How this instrument works
Graham's Law of diffusion (also called Graham's Law of effusion) states that the rate at which a gas effuses through a tiny opening, or diffuses through another gas, is inversely proportional to the square root of its molar mass. Put two gases side by side and the lighter one always wins the race, and this formula says by exactly how much: the ratio of their rates equals the square root of the inverse ratio of their molar masses. Double a gas's molar mass and its rate falls not by half, but by a factor of the square root of two — about 0.71 times as fast.
The square root falls straight out of kinetic theory rather than being an empirical curve fit. At a given temperature, every gas molecule in a mixture shares the same average kinetic energy, ½mv² — heavier molecules simply move slower to keep that product constant. Solve ½m₁v₁² = ½m₂v₂² for the velocity ratio and the masses land under a square root, which is the entire derivation behind Thomas Graham's 1848 observation, made decades before anyone could measure a molecular mass directly.
The law is exact for effusion — molecules leaking one at a time through an opening much smaller than their mean free path, so they never collide with each other on the way out. Diffusion, where a gas migrates through another gas or through air, follows the same square-root trend but is a looser approximation, since molecules collide constantly with whatever they are diffusing through. That distinction matters industrially: uranium enrichment by gaseous diffusion of UF6 relies on effusion-like separation, and it takes thousands of stages precisely because ²³⁵U and ²³⁸U differ in molar mass by only about 1%, giving a rate ratio barely above 1.004.
- Enter the molar mass of your reference gas into Molar mass of gas 1 (g/mol) — hydrogen's 2.016 g/mol loads by default.
- Enter the second gas's molar mass into Molar mass of gas 2 (g/mol) — oxygen's 32 g/mol loads by default.
- Read Rate1 / Rate2 (effusion/diffusion ratio): a value above 1 means gas 1 moves faster than gas 2; below 1 means gas 1 moves slower.
- Swap the two molar masses to flip the comparison — Rate1/Rate2 and Rate2/Rate1 are exact reciprocals of one another.
- Molar mass of gas 1 (g/mol) must stay above zero -- it's the denominator, so zero or negative values aren't accepted. Molar mass of gas 2 (g/mol) may be zero, which correctly returns a ratio of 0.
Worked example — hydrogen effusing against oxygen
Enter 2.016 into Molar mass of gas 1 (g/mol) — hydrogen's standard atomic-weight-based molar mass — and 32 into Molar mass of gas 2 (g/mol) for oxygen. Rate1 / Rate2 (effusion/diffusion ratio) reads 3.984095: the instrument forms the ratio 32 / 2.016 = 15.873 and takes its square root.
That result matches the textbook shorthand that hydrogen effuses roughly four times faster than oxygen, and it follows exactly from the two molar masses entered — no measured leak time or pressure drop is needed once the identities of both gases are known.
Questions
Why does the lighter gas always diffuse or effuse faster?
Because at any shared temperature, every gas molecule carries the same average kinetic energy, ½mv². For that product to stay constant, a molecule with smaller mass m must have a larger average speed v — there's no way around it. Graham's Law simply restates that physical fact as a ratio: rate is proportional to 1/sqrt(M), so halving the molar mass multiplies the rate by sqrt(2), about 1.41 times faster, not twice as fast.
What's the real difference between effusion and diffusion?
Effusion is a gas leaking through an opening so small that molecules pass through one at a time without colliding with each other near the hole — into a vacuum, ideally. Diffusion is a gas spreading through another gas or through air, where constant collisions slow the process down. Graham's Law is derived exactly for effusion; it still describes diffusion's trend accurately, but diffusion rates are also affected by the medium the gas is moving through, so the match is closer for effusion.
Does it matter which gas goes in field 1 versus field 2?
No information is lost either way, but the number changes. Putting the lighter gas in field 1 gives a ratio greater than 1 (gas 1 is faster); swapping the two molar masses gives the reciprocal, a ratio less than 1. Rate1/Rate2 and Rate2/Rate1 always multiply to exactly 1, so pick whichever arrangement answers the question you're actually asking.
What is Graham's Law used for outside the classroom?
Its best-known industrial use is uranium enrichment: natural uranium is converted to gaseous UF6 and passed through thousands of membrane stages, each one nudging the mixture slightly richer in the lighter, fissile ²³⁵U isotope, because ²³⁵UF6 and ²³⁸UF6 differ in molar mass by only about 3 g/mol out of roughly 349. It is also used more simply in the lab to identify an unknown gas by timing its effusion rate against a known reference gas.
Can this formula compare isotopes of the same element?
Yes — Graham's Law depends only on molar mass, so it applies equally well to two isotopologues of the same compound, such as ²³⁵UF6 versus ²³⁸UF6. The catch is that isotopes are usually close in mass, so the resulting ratio sits just barely above 1 (around 1.0043 for uranium hexafluoride), which is why separating isotopes this way takes many repeated stages rather than one pass.
Why does the instrument reject a molar mass of zero or less?
Molar mass is a physical quantity that can't be zero or negative for a real substance, and the formula divides by M1, so a non-positive entry would make the ratio undefined or nonsensical rather than simply wrong. The instrument blanks the reading and explains why on the formula line instead of returning a meaningless number.