# Grahams law of diffusion: Worked example and importance

### Graham’s Law of Diffusion:

Diffusion involves the movement of gaseous, liquid and solid particles from a region of higher concentration to that of lower concentration. Consider the analogy below:

When a perfume is sprayed in a room, the particles of the perfume will move from a region of higher concentration (region where the perfume was sprayed) to a region of lower concentration (other regions in the room where the perfume wasn’t sprayed). As a result, anyone present in this region of lower concentration will also perceive the scent of the perfume.

This movement of particles from a region of higher concentration to a region of lower concentration is termed diffusion.

The particles undergoing diffusion can be gaseous, liquid or solid. Importantly, the movement of such particles must be from a region of higher concentration to a region of lower concentration.

Diffusion is fastest in gaseous particles. This is due to the fact that gaseous particles are able to move freely since the cohesive or binding forces between them are extremely small or negligible.

It is noteworthy to state that diffusion is slowest in solid particles.

Diffusion is technically not the same thing as effusion Diffusion always accompanies effusion; with both happening almost simultaneously.

Effusion is the movement of gaseous particles through a very small opening; and is always accompained by diffusion.

For instance, when an inflated balloon is pierced with a needle, both effusion and diffusion will occur.

When diffusion occurs, it continues until equilibrium is attained. With reference to our previous instance, the scent of the perfume will continue to move from the region of higher concentration to the region of lower concentration until every part of the room becomes equally filled, assuming all external openings have been shut.

When a gas has a high rate of diffusion, it will take a shorter time for diffusion to occur in such gas. This implies that the gas will diffuse rapidly.

Conversely, diffusion will take a longer time for gases with lower rate of diffusion: that is, such gases will diffuse slowly.

The Scottish chemist Thomas Graham (1805-1869) studied the rate at which gases diffuse. He established a law of diffusion which was named after him. This is the Graham’s law of diffusion.

Graham’s law of diffusion states that the rate (r) of diffusion in a gas at a given temperature is inversely proportional to the square root of its density or molecular mass (m).

Below is an equation on Graham's law of diffusion.

Note that the vapour density (v.d) of a gas is equal to half its relative molecular mass(r.m.m).

Therefore: r.m.m. = 2 x v.d

#### Worked example of Graham's law

30cm3 of a gas with an empirical formula of CH3 diffuses through a porous partition in 45.2s. If 30 cm3 of hydrogen diffused in 11.7s under the same conditions. Calculate:

1. The vapour density of the CH3 gas

2. The molecular formula of the gas CH3

(Mass of H2 = 2)

Solution:

(i) Vapour density of the CH3 gas.

Using the formula derived from the above diagram:

• tx/tH= √mx /√mH

• t = time, m = molar mass, x = CH3 and H = hydrogen

• Time taken for CH3 to diffuse = 45.2s

• Time taken for H2 to diffuse = 11.7s

• Mass of H2 = 2

• Mass of CH3 = ?

Let us substitute these values into the above formula:

• 45.2/11.7 = √(mx/2)

• 2(45.2/11.7)2 = mx

• mx = 2 x 14.92

• mx = 29.84g

• Mass of CH3 = 29.84g

• r.m.m = 2 x v.d

Where r.m.m = Relative Molecular Mass and v.d = Vapour Density

• v.d = r.m.m./2

• v.d = 29.84/2

• v.d = 14.92

(ii) Molecular formula of the gas CH3

• xCH3 = 30

• 15x = 30

• x = 2

Therefore, molecular formula = C2H6. This is ethane

### Importance of Graham's law:

1. It is applied in the separation of gases with different densities.
2. It is applied in the separation of elements with the same atomic number but different mass number (isotopes).
3. To determine the densities and molecular masses of unknown gases by comparing their rates of diffusion with known gases.

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