Graham's law of effusion

Result

Graham's law of effusion

T-022Home CU-104Threads thermo

Statement

Effusion — the escape of gas molecules, one at a time, through a hole small enough that molecules pass through individually without colliding near it — occurs at a rate proportional to the molecules' mean speed, which maxwell-boltzmann-speeds established scales as \(1/\sqrt{M}\) at fixed temperature. Comparing two gases at the same temperature and pressure, \(\dfrac{\text{rate}_1}{\text{rate}_2}=\sqrt{\dfrac{M_2}{M_1}}\): the lighter gas always effuses faster, by a factor equal to the square root of the molar mass ratio.

Why it matters

This result is the direct, practical capstone of the unit's kinetic-theory thread: kinetic-theory-pressure and maxwell-boltzmann-speeds established that lighter molecules move faster at a given temperature, and Graham's law is exactly that fact made experimentally testable and practically exploitable — both as a historical technique for estimating unknown molar masses before precise mass spectrometry existed, and, far more consequentially, as the working principle behind gaseous diffusion enrichment of uranium hexafluoride, the method used to separate fissile \(^{235}\text{U}\) from \(^{238}\text{U}\) during the Manhattan Project and for decades of subsequent nuclear fuel production.

Hypotheses

The hole is small compared to the gas's mean free path (the average distance a molecule travels between collisions), so molecules pass through individually rather than as a bulk, collective flow.This is the defining condition of the effusion (or "Knudsen") regime, as opposed to ordinary hydrodynamic flow through a larger opening; if the hole is not small in this sense, gas escapes more like a fluid jet, governed by pressure-difference-driven flow physics entirely different from the molecular, one-at-a-time picture this result depends on (Fails without). The region beyond the hole is at negligible pressure (effectively vacuum), so there is no significant counter-flow of gas back through the hole.This simplifies the calculation to a one-directional flux; effusion into a region of comparable or higher pressure would require separately accounting for molecules crossing the hole in both directions.

Proof

\Phi = \frac14 n\langle v\rangle

The effusion flux (molecules crossing unit area of the hole per unit time) is a standard kinetic-theory result obtained by integrating the one-directional molecular flux over the Maxwell-Boltzmann velocity distribution (maxwell-boltzmann-speeds): only molecules with a velocity component toward the hole contribute, and integrating this contribution over the full distribution yields exactly one-quarter of the number density \(n\) times the mean speed \(\langle v\rangle\). A

\langle v\rangle = \sqrt{\frac{8RT}{\pi M}}

The mean speed, already derived in maxwell-boltzmann-speeds, depends on temperature and molar mass in exactly this way. A

\Phi \propto \frac{1}{\sqrt{M}} \quad (\text{fixed } n, T)

Substituting Step 2 into Step 1, and holding number density and temperature fixed (comparing two gases under identical conditions), the effusion flux depends on molar mass only through the \(1/\sqrt{M}\) factor inherited from the mean speed. A

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

Taking the ratio of Step 3 for two different gases at the same temperature and pressure (hence, by the ideal gas law, the same number density \(n\)) cancels every factor except the molar masses, giving Graham's law in its standard comparative form. A

Result

\Phi = \frac14n\langle v\rangle \propto \frac{1}{\sqrt{M}}, \qquad \frac{\text{rate}_1}{\text{rate}_2}=\sqrt{\frac{M_2}{M_1}}

Reading. Whichever gas has the smaller molar mass effuses faster, by a factor equal to the square root of the ratio of molar masses — a direct, quantitative, experimentally testable consequence of the kinetic-theory result that lighter molecules move faster at a given temperature.

Scope. Strictly an effusion (small-hole, Knudsen-regime) result; the comparative ratio form requires both gases to be compared at the same temperature and pressure. Thomas Graham's original 1830s–40s experiments were actually on diffusion (movement through a porous barrier or through another gas, a related but not identical process); the precise \((1/4)n\langle v\rangle\) effusion-flux derivation used here came later, once Maxwell's speed distribution made it possible.

