Effusion and Diffusion

Open a bottle of ammonia in one corner of a room. Open a bottle of hydrochloric acid in another corner. A faint white ring of ammonium chloride will eventually appear in the air between them, closer to the acid. This everyday observation — gases spreading and mixing at different speeds — is governed by a strikingly simple rule called Graham's law. It connects something you can measure with a stopwatch (how fast a gas escapes or spreads) to something fundamental about the molecule itself (its mass). Graham's law is a direct, testable consequence of the Kinetic-Molecular Theory of Gases. That theory tells us that all gas molecules at a given temperature share the same average kinetic energy. This holds true regardless of identity. Because kinetic energy depends on both mass and speed, lighter molecules must move faster than heavier ones. This keeps their average kinetic energy the same. Graham's law simply turns that idea into an equation. You can use it to predict rates, compare isotopes, or estimate an unknown molar mass from a timed experiment. This article develops the physical picture behind effusion and diffusion. It derives Graham's law rigorously from kinetic theory. It also works through fully numeric examples. These include the classic method for finding an unknown gas's molar mass from its effusion rate relative to a known gas.

Key takeaways

Effusion is escape through a tiny hole into a vacuum; diffusion is spreading through another gas — both are governed by the same mass-dependence.
Graham's law: the rate of effusion is inversely proportional to the square root of molar mass.
Rates compare as $\text{rate}_1/\text{rate}_2 = \sqrt{M_2/M_1}$ — the lighter gas always effuses faster.
Graham's law follows directly from the Kinetic-Molecular Theory of Gases: equal average kinetic energy at a given temperature forces lighter molecules to move faster.
It is used practically to estimate unknown molar masses and to separate isotopes (historically, uranium-235 from uranium-238 via UF₆ gas).
Key idea: at a fixed temperature, effusion/diffusion rate scales as $1/\sqrt{M}$ , because all gases share the same average kinetic energy but not the same speed.

The core insight

Same temperature means same average kinetic energy per molecule, not the same speed — so heavier molecules must crawl while lighter ones sprint, and this speed gap is exactly what Graham's law quantifies.

The physical picture

Picture a rigid container of gas with one tiny pinhole in its wall, so small that a molecule wandering near it escapes into the vacuum outside without colliding with other molecules on the way out. This leaking-through-a-pinhole process is effusion. Contrast it with diffusion, where a gas is not escaping into empty space but slowly threading its way through the "traffic" of other gas molecules already filling a space — the ammonia and HCl scenario above is diffusion, not effusion. Despite this difference in setting, both processes have the same molecular root cause: molecular speed. A molecule can only effuse through the hole, or nudge its way through diffusion, at a rate tied to how fast it moves and how often it heads in the right direction. The Kinetic-Molecular Theory of Gases states that at a fixed temperature $T$ , every gas — regardless of molar mass — has the same average translational kinetic energy per molecule:

\bar{KE} = \tfrac{3}{2}k_BT

Since $\bar{KE} = \tfrac{1}{2}m\overline{v^2}$ is the same number for every gas at temperature $T$ , a heavier molecule ( $m$ large) must have a smaller mean-square speed $\overline{v^2}$ , and a lighter molecule must have a larger one. This is the entire physical content of Graham's law: light molecules move faster, so they effuse and diffuse faster. The rate of effusion through the pinhole is proportional to how many molecules strike the hole per second, which in turn is proportional to the molecules' average speed (faster molecules arrive at the hole more often). This lets us translate a statement about molecular speeds directly into a statement about macroscopically measurable escape rates.

Figure

Effusion: light gas (He) escapes faster than heavy gas (Ar) through the same pinhole

The core relations

\text{rate}_1 \propto \bar{v}_1, \qquad \bar{v} = \sqrt{\dfrac{3RT}{M}}

where: rate is the number of molecules effusing (or diffusing) per unit time, $\bar{v}$ is a characteristic molecular speed (root-mean-square speed), $R$ is the gas constant, $T$ is absolute temperature, and $M$ is molar mass.

\frac{\text{rate}_1}{\text{rate}_2} = \sqrt{\frac{M_2}{M_1}} \qquad (\text{same } T)

where: subscripts 1 and 2 label two different gases compared at the same temperature and, for effusion, the same pinhole geometry and pressure difference.

\frac{t_2}{t_1} = \sqrt{\frac{M_2}{M_1}}

where: $t_1, t_2$ are the times required for equal amounts (equal moles, or equal volumes at the same $P,T$ ) of gases 1 and 2 to effuse — time is inversely related to rate, so the ratio flips relative to the rate equation.

Derivation ADeriving Graham's Law from Kinetic-Molecular Theory

Goal. Show that the ratio of effusion rates of two gases at the same temperature equals the square root of the inverse ratio of their molar masses.

