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Rate of Effusion Calculator (Graham's Law)

Use Graham's Law to compare how fast two gases effuse through a small opening. Choose a solve mode: find the rate ratio from two molar masses, determine an unknown molar mass from observed rates, or compare the time each gas takes to pass through an aperture. The common gas table below gives you molar masses to look up instantly.

Your details

Select which quantity you want to find. Graham's Law can be rearranged to solve for any variable.
Molar mass of the first gas in grams per mole. H₂ = 2.016, He = 4.003, O₂ = 32.00.
g/mol
Molar mass of the second gas in grams per mole. O₂ = 32.00, N₂ = 28.01, CO₂ = 44.01.
g/mol
Rate ratio (r₁ / r₂)
3.984

How many times faster Gas 1 effuses compared to Gas 2

Time ratio (t₁ / t₂)0.251
Gas 1 speed vs Gas 2298.4%
Unknown molar mass-
Gas 1 relative speed3.98397085952242
Gas 2 relative speed1

Gas 1 effuses 298.4% faster than Gas 2.

  • The rate ratio r₁/r₂ = 3.9840: for every molecule of Gas 2 that passes through the aperture, 3.98 molecules of Gas 1 pass through.
  • The time ratio t₁/t₂ = 0.2510: if Gas 2 takes a reference time, Gas 1 takes 0.2510 times as long to effuse the same volume.
  • Gas 2 is heavier (31.998 g/mol vs 2.016 g/mol) and so effuses more slowly, consistent with kinetic-molecular theory.

Next stepGraham's Law assumes ideal-gas behavior at the same temperature and pressure. At high pressures or very low temperatures, intermolecular forces cause real-gas deviations.

Formula

r1r2=M2M1,M1=M2(r2r1)2,t1t2=M1M2\dfrac{r_1}{r_2} = \sqrt{\dfrac{M_2}{M_1}}, \quad M_1 = M_2 \left(\dfrac{r_2}{r_1}\right)^{2}, \quad \dfrac{t_1}{t_2} = \sqrt{\dfrac{M_1}{M_2}}

Worked example

Hydrogen (H₂, M = 2.016 g/mol) vs. oxygen (O₂, M = 32.00 g/mol): rate ratio = √(32.00 / 2.016) = √15.873 ≈ 3.984. So hydrogen effuses about 3.98 times faster than oxygen. If oxygen takes 100 s to effuse a fixed volume, hydrogen takes 100 / 3.984 ≈ 25.1 s.

What is Graham's Law of Effusion?

Graham's Law, stated by Thomas Graham in 1848, describes how fast a gas escapes through a tiny opening (effusion) or spreads through space (diffusion). The law says the rate of effusion of a gas is inversely proportional to the square root of its molar mass: lighter gases move faster because, at the same temperature, all gases have the same average kinetic energy (KE = ½ m v²). Since lighter molecules need higher speeds to carry the same kinetic energy as heavier ones, they escape through an aperture more quickly. The ratio form r₁/r₂ = √(M₂/M₁) lets you compare any two gases directly without knowing their absolute speeds.

How to use this calculator

Choose a solve mode from the dropdown. In Rate ratio mode, enter the molar masses of both gases and the calculator returns how many times faster Gas 1 effuses compared to Gas 2. In Unknown molar mass mode, enter the known molar mass plus the observed rates of both gases to back-calculate the unknown molar mass - useful in analytical chemistry for identifying an unknown gas. In Time ratio mode, enter both molar masses to find how long Gas 1 takes to effuse relative to Gas 2 (the inverse of the rate ratio). The reference table lists molar masses for eleven common gases so you do not need to look them up separately.

Effusion vs. diffusion

Effusion is the escape of gas through a single small hole into a vacuum or lower-pressure region. Diffusion is the gradual mixing of one gas into another across a pressure gradient. Graham's Law applies strictly to effusion under ideal conditions. For diffusion in a real gas mixture, the effective transport is also influenced by molecular collisions (the mean free path), which is captured by the more complete Chapman-Enskog theory. For the purposes of most chemistry courses and laboratory work, Graham's Law gives very good agreement and is the standard tool for comparing effusion rates.

