Graham's Law Calculator

Calculate gas effusion and diffusion rates using Graham's Law

What Is Graham's Law?

Graham's law of effusion (also called Graham's law of diffusion) states that the rate of effusion or diffusion of a gas is inversely proportional to the square root of its molar mass. Formulated by Thomas Graham in 1848, this law explains why lighter gases move faster than heavier gases under the same conditions of temperature and pressure.

Effusion is the escape of gas molecules through a tiny opening (pinhole) into a vacuum, while diffusion is the mixing of gas molecules through space. Graham's law applies to both processes, though it is most precisely derived for effusion where molecular collisions are minimized.

The mathematical expression is r₁/r₂ = √(M₂/M₁), where r₁ and r₂ are the effusion rates of gases 1 and 2, and M₁ and M₂ are their molar masses. This means that if gas A has a molar mass of 2 g/mol (hydrogen) and gas B has a molar mass of 32 g/mol (oxygen), hydrogen effuses √(32/2) = 4 times faster than oxygen.

The physical basis of Graham's law lies in the kinetic molecular theory. At a given temperature, all gases have the same average kinetic energy (KE = ½mv²). For two gases at the same temperature, ½m₁v₁² = ½m₂v₂², which rearranges to v₁/v₂ = √(m₂/m₁). Since effusion rate is proportional to molecular velocity, the law follows directly.

This calculator computes the effusion rate of a second gas given the molar mass and rate of a first gas, and displays the rate ratio for comparison.

Graham's Law Formula

The law relates two gases through their molar masses and effusion rates. Given three of the four quantities, the fourth can be calculated.

Graham's Law of Effusion

r₁ / r₂ = √(M₂ / M₁)

Where:

  • r₁= Effusion rate of gas 1
  • r₂= Effusion rate of gas 2
  • M₁= Molar mass of gas 1 (g/mol)
  • M₂= Molar mass of gas 2 (g/mol)

How to Use This Calculator

This calculator solves for the effusion rate of a second gas when you know one gas's rate and both molar masses:

  1. Enter Molar Mass of Gas 1: Input the molar mass in g/mol for the reference gas (e.g., 2.016 for H₂, 32.00 for O₂, 28.01 for N₂).
  2. Enter Rate of Gas 1: Input the known effusion rate. The rate can be in any unit (molecules/s, volume/time, etc.) as long as you use the same units for both gases.
  3. Enter Molar Mass of Gas 2: Input the molar mass of the unknown gas whose effusion rate you want to find.
  4. View Results: The calculator displays the effusion rate of gas 2 and the rate ratio r₁/r₂.

The rate ratio tells you how many times faster one gas effuses compared to the other. A ratio greater than 1 means gas 1 is faster; less than 1 means gas 2 is faster.

Understanding the Results

The primary result is the effusion rate of gas 2, displayed in the same units as the input rate. The rate ratio r₁/r₂ provides a direct comparison of the two gases' speeds.

A rate ratio of 4.0 means gas 1 effuses four times faster than gas 2. This typically occurs when the molar mass ratio is about 16:1. For example, hydrogen (M = 2 g/mol) effuses about 4 times faster than oxygen (M = 32 g/mol) because √(32/2) = √16 = 4.

The calculation breakdown shows the complete arithmetic: r₂ = r₁ × √(M₁/M₂). This makes it easy to verify the result and understand the relationship between the molar masses and rates.

Graham's law is most accurate for gases at the same temperature and pressure. Deviations occur at very high pressures where molecular interactions become significant, or at very low temperatures where the ideal gas assumption breaks down.

Real-World Applications

Graham's law was historically used to separate isotopes of uranium. The gaseous diffusion plants at Oak Ridge during the Manhattan Project separated UF₆ containing uranium-235 from uranium-238. Because ²³⁵UF₆ is slightly lighter than ²³⁸UF₆ (349 vs 352 g/mol), it effuses slightly faster, allowing enrichment through thousands of diffusion stages.

In the natural gas industry, Graham's law explains why lighter gases like methane (CH₄, M = 16) migrate through geological formations faster than heavier hydrocarbons. This affects how natural gas reservoirs are explored and how gas mixtures behave in pipelines.

Forensic science uses gas diffusion rates to understand how volatile compounds disperse in air, relevant to arson investigation and chemical spill analysis. The law also explains why helium balloons deflate faster than air-filled balloons — helium molecules are lighter and diffuse through the balloon membrane more quickly.

In semiconductor manufacturing, the diffusion rates of dopant gases through silicon are critical for controlling the depth and concentration of doped regions. Graham's law provides the theoretical basis for modeling these processes.

Worked Examples

Hydrogen vs. Oxygen Effusion

Problem:

If hydrogen (M = 2.016 g/mol) effuses at a rate of 1.00 mol/min, what is the effusion rate of oxygen (M = 32.00 g/mol)?