Corollaries & converses

Time to effuse a fixed amount of gas is inversely related to rate, so \(\dfrac{t_1}{t_2}=\sqrt{\dfrac{M_1}{M_2}}\) — the molar-mass ratio inside the square root is inverted compared to the rate formula, a frequent and important source of sign confusion (Common errors).
Effusion provides only a small enrichment factor per single pass when the two molar masses are close (as for isotopically distinct forms of the same compound); achieving substantial separation requires cascading many effusion stages in sequence, with the overall enrichment factor equal to the single-stage factor raised to the power of the number of stages (Worked examples, Problems).
Converse: measuring the ratio of effusion rates (or times) for an unknown gas against a gas of known molar mass allows the unknown molar mass to be determined directly from Step 4, rearranged — a historically important technique for molar-mass determination before mass spectrometry became routine.

Fails without

Apply Graham's law to a large opening rather than a small pinhole (violate Hypotheses' small-hole condition): gas escaping through a wide-open valve or large opening flows collectively as a fluid, governed by pressure-difference-driven hydrodynamic flow physics (more akin to Bernoulli's principle or Poiseuille flow) rather than by individual molecular effusion — the simple \(1/\sqrt{M}\) dependence derived here does not apply to that regime at all.
Compare two gases at different temperatures or pressures without accounting for the difference: Step 4's simple ratio form specifically assumed matched \(T\) and \(P\) between the two gases (so that \(n\) cancels); comparing effusion rates measured under different conditions without correcting for this would incorrectly attribute a temperature- or pressure-driven rate difference entirely to the molar-mass ratio.

Common errors

Confusing rate with time and using the wrong molar-mass ratio orientation — rate is proportional to \(1/\sqrt{M}\) (lighter gas, faster rate), while time is proportional to \(\sqrt{M}\) (lighter gas, shorter time) — directly the Corollaries' first bullet, and one of the single most common errors in Graham's law problems.
Applying the law to a large opening rather than genuine effusion through a small hole (Fails without, first bullet).
Forgetting that the comparative ratio form (Step 4) requires both gases to be at the same temperature and pressure; applying it across mismatched conditions without correction (Fails without, second bullet).

Discussion

Thomas Graham established the empirical relationship between gas diffusion/effusion rate and (inverse square root of) density or molar mass through careful experimentation in the 1830s and 1840s, decades before kinetic theory existed to explain why it should hold; this is the same recurring historical pattern seen throughout this network's content — ideal-gas-law's empirical assembly by Clapeyron, the VSEPR and molecular-orbital-theory results in the bonding unit — where a robust empirical regularity was discovered and used practically long before its underlying molecular mechanism was understood. That mechanism arrived specifically once Maxwell's 1860 speed distribution (maxwell-boltzmann-speeds) made the precise \(\Phi=\tfrac14n\langle v\rangle\) effusion-flux derivation possible, roughly thirty years after Graham's original experimental work.

The most consequential real-world application of Graham's law is gaseous diffusion enrichment of uranium hexafluoride (\(\text{UF}_6\)) gas, used industrially from the Manhattan Project through the late 20th century to separate the fissile isotope \(^{235}\text{U}\) from the far more abundant \(^{238}\text{U}\). Because both isotopes form chemically identical \(\text{UF}_6\) molecules differing only in the uranium atom's mass, the molar masses (\(\approx349.03\) and \(\approx352.04\,\text{g/mol}\) respectively) differ by less than \(1\%\), giving a single-stage enrichment factor of only about \(1.0043\) (Worked example 2) — explaining why real enrichment facilities required cascades of many hundreds to thousands of effusion stages in series to achieve weapons- or reactor-grade enrichment levels from natural uranium's \(0.7\%\) \(^{235}\text{U}\) abundance.

Common misconception: that effusion and diffusion are simply two names for the same process. Effusion specifically describes molecules passing individually through a small hole into a vacuum or near-vacuum; diffusion describes the (generally slower, collision-mediated) spreading of one gas through another gas already present, or through a porous medium — related processes with similar mass dependence, but not physically identical, a distinction Graham's own historical terminology sometimes blurred (Result's scope note).

