Question

If oxygen effuses through an opening at a rate of $67$ particles per second, how fast would Hydrogen gas effuse through the same opening?

Answer 1

Hint : Graham’s Law shows an empirical relationship that states that the ratio of the rates of diffusion or effusion of two gases is square root of the inverse ratio of their molar masses. This relationship is based on the postulate which states that all gases at constant temperature have the same kinetic energy.

Complete Step By Step Answer:
We know that rate of effusion of gases is inversely proportional to square root of molecular mass of gases, that is, according to Graham’s Law, $Rate\alpha \sqrt{\dfrac{1}{M}}$ here $M$ is the molar mass. From this we get that, the smaller the value of molecular mass of gas, the greater will be the rate of effusion of same volume, we can write as;
${{\dfrac{{{R}_{1}}}{{{R}_{2}}}}_{{}}}=\sqrt{\dfrac{{{M}_{2}}}{{{M}_{1}}}}$ where $R$ is the rate of effusion and $M$ is the mass. The subscript number refers to either oxygen or hydrogen; it's a relative equation.
Here if we say that substance one is hydrogen and two is oxygen, then
${{R}_{1}}=?,{{R}_{2}}=67$ and ${{M}_{1}}=1,{{M}_{2}}=16$
Rewriting the equation we get;
${{R}_{1}}={{R}_{2}}\times \sqrt{\dfrac{{{M}_{2}}}{{{M}_{1}}}}$
By substituting in the above equation we realize;
${{R}_{1}}=67\times \sqrt{\dfrac{16}{1}}=67\times 4=268$
Therefore, $268$ particles per second effuses Hydrogen gas effuse through the same opening.

Note :
Diffusion or effusion is two different entities. When gas molecules disperse throughout a container, it is diffusion. Effusion occurs when gas passes through a small opening that is smaller than the mean free path of the particle, that is nothing but the average distance travelled between collisions between the molecules.