Graham's Law of Diffusion: Diffusion, Formula & Solved Examples

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Jasmine Grover

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Graham’s law of diffusion states that the rate of diffusion of a gas or its effusion is inversely proportional to the square root of its molecular weightThe principle of the law of diffusion states that at any given temperature and pressure the rate of diffusion of any given gas is inversely proportional to the square root of its density.

  • Diffusion is the process through which particles from one gas move to another.
  • It is caused when the particles of a material are in low concentration and they move to areas with high area of concentration.
  • An example of diffusion in the spreading of scent of perfume in the entire room when it’s sprayed in one part.
  • Effusion is the process through which a gas escapes the container’s wall through a very small hole on the surface. 
  • The rate of diffusion refers to the change in diffusing molecules over time.

Read More: Difference between Effusion and Diffusion

Key terms: Effusion, Square root, Temperature, Density, Pressure, Collision, temperature, Molecule, Law of Diffusion


Rate of Diffusion

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Diffusion of a gas is the random motion of a particle involved in the net movement of the substance from the higher concentration area to the lower concentration area.

  • Each of the particles inside the gas starts colliding with each other.
  • The rate of diffusion of a gas is directly proportional to the square root of the density of a gas molecule.
  • The density of a gas molecule is equal to the mass of a gas that is divided by the volume of the gas molecule.
  • If the volume of the gas molecule is held constant one gas can be compared with another gas.
  • Graham's Law of Diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molecular weight.

Diffusion of Gases
Diffusion of Gases

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Rate of Effusion

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The effusion of a gas is the process in which the air particles escape or leak through a hole whose diameter is considerably smaller than the mean free path of the molecules.

  • In this process, all the particles and the molecules that reach the hole will pass through, as due to the collisions between the molecules in these places are negligible in number.
  • The rate of effusion of a gas can be defined as the process by which particles of a material from the closed space start escaping with time.
  • Suppose when we make an opening in the balloon, the gas present inside it starts escaping into the atmosphere and the balloon starts deflating from inside.
  • This is called the effusion of a gas into the atmosphere.
  • Hence, we can say that the rate of effusion of a gas is inversely proportional to the density of a gas and its molar mass.

Rate of Effusion 1/√density 1/√M

  • Here M is the molar mass of a gas molecule.

Read More: Difference between Diffusion and Osmosis


Graham's Law of Diffusion

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Graham's Law was first discovered by a scientist named Thomas Graham in the year 1848. He experimented with the effusion process of a gas and formulated an important aspect that the gas molecules are lighter in weight and will travel faster in air than the heavier gas molecule. This law is popularly known as Graham's law of Effusion.

According to Graham’s law of diffusion and effusion, the atoms and molecules with lower molecular mass will diffuse faster in air than the molecules and atoms that are heavier, at constant temperature and pressure of a gas. It states that the rate of effusion of a gas molecule is inversely proportional to the square root of the density of a gas or its molecular mass.

Graham's law of Diffusion Formula and Examples

Graham’s Law Formula = \({Rate_1 \over Rate_2} = {\sqrt {M2 \over M1}}\)

  • Here, Rate 1 = rate of effusion of the first gas
  • Rate 2 = rate of effusion of the second gas.
  • M1 = Molar mass of first gas
  • M2 = Molar mass of second gas.

It states that the rate of diffusion or effusion is inversely proportional to its molecular mass. Graham's law of diffusion, also known as Graham's law of effusion, states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molecular weight. 

According to Graham's law, lighter gases diffuse or effuse quickly than heavier gases at the same temperature and pressure.

Applications of Graham’s law of diffusion

Graham's law has several applications such as –

  • The separation of gases in industries such as petrochemicals, air separation, and natural gas processing.
  • It is also used in fields such as atmospheric science, where it is used to understand the mixing of gases in the atmosphere.

Importance of Graham's law of diffusion

Graham's law of diffusion is important as:

  • It provides a basis for separating gases of different molecular weights. So, the lighter gases can be separated from heavier gases using gas chromatography.
  • Graham's law of diffusion helps to understand the mixing of gases in the atmosphere. It explains why lighter gases rise to the upper atmosphere while heavier gases remain closer to the surface.
  • The law of diffusion is also used in chemical processes such as distillation.
  • Graham's law of diffusion is a fundamental law of physics to understand the behavior of gases and how they interact with their environment.

Graham’s Law of Diffusion Question

Suppose we have two gases, gas A with a molecular weight of 16 g/mol and gas B with a molecular weight of 64 g/mol. If both gases are at the same temperature and pressure, what is the ratio of their diffusion rates?

Using Graham's law, we can calculate the ratio of the diffusion rates as follows:

Ratio of diffusion rates = (Rate of diffusion of gas A) / (Rate of diffusion of gas B)
Ratio of diffusion rates = √(Molecular weight of gas B) / √(Molecular weight of gas A)
Ratio of diffusion rates = √(64 g/mol) / √(16 g/mol)
Ratio of diffusion rates = 8 / 4
Ratio of diffusion rates = 2

Therefore, the ratio of the diffusion rates of gas A to gas B is 2:1. This means that gas A will diffuse twice as quickly as gas B at the same temperature and pressure.

Read More: Bulk Modulus


Things to Remember

  • Diffusion is a phenomenon in which gas moves from higher concentration to lower concentrations.
  • In effusion, the gas moves from lower to higher concentrations.
  • The diffusion of a gas leads to a significant disorder in the entire system of the gas. It acts at slower rates in liquid as well as solid gasses. 
  • Graham's law of diffusion is used in determining the molecular masses and the vapor density of a gas molecule.
  • Mathematically, Graham’s law can be expressed as Rate of diffusion/effusion ∝ 1/√(molecular weight)
  • This is also used in separation of different gasses from the mixture of gasses and in separating the isotopes of same elements.