Worked examples

\text{H}_2\ (M=2.016)\ \text{vs.}\ \text{O}_2\ (M=32.00),\ \text{same } T,P: \quad \frac{\text{rate}(\text{H}_2)}{\text{rate}(\text{O}_2)}=\sqrt{\frac{32.00}{2.016}}\approx3.98

Hydrogen effuses roughly \(4\) times faster than oxygen at the same temperature and pressure — the classic textbook illustration of Graham's law, with the ratio close to (but not exactly) \(4\) since \(M(\text{H}_2)=2.016\), not exactly \(2\). A

^{235}\text{UF}_6\ (M\approx349.03)\ \text{vs.}\ ^{238}\text{UF}_6\ (M\approx352.04): \quad \frac{\text{rate}(^{235}\text{UF}_6)}{\text{rate}(^{238}\text{UF}_6)}=\sqrt{\frac{352.04}{349.03}}\approx1.0043

The historically real single-stage enrichment factor for uranium hexafluoride gaseous diffusion, directly reproducing the well-documented figure; the smallness of this factor (well under half a percent) is exactly why isotope separation by this method required so many cascaded stages in practice. A

\text{rate}(\text{H}_2)/\text{rate}(\text{O}_2)\approx3.98; \qquad \text{single-stage }^{235}\text{UF}_6/^{238}\text{UF}_6\text{ enrichment}\approx1.0043

Reading. The same formula spans an enormous range of practical significance, from a dramatic, easily observed factor-of-4 laboratory demonstration to an extremely small but industrially decisive isotope-separation factor.

Scope. Both calculations require only the two molar masses, at matched temperature and pressure, per the Result's scope note.

Problems

An unknown gas is found to effuse at \(0.25\times\) the rate of hydrogen gas (\(M=2.016\,\text{g/mol}\)) under identical conditions. Determine the unknown gas's molar mass and suggest a plausible identity.
Solution
From \(\dfrac{\text{rate}_{\text{unk}}}{\text{rate}_{\text{H}_2}}=\sqrt{\dfrac{M_{\text{H}_2}}{M_{\text{unk}}}}\), rearranging: \(M_{\text{unk}}=\dfrac{M_{\text{H}_2}}{(\text{ratio})^2}=\dfrac{2.016}{(0.25)^2}\approx32.3\,\text{g/mol}\) — closely matching the molar mass of \(\text{O}_2\) (\(32.00\,\text{g/mol}\)), a plausible identification.
Gas A (\(M=16.0\,\text{g/mol}\)) takes \(4.0\,\text{min}\) to effuse a given volume through a small hole. Gas B has molar mass \(M=64.0\,\text{g/mol}\) (four times that of A). Find how long gas B takes to effuse the same volume under identical conditions.
Solution
Using the Corollaries' time relationship (inverse of the rate ratio): \(\dfrac{t_B}{t_A}=\sqrt{\dfrac{M_B}{M_A}}=\sqrt{4}=2\). So \(t_B=2\times4.0=8.0\,\text{min}\) — note that gas B, being heavier, takes longer (not shorter) to effuse the same amount, exactly the inverse of the rate relationship, and precisely the distinction flagged in Common errors.
Using the single-stage \(^{235}\text{UF}_6\) enrichment factor \(\approx1.0043\) from Worked example 2, estimate how many cascaded effusion stages would be needed to achieve an overall enrichment factor of \(1.5\times\) (i.e. increasing the \(^{235}\text{U}\) fraction relative to \(^{238}\text{U}\) by a factor of \(1.5\)).
Solution
Each stage multiplies the enrichment by the single-stage factor, so after \(n\) stages the overall factor is \((1.0043)^n\). Setting \((1.0043)^n=1.5\) and solving: \(n=\dfrac{\ln(1.5)}{\ln(1.0043)}\approx95\) stages — and real historical enrichment facilities, needing far more than a \(1.5\times\) enrichment to reach weapons- or reactor-grade material from natural uranium, required proportionally many more stages, consistent with the historically documented scale (hundreds to thousands of stages) of actual gaseous diffusion plants.
Explain, without further calculation, why Graham's law would not correctly predict the relative rates at which two gases escape from a container through a hole roughly the size of a coin (rather than a microscopic pinhole).
Solution
Graham's law depends on the effusion (Knudsen) regime specifically, requiring the hole to be small compared to the gas's mean free path so that molecules pass through individually without colliding with one another near the opening (Hypotheses). A coin-sized hole is vastly larger than any ordinary gas's mean free path, so gas escaping through it behaves as a bulk, collective hydrodynamic flow (driven by the pressure difference across the opening, more akin to air rushing out of a punctured tyre) rather than as individual molecular effusion events — a physically different process with a different (and generally much weaker or entirely different) dependence on molar mass, exactly the scope restriction flagged in Fails without.