This Graham’s law calculator compares how fast two gases effuse (or diffuse): rate₁/rate₂ = √(M₂/M₁). Lighter molecules move faster at the same temperature — enter two molar masses for the rate ratio, or one mass against a known partner to solve for an unknown gas.

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Why the Square Root Rules Gases

At a given temperature, average molecular kinetic energy is fixed (½mv² = ³⁄₂RT per mole) — so velocity must fall as mass rises, specifically as 1/√M: v(rms) = √(3RT/M). Effusion rate is proportional to molecular speed (how often molecules find the hole), and Graham’s law is that statement as a ratio: rate₁/rate₂ = √(M₂/M₁). The defaults pit helium (4.003) against oxygen (31.998): √7.99 = 2.83× — helium effuses nearly three times faster, the physics behind helium balloons deflating overnight through “sealed” latex (the pores are effusion holes at molecular scale) while oxygen-filled balloons keep their shape for days. The square root cuts both ways as a design constraint: a 4× mass ratio buys only 2× speed — gases are hard to separate by speed alone.

The law’s industrial biography is uranium enrichment: UF₆ gas from ²³⁵U (M ≈ 349) vs ²³⁸U (M ≈ 352) effuses at ratio √(352/349) = 1.0043 per pass — a 0.43% enrichment per stage, cascaded through thousands of stages at Oak Ridge. The same ratio, read backwards, identifies unknown gases: measure a gas’s effusion time against a known standard and M(unknown) = M(known) × (t(unknown)/t(known))² — the classical molar-mass measurement this calculator’s time input supports. Diffusion in open space (perfume across a room) follows the same mass dependence approximately, muddied by collisions with air; true Graham conditions are effusion through a pinhole into vacuum, where the law is exact.

Worked Examples (Verified)

  • He vs O₂ (defaults): ratio 2.827; if O₂ takes 100 s, He takes 35.4 s.
  • Unknown gas ID: a gas effuses 2.00× slower than O₂ → M = 31.998 × (2.00)² = 128.0 g/mol — SO₂? No (64.1); Xe (131.3) is the near match the method points to, tightened by better timing.
  • Hydrogen vs nitrogen: √(28.02/2.016) = 3.73× — why hydrogen leaks are so mobile, and hydrogen sensors sit at ceilings.

Graham’s Law Discipline

  • Same temperature is a precondition: the ratio form assumes both gases share T (kinetic energies equal); comparing a hot gas to a cold one needs the full √(T/M) form.
  • Times invert rates: slower gas, LONGER time — t₁/t₂ = √(M₁/M₂); flipping this is the standard homework error, and the calculator labels which time belongs to which gas.
  • Effusion ≠ bulk flow: Graham conditions mean molecular flow through holes smaller than the mean free path; a punctured tire (pressure-driven bulk flow) does not obey the ratio.
  • Molar masses of the GAS SPECIES as they exist: water vapor is 18.015 (not atoms), UF₆ is 352 (not uranium’s 238) — use the traveling molecule’s mass.
  • Diffusion through another gas is only approximately Graham: collision cross-sections intrude, so treat room-air diffusion numbers as estimates.

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Frequently Asked Questions

How do you use Graham’s law?

rate₁/rate₂ = √(M₂/M₁): the lighter gas’s rate over the heavier’s is the square root of the inverse mass ratio. Helium vs oxygen: √(31.998/4.003) = 2.83 — helium effuses 2.83 times faster. This calculator returns the ratio and converts a known effusion time for one gas into the other’s time for the same amount.

Why do lighter gases effuse faster?

Temperature fixes average kinetic energy (³⁄₂RT per mole), not speed. With ½Mv² equal across gases, v must scale as 1/√M. Molecules strike the pinhole in proportion to their speed, so effusion inherits the 1/√M law. It’s statistical, not per-molecule: every gas has a Maxwell distribution of speeds, but the whole distribution shifts with M exactly as Graham found empirically in the 1830s–40s.

How do you find an unknown molar mass by effusion?

Time the unknown and a known gas effusing the same amount under identical conditions: M(unknown) = M(known) × (t(unknown)/t(known))². A gas taking twice oxygen’s time has M = 32 × 4 = 128 g/mol. Precision timing of this ratio was a standard 19th-century molar-mass method — and still appears wherever a mass spectrometer isn’t handy.

Graham’s law vs diffusion — same thing?

Kin: effusion is escape through a pinhole into vacuum (the exact domain); diffusion is spreading through space or another medium, where intermolecular collisions modify the pure mass law into an approximation. Textbooks borrow the ratio for both; the calculator’s numbers are exact for effusion and indicative for free diffusion. Graham himself studied both and stated the rate law for each.

How does Graham’s law enrich uranium?

²³⁵UF₆ (M ≈ 349.04) effuses √(352.04/349.04) = 1.0043× faster than ²³⁸UF₆ — passing the gas through a porous barrier enriches the light isotope by 0.43% per stage. Reaching reactor-grade (~3–5%) takes hundreds of stages in cascade; weapons-grade, thousands. Gaseous diffusion plants were among the largest industrial buildings ever built precisely because the square root is so stingy.

Why do helium balloons deflate so fast?

Latex is porous at molecular scale, and helium is the second-lightest, smallest molecule there is: it effuses through the pores ~(2.83× faster than the O₂ trying to replace it, and far faster relative to N₂), while air diffuses inward more slowly — net volume loss overnight. Metallized mylar’s tighter structure slows (doesn’t stop) the same physics; the float-time difference between latex and foil balloons is Graham’s law as party trivia.

Does temperature change the ratio?

Only if the two gases are at DIFFERENT temperatures: the general form is rate₁/rate₂ = √(T₁M₂/(T₂M₁)). At equal T it cancels, which is why the standard law shows no T. Individually, each gas’s absolute rate scales as √T — heating speeds both, but their ratio stands still.

Is this Graham’s law calculator free?

Yes — free, instant, no sign-up. Enter two molar masses (and optionally one gas’s effusion time) for the rate ratio and the matched time.