Graham's Law of Effusion Calculator
How many times faster one gas escapes through a tiny hole than another, from their molar masses, and how long the second gas takes for the same amount.
Results
What this tool does
Thomas Graham found in 1846 that a gas leaking through a pinhole does so at a rate inversely proportional to the square root of its molar mass. Kinetic theory explains why: at the same temperature every gas has the same average kinetic energy, so lighter molecules have to move faster. Hydrogen (2 g/mol) therefore escapes about four times faster than oxygen (32 g/mol). The same law ran the gaseous diffusion plants that enriched uranium: the two forms of uranium hexafluoride differ in speed by only about 0.43 %, so the gas had to pass through a very long cascade of stages. Strictly, it holds for holes much smaller than the distance a molecule travels between collisions.
Formula
rate₁ ÷ rate₂ = √(M₂ ÷ M₁) · t₂ = t₁ × √(M₂ ÷ M₁)
Variables
| Symbol | Meaning | Unit |
|---|---|---|
m1 | Molar mass of gas 1 | g/mol |
m2 | Molar mass of gas 2 | g/mol |
t1 | Time gas 1 takes to effuse | s |
R | Gas 1 escapes this many times faster | × |
T2 | Time gas 2 takes for the same amount | s |
PF | Gas 1 is faster by | % |
Worked example
- Molar mass of gas 12.016 g/mol
- Molar mass of gas 232 g/mol
- Time gas 1 takes to effuse10 s
- Gas 1 escapes this many times faster3.9841 ×
- Time gas 2 takes for the same amount39.841 s
- Gas 1 is faster by298.41 %
Limitations
- The formula assumes ideal conditions: no friction losses, no air resistance and no efficiency losses unless you enter them.
- The default values are typical reference figures, not measurements of your situation. Replace them with your own data whenever you have it.