Graham's law of effusion, formulated by Scottish chemist Thomas Graham in 1848, describes a simple but powerful relationship: lighter gas molecules move faster than heavier ones at the same temperature, and effuse or diffuse proportionally faster. This calculator applies that relationship to compare two gases, solve for an unknown rate, or back out an unknown molar mass.
How Graham's law works
At a given temperature, all gas molecules share the same average kinetic energy: KE = ½mv². Because kinetic energy is fixed, a lighter molecule (smaller m) must move with a higher average speed (v) to balance the equation. Rearranging the kinetic-energy relationship for two gases at the same temperature gives Graham's law: rate1/rate2 = √(M2/M1). The rate ratio depends on the square root of the inverse mass ratio, not the mass ratio itself — a gas that is 4× lighter effuses only 2× faster, not 4× faster.
Inputs and what they mean
The Rate Ratio tab takes the molar masses of two gases in grams per mole (g/mol) and returns how many times faster or slower Gas 1 moves relative to Gas 2. The Solve Rate tab adds a known rate for Gas 2, letting you calculate Gas 1's actual rate rather than just the ratio. The Molar Mass tab works in reverse: given one known molar mass and a measured rate ratio, it solves for the unknown gas's molar mass — a common technique for identifying an unknown gas in a lab setting.
Limits and edge cases
Graham's law strictly applies to effusion through a small orifice into a vacuum and, as an approximation, to diffusion under similar conditions — it assumes ideal-gas behavior and that both gases are at the same temperature and pressure. It breaks down at very high pressures or low temperatures where real-gas effects (intermolecular forces, molecular volume) become significant. The law compares only the relative rates of two gases; it does not by itself give an absolute effusion rate in physical units like moles per second — for that, a rate for one gas must be measured or given.