# Graham’s Law: Diffusion-Effusion and Its Applications

Diffusion is the random movement of particles from a higher concentration region to an area of lower concentration. Graham’s law of diffusion states that the rate of diffusion of gases is inversely proportional to the square root of their molecular weight.

Light gases (low molecular weight) have a higher diffusion rate as compared with heavier ones (high molecular weight) provided that the temperature, pressure, and density remain constant. For example, hydrogen with a molecular weight of 2 amu has a 4 times greater rate of diffusion than oxygen with a molecular weight of 16 amu. Thomas Graham (1805-1869), a Scottish chemist after observing effusion and diffusion through a thick plug of plaster of paris, formulated this law in 1829.

## Relationship between Entropy and Diffusion of gases

Entropy is the randomness of gas particles and the diffusion of gases is based on entropy law. The gas particles diffuse in the surrounding towards an increased state of entropy. As this process is spontaneous it occurs on its own. ## Equation of Graham’s law

Consider two different gases with Molecular masses M1 and M2. The rate of their relative diffusion is given by:

R1 ∝ 1/M1

R2 ∝ 1/M2

### Derivation of equation for Graham law of diffusion with KMT

In1857, Clausius proved Graham’s law with the Kinetic molecular theory (KMT) of gas. If two gases A and B are with masses m1 and m2, and the velocities of their molecules are CA and CB:

From the Kinetic Molecular Theory of Gas;

Kinetic energy of gas 1= 1/3 m1NACA2 → (i)

Kinetic energy of gas 2 = 1/3 m2NBCB2 → (ii)

From equation (i) and equation (ii),

1/3 m1NACA2= 1/3 m2NBCB2 → (iii)

As we know that by Avagadro’s Law;

NA = NB → (iv)

Equation (iii) can be written after eliminating equation (iv),

mACA2 = mB CB2 → (v)

If we suppose m1 and m2 to be the molecular weights such as MA and MB, equation (v) can be written as;

MACA2= MBCB2

or,

(CA/CB)2 = MB/MA

By taking square root on both sides, we get;

CA/CB = √MB/MA

The rate of diffusion of a gas is directly proportionated to its velocity, hence it is concluded that:

r1/r2 = √MB/MA

The above equation is known as Graham’s law.

## Calculation of molecular mass (Mr)

Graham’s law is used to accurately calculate the unknown molecular mass of a gas. For this purpose, the rates of diffusion of these gases are to be measured carefully. The temperature and pressure should be kept constant during this analysis.

Example: The rate of diffusion of a gas is twice as compared with carbon dioxide. what is the molecular mass of unknown gas?

Solution:

Ratio of the rate of gases= r1 : r2 = 2 : 1

Molecular mass of carbon dioxide = M2= 44 g/mol

The molecular mass of unknown gas = M1 =?

Using Graham’s Law of Diffusion,

r1/r2 = √M2/M1

2/1 = √ 44/M1

4 = 44/M1

M1 = 11

Hence the molecular mass of unknown gas is 11 g/mol.

Practical demonstration of the rate of diffusion

Two pieces of the cotton pack are taken and dipped in ammonia and hydrochloric acid solutions. These pieces are then placed at positions A and B, the ends of the cylindrical tube. Now the vapors of these gases start moving towards each other. Eventually, the tube gets divided into 3 sections. The white fumes of ammonium chloride appear in the section near HCl cotton. Because HCl has a greater molecular mass hence diffuses slowly. While ammonia with lighter Mr travels at a higher speed to reach point C.

## Graham’s Law of Effusion

The movement of molecules from an area of higher concentration to an area of lower concentration through a small hole is called effusion. The rate of diffusion depends upon the molecular mass of the gas. It has been observed that the rate of effusion of a gas is much easier to calculate, hence the rate of effusion is determined and employed to find the molecular mass of unknown gas.

The rate of effusion is given by;

⇒ Pressure and temperature are kept constant while working in Graham’s law.

Example

A 20 cm3 of an unknown gas effuses through a pin-hole in 100 seconds, The same volume of oxygen O2 effuses in 140 seconds. Calculate the Mr of the unknown gas.

