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Rate of Effusion Calculator

Chemistry

Calculate relative rates of effusion using Graham's law: rate₁/rate₂ = √(M₂/M₁). Compare any two gases by molar mass for gas identification.

Reviewed by the thecalcu.com team · Last updated April 18, 2026

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Rate Ratio (r₁/r₂)

4
Rate of Gas 2 (same units)
0.25
Relative Rate of Gas 1 (%)
80

This calculator computes your Rate Ratio (r₁/r₂), Rate of Gas 2 (same units), Relative Rate of Gas 1 (%) from the values you enter.

Inputs
Molar Mass of Gas 1 (g/mol)Molar Mass of Gas 2 (g/mol)Rate of Gas 1 (any unit)
Outputs
Rate Ratio (r₁/r₂)Rate of Gas 2 (same units)Relative Rate of Gas 1 (%)

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.

Why Use a Rate of Effusion Calculator?

The square root in Graham's law is the source of errors: students often compute M₁/M₂ instead of the inverse ratio M₂/M₁, or forget to take the square root. For the rate ratio of Gas 1 to Gas 2, the heavier gas denominator (M₂) goes in the numerator of the molar mass ratio under the square root, the calculation is counter-intuitive. This calculator explicitly shows the rateRatio = √(M₂/M₁) for Gas 1 relative to Gas 2.

For NCERT Class 11 (Chapter 5, States of Matter) and JEE problems, Graham's law calculations appear in both direct (find rate ratio) and inverse (find molar mass from rate ratio) forms.

Who Should Use This Calculator?

Class 11 and undergraduate chemistry students studying kinetic molecular theory and Graham's law in the States of Matter chapter, where effusion and diffusion calculations are standard examination topics.

JEE Main and JEE Advanced aspirants solving Graham's law problems that require comparing effusion rates or determining molar masses from rate ratios.

Analytical chemists estimating relative effusion rates through capillary or porous membrane systems for qualitative gas identification.

Nuclear fuel cycle engineers and researchers computing theoretical isotopic separation factors for gaseous diffusion processes using UF₆ or other isotope-specific gases.

Environmental scientists computing the relative rate at which different greenhouse gases (CO₂, CH₄, N₂O) permeate through atmospheric boundaries.

What Insights Does the Rate of Effusion Calculator Give You?

Rate Ratio (r₁/r₂) is the primary output, how much faster Gas 1 effuses compared to Gas 2. If the ratio is greater than 1, Gas 1 (the lighter gas, since M₁ < M₂) is faster. The ratio is always √(M₂/M₁): lighter gas in the denominator of the mass ratio under the square root.

Rate of Gas 2 is the effusion rate of the heavier gas in the same units as Gas 1's input rate. If Gas 1 effuses at 1 L/min, Gas 2 at √(M₁/M₂) × 1 L/min.

Relative Rate of Gas 1 (%) shows what fraction of the total effusing gas (per unit time) is Gas 1, relevant for separations and composition analysis.

How to use this Effusion Rate calculator

  1. Enter Molar Mass of Gas 1 (g/mol), for example, H₂ = 2, He = 4, N₂ = 28, O₂ = 32, CO₂ = 44, UF₆ ≈ 349.
  2. Enter Molar Mass of Gas 2 (g/mol), the comparison gas.
  3. Enter Rate of Gas 1, the measured or reference effusion rate in any consistent units (mL/min, L/s, relative value, etc.).
  4. Read Rate Ratio (r₁/r₂), how much faster Gas 1 effuses relative to Gas 2.
  5. Use Rate of Gas 2 for the absolute rate if Gas 1's rate is known.
Show formula & methodology ↓Show less ↑

