrt{2}$$ and the mole fraction of hydrogen is: $$x(\ce{H2}) = \frac{n(\ce{H2})}{\text{total amount of substance}} = \frac{1}{3\sqrt{2} + 1}$$