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

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

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

  • 19 Questions around this concept.

Solve by difficulty

Equal moles of hydrogen and oxygen gases are placed in a container with a pinhole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape?

Two gases A and B having the same volume diffuse through a porous partition in 20 and 10 second respectively. The molecular mass of A is 49u. Molecular mass of B will be

50 mL of each gas A and gas B takes 150 and 200 seconds respectively for effusing through a pinhole under similar conditions. If the molecular mass of gas B is 36, the molecular mass of gas A will be :

A certain gas takes three times as long to effuse out as helium. Its molecular mass will be:

The composition (by moles) of effused mixture which initially had equal moles of H2, F2 and Heafter 6 effusion steps is  x: 1: y, respectively. Find the value of (xy)

The diffusion coefficient of on ideal gas is Proportional to its mean free path and mean speed The Absolute temperature of an ideal gas is increased 8 atm and its pressure 4 times. As a Result, the diffusion coefficient ot this gas Increase A times. The value at A is?

A large cylinder of helium filled at 2000 mm of Hg had a small Orifice through which helium escaped into evacuated space at the rate of 6.4 moles/ hours. How long would it take for 10 mole of CO to leak through a similar orolfice if the co were confined at the same pressure?

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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 distance travelled by a particular gas or liquid.

Differentiation Between Diffusion and Effusion

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

Effusion
It is 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}}

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