NIT Jamshedpur Seat Matrix 2024 - Check Previous Year Matrix here

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

Updated on Sep 18, 2023 18:35 AM | #JEE Main

Quick Facts

  • Graham’s Law of Diffusion is considered one the most difficult concept.

  • 13 Questions around this concept.

Solve by difficulty

100 cm3 of N2O diffuse through a porous plug in 100sec. What volume of CO2 will diffuse under similar conditions in 50 sec.

The gas that has the slowest rate of diffusion among O2 , H2 , CO2 and CH4 is

A specific volume of $\mathrm{H}_2$ requires 24 s to diffuse out of a container. The time requires by an equal volume of $O_2$ to diffuse out under identical conditions, is 

According to Graham's Law , the rate of diffusion of $\mathrm{CO}, \mathrm{O}_2, \mathrm{~N}_2$ and $\mathrm{CO}_2$ follows the order :

An air bubble of radius 0.1 cm lies at a depth of 20 cm below the free surface of a liquid of density $1000 \mathrm{~kg} / \mathrm{m}^3$. If the pressure inside the bubble is $2100 \mathrm{~N} / \mathrm{m}^2$ greater than the atmosphet ic pressure, then the surface tension of the liquid m . SI unit is (use $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )
 

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}}

"Stay in the loop. Receive exam news, study resources, and expert advice!"

Get Answer to all your questions

Back to top