Rate of effusion of unknown gas = volume of gas/ time taken

= 20/100

Rate of Unknown gas = 0.2 ml/s

Similarly ;

Rate of dissusion of O2 gas

=20/140

= 0.14 ml/s

Putting r1, M1 and r2 in Graham’s law, we get;

0.14/0.2 = √M/ 32

Taking squares on both sides

0.71 = M/ 32

⇒ M = 0.71 x 32

= 22. 8 g/mol

Hence, 22.8 g/mol is the molecular mass of the unknown gas.

## Applications of Graham’s law

Graham’s law has many applications such as:

1. Separation of isotopes
2. Dialyser
3. Heterogeneous catalysis
4. Determination of molecular masses of unknown and newly discovered gases, etc

### Separation of Isotopes

Isotopes are the species with same atomic numbers but different atomic masses. They can be separated easily by using diffusion principles. For example, isotopes of uranium are separated through a diffusion process known as the Uranium enrichment process.

Uranium-238 and U-235 are two (of a total of 5) isotopes of uranium found in nature. However, uranium-235 is a fissionable isotope, so it is first separated from U-238 to get pure U-235. For this purpose, U is first reacted with Flourione to form UF6. Actually, a mixture of 235UF6 and 238UF6 is produced and these isotopes have significantly different molecular weights. This mixture is volatilized and separated on the basis of differences in the rate of their diffusions. This process is done in the Enrichment plant as shown below. ### Diffusion of ions in liquids: Dialyser

The transport of ions in liquids is somewhat the same in all gases. The rate of diffusion in liquid also depends upon their ionic masses. Its best application is the dialysis chamber which works on the principle of diffusion.

In this method, the unwanted ions (nitrogenous wastes of the body) are removed from the solution (blood). When two solutions are separated through a perforated filter, which allows smaller ions to pass through, are placed together, the large molecular ion tends to stay in the solution due to a lower diffusion rate whereas the lighter one diffuses across the solution. ### Heterogeneous catalysis

Gases are diffused on the metal surfaces where they adsorb and react with each other. The rate of diffusion of these gases controls reaction kinetics when in a reaction.

For example, in heterogeneous catalysis, particular conditions of temperature and pressure are required to form ammonia from nitrogen and hydrogen. The reaction is catalyzed by iron.

The following steps are involved in this reaction.

• Diffusion

Nitrogen gas and hydrogen gas diffuse to the surface of the iron.

The reactants molecules are chemically adsorbed onto the surface of the iron.

• Reaction

The adsorbed nitrogen and hydrogen atoms react together to form ammonia.

• Desorption

The bonds between the ammonia and the surface of iron weaken and are eventually broken.

• Diffusion

Ammonia diffuses away from the surface of the iron.

Hydrogenation of Vegetable oil

A similar process goes for the hydrogenation of edible oils. The liquid oil is adsorbed along with hydrogen, for the reaction to occur.

Pharmaceuticals

Drugs used to treat several tumors and cysts are first investigated by using colored compounds, e.g gelatin through their rates of diffusion. This phenomenon is known as the Drug Modeling Technique.

## Concepts Berg

Can you use Graham’s law of effusion on diffusion?

Both have the same equation but different working principles. Effusion is measured by using vacuumed container and no molecular collision is observed (small hole) whereas, whereas diffusion is the movent of molecules due to a concentration gradient through a large area.

What is the concept of Graham’s law of effusion?

Effusion, Work, and The principle of entropy are the basis of Graham’s law of discussion. When a gas is trapped in a container and is allowed to expand in a vacuumed container, It rapidly moves towards less concentration to increase its entropy.

What gases diffuse the fastest, O2, CH4, CO2, or Cl2?

The rates of diffusion of these gases are as:

CH4 > O2 > CO2 > Cl2

Methane diffuses at the fastest rate because it has the least molecular weight. Similarly, chlorine has the most molecular weight here.

What is the real formula of Graham’s law?

The real formula of Graham’s law is its mathematical form.

r = 1/√M

Which gas diffuses faster hydrogen, oxygen, carbon monoxide, or nitrogen?

Rate of diffusion;

Hydrogen > Nitrogen > Oxygen > Carbon monoxide

The rate of diffusion is inversely proportional to the molecular weight of the given gas. Carbon monoxide being the heaviest gas in these examples has the lowest diffusion rate followed by oxygen, and nitrogen, and then hydrogen gas has the fastest rate of diffusion among them.

Reference Books

• Essential of Physical Chemistry: 2nd edition By B.S Bahl and Arun Bahl
• Cambridge International AS A level Chemistry by Lawrie Ryan and Roger Norris