Effusion and Graham's Law

In 1945, the Manhattan Project needed a way to separate uranium-235 from the far more abundant uranium-238. The two isotopes are chemically identical, so no reaction can tell them apart. Engineers instead pumped uranium hexafluoride gas through thousands of porous barriers. This process, called effusion, exploits a tiny speed difference between molecules of slightly different mass. Effusion is the escape of gas molecules through a hole so small that molecules leave one at a time, without colliding on the way out. Graham's law tells us the escape rate depends only on molar mass and temperature. It follows directly from the kinetic theory of gases, using the same molecular speed distribution that gives us the mean and root-mean-square speeds. This article derives the effusion rate from the Maxwell-Boltzmann distribution. It states Graham's law precisely. It shows how tiny mass differences compound over many stages into useful separation, using real isotope-enrichment numbers you can check by hand.

Key takeaways

Effusion is molecular escape through a hole smaller than the mean free path, one molecule at a time.
The effusion flux is proportional to the mean speed and the number density: $Z_W = \tfrac14 n\bar{c}$ .
Graham's law: effusion rate is inversely proportional to the square root of molar mass, at fixed $T$ and $p$ .
Separating isotopes needs many effusion stages because the single-stage separation factor is close to 1.
Effusion (through a pinhole) differs sharply from bulk flow (through a wide pipe), which depends on viscosity, not molar mass.
Key idea: the rate a gas leaks through a tiny hole is set by how many molecules arrive per second, and that traffic rate scales as $1/\sqrt{M}$ .

The core insight

Effusion rate is controlled by the flux of molecules hitting an area from one side, a quantity fixed by kinetic theory as $\tfrac14 n\bar c$ ; since $\bar c \propto 1/\sqrt M$ , lighter molecules stream through a small hole faster than heavier ones at the same temperature and pressure, which is exactly Graham's law.

The physical picture

Imagine a rigid box of gas with a tiny hole in one wall — small compared with the mean free path of the molecules, and small compared with the container itself. Molecules move in straight lines between collisions, following the Maxwell-Boltzmann speed distribution. Some of them, purely by the geometry of random motion, are heading toward the hole. If the hole is small enough, a molecule that happens to be aimed at it exits without ever colliding with another molecule near the opening. This is the effusive regime, and it applies whenever the hole diameter is much smaller than the mean free path $\lambda$ . Contrast this with the everyday leak of air from a punctured tire, where the hole is large compared to $\lambda$ . There, gas gushes out as a bulk flow — a fluid current governed by pressure gradients and viscosity, not by individual molecular trajectories. Bulk flow does not obey Graham's law: the mass dependence is different, and viscosity (which depends weakly on molar mass and more on molecular size and intermolecular forces) enters instead. The key quantity in effusion is the collision flux — the number of molecules striking a unit area of wall per unit time from one side. Kinetic theory gives this a clean form, $\tfrac14 n\bar c$ , derived by integrating over all molecules with a velocity component toward the wall, weighted by that velocity component, across the full three-dimensional Maxwell-Boltzmann distribution. Because the mean speed $\bar c$ depends on $\sqrt{T/M}$ , so does the effusion rate. Heavier molecules move more sluggishly on average, so fewer of them reach the hole per second.

Figure

Effusion versus bulk flow through a wall

Figure

Effusion rate ratio versus molar mass, at fixed T

The core relations

Z_W = \frac{1}{4} n \bar{c}

where: $Z_W$ is the effusive flux (collisions per unit area per unit time), $n$ is the number density, $\bar{c}$ is the mean speed.

\bar{c} = \sqrt{\frac{8RT}{\pi M}}

where: $R$ is the gas constant, $T$ is absolute temperature, $M$ is molar mass.

