Effusion Rate Calculator | Graham's Law of Effusion

What is an effusion rate calculator?

An effusion rate calculator finds how fast one gas escapes through a tiny hole compared to another gas. It uses Graham's law of effusion, which links the escape speed of a gas to its molar mass and temperature. Effusion happens when gas molecules pass one at a time through an opening so small that the molecules do not collide with each other as they go through.

Graham's law of effusion formula

Graham's law compares the effusion rates of two gases, gas 1 and gas 2:

$\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}} \times \sqrt{\frac{T_1}{T_2}}$

• Rate₁ and Rate₂ are the effusion rates of gas 1 and gas 2.
• M₁ and M₂ are the molar masses of gas 1 and gas 2, in grams per mole (g/mol).
• T₁ and T₂ are the absolute temperatures of gas 1 and gas 2, in kelvin (K).

A result above 1 means gas 1 effuses faster than gas 2. A result below 1 means gas 2 is faster. A result of exactly 1 means both gases effuse at the same rate.

Why the formula works

The formula comes from the kinetic theory of gases. All gases at the same temperature have the same average kinetic energy, ½mv² = (3/2)kT, where m is the mass of one molecule, v is its average speed, k is the Boltzmann constant, and T is the absolute temperature. Solving for speed gives v = √(3kT/m). A lighter molecule must move faster than a heavier one to hold the same kinetic energy. Since the rate of effusion rises with molecular speed, and molar mass scales with molecular mass, dividing the speed of gas 1 by the speed of gas 2 gives the formula above.

Two simpler cases

• Same temperature. If T₁ equals T₂, the temperature terms cancel and the formula becomes Rate₁/Rate₂ = √(M₂/M₁). Only the molar masses matter.
• Same molar mass. If M₁ equals M₂, the formula becomes Rate₁/Rate₂ = √(T₁/T₂). Only the temperatures matter.

How to use the effusion rate calculator

1. Enter the molar mass and temperature of gas 1.
2. Enter the molar mass and temperature of gas 2.
3. Read the relative effusion rate. It shows how many times faster gas 1 effuses than gas 2.

Both molar mass and temperature must be positive numbers. Temperature must be entered in kelvin, not Celsius or Fahrenheit, because the formula depends on absolute temperature.

Common gas molar masses

Worked examples

Example 1: helium versus methane, same temperature

Helium (He) has a molar mass of 4.00 g/mol. Methane (CH₄) has a molar mass of 16.00 g/mol. Both are at 298 K.

$\frac{\text{Rate}_{He}}{\text{Rate}_{CH_4}} = \sqrt{\frac{16.00}{4.00}} \times \sqrt{\frac{298}{298}} = \sqrt{4} \times 1 = 2.00$

Helium effuses 2.00 times faster than methane.

Example 2: hydrogen versus oxygen, different temperatures

Hydrogen (H₂) has a molar mass of 2.02 g/mol and a temperature of 400 K. Oxygen (O₂) has a molar mass of 32.00 g/mol and a temperature of 300 K.

$\sqrt{\frac{32.00}{2.02}} \approx 3.9801 \qquad \sqrt{\frac{400}{300}} \approx 1.1547$

$\frac{\text{Rate}_{H_2}}{\text{Rate}_{O_2}} = 3.9801 \times 1.1547 \approx 4.5959$

Hydrogen at 400 K effuses about 4.60 times faster than oxygen at 300 K. Rounding each square root to two decimal places before multiplying, as some worked examples online do, gives 3.98 × 1.15 = 4.58, which is slightly off. Keeping more decimal places until the final step avoids that error.

Applications of Graham's law

• Isotope separation. The Manhattan Project used gaseous diffusion to separate uranium-235 from uranium-238. The two isotopes form uranium hexafluoride gas with slightly different molar masses, so they effuse at slightly different rates.
• Leak detection. Helium is a common tracer gas for finding leaks in vacuum systems and pressure vessels because its low molar mass lets it effuse through small leaks quickly.
• Gas chromatography. Molecules of different mass move through a chromatography column at different rates, which helps separate and identify compounds in a gas mixture.
• Membrane gas separation. Industrial membranes that purify or separate gas mixtures rely partly on the different effusion rates of the component gases.

History

The Scottish chemist Thomas Graham first described this relationship in 1846. He measured how fast different gases escaped through small openings and found that the rate was inversely proportional to the square root of the gas density. His results later helped support the kinetic theory of gases, developed further by James Clerk Maxwell and Ludwig Boltzmann in the 1860s and 1870s.

Frequently asked questions

What is the difference between effusion and diffusion? Effusion is the escape of gas molecules through a small hole into a vacuum or region of lower pressure, where the hole is smaller than the average distance a molecule travels between collisions. Diffusion is the spreading of gas molecules through another gas or medium, driven by differences in concentration, and it involves molecules colliding with each other.

Why must temperature be in kelvin? The kinetic energy of gas molecules is proportional to absolute temperature. Celsius and Fahrenheit scales do not start at absolute zero, so using them would give the wrong ratio.

Does pressure affect the relative effusion rate? No, as long as both gases are at the same pressure. Pressure changes the absolute effusion rate of each gas, but it affects both gases equally, so the ratio between them stays the same.

Can Graham's law find the molar mass of an unknown gas? Yes. If the effusion rate of an unknown gas is measured against a gas of known molar mass, the formula can be rearranged:

$M_{unknown} = M_{known} \times \left(\frac{\text{Rate}_{known}}{\text{Rate}_{unknown}}\right)^2 \times \frac{T_{unknown}}{T_{known}}$

Does Graham's law apply to liquids? No. It applies only to gases. Liquid molecules are packed much closer together and interact far more strongly, so the same formula does not describe how liquids move through small openings.

How much faster does a gas effuse if its temperature doubles? The effusion rate is proportional to the square root of the absolute temperature. Doubling the temperature multiplies the rate by √2, or about 1.41, if the molar mass stays the same.