Q.An unknown gas $X$ diffuses twice as fast as sulfur dioxide ($\text{SO}_2$, $M = 64\ \text{g mol}^{-1}$) under identical conditions. Calculate the molar mass of $X$ and suggest what it might be.

Graham's Law of Diffusion – From Intuition to Precision

Imagine you're in a room where someone opens a bottle of perfume at one end. You don't smell it instantly — it takes time for the perfume molecules to wander across the room. Now imagine the same experiment with a bottle of ammonia. You'd smell the ammonia much faster. Why? The ammonia molecules are lighter.

That's the core physical idea: lighter gas molecules move faster, on average, than heavier ones at the same temperature. Since diffusion and effusion are processes driven by molecular motion, a lighter gas will spread out (diffuse) or escape through a tiny hole (effuse) more quickly than a heavier gas.

The Precise Statement

Graham's Law of Diffusion/Effusion states:

At constant temperature and pressure, the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass (or density).

Mathematically, for two gases A and B:

where:

• $r$ = rate of diffusion/effusion (volume or moles per unit time)
• $M$ = molar mass
• $\rho$ = density (at same T and P)

Why the Square Root? (The Physics)

The reason comes from kinetic molecular theory. At a given temperature, the average kinetic energy of gas molecules is the same for all gases:

Here $m$ is the mass of one molecule and $v$ is its speed. Rearranging:

Since molar mass $M$ is proportional to molecular mass $m$, the average molecular speed is inversely proportional to $\sqrt{M}$. And since the rate of diffusion/effusion is directly proportional to this average speed, you get Graham's Law.

Worked Example

Problem: Hydrogen ($M = 2\ \text{g/mol}$) and oxygen ($M = 32\ \text{g/mol}$) are allowed to effuse through identical pinholes. How much faster does hydrogen effuse?

Solution:

Hydrogen effuses 4 times faster than oxygen. …