\text{Rate of effusion} \; \propto \; \frac{p}{\sqrt{MT}} \times T = \frac{p}{\sqrt{M}}\sqrt{\frac{1}{T}}\, ,\qquad \frac{dN}{dt} = -\frac{Z_W A}{1}

where: $p$ is pressure, $A$ is the hole area; this combines $n = p/k_BT$ with $\bar c$ .

\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}} \qquad (\text{Graham's law, same } T, p)

where: subscripts 1, 2 label two gases (or isotopologues) effusing under identical conditions.

\alpha = \sqrt{\frac{M_2}{M_1}}

where: $\alpha$ is the single-stage separation factor for isotopic enrichment, with $M_2 > M_1$ .

The core relations

Derivation AFrom the Maxwell-Boltzmann distribution to Graham's law

Goal: derive the effusive flux $Z_W$ and hence the mass dependence of the effusion rate, starting from the three-dimensional Maxwell-Boltzmann speed distribution.

Step 1. Set up the flux integral. Consider a small hole of area $A$ in a container wall, with the $x$ -axis normal to the wall. A molecule with velocity component $v_x > 0$ moving toward the hole will exit within time $dt$ if it lies within a slab of thickness $v_x\,dt$ behind the hole. The number of such molecules is $n f(v_x)\,dv_x \cdot A v_x\,dt$ , where $f(v_x)$ is the one-dimensional Maxwell-Boltzmann velocity distribution,

f(v_x) = \left(\frac{m}{2\pi k_BT}\right)^{1/2} \exp\!\left(-\frac{mv_x^2}{2k_BT}\right).

Step 2. Integrate over all outward-moving molecules. The flux (molecules per unit area per unit time) is obtained by integrating $v_x f(v_x)$ over $v_x$ from 0 to $\infty$ (only positive $v_x$ contributes to escape):

Z_W = n\int_0^\infty v_x f(v_x)\,dv_x = n\left(\frac{m}{2\pi k_BT}\right)^{1/2}\int_0^\infty v_x \exp\!\left(-\frac{mv_x^2}{2k_BT}\right)dv_x.

The integral is a standard Gaussian-type integral equal to $k_BT/m$ , giving

Z_W = n\left(\frac{k_BT}{2\pi m}\right)^{1/2}.

Step 3. Recognize the mean speed. The mean speed from the full three-dimensional distribution is

\bar{c} = \left(\frac{8k_BT}{\pi m}\right)^{1/2} = \sqrt{\frac{8RT}{\pi M}},

so that $Z_W$ can be rewritten compactly using $\bar c$ :

Z_W = \frac{1}{4}n\bar{c}.

This is the central result: the effusive flux is exactly one-quarter of the product of number density and mean speed — a purely geometric factor arising from averaging $v_x$ over a hemisphere of directions.

Step 4. Convert to a rate expressed via pressure. Using the ideal gas relation $n = p/k_BT$ , the rate of molecules effusing per unit time through hole area $A$ is

\frac{dN}{dt} = Z_W A = \frac{pA}{\sqrt{2\pi m k_BT}} = \frac{pA N_A}{\sqrt{2\pi M R T}}.

The rate of effusion, measured as moles per second, is therefore proportional to $p/\sqrt{MT}$ .

Step 5. Compare two gases at identical $T$ and $p$ . For two gases (or two isotopologues of the same compound, so $p$ , $T$ , and hole geometry are identical), take the ratio:

\frac{(\text{rate})_1}{(\text{rate})_2} = \frac{p A/\sqrt{2\pi M_1 R T}}{p A/\sqrt{2\pi M_2 R T}} = \sqrt{\frac{M_2}{M_1}}.

All shared factors — $p$ , $A$ , $T$ , $R$ — cancel exactly, leaving only the mass ratio.

