Question

The time taken for effusion of 64mL of oxygen will be best as the time taken for the effusion of which of the following gases under identical conditions?
(A) ${\text{64mL of }}{{\text{H}}_{\text{2}}}$
(B) ${\text{100mL of }}{{\text{N}}_{\text{2}}}$
(C) ${\text{64mL of C}}{{\text{O}}_{\text{2}}}$
(D) ${\text{45}}{\text{.24mL of S}}{{\text{O}}_{\text{2}}}$

Answer 1

There are two important properties of gases which are diffusion and effusion. Diffusion and effusion can be considered similar only in some cases. In this question both the properties are related to a particular quantity following a particular law.

Complete step by step answer:
Diffusion: The phenomenon where there is movement of particles from the area of high concentration to the area of low concentration. It can be also defined as the movement of molecules from high pressure to low pressure. Here the molecules spread through the medium. For example, if we spray at one corner of the room, we can smell the same at the other corner. Here diffusion of spray particles takes place.

Effusion: Effusion is a process where air escapes through a hole. The diameter of the hole is smaller than the mean free path of the molecules. Due to these circumstances all the molecules which reach the holes will pass through. The collision between the molecules is negligible.
Effusion and Diffusion are two different properties of gas. During effusion particles move faster as compared to diffusion because there is no collision between molecules.
Graham's Law was formulated by Thomas Graham in the year 1848. He discovered that gas molecules which are lighter will travel faster than the heavier gas molecules.
According to Graham’s law, at constant temperature and pressure molecules or atoms with lower molecular mass will effuse faster than the higher molecular mass molecules atoms. He found out the rate at which they escape through diffusion.
The rate of effusion of gas is inversely proportional to the square root of the molecular mass. ${\text{Rate}} \propto \dfrac{{\text{1}}}{{\sqrt {{\text{Molecular mass}}} }}$

The relation can be used to compare the rates of two different gases at constant pressure and temperature.
$\dfrac{{{\text{Rat}}{{\text{e}}_{\text{1}}}}}{{{\text{Rat}}{{\text{e}}_{\text{2}}}}}{\text{ = }}\sqrt {\dfrac{{{{\text{M}}_{\text{2}}}}}{{{{\text{M}}_{\text{1}}}}}}$
${{\text{M}}_{\text{1}}}{\text{ = molar mass of gas 1 which is }}{{\text{O}}_{\text{2}}}{\text{ = 32mL}}$
${{\text{M}}_{\text{2}}}{\text{ = molar mass of gas 2 = 64mL}}$
${\text{Rat}}{{\text{e}}_{\text{1}}}{\text{is the rate of effusion of the first gas = }}{{\text{r}}_1}{\text{ = 64mL}}$
${\text{Rat}}{{\text{e}}_2}{\text{is the rate of effusion of the second gas = }}{{\text{r}}_2}$

Substituting the values in the relation:
$\dfrac{{64}}{{{{\text{R}}_{\text{2}}}}}{\text{ = }}\sqrt {\dfrac{{64}}{{32}}}$
$\dfrac{{64}}{{{{\text{R}}_{\text{2}}}}}{\text{ = }}\sqrt 2$
${{\text{R}}_2} = \dfrac{{64}}{{\sqrt 2 }} = 45.25mL$
So, the correct answer is “Option D”.

Note: Diffusion does not require a barrier for the movement while effusion requires a barrier for the movement of particles. Though diffusion and effusion are two different properties of gases but in terms of speed they are considered similar.