Question

X mL of hydrogen gas effuses through a hole in a container in 5s. The time taken for the effusion of the same volume of the gas specified below, under identical conditions, is
(a) 10s, He
(b) 20s, \[{{O}_{2}}\]
(c) 25s, CO
(d) 55s, \[C{{O}_{2}}\]

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 solution:
> 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,
\[Rate\propto \sqrt{\dfrac{1}{M}}\]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.
So, for effusion of same volume, we can write,
\[\dfrac{{{t}_{1}}}{{{t}_{2}}}=\sqrt{\dfrac{{{M}_{1}}}{{{M}_{2}}}}\] or
\[\dfrac{{{R}_{1}}}{{{R}_{2}}}=\sqrt{\dfrac{{{M}_{1}}}{{{M}_{2}}}}\]
> The above equation in terms of volume \[{{V}_{1}}\] and \[{{V}_{2}}\]
\[\sqrt{\dfrac{{{M}_{1}}}{{{M}_{2}}}=}\dfrac{\dfrac{{{V}_{2}}}{{{t}_{2}}}}{\dfrac{{{V}_{1}}}{{{t}_{1}}}}\], here it is given same volume is effusion occurs so \[{{V}_{1}}\]=\[{{V}_{2}}\]=X and \[{{t}_{1}}\]=5sec and \[{{M}_{1}}\]of hydrogen is 2g per mol.
> Therefore, substituting these values we get
\[\sqrt{\dfrac{2}{{{M}_{2}}}}=\dfrac{5}{{{t}_{2}}}\]
> Now let us consider each options in the question,
-By substituting the molar mass of helium(4 g per mol) and time 10s in the above equation we realise
\[\sqrt{\dfrac{2}{4}}\ne \dfrac{5}{10}\]. So option (a) is incorrect.
-Substituting the values of molar mass of oxygen(32 g per mol) and time 20 sec. We get
\[\sqrt{\dfrac{2}{32}}=\dfrac{5}{20}\]So, option (b) is correct.
-For CO molar mass is 28 g per mol and time here is 25s,
\[\sqrt{\dfrac{2}{28}}\ne \dfrac{5}{25}\], option (c) is incorrect.
-For carbon dioxide, molar mass is 44 g per mol, time is 55s, we get
\[\sqrt{\dfrac{2}{44}}\ne \dfrac{5}{55}\]
Therefore, the correct answer to this question is option (b).

Note: Diffusion and 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.