Graham’s Law of Diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its density or molecular mass. This means that lighter gases diffuse faster than heavier gases under the same conditions of temperature and pressure.
Mathematically, the law is expressed as:
$$\text{Rate of diffusion} \propto \frac{1}{\sqrt{M}}$$
or comparing two gases:
$$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$$
where:
\(r_1\) and \(r_2\) = rates of diffusion of gases 1 and 2 respectively, and
\(M_1\) and \(M_2\) = their respective molecular masses (or densities).
Illustration:
If hydrogen (H2) and oxygen (O2) are allowed to diffuse under the same conditions, their rates of
diffusion can be compared using the formula:
$$\frac{r_{\text{H}_2}}{r_{\text{O}_2}} = \sqrt{\frac{M_{\text{O}_2}}{M_{\text{H}_2}}} = \sqrt{\frac{32}{2}} = 4$$
This means hydrogen diffuses four times faster than oxygen.
Example 1
If the rate of diffusion of gas X is half that of hydrogen under the same conditions, calculate the relative molecular mass (RMM) of gas X.
Solution
Therefore, the molecular mass of gas X = 8.
Example 2
A gas Y diffuses three times faster than sulphur dioxide (SO₂). Find the molecular mass of Y. (MSO₂ = 64)
Solution
Hence, the molecular mass of Y is 7.11.
Example 3
The rate of diffusion of an unknown gas Z is 1.414 times that of nitrogen gas (N₂). Calculate the molecular mass of Z.
Solution
Therefore, the molecular mass of Z = 14.
Example 4
If oxygen diffuses at a rate of 10 cm³/s, find the rate of diffusion of hydrogen under the same conditions. (MO₂ = 32, MH₂ = 2)
Solution
Example 5
If equal volumes of methane (CH₄) and ethene (C₂H₄) diffuse through a porous plug, calculate the ratio of their rates of diffusion.
Solution
Example 6
A gas takes 64 seconds to diffuse through a porous plug, while nitrogen takes 48 seconds. Calculate the molecular mass of the gas.
Solution
Since rate is inversely proportional to time, $$\frac{r_1}{r_2} = \frac{t_2}{t_1} = \sqrt{\frac{M_2}{M_1}}$$ $$\frac{48}{64} = \sqrt{\frac{28}{M_1}}$$ $$\frac{3}{4} = \sqrt{\frac{28}{M_1}}$$ $$\frac{9}{16} = \frac{28}{M_1}$$ $$M_1 = \frac{28 \times 16}{9} = 49.8$$ Molecular mass of the gas ≈ 50.
Example 7
A gas diffuses four times slower than hydrogen. What is its molecular mass?
Solution
Hence, the molecular mass of the gas is 32.
Example 8
The rate of diffusion of chlorine gas is 0.75 times that of oxygen. Calculate the molecular mass of chlorine. (MO₂ = 32)
Solution
Therefore, the molecular mass of chlorine ≈ 57.