Real-world applications

Graham's Law has important practical uses. The most famous is uranium enrichment: U-235 and U-238 are separated by passing uranium hexafluoride (UF₆) gas through many porous barriers. Because UF₆ containing U-235 (molar mass 349.03 g/mol) is slightly lighter than that containing U-238 (molar mass 352.04 g/mol), the rate ratio is √(352.04/349.03) ≈ 1.0043 - a tiny but exploitable difference amplified over thousands of stages. Graham's Law is also used in leak detection, respiratory physiology (comparing oxygen and carbon dioxide transport), and industrial membrane separations.

Common gas molar masses and effusion rates relative to O₂

GasFormulaMolar mass (g/mol)Rate vs O₂
HydrogenH₂2.0163.984
HeliumHe4.0032.827
MethaneCH₄16.0431.412
AmmoniaNH₃17.0311.371
Water vaporH₂O18.0151.333
NitrogenN₂28.0141.069
OxygenO₂31.9981.000
ArgonAr39.9480.895
Carbon dioxideCO₂44.0090.853
Sulfur dioxideSO₂64.0640.707
Uranium hexafluorideUF₆352.0190.301

Molar masses from IUPAC 2021 atomic weights. Effusion rate relative to O₂ calculated via Graham’s Law: rate = √(M(O₂) / M(gas)).

Frequently asked questions

What is the formula for Graham's Law of Effusion?

The rate ratio form is r₁/r₂ = √(M₂/M₁), where r₁ and r₂ are the effusion rates of Gas 1 and Gas 2, and M₁ and M₂ are their molar masses in g/mol. The law can be rearranged to solve for an unknown molar mass: M₁ = M₂ × (r₂/r₁)². The time ratio (for equal volumes) is t₁/t₂ = √(M₁/M₂), the inverse of the rate ratio.

Why does a lighter gas effuse faster?

At the same temperature, all gas molecules have the same average kinetic energy (KE = ½ m v²). A lighter molecule must move faster to carry the same kinetic energy as a heavier one. Since effusion rate depends on molecular speed, lighter gases escape through an aperture more quickly. Hydrogen (2 g/mol) is about four times faster than oxygen (32 g/mol) because √(32/2) = 4.

What units should I use for the effusion rates?

Any rate unit works as long as both gases use the same unit, because Graham's Law uses the ratio r₁/r₂ and the units cancel. Common choices are mL/s, cm³/s, mol/s, or number of molecules per second. If you are comparing times rather than rates, convert the time ratio to a rate ratio by taking its reciprocal.

Can I use this calculator to identify an unknown gas?

Yes. Select 'Molar mass of Gas 1 (or Gas 2) from rates', enter the known molar mass of the reference gas and the observed effusion rates of both gases, and the calculator returns the molar mass of the unknown gas. Compare this value against the common gas reference table or a periodic table to identify the compound. For example, if the unknown molar mass is about 44 g/mol, the gas is likely carbon dioxide or propane.

Does Graham's Law apply at all temperatures and pressures?

Graham's Law assumes ideal-gas behavior: the molecules are point masses that do not interact with each other. It is most accurate at low pressures and moderate temperatures. At high pressures, intermolecular forces cause deviations, and the lighter gas may not effuse as much faster as the law predicts. The law also assumes the aperture is smaller than the mean free path of the gas (Knudsen effusion regime); if the hole is large, viscous flow dominates and the simple ratio no longer holds.

How is Graham's Law used in uranium enrichment?

Uranium is processed as uranium hexafluoride gas (UF₆). The two main isotopes give slightly different molar masses: ²³⁵UF₆ is about 349.03 g/mol and ²³⁸UF₆ is about 352.04 g/mol. Graham's Law gives a rate ratio of √(352.04/349.03) ≈ 1.0043 - only 0.43% faster. A single stage barely separates them, so thousands of porous membrane stages are cascaded to enrich U-235 to reactor or weapons grade.

Sources

Written by Grace Mbeki, MSc Data Scientist & Educator · Nairobi, Kenya

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