Skip to main content

Graham's Law Calculator

Calculate the relative rate of effusion of two gases using Graham's law: rate1/rate2 = sqrt(M2/M1). The lighter gas effuses faster.

Common Gas Molar Masses
H2 (Hydrogen)2.016
He (Helium)4.003
CH4 (Methane)16.043
NH3 (Ammonia)17.031
Ne (Neon)20.180
N2 (Nitrogen)28.014
O2 (Oxygen)32.000
Ar (Argon)39.948
CO2 (Carbon dioxide)44.009
SO2 (Sulfur dioxide)64.066
Cl2 (Chlorine)70.906
SF6 (Sulfur hexafluoride)146.055

Graham’s law: effusion and molar mass

Graham’s law connects how fast a gas escapes through a small aperture to its molar mass. The relationship is rate1/rate2 = sqrt(M2/M1) — the lighter gas effuses faster, and the speed advantage scales as the square root of the mass ratio. Halve the molar mass and effusion gets 1.41× faster. Compare hydrogen (M = 2) to oxygen (M = 32) and you get sqrt(16) = 4×.

The square root comes straight out of kinetic molecular theory. At a given temperature every gas has the same average translational kinetic energy, so a lighter molecule has to move proportionally faster to carry that same energy at lower mass — and average speed goes as 1/sqrt(m). Effusion rate tracks the speed of the molecules hitting the aperture, so the rate ratio inherits the same square-root dependence.

This calculator takes two molar masses and returns the effusion-rate ratio along with which gas is faster. The math is one line, but the consequences run from helium leak detection to the entire architecture of the Manhattan Project’s enrichment cascades.

What the calculator does

  1. Enter molar masses of Gas 1 and Gas 2 in g/mol (gas names are optional, just for labeling).
  2. The calculator returns rate1/rate2 = sqrt(M2/M1) and identifies which gas is faster.
  3. A reference table lists common gases for quick comparison.

Worked examples

Hydrogen vs. oxygen.

  • M(H2) = 2.016, M(O2) = 32.000
  • rate(H2)/rate(O2) = sqrt(32.000/2.016) = sqrt(15.87) = 3.98
  • H2 effuses about 4× faster — the textbook benchmark case.

Helium vs. nitrogen (leak detection).

  • M(He) = 4.003, M(N2) = 28.014
  • rate(He)/rate(N2) = sqrt(28.014/4.003) = 2.65
  • He effuses 2.65× faster than air, which is exactly why leak detectors flood a vacuum chamber with He and watch the residual-gas spectrum.

UF6 isotopes (uranium enrichment).

  • M(235-UF6) = 349.034, M(238-UF6) = 352.041
  • rate(235)/rate(238) = sqrt(352.041/349.034) = 1.0043
  • A 0.43% separation per stage. That tiny number is why gaseous-diffusion plants needed thousands of cascaded stages and why centrifuges (which exploit a different mass-dependence) eventually replaced them.

Frequently Asked Questions

What is Graham's law of effusion?
Graham's law says the rate of effusion is inversely proportional to the square root of molar mass. For two gases at the same temperature, rate1/rate2 = sqrt(M2/M1). The lighter gas wins because at equal temperatures all gases have the same average kinetic energy, and lighter molecules have to move faster to carry that energy at lower mass.
What is the difference between effusion and diffusion?
Effusion is escape through a hole small enough that molecules pass one at a time — smaller than the mean free path of the gas. Diffusion is the bulk mixing of gases through one another. Graham's law is exact for effusion under ideal-gas assumptions; for diffusion it gives a reasonable first-order answer but real diffusion is complicated by molecule-molecule collisions and concentration gradients.
Why do lighter gases effuse faster?
Equipartition: at the same temperature, every gas has the same average translational kinetic energy, (1/2)mv^2 = (3/2)kT. Same KE, smaller mass, higher speed — and v scales as 1/sqrt(m). Faster molecules hit the aperture more often, so the effusion rate scales the same way. The square-root in Graham's law comes directly from this kinetic-theory result.
How was Graham's law used in uranium enrichment?
The Manhattan Project separated U-235 from U-238 by converting both to UF6 gas (the only volatile uranium compound) and running it through gaseous-diffusion barriers. The molar masses are 349 vs 352 g/mol, so the per-stage separation factor is sqrt(352/349) = 1.0043 — only 0.43% enrichment per pass. Achieving weapons-grade required thousands of stages in cascade.
Does Graham's law apply to gas mixtures?
Each component effuses independently at the rate set by its own molar mass; the law treats one gas at a time. This independence is the same idea that underlies Dalton's law of partial pressures: in an ideal-gas mixture each species behaves as if the others were not there. Identifying the limiting reagent in a mixture experiment uses this property.