Graham's Law: Diffusion And Effusion MCQ - Practice Questions & Answers

Quick Facts

Graham’s Law of Diffusion is considered one the most difficult concept.
12 Questions around this concept.

Concepts Covered - 0

Graham’s Law of Diffusion

Graham’s Law of Diffusion
According to it "At constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density or molecular weight". It is applicable only at low pressure.

$\mathrm{r} \propto \frac{1}{\sqrt{\mathrm{M}}} \text { or } \frac{1}{\sqrt{\mathrm{d}}}$
Here r = rate of diffusion or effusion of a gas or liquid. M and d are the molecular weight and density respectively.

For any two gases, the ratio of the rate of diffusion at constant pressure and temperature can be shown as

$\mathrm{r}_{1} / \mathrm{r}_{2}=\sqrt{\mathrm{M}_{2} / \mathrm{M}}_{1} \text { or } \sqrt{\mathrm{d}_{2} / \mathrm{d}_{1}}$
Hence diffusion or effusion of a gas or gaseous mixture is directly proportional to the pressure difference of the two sides and is inversely proportional to the square root of the gas or mixture effusing or diffusing out.

Some Other Relation Based on Graham’s law
As r = V/t = Volume/time, thus:

$\\\mathrm{\frac{V_{1} t_{2}}{V_{2} t_{1}}=\sqrt{M_{2} / M_{1}}}\\\\\mathrm{\text { As } r=\frac{n}{t}=\frac{d}{t}=\frac{w}{t}}\\\\\mathrm{Thus,\: \frac{n_{1} t_{2}}{n_{2} t_{1}}=\sqrt{M_{2} / M_{1}}}\\\\\mathrm{\frac{w_{1} t_{2}}{w_{2} t_{1}}=\sqrt{M_{2} / M_{1}}}\\\\\mathrm{\frac{\mathrm{d}_{1} \mathrm{t}_{2}}{\mathrm{d}_{2} \mathrm{t}_{1}}=\sqrt{\mathrm{M}_{2} / \mathrm{M}_{1}}}$
Here n represents the number of moles, w represents weight in gram and d represents the distance traveled by a particular gas or liquid.

Differentiation Between Diffusion and Effusion

Diffusion
It is the movement of gaseous or liquid molecules without any porous bars that are, the spreading of molecules in all directions.

Effusion
It is the movement of gases molecules or liquid molecules through a porous bar that is, a small hole or orifice.

Uses of Graham’s Law

Detecting the presence of Marsh gas in mines.
Separation of isotopes by different diffusion rates. For example, U-235 and U-238
Detection of molecular weight and vapour density of gases using this relation.
$\mathrm{\frac{r_{1}}{r_{2}}=\left(\frac{m_{2}}{m_{1}}\right)^{1 / 2} \text { or }\left(\frac{d_{2}}{d_{1}}\right)^{1 / 2}}$