Step 1. Start from equal average kinetic energy. The Kinetic-Molecular Theory of Gases asserts that at temperature $T$ , every gas has the same average translational kinetic energy per mole:

\tfrac{1}{2}M_1\overline{v_1^2} = \tfrac{1}{2}M_2\overline{v_2^2} = \tfrac{3}{2}RT

Here $M_1, M_2$ are molar masses and $\overline{v^2}$ is the mean-square speed, so that $M\overline{v^2}$ has units of energy per mole.

Step 2. Solve each gas's mean-square speed for $M$ and $T$ .

\overline{v_1^2} = \frac{3RT}{M_1}, \qquad \overline{v_2^2} = \frac{3RT}{M_2}

Both expressions contain the same $3RT$ factor because $T$ is identical for the two gases being compared.

Step 3. Take the ratio to eliminate $T$ .

\frac{\overline{v_1^2}}{\overline{v_2^2}} = \frac{3RT/M_1}{3RT/M_2} = \frac{M_2}{M_1}

The temperature and the gas constant cancel completely, leaving a ratio that depends only on molar masses.

Step 4. Take square roots to get root-mean-square speed.

\frac{\sqrt{\overline{v_1^2}}}{\sqrt{\overline{v_2^2}}} = \sqrt{\frac{M_2}{M_1}}

The root-mean-square speed $v_{rms} = \sqrt{\overline{v^2}}$ is the natural speed scale to use here.

Step 5. Connect speed to effusion rate. The number of molecules striking a small hole per unit time is proportional to the molecule's mean speed (more precisely, to the mean speed component perpendicular to the hole, but this still scales the same way with $M$ as $v_{rms}$ does, since all these characteristic speeds carry the same $\sqrt{T/M}$ dependence). Writing rate $\propto v$ :

\frac{\text{rate}_1}{\text{rate}_2} = \frac{v_1}{v_2} = \sqrt{\frac{M_2}{M_1}}

This is Graham's law. Note the inversion: the lighter gas (smaller $M_2$ becomes a numerator inside the root when it is gas 2, or conversely gas 1's rate grows when $M_1$ is small) effuses faster — light molar mass in the denominator position under the root always corresponds to the faster gas.

Step 6. Convert to a time-ratio form. If equal moles of the two gases are allowed to effuse, and gas 1 takes time $t_1$ while gas 2 takes time $t_2$ , then rate $\propto 1/t$ , so

\frac{1/t_1}{1/t_2} = \frac{t_2}{t_1} = \sqrt{\frac{M_2}{M_1}}

This is often the more directly measurable version: you time how long it takes a fixed amount of gas to leak out, rather than measuring molecular speeds directly.

Worked examples

Example 1 — Comparing effusion rates of two named gases. Compare the effusion rates of hydrogen gas ( $H_2$ , $M = 2.016\ \text{g/mol}$ ) and methane ( $CH_4$ , $M = 16.04\ \text{g/mol}$ ) at the same temperature and pressure.

Step 1. Assign gas 1 = $H_2$ (lighter, expected faster) and gas 2 = $CH_4$ .

Step 2. Apply Graham's law:

\frac{\text{rate}_{H_2}}{\text{rate}_{CH_4}} = \sqrt{\frac{M_{CH_4}}{M_{H_2}}} = \sqrt{\frac{16.04}{2.016}}

Step 3. Compute the ratio inside the root:

\frac{16.04}{2.016} = 7.957

Step 4. Take the square root:

\sqrt{7.957} = 2.821

So hydrogen effuses about 2.82 times faster than methane at the same temperature. This is a genuinely asymmetric, non-round comparison — not $H_2$ vs. $D_2$ or some contrived 4:1 mass ratio — and the number checks against intuition: $H_2$ is roughly 8 times lighter, and speed scales as the square root of that mass ratio, so a factor near 2.8 (since $\sqrt{8} \approx 2.83$ ) is exactly what should appear.

Limiting-case check. If the two gases had identical molar mass ( $M_1 = M_2$ ), the ratio under the root becomes 1, giving rate $_1$ /rate $_2 = 1$ — the gases effuse at identical rates, as physically required, since kinetic theory gives them the same rms speed at equal $T$ and equal $M$ .

Example 2 — Finding an unknown molar mass from a timed effusion experiment. A sample of pure $N_2$ gas ( $M = 28.02\ \text{g/mol}$ ) takes 84.0 seconds to effuse completely through a pinhole. Under otherwise identical conditions (same apparatus, temperature, pressure, amount of gas), an unknown gas takes 142.0 seconds to effuse. Find the molar mass of the unknown gas.

Step 1. Assign gas 1 = $N_2$ (known, $t_1 = 84.0\ \text{s}$ , $M_1 = 28.02$ ), gas 2 = unknown ( $t_2 = 142.0\ \text{s}$ , $M_2 = ?$ ).

