Graham Law Calculator
Compare gas effusion rates and times from molar masses with Graham law.
What it does
Small molecules move fast and escape first. When two gases sit at the same temperature, the lighter one carries more speed per molecule and slips through pinholes sooner. Graham law turns that picture into one square root: the rate ratio equals the inverse square root of the mass ratio.
The tool above applies the law directly. Enter both molar masses plus a measured time for gas 2 to get the rate ratio, the predicted time for gas 1, and the time ratio. Defaults compare hydrogen against oxygen with 120 s for oxygen, so the page opens solved at 3.984 x and 30.12 s.
The famous factor of four falls out at once: oxygen is nearly sixteen times heavier than hydrogen, and the square root of sixteen is four. Helium against air gives 2.690 x, while neon against xenon gives 2.551 x with 95.27 s, matching the OpenStax worked example to the decimal the page keeps.
Conditions decide trust. Both gases must share temperature and pressure, behave nearly ideally, and pass through an opening fine enough for molecular flow. Change any of these and the measured ratio drifts from the square root, which is why enrichment plants chain many stages instead of trusting one.
Use the table for common pairs, then run your gases. The page assumes equal amounts transferred and reports times accordingly. Within range it answers lab questions fast: which gas wins, by how much, and how long the slow one needs.
How to use
- Enter the molar mass of gas 1.
- Enter the molar mass of gas 2.
- Enter the measured time for gas 2.
- Read the rate ratio first.
- Read the predicted time for gas 1.
- Flip the masses to check the inverse case.
Formula
- M1
- Molar mass of gas 1 in g/mol.
- M2
- Molar mass of gas 2 in g/mol.
- Time2
- Measured effusion time of gas 2 in seconds.
- Ratio
- Effusion rate ratio of gas 1 to gas 2.
M1 and M2 are molar masses in g/mol and the reference time belongs to gas 2. The square root form assumes ideal gases at equal temperature and pressure with molecular flow through the opening. Times scale inversely with rates for equal amounts transferred.
Worked example
Example: hydrogen versus oxygen
Oxygen (32.00 g/mol) needs 120 s to effuse. How fast does hydrogen (2.016 g/mol) effuse under the same conditions?
- Ratio = sqrt(M2/M1) = sqrt(32.00/2.016) = sqrt(15.873) = 3.984 x.
- Hydrogen effuses about 4 times as fast as oxygen at the same temperature.
- Time for H2 = 120 / 3.984 = 30.12 s against 120 s for O2.
- Time ratio = 1/3.984 = 0.251 x.
- Lighter gas always wins when temperature and pressure match.
- Heavier gases lag by the square root of the mass ratio.
Reference table
| Pair | Ratio | Faster gas |
|---|---|---|
| H2 vs O2 | 3.984 x | H2 |
| He vs air | 2.69 x | He |
| Ne vs Xe | 2.551 x | Ne |
| CH4 vs CO2 | 1.656 x | CH4 |
| O2 vs CO2 | 1.173 x | O2 |
Limits
- Ideal gases at equal T and P only.
- Molecular flow through small openings.
- Equal amounts transferred assumption.
- No porous barrier or pressure gradient modeling.
Common mistakes
- Inverting which mass sits on top.
- Using atomic mass instead of molecular mass.
- Comparing gases at different temperatures.
- Entering times for different amounts.
- Expecting exact plant scale agreement.
Try it yourself
Quick checks. Answers sit tucked under each question.
Helium (4.003) versus air (28.97) with 100 s for air?
Ratio 2.69 x. Time 37.17 s. Time ratio 0.372 x.
Neon (20.18) versus xenon (131.3) with 243 s for xenon?
Ratio 2.551 x. Time 95.27 s. Time ratio 0.392 x.
Methane (16.04) versus CO2 (44.01)? Find the ratio by hand.
Ratio = sqrt(44.01/16.04) = sqrt(2.744) = 1.656 x. Unknown gas (methane) wins.
Questions
What is Graham law?
The rule that a gas effuses at a rate inversely proportional to the square root of its molar mass. Light gases escape faster.
Effusion versus diffusion?
Effusion is escape through a tiny hole into vacuum or low pressure. Diffusion is mixing through space. Both follow the same mass ratio.
What conditions must hold?
Same temperature and pressure for both gases, ideal behavior, and a hole small enough that molecules pass without colliding.
How do times relate to rates?
The ratio of rates inverts into a ratio of times: t1/t2 = sqrt(M1/M2). The tool reports both.
Does it match textbook examples?
Yes. Enter 131.3 and 20.18 with 243 s for xenon to get neon at 95.27 s, matching the textbook example.
Is it exact for real gases?
No. Real gases deviate at high pressure, and porous barriers add Knudsen subtleties. Treat plant scale results as estimates.