Graham's Law of Diffusion- Diffusion, Effusion, Formula, Factors Affecting Rate of Effusion of Gases, Practice problems and FAQs

Have you ever noticed why you get the aroma of hot food from a far distance when it is getting prepared in your neighbour’s house? Did you observe when you burn incense sticks, the smell of burning incense sticks can even be identified from the other part of the room? Well! All these are examples of diffusion which takes place in our daily life. In the year 1848, Thomas Graham studied the diffusion and effusion of gases. So, let’s dig in a bit deeper into the concept and try to understand what diffusion and effusion are and what Thomas Graham proposed.

Table of contents

Diffusion
Effusion
Graham’s Law of effusion/diffusion
Factors affecting rate of effusion of gases
Practice problems
Frequently asked questions-FAQs

Diffusion

It is defined as the movement of gas from the region of high concentration to the region of low concentration without any physical aid or its just intermixing of different gases on its own due to pressure difference.

Diffusion is an important phenomenon occurring in our daily life. Diffusion is maximum in the case of gases and then followed by liquids and solids.

Let us consider the above example in which when the valve is opened and the gases present in both the containers start mixing due to the concentration or pressure difference and get distributed uniformly in the beaker till the equilibrium is reached.

Effusion

The movement of gas molecules through a small orifice due to the difference in pressure is known as effusion. It occurs when the size of the orifice is smaller than the mean free path of the particles of the gas.

Graham’s Law of effusion/diffusion

In the year 1848, Thomas Graham through his experiments observed that the rate of effusion/diffusion of a lighter gas is more than the rate of effusion/diffusion of heavier gas and stated a law known as Graham’s law of effusion/diffusion.

According to Graham's law “Rate of effusion/diffusion of a gas at a constant temperature and pressure is inversely proportional to the square root of its molar mass”.

Mathematically,

Where, r is the rate of the effusion/diffusion and M is the molar mass of the gas.

From the ideal gas equation, we can say;

density (d)

Where d is the density of the gas, P is the pressure of the gas, M is the molar mass of the gas, R is the universal gas constant and T is the temperature of the gas.

For an ideal gas at constant pressure and temperature;

d ∝ M

So, it can be said that;

r ∝ 1d

Rate of effusion/diffusion can be defined in multiple ways.

Rate of effusion/diffusion can be defined as the no. of moles of gas diffused.

Rate of effusion/diffusion is also defined as the volume of the gas diffused per unit time.

Rate of effusion/diffusion can also be defined as the distance travelled by the gas per unit time.

Factors affecting rate of effusion of gases

In fact, apart from molar mass and density of the gas, the rate of effusion/diffusion of a gas depends on other factors as well:

Pressure: Rate of effusion/diffusion is directly proportional to the pressure of the gas. The higher the pressure of the gas in the container, the more will be the rate with which the gas will effuse or diffuse.

Mathematically,

r ∝ P

Where, r is the rate of the effusion/diffusion and P is the pressure of the gas in the container.

Area of the orifice: The rate of effusion/diffusion is directly proportional to the area of the orifice through which a gas is effusing. The bigger the area of the orifice, the higher will be the rate of effusion/diffusion.

Mathematically,

r ∝ A

Where, r is the rate of the effusion/diffusion and A is the area of the orifice.

Temperature: The rate of effusion/diffusion is inversely proportional to the temperature of the gas. More the temperature of the gas, the lower will be the rate with which the gas will effuse or diffuse.

Mathematically,

r ∝ 1T

Where, r is the rate of the effusion/diffusion and T is the temperature of the gas.

Number of moles: The rate of effusion/diffusion is directly proportional to the number of moles of gas which were present initially in the container.

Mathematically,

r ∝ n

Where, r is the rate of the effusion/diffusion and n is the number of moles of gas present initially in the container.

Suggested video link

https://www.youtube.com/watch?v=3QcH9-rEB_c

Practice problems

Q 1. If two gases A and B are present in two different containers which are connected with a cylindrical tube of a certain length that is allowed to diffuse. Select the correct option for the gas, where the first time these two gases are meeting.
Assume that the molar mass of gas A = 64 gmol^-1 and molar mass of gas B=32 gmol^-1 respectively.

a. Closer to the container containing gas A
b. Closer to the container containing gas B
c. At an equal distance from both the containers
d. None of the above

Answer: (A)
According to the given data,

Molar mass of gas A = 64 gmol^-1

Molar mass of gas B=32 gmol^-1

Using Graham’s law of diffusion we know,

Where, r is the rate of the effusion/diffusion and M is the molar mass of the gas.

Rate of diffusion can be defined as the distance travelled by the gas per unit time.

Comparing equation (i) and (ii) we get,

Higher the molar mass of the gas, the lesser the distance it will travel. Therefore, the first intermixing will take place closer to the container containing gas A.

Q 2.Select the correct option in which the rate of effusion of the given gas will be maximum at a constant temperature and pressure.

a. SO2
b. O2
c. CH4
d. CO2

Answer: C
According to the Graham’s law (At constant temperature and pressure),

r1M

Molar mass of SO2=64 g mol^-1

Molar mass of O2=32 g mol^-1

Molar mass of CH4= 16 g mol^-1

Molar mass of CO2= 44 g mol^-1

As the molar mass of CH4 is minimum. Therefore, the rate of effusion will be maximum.

Q 3. If the rate of diffusion of CH₄ and some unknown gas is 2:1, measured under constant temperature and pressure. The molecular mass of unknown gas is:

Answer: (D)
Molar mass of CH4 =16 gmol^-1

Let the rate of diffusion of unknown gas be r_dX.

Let the molar mass of unknown gas be M_X g mol^-1.

According to Graham’s law at constant temperature and pressure;

Putting the value in the equation (i) we get,

Squaring both the sides, we get;

M_X=2 x 16 =32 gmol^-1

Q 4. What will be the ratio of the rate of diffusion of SO2:SO3 if the initial moles of SO2 and SO3 2 mol and 3 mol respectively?

a. 2:3
b. 5:2
c. 5 :3
d. 5:3

Answer: (C)
Initial no. of moles of SO₂=2 mol

Initial no. of moles of SO₃=3 mol

Molar mass of SO₂=64 g mol^-1

Molar mass of SO₃=80 g mol^-1

According to Graham's law;

Putting the values in the above equation;

Frequently Asked Questions-FAQs

Q 1. What is the difference between osmosis and diffusion?
Answer: Osmosis is the process of movement of solvent molecules through a semipermeable membrane from the region of lower concentration to the region of higher concentration. Diffusion is defined as the spontaneous movement of gas molecules from the high concentration region to the low concentration region without any physical aid.

Q 2. Why is diffusion in gases faster as compared with solids?
Answer: Diffusion is faster in the case of gases as compared with liquids or solids because in case of gases, interparticle forces of attraction between the particles are minimum and particles possess higher kinetic energy but in the case of solids interparticle forces of attraction between particles are maximum and kinetic energy is minimum therefore diffusion in case of solids is almost negligible.

Q 3. State some examples of effusion in daily life.
Answer: Some of the examples of effusion are:

Air coming out of the balloon when it is pricked.
Air coming out of a tyre when got punctured.
Water vapours coming out of a pressure cooker.

Q 4. What are the applications of Graham’s law of diffusion?
Answer: There are some important applications of Graham’s law of diffusion which include: