Rate of Effusion Calculator
ChemistryCalculate relative rates of effusion using Graham's law: rate₁/rate₂ = √(M₂/M₁). Compare effusion rates of any two gases by molar mass. Useful for gas identification.
Rate Ratio (r₁/r₂)
What is a Effusion Rate?
The Rate of Effusion Calculator computes the relative rates of effusion of two gases using Graham's law: r₁/r₂ = √(M₂/M₁). Enter the molar masses of both gases and the known rate of gas 1 to get the rate ratio, the rate of gas 2, and the relative rate of gas 1.
Graham's law of effusion — derived by Thomas Graham in 1848 and later explained by kinetic molecular theory — states that the rate at which a gas escapes through a small orifice is inversely proportional to the square root of its molar mass. This square-root relationship arises because molecular speed is proportional to 1/√M (from the kinetic energy equation ½mv² = 3/2 kT), and effusion rate is proportional to molecular speed.
The most historically significant application is uranium isotope separation: ²³⁵UF₆ (M = 349.03) effuses slightly faster than ²³⁸UF₆ (M = 352.04), with a rate ratio of √(352.04/349.03) = 1.00431 — a mere 0.43% difference per stage. The calculation of this ratio using Graham's law is one of the most consequential applications of a simple formula in the history of science.
For the Molar Mass of Gas Calculator (which finds molar mass from gas density), and for gas diffusion at STP (see STP Calculator), related calculations are available.
How to use this Effusion Rate calculator
- Enter Molar Mass of Gas 1 (g/mol) — for example, H₂ = 2, He = 4, N₂ = 28, O₂ = 32, CO₂ = 44, UF₆ ≈ 349.
- Enter Molar Mass of Gas 2 (g/mol) — the comparison gas.
- Enter Rate of Gas 1 — the measured or reference effusion rate in any consistent units (mL/min, L/s, relative value, etc.).
- Read Rate Ratio (r₁/r₂) — how much faster Gas 1 effuses relative to Gas 2.
- Use Rate of Gas 2 for the absolute rate if Gas 1's rate is known.
Formula & Methodology
Graham's law of effusion:r₁/r₂ = √(M₂/M₁) r₂ = r₁ / √(M₂/M₁) = r₁ × √(M₁/M₂)Inverse application — finding unknown molar mass:M_unknown = M_ref × (r_ref/r_unknown)² = M_ref × (t_unknown/t_ref)² [using time inversely proportional to rate]Worked example — separation factor for ²³⁵UF₆/²³⁸UF₆: M(²³⁵UF₆) = 235.04 + 6×19.00 = 349.04 g/mol M(²³⁸UF₆) = 238.05 + 6×19.00 = 352.05 g/molr(²³⁵UF₆)/r(²³⁸UF₆) = √(352.05/349.04) = √1.00863 = 1.00431²³⁵UF₆ effuses only 0.43% faster per stage. To enrich ²³⁵U from 0.71% (natural) to 3.5% (reactor fuel grade): ln(3.5/0.71) / ln(1.00431) ≈ 1,584/0.00430 ≈ 368 stages minimum (ideal cascade). The gaseous diffusion plant at Paducah, Kentucky used over 1,000 stages; modern centrifuge cascades achieve much higher per-stage enrichment.
Frequently Asked Questions