\boxed{\dfrac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\dfrac{M_2}{M_1}}}

Worked examples

Example 1 — Comparing hydrogen and oxygen effusion (foundational). Question: at 298 K, how much faster does H₂ effuse than O₂ through the same pinhole, at the same pressure? Molar masses: $M(\text{H}_2) = 2.016\ \text{g mol}^{-1}$ , $M(\text{O}_2) = 32.00\ \text{g mol}^{-1}$ . Apply Graham's law directly:

\frac{\text{Rate(H}_2)}{\text{Rate(O}_2)} = \sqrt{\frac{M(\text{O}_2)}{M(\text{H}_2)}} = \sqrt{\frac{32.00}{2.016}} = \sqrt{15.87} = 3.984.

So H₂ effuses about 3.98 times faster than O₂ at the same temperature and pressure. Sanity check: mean speeds scale the same way since $\bar c \propto 1/\sqrt M$ ; $\bar c(\text{H}_2) = \sqrt{8(8.314)(298)/(\pi \times 0.002016)} = 1782\ \text{m s}^{-1}$ , and $\bar c(\text{O}_2) = \sqrt{8(8.314)(298)/(\pi \times 0.03200)} = 447\ \text{m s}^{-1}$ . The ratio $1782/447 = 3.99$ , matching the mass-ratio calculation within rounding. This mass sensitivity is why helium balloons deflate faster than air-filled ones — He (4.00 g/mol) effuses through microscopic pores in rubber much faster than N₂ (28.0 g/mol), with a rate ratio of $\sqrt{28.0/4.00}=2.65$ .

Example 2 — Single-stage and cascade enrichment of uranium-235 (applied, harder). Uranium hexafluoride gas contains ²³⁵UF₆ (natural abundance 0.720%) and ²³⁸UF₆ (99.275%, ignoring trace ²³⁴U). Molar masses, using F = 18.998 g/mol (6 × 18.998 = 113.99 for the fluorine contribution):

M(^{235}\text{UF}_6) = 235.04 + 113.99 = 349.03\ \text{g mol}^{-1},

M(^{238}\text{UF}_6) = 238.05 + 113.99 = 352.04\ \text{g mol}^{-1}.

Step 1: single-stage separation factor.

\alpha = \sqrt{\frac{352.04}{349.03}} = \sqrt{1.00863} = 1.00431.

So a single porous-membrane stage enriches the light isotopologue by only 0.431%. This is a strikingly small number — it is why isotope separation by effusion is notoriously slow and energy-intensive. Step 2: the isotopic ratio after one stage. Starting ratio $R_0 = 0.00720/0.99275 = 7.253\times10^{-3}$ (mole ratio ²³⁵U to ²³⁸U). After one ideal stage, the exiting (lighter, permeate) stream has ratio

R_1 = \alpha R_0 = 1.00431 \times 7.253\times10^{-3} = 7.284\times10^{-3}.

Step 3: cascading $n$ stages. Since each ideal stage multiplies the ratio by $\alpha$ ,

R_n = \alpha^n R_0.

To reach a typical reactor-grade enrichment of about 4% ²³⁵U (mole ratio $R_n = 0.0400/0.9600 = 0.04167$ ), solve for $n$ :

n = \frac{\ln(R_n/R_0)}{\ln \alpha} = \frac{\ln(0.04167/0.007253)}{\ln(1.00431)} = \frac{\ln(5.746)}{0.004301} = \frac{1.749}{0.004301} \approx 407.

So roughly 407 ideal effusion stages are needed to go from natural to 4% enrichment — consistent with historical gaseous-diffusion cascades, which used well over a thousand real (non-ideal) stages because actual stage efficiencies fall short of the ideal $\alpha$ . Sanity check: since $\alpha$ is so close to 1, $\ln\alpha \approx (\alpha - 1) = 0.00431$ , and $1.749/0.00431 = 406$ , matching our more precise log calculation.