Read More: Young's Modulus


Sample Questions

Ques. What is the equation for Graham's law of effusion? (2 marks)

Ans. Graham's law of effusion can be used for calculating the rate of diffusion or effusion of a gas and its molar masses. The equation of the law of effusion can be expressed in the equation: r 1 / r 2 = √ (m 2 / m 1)

r1 = rate of effusion of the first gas

r2 = rate of effusion of the second gas.

m1 = Molar mass of first gas

m2 = Molar mass of second gas.

Ques. Which gas Effuses the fastest? (3 marks)

Ans. As we know, the rate of effusion of a gas is inversely proportional with the square root of the molecular masses of a gas as per Graham's law of effusion. Hence, the gas with the lowest molecular weight will effuse faster than the heavier gas. Therefore we can say the lightest and fastest gas is helium.

Ques. Which gas is more ideal at STP? (3 marks)

Ans. The real gas which acts most likely as an ideal gas is helium. It is because helium, not at all like most gases, exists as a solitary atom, which makes the van der Waals scattering powers as low as could really be possible. Another element is that helium, as other noble gases, has a totally filled valence electron shell.

Ques. What is the value of R in an ideal gas equation? (2 marks)

Ans. The numerical value of R on the ideal gas equation will depend upon the units involved in the equation, but it is usually stated with the S.I units as R: 8.314 J/mol K. The R is known as the Universal Gas Constant of the equation. 

Ques. Calculate the molar mass of a gas whose rate of diffusion is 2.92 times the diffusion rate of Ammonia NH3? (4 marks)

Ans. As we know that the rate of diffusion of a gas is 2.92 times the ammonia gas, hence we can say that the ratio of rates of diffusion of the given gas should be 1/ 2.92. So it can be written as r1/r2 = 1/ 2.92.

It is known that the molar mass of the Ammonia gas is 17.03. Them according to Graham's law of effusion 

r 1 / r 2 = √ (m 2 / m 1)

After substituting the value given above we get

1/ 2.92 = √ (mass 2 / 17.03 )

After squaring both sides 

0.11728 = (mass 2 / 17.03 )

Hence, mass 2 = 2.0 g per mol. 

Ques. Find the relative diffusion rates of a water H20 as compared to the hard water whose molar mass is 20.02? (5 marks)

Ans. After calculating the molar masses of water and hard water we get

The molar mass of H20 (mass 1) = 18.01

The molar mass of Hard Water (mass 2) = 20.02 

Now let's assume that the rate of diffusion formula of a heavy water as one as it has a slower rate of diffusion. 

r2 = 1

According to Graham's law of effusion 

r 1 / r 2 = √ (m 2 / m 1)

After squaring both sides we get 

( rate 1 / rate 2) ² = (mass 2 / mass 1)

After substituting the values given above in the equation we get

( rate 1)² / 1 = 20.02 / 18.01

( rate 1)² = 1.11

rate 1 = 1.05

Hence the relative diffusion rate of a gas is 1.05.

Ques. A vessel having a capacity of 120 ml contains some amount of gas at 35°C and q pressure of 1.2 bar. The gas is then transferred into another vessel of 180 mL at a temperature of 35°C. What would be its pressure? (3 marks)

Ans. It is given in the questions 

V1 = 120 ml

V2 = 180 ml

P1 = 1.2 bar

P2 = ?

As we know, the temperature remains constant at 35° Celsius. 

P1 V1 = P2 V2

1.2 bar × 120 ml = P2 × 180 ml

144 bar ml = P2 × 180 ml

144 bar ml / 180 ml = P2

P2 = ? bar

P2 = 0.8 bar

Ques. Calculate the molar mass of a gas whose rate of diffusion is 4 times the diffusion rate of water? (5 marks)

Ans. As we know that the rate of diffusion of a gas is 4 times the water gas, hence we can say that the ratio of rates of diffusion of the given gas should be 1/ 4. So it can be written as r1/r2 = 1/ 4.

It is known that the molar masses of the water gas is 18. Them according to Graham's law of effusion 

r 1 / r 2 = √ (m 2 / m 1)

After substituting the value given above we get

1/ 4 = √ (mass 2 / 18 )

After squaring both sides 

1/ 16 = (mass 2 / 18 )

Hence, mass 2 = 1.125 g per mol

Ques. Find the relative diffusion rates of a hydrogen as compared to the nitrogen whose molar masses is 14? (5 marks)

Ans. After calculating the molar masses of hydrogen and nitrogen we get

The molar mass of water (mass 1) = 2

The molar mass of nitrogen (mass 2) = 14 

Now let's assume that the rate of diffusion formula of hydrogen as one as it has a slower rate of diffusion. 

r2 = 1

According to Graham's law of effusion 

r 1 / r 2 = √ (m 2 / m 1)

After squaring both sides we get 

( rate 1 / rate 2) ² = (mass 2 / mass 1)

After substituting the values given above in the equation we get

( rate 1)² / 1 = 14 / 2

( rate 1)² = 7

rate 1 = 2.64

Hence the relative diffusion rate of a gas is 2.64.

Ques. What will be the minimum pressure that is required to compress 500 dm³ of air at 1 bar of pressure to 200 dm³ at a temperature of 30° C? (3 marks)

Ans. It is given in the questions 

V1 = 500 dm³

V2 = 200 dm

P1 = 1 bar 

P2 = ?

As we know, the temperature remains constant at 30° Celsius. 

P1 V1 = P2 V2

1 bar × 500 dm³ = P2 × 200 dm³

500 dm³ = P2 × 200 dm³

500 dm³ / 200 dm³ = P2

P2 = 5/2 bar

P2 = 2.5 bar


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