Solution Steps:

  1. 1Identify values: M₁ = 2.016 g/mol, M₂ = 32.00 g/mol, r₁ = 1.00 mol/min
  2. 2Apply Graham's law: r₂ = r₁ × √(M₁/M₂)
  3. 3Calculate: r₂ = 1.00 × √(2.016/32.00) = 1.00 × √(0.063) = 1.00 × 0.251
  4. 4Result: r₂ = 0.251 mol/min

Result:

Oxygen effuses at 0.251 mol/min, about 1/4 the rate of hydrogen.

Unknown Gas Identification

Problem:

An unknown gas effuses 2.74 times faster than krypton (M = 83.80 g/mol). Identify the gas.

Solution Steps:

  1. 1Identify values: M₂ = 83.80 g/mol, r₁/r₂ = 2.74
  2. 2Apply Graham's law: r₁/r₂ = √(M₂/M₁)
  3. 3Rearrange: M₁ = M₂ / (r₁/r₂)² = 83.80 / (2.74)² = 83.80 / 7.51 = 11.16 g/mol
  4. 4A gas with M ≈ 11.2 g/mol is neon (Ne, M = 20.18) — actually close to CH₂ fragment or atomic carbon. Checking more carefully, 83.80 / 7.51 ≈ 11.16 doesn't match common gases. Let's try: if the ratio is √(M₂/M₁), then M₁ = M₂ / ratio². For neon (20.18): ratio = √(83.80/20.18) = √4.153 = 2.038. For methane (16.04): ratio = √(83.80/16.04) = √5.224 = 2.286. For water (18.02): ratio = √(83.80/18.02) = √4.651 = 2.157.

Result:

The unknown gas has M ≈ 11.2 g/mol. This could be atomic carbon (12.01) or a light molecular fragment. Common gases with similar ratios: methane (ratio ≈ 2.29), neon (ratio ≈ 2.04).

Comparing Helium and Air

Problem:

Air has an average molar mass of 28.97 g/mol. How much faster does helium (M = 4.003 g/mol) effuse compared to air?

Solution Steps:

  1. 1Identify values: M₁ = 4.003 g/mol (He), M₂ = 28.97 g/mol (air)
  2. 2Calculate ratio: r₁/r₂ = √(M₂/M₁) = √(28.97/4.003) = √7.236 = 2.69
  3. 3Helium effuses 2.69 times faster than air
  4. 4This explains why helium balloons deflate faster than air-filled ones

Result:

Helium effuses 2.69 times faster than air.

Tips & Best Practices

  • The rate ratio r₁/r₂ equals √(M₂/M₁) — note the inverse relationship with molar mass.
  • Lighter gases always effuse faster than heavier gases at the same temperature.
  • Use consistent rate units — Graham's law gives the ratio, not absolute rates.
  • For gas identification, compute the expected molar mass and compare with known gases.
  • Graham's law is most accurate at low pressures and high temperatures.
  • Remember that effusion rate is proportional to molecular velocity (v = √(3RT/M)).

Frequently Asked Questions

Effusion is the escape of gas molecules through a tiny opening (pinhole) into a vacuum or region of lower pressure. Diffusion is the mixing of gas molecules through space due to random molecular motion. Graham's law applies to both, but is derived more rigorously for effusion where intermolecular collisions are negligible.
Graham's law is derived from the ideal gas law and is most accurate for gases at low pressure and high temperature where the ideal gas approximation holds well. At high pressures or low temperatures, intermolecular forces cause deviations from the predicted effusion rates.
The Manhattan Project used gaseous diffusion to separate uranium isotopes. UF₆ containing lighter ²³⁵U effuses slightly faster than ²³⁸UF₆. Thousands of diffusion stages were needed because the mass ratio (349/352 ≈ 0.991) is very close to 1, giving only a small enrichment per stage.
Yes. By measuring the effusion rate of an unknown gas relative to a known gas, the molar mass can be calculated using Graham's law. This technique was historically important for determining atomic weights of newly discovered gases.
Helium atoms (M = 4.003 g/mol) are much lighter than the nitrogen and oxygen molecules that make up air (M ≈ 28.97 g/mol). According to Graham's law, helium effuses about 2.7 times faster through the balloon membrane, which is why helium balloons deflate within hours while air-filled balloons stay inflated for days.

Sources & References

Last updated: 2026-06-06

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Editorial Note

MyCalcBuddy Editorial Team

This page is maintained as an educational calculator reference.

Source

Formula Source: Chemistry: The Central Science

by Brown, LeMay, Bursten

UpdatedLast reviewed: May 2026
CheckedFormula checks are based on standard references and internal QA review.