Limiting cases & common traps

Situation	What to remember
Hole size comparable to or larger than $\lambda$	Not effusion — this is viscous/bulk flow; Graham's law does not apply
Comparing gases at different $T$	Must include the $\sqrt{T}$ factor; Graham's law as stated assumes equal $T$
Comparing amount effused (moles) vs. rate	Effusion rate (mol/s) uses $1/\sqrt{M}$ ; total moles effused over finite time also needs $p(t)$ , which itself decays as gas leaves
Mixtures	Each component effuses independently, proportional to its own partial pressure and $1/\sqrt{M_i}$ — lighter components are preferentially depleted from the container over time
Isotope separation	Single stage factor $\alpha=\sqrt{M_2/M_1}$ is very close to 1 for heavy isotopes; cascades of many stages are required

The chemist behind it

Thomas Graham, a Scottish chemist working in Glasgow and later London in the 1830s–1840s, carried out careful quantitative experiments on gases escaping through fine tubes and plaster plugs. He measured how long it took equal volumes of different gases to effuse and found the times were proportional to the square root of density — equivalent, at fixed temperature, to molar mass. Graham published this relationship in 1846, decades before Maxwell and Boltzmann supplied the microscopic theory of molecular speeds that explains why it holds. Graham also pioneered the study of colloids and dialysis, coining the term "colloid," and his effusion tubes were a direct ancestor of the porous-barrier cascades used in 20th-century uranium enrichment.

Origins · 1 minThomas Graham and how this was discovered

Thomas Graham

1805-1869 · Scottish chemist who discovered the law of gaseous effusion decades before kinetic theory explained it

Ask the tutor: How would the required number of stages change if we wanted weapons-grade enrichment near 90% instead of 4%?

Why does Graham's law use molar mass rather than molecular size?

Effusion rate depends on how fast molecules travel toward the hole, not on how bulky they are. The mean speed from kinetic theory, $\bar c=\sqrt{8RT/\pi M}$ , depends only on temperature and mass — not on molecular diameter or shape. Diameter matters for collision frequency and mean free path (which set whether effusion or bulk flow applies), but once you are safely in the effusive regime, the escape rate itself is a pure mass effect.

Can Graham's law be used to separate isotopes of light elements like hydrogen or carbon more easily than uranium?

Yes — the separation factor $\alpha=\sqrt{M_2/M_1}$ is larger when the relative mass difference is bigger. For H2 versus D2, $\alpha=\sqrt{4.028/2.016}=1.414$ , dramatically larger than uranium's 1.0043. This is why early heavy-water production used gaseous diffusion-like methods effectively, while uranium enrichment needed hundreds of cascaded stages to achieve the same overall separation.

Does effusion rate depend on the size of the hole?

Yes, linearly: the total molar rate is $Z_W A$ , proportional to hole area $A$ , as long as the hole remains small compared to the mean free path so each molecule crosses independently. Doubling the hole area doubles the escaping flux, but the mass dependence embodied in Graham's law is unaffected because $A$ cancels when you take the ratio of two gases through the same hole.

How is effusion different from ordinary diffusion of one gas into another?

Effusion is escape into a vacuum or through an isolated small aperture, with no collisions right at the hole. Diffusion is the random-walk spreading of one gas through a background of other gas molecules, dominated by frequent intermolecular collisions. Both happen to show a $1/\sqrt M$ -like trend, but diffusion coefficients also depend on collision cross-sections and the identity of the surrounding gas, which effusion completely ignores.

Why did gaseous diffusion plants need over a thousand stages instead of the 407 predicted for ideal stages?

The 407-stage estimate assumes every stage achieves the full ideal separation factor $\alpha=\sqrt{M_2/M_1}$ with no mixing losses. Real membrane stages have imperfect cut fractions, back-mixing, and pressure losses, so their effective separation per stage is smaller than ideal. Historical cascades (like K-25 at Oak Ridge) needed thousands of real stages, plus enormous pumping power, to compensate for this inefficiency.

Effusion and Graham's Law

The physical picture

The core relations

The core relations

Worked examples

Limiting cases & common traps

The chemist behind it

Before this