Graham's Law Calculator
Relative effusion rate of two gases from molar masses
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Graham's Law Calculator turns the numbers behind relative effusion rate of two gases from molar masses into a result, step by step. You supply Molar Mass of Gas 1 (g/mol) and Molar Mass of Gas 2 (g/mol) to Graham's Law; the answer follows at once. Every Graham's Law answer can be reproduced by hand. Graham's Law replaces guesswork. Graham's Law is privacy-first: Science inputs stay offline. Graham's Law handles grahams law calculator effusion diffusion rate gas molar mass the same way as free online graham's law calculator — one of the Science tools on CalcProMaster. Graham's Law has no learning curve. The fields are labeled. Start with Graham's Law; the first number takes seconds.
What the defaults give
With molar masses of 2 and 32 g/mol, gas 1 effuses 4.00 times faster than gas 2.
Four is the square root of the 16-to-1 mass ratio: lighter molecules move faster, and the relationship is a square root, not a straight line.
The formula
rate₁ ÷ rate₂ = √(M₂ ÷ M₁)
M is molar mass in grams per mole. The heavier gas sits in the numerator because a larger M₂ means a bigger ratio — gas 1 escapes faster. Alternatively, rate ∝ 1/√M, so doubling the mass slows effusion by a factor of √2, about 1.41.
Worked examples
H₂ (2) vs O₂ (32) → 4.00×
H₂ (2) vs helium (4) → 1.41×
He (4) vs O₂ (32) → 2.83×
H₂ (2) vs methane (16) → 2.83×
H₂ (2) vs krypton (84) → 6.48×
O₂ (32) vs UF₆ (352) → 3.32×
The hydrogen-oxygen comparison is the one worth remembering: a 16-fold mass difference shows up as only 4-fold in rate. That square root is also why uranium enrichment is expensive — separating ²³⁵UF₆ from ²³⁸UF₆ is a mass ratio of 352 to 349, a rate ratio of about 1.004, so a single stage barely separates them at all.
Assumptions the number rests on
Graham’s law treats effusion as molecules passing through a pinhole one at a time, with no collisions in the hole. Real leaks through a crack are closer to viscous flow, where the ratio behaves differently and the lighter gas advantage shrinks.
Temperature must be the same on both sides. The derivation starts from kinetic energy, ½mv², being equal for both gases at a given temperature; heat one side and the comparison is no longer about mass. Pressure and hole size are assumed identical too.
Molar mass, not molecular mass, is what goes in. For a mixture, use the average molar mass; for a gas that dimerises, such as acetic acid vapour, the effective mass is larger than the formula suggests.
The law is a good approximation for small molecules at low pressure and a poor one for large ones, where the mean free path is short compared with the opening. It is still the right first estimate for "which of these two gases leaks out faster, and roughly by how much".
Frequently Asked Questions
How much faster does hydrogen effuse than oxygen?
Four times. √(32/2) = √16 = 4, so every oxygen molecule that escapes is matched by four hydrogen molecules.
Why a square root and not a direct ratio?
Kinetic energy is ½mv². At equal temperature the ½mv² values are equal, so speed scales with 1/√m rather than 1/m.
Does temperature affect the ratio?
Not if both gases are at the same temperature. Raise one side and the comparison no longer isolates mass.
What if I swap the two gases?
The ratio inverts. Oxygen effuses 0.25 times as fast as hydrogen, the reciprocal of 4.00.
Does this apply to a leak in a tyre?
Only roughly. A macroscopic puncture is closer to viscous flow, where the mass dependence is weaker than Graham’s law predicts.
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