Step 2. Use the time-ratio form of Graham's law:

\frac{t_2}{t_1} = \sqrt{\frac{M_2}{M_1}}

Step 3. Substitute known numbers:

\frac{142.0}{84.0} = \sqrt{\frac{M_2}{28.02}}

1.690 = \sqrt{\frac{M_2}{28.02}}

Step 4. Square both sides to remove the root:

1.690^2 = \frac{M_2}{28.02}

2.857 = \frac{M_2}{28.02}

Step 5. Solve for $M_2$ :

M_2 = 2.857 \times 28.02 = 80.06\ \text{g/mol}

The unknown gas has a molar mass of about 80.1 g/mol — close to the molar mass of $SO_3$ (80.06 g/mol) or $Br$ atomic contexts, though identifying the actual substance would need further chemical evidence (only the mass is pinned down by this measurement).

Limiting-case check. As the unknown gas becomes very heavy ( $M_2 \to \infty$ ), the predicted time $t_2 \to \infty$ as well, since $t_2 = t_1\sqrt{M_2/M_1}$ grows without bound — a heavier and heavier gas takes longer and longer to effuse, consistent with sluggish, slow-moving heavy molecules rarely finding the pinhole.

Limiting cases & common traps

Situation	What happens	Why
Equal molar masses ( $M_1 = M_2$ )	rate ratio = 1	No mass advantage; both gases have identical $v_{rms}$ at same $T$
$M_2 \gg M_1$	rate $_1$ /rate $_2 \to$ very large	Gas 1 is overwhelmingly faster; heavy gas 2 crawls
Comparing at different temperatures	Simple Graham's law ratio invalid	The derivation explicitly cancels $T$ only because $T$ was assumed equal
Confusing rate ratio with time ratio	Sign/placement of $M_1, M_2$ flips	rate $\propto 1/t$ , so the fraction inverts between the two forms
Using molecular mass instead of molar mass in mixed units	Wrong numeric answer	Both gases' $M$ must be in the same units (g/mol is standard)

The chemist behind it

Thomas Graham was a Scottish chemist working in the 1830s and 1840s who noticed, through careful measurement, that the rate at which different gases escaped through porous plugs or fine capillary openings tracked systematically with something related to their densities — and since density at fixed conditions is proportional to molar mass, this was really a molar-mass effect. He formulated his law purely empirically, from painstaking timing experiments, well before the kinetic-molecular theory existed to explain why it worked. It was only decades later, once the kinetic theory of gases matured through the work of Maxwell, Boltzmann, and others, that Graham's empirical rule was shown to fall directly out of the equal-kinetic-energy postulate — turning a useful measurement trick into a predictable consequence of molecular motion.

Origins · 1 minThomas Graham and how this was discovered

Thomas Graham

1805–1869 · Scottish chemistry, gas diffusion/effusion studies and colloid chemistry

Ask the tutor: If I mix helium and neon in a closed flask with a pinhole and let them effuse together, does the gas that escapes first get enriched in helium, and by roughly what factor compared to the original mixture?

Is effusion the same thing as diffusion?

No. Effusion is escape through a small hole into a vacuum (or a region with essentially no gas to bump into), while diffusion is the spreading of one gas through a region already occupied by another gas. Both obey the same $1/\sqrt{M}$ mass dependence because both rates trace back to molecular speed, but diffusion is also slowed by molecular collisions along the way, making its treatment more complex in detail.

Why does the lighter gas always win the race?

Because the Kinetic-Molecular Theory of Gases requires all gases at a given temperature to share the same average kinetic energy. Since kinetic energy involves both mass and the square of speed, a molecule with smaller mass must compensate with a larger speed to keep the energy the same as a heavier molecule's.

Can Graham's law be used to separate isotopes?

Yes — historically this was one of its most important applications. Uranium hexafluoride gas made from the lighter isotope $^{235}U$ effuses very slightly faster than the version made from $^{238}U$ , because the molar masses differ only slightly. Repeating the effusion process through many stages amplifies this tiny speed difference into a usable separation, which was the basis of gaseous diffusion enrichment plants.

Does Graham's law depend on pressure?

The ratio of rates for two gases compared under the same conditions does not depend on the absolute pressure, since pressure affects both gases' rates in the same way and cancels in the ratio. However, the individual (absolute) effusion rate of a single gas does increase with the pressure difference driving molecules through the hole.

What if the two gases are compared at different temperatures?

Then the simple ratio $\sqrt{M_2/M_1}$ no longer applies, because the derivation explicitly relied on $T$ being identical for both gases so that the $3RT$ factor could cancel. In that case you must go back to $v \propto \sqrt{T/M}$ and keep both temperature and mass in the ratio.