Graham's Law of Diffusion

Graham's law of diffusion describes how quickly different gases spread out and mix — diffusion — or escape

through a tiny pinhole into a vacuum — effusion. It states that, 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:

Comparing two gases under the same conditions gives the more directly useful form:

A lighter gas (smaller $M$) diffuses faster than a heavier gas (larger $M$) — this follows directly from the

kinetic theory, since at a given temperature all gases have the same average kinetic energy

($\tfrac{1}{2}Mv^2_{\text{rms}} = \tfrac{3}{2}RT$, the same for every gas), so a smaller molar mass $M$ must be

compensated by a larger root-mean-square speed $v_{\text{rms}}$ for the kinetic energy to stay the same — and a

faster average speed means faster diffusion.

Worked example — comparing $\text{H}_2$ and $\text{O}_2$. Hydrogen ($M = 2\ \text{g mol}^{-1}$) and oxygen

($M = 32\ \text{g mol}^{-1}$), at the same temperature and pressure:

Hydrogen diffuses four times faster than oxygen — a large, easily measurable difference that follows purely

from the ratio of their molar masses, sixteen-fold, being a perfect square.

Worked example — identifying an unknown gas. An unknown gas $X$ diffuses twice as fast as sulfur dioxide

($\text{SO}_2$, $M = 64\ \text{g mol}^{-1}$) under identical conditions, so $\dfrac{r_X}{r_{\text{SO}_2}} = 2$.

Using Graham's law: $2 = \sqrt{\dfrac{64}{M_X}}$. Squaring both sides: $4 = \dfrac{64}{M_X}$, so

$M_X = \dfrac{64}{4} = 16\ \text{g mol}^{-1}$ — the molar mass of methane, $\text{CH}_4$ ($12 + 4\times1 = 16$),

identifying $X$ as methane (or another gas of the same molar mass).

Graham's law is exploited industrially in isotope enrichment — for instance, separating the fissile isotope …