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/mol

r(²³⁵UF₆)/r(²³⁸UF₆) = √(352.05/349.04) = √1.00863 = 1.00431

²³⁵UF₆ 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

What is effusion and how does it differ from diffusion?
Effusion is the escape of gas molecules through a tiny orifice (hole much smaller than the mean free path of the gas) into a vacuum or lower-pressure region. Diffusion is the spreading of gas molecules through another gas or through a medium due to concentration gradients. Both are governed by molecular speeds, but effusion occurs through a physical opening while diffusion occurs through bulk gas mixing. Graham's law applies specifically to effusion; for diffusion through a gas, the relative rates are also approximately √(M₂/M₁) but the process is more complex.
What is Graham's law of effusion?
Graham's law of effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass: rate ∝ 1/√M. For two gases at the same temperature and pressure, the ratio of effusion rates is: r₁/r₂ = √(M₂/M₁). This relationship follows from the kinetic molecular theory: average molecular speed u = √(3RT/M), so a lighter gas moves faster and effuses more rapidly.
How does molar mass affect the rate of effusion?
Lighter gases effuse faster than heavier gases. The rate ratio is the square root of the inverse mass ratio. For example: H₂ (M = 2 g/mol) vs. O₂ (M = 32 g/mol): rate(H₂)/rate(O₂) = √(32/2) = √16 = 4. Hydrogen effuses 4 times faster than oxygen. For He (M = 4) vs. N₂ (M = 28): rate(He)/rate(N₂) = √(28/4) = √7 ≈ 2.65. Helium is 2.65 times faster.
How do I use the Rate of Effusion Calculator?
Enter the Molar Mass of Gas 1 and Molar Mass of Gas 2 in g/mol. Enter the Rate of Gas 1 in any unit (mL/s, L/min, relative value, etc.). The calculator applies Graham's law to return the rate ratio r₁/r₂ = √(M₂/M₁), the rate of Gas 2 in the same units, and the relative rate of Gas 1 as a percentage of the total.
What is Graham's law used for in practice?
Graham's law has several practical applications: (1) Gas separation by effusion, uranium enrichment for nuclear fuel uses gaseous UF₆ effusion through porous barriers. ²³⁵UF₆ (M=349) effuses slightly faster than ²³⁸UF₆ (M=352), and thousands of cascade stages achieve the needed enrichment. (2) Identifying unknown gases, measuring effusion rate relative to a known gas allows calculation of the unknown gas's molar mass. (3) Predicting gas leak rates through small cracks, relevant to pressure vessel safety and gas distribution systems.
How is Graham's law applied to uranium isotope separation for nuclear fuel?
India's nuclear fuel cycle (under DAE/NFC) uses uranium enrichment for PHWRs (natural uranium, no enrichment needed) and for advanced reactors. International uranium enrichment is done via gaseous diffusion or centrifuge cascade using UF₆ gas. Graham's law gives the theoretical separation factor for a single stage: α = r(²³⁵UF₆)/r(²³⁸UF₆) = √(352.04/349.03) = √1.00863 = 1.00431, a separation factor of only 1.004 per stage, requiring thousands of stages to achieve weapons-grade enrichment (>90% ²³⁵U). Modern gas centrifuges achieve much higher separation factors per stage.
What is the average speed of gas molecules according to kinetic theory?
Kinetic molecular theory gives three molecular speed statistics: mean speed (c̄) = √(8RT/πM); root mean square speed (u_rms) = √(3RT/M); most probable speed (u_mp) = √(2RT/M). All are inversely proportional to √M, consistent with Graham's law. At 25°C: u_rms(H₂) = √(3×8.314×298/0.002) = 1,920 m/s; u_rms(O₂) = √(3×8.314×298/0.032) = 483 m/s. The ratio 1920/483 = 3.97 ≈ 4, consistent with Graham's law: √(32/2) = 4.
Can Graham's law be used to determine molar mass of an unknown gas?
Yes, this is one of its most important analytical applications. Measure the time for a known volume of the unknown gas and a reference gas (usually H₂ or O₂) to effuse through the same orifice under the same conditions. Rate is inversely proportional to time at constant volume: r ∝ 1/t. So M_unknown = M_ref × (t_unknown/t_ref)². For example, if the unknown gas effuses in 3× the time of H₂ (M = 2): M_unknown = 2 × (3)² = 18 g/mol, consistent with water vapour or neon.
Does effusion rate depend on temperature or pressure?
At the same temperature and pressure (or equal temperature and pressure for both gases), Graham's law depends only on molar mass. The absolute rate of effusion (not the ratio) does depend on temperature (rate ∝ √T) and on the pressure difference across the orifice. Graham's law gives the rate ratio between two gases at the same conditions, so temperature and pressure cancel out in the ratio. This makes the law particularly useful for comparison experiments at ambient conditions.
How does Graham's law relate to the Maxwell-Boltzmann distribution?
The Maxwell-Boltzmann distribution gives the probability distribution of molecular speeds: f(v) ∝ v² × exp(−Mv²/2RT). The mean speed c̄ = √(8RT/πM) ∝ 1/√M, directly showing the inverse square root relationship with molar mass. Graham's law is a macroscopic consequence of this microscopic speed distribution: since lighter molecules have higher average speeds, they strike the orifice more frequently and effuse faster. The full Maxwell-Boltzmann analysis was developed by James Clerk Maxwell (1859) and Ludwig Boltzmann (1872).