Diffusion Of Gases

1.0What is Diffusion?

Diffusion is the movement of a substance from an area of high concentration to an area of low concentration. This process occurs faster in gases and liquids than in solids because their particles move more freely and randomly.

The word "diffusion" comes from the Latin word diffundere, meaning "to spread out."

Increasing temperature or pressure speeds up the rate of diffusion.

Diffusion of Gases Gases have a remarkable property of spreading out and intermixing throughout the space available to them, even against gravity.

For example, if a bottle of ammonia is opened in one corner of a room, its smell can be detected in the opposite corner.

Let’s understand this with an example: 

Suppose we have two gas jars — one containing air (colourless) and the other containing reddish-brown bromine gas. If we invert the air over the bromine jar and remove the cover between them, the gases begin to intermix. Over time, both jars appear reddish-brown.

Even though bromine is heavier than air, it travels upward while the lighter air moves downward. This happens due to the rapid movement of gas molecules and the presence of intermolecular spaces in gases. The high velocities of gas molecules allow them to overcome gravitational forces and mix with each other.

Thus, diffusion is defined as: The spontaneous intermixing of two or more gases, even against gravity.

2.0Examples Of Diffusion of Gases in Daily Life

• The fragrance of perfume can be smelled from a distance.
• The smell of a burning incense stick spreads throughout the room.
• The aroma of food cooking travels through the house.

3.0Graham’s Law of Diffusion 

Thomas Graham studied the diffusion of gases and formulated a relationship between the rate of diffusion and the density of a gas. This is known as Graham’s Law of Diffusion, which states:

At constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density.

Mathematically:

$r\propto \frac{1}{\sqrt{d}}$

Where:

• r = rate of diffusion
• d = density of the gas

If r₁ and r₂ are the rates of diffusion of two gases with densities d₁ and d₂ respectively, then:

$\frac{r_1}{r_2}=\sqrt{\frac{d_2}{d_1}}$

Since the molecular mass (M) of a gas is twice its vapor density:

$\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}$

Where M₁ and M₂ are the molecular masses of the two gases.

4.0Rate Of Diffusion

The rate of diffusion of a gas is the ratio of the volume of the gas diffused to the time taken for diffusion:

$r=\frac{v}{t}$

Where V is the volume diffused and t is the time taken. If two gases diffuse for the same time, their relative rates of diffusion can be compared.

Effusion is a specific type of diffusion in which gas particles pass through a fine orifice or pinhole without colliding with each other.

5.0Applications of Gaseous Diffusion

• Fractional diffusion is used to separate gases with different densities, such as separating isotopes in a gaseous mixture.
• Graham’s law helps determine the density or molecular mass of an unknown gas by comparing its rate of diffusion to that of a known gas.
• Diffusion helps disperse harmful gases into the atmosphere, diluting their effects and keeping the air homogeneous.
• An example of this is how carbon dioxide and oxygen diffuse into water, providing essential gases for aquatic plants and animals. Plants use dissolved CO₂ for photosynthesis, while animals rely on dissolved oxygen for respiration.

6.0Solved Examples

Example 1: The rate of diffusion of hydrogen gas (H₂) and an unknown gas is 3:1, measured under the same temperature and pressure conditions. What is the molecular mass of the unknown gas?

Given:

• Rate of diffusion of H₂ : Rate of diffusion of unknown gas = 3 : 1
• Molecular mass of H₂ = 2 g/mol

Using Graham’s law of diffusion: $\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}$

Where:

• r₁ and r₂​ are the rates of diffusion of the gases.
• M₁​ is the molecular mass of H_2​.
• M_2​ is the molecular mass of the unknown gas.

Substitute the given values: $\frac{3}{1}=\sqrt{\frac{M_2}{2}}$

Square both sides: $9=\frac{M_2}{2}$

Solve for M_2​:$M_2=9\times 2=18g/mol$

Answer: The molecular mass of the unknown gas is 18 g/mol.

Example 2: What will be the ratio of the rate of diffusion of CO₂:CO if the initial moles of CO₂ and CO are 4 mol and 5 mol, respectively?

Solution:
Graham’s Law of Diffusion, states: $\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}$

Where:

• r₁ and r₂​ are the rates of diffusion of the gases.
• M₁ and M_2​ are the molar masses of the gases.

Given:

• Molar mass of CO₂=44 g/mol
• Molar mass of CO=28 g/mol

Using Graham’s law:

$\frac{r_{CO2}}{r_{CO}}=\sqrt{\frac{28}{44}}=\sqrt{0.636}\approx 0.797$

Answer: The ratio of the rates of diffusion:

$r_{CO2}:r_{CO}=0.797:1\approx 4:5$