Rate of Effusion Calculator

Compare gas effusion rates with Graham's law using molar mass and one known rate.

Graham's law calculator
Enter a known rate and the molar masses of two gases.

About gas effusion rates

Effusion is the movement of gas particles through a very small opening into a vacuum or a region of much lower pressure. The rate depends on how quickly the particles move, and molecular speed depends strongly on molar mass when gases share the same temperature. Graham's law captures this relationship: the ratio of two effusion rates equals the square root of the inverse ratio of their molar masses. A lighter gas therefore passes through the opening faster than a heavier gas. This calculator finds the rate of a second gas from one measured or assigned rate and the molar masses of both gases. It evaluates Rate 2 = Rate 1 × square root of (Molar mass 1 / Molar mass 2). The known rate may use any consistent unit, such as milliliters per minute, moles per second, or a relative rate. The answer uses the same rate unit because the molar-mass ratio is dimensionless. Both molar masses must use the same unit, normally grams per mole. Graham's law follows from the kinetic molecular model. At equal temperature, gases have the same average translational kinetic energy. Since kinetic energy depends on mass and speed squared, particles with lower mass must have a higher characteristic speed. Effusion rate is proportional to that speed under ideal conditions. The square-root relationship is important: a gas with one quarter of another gas's molar mass effuses twice as fast, not four times as fast. The law works best for dilute gases, tiny openings, equal temperatures, and pressures low enough that collisions near the opening do not dominate. Real gases can deviate when intermolecular forces become important, while diffusion through another gas is more complicated because particles collide repeatedly. Effusion and diffusion are related ideas but are not interchangeable in careful laboratory work. Use molecular molar masses rather than atomic masses for gases such as hydrogen, nitrogen, and oxygen, because those elements normally occur as diatomic molecules. Keep temperatures equal when comparing experimental rates and make sure the opening and pressure conditions are unchanged. With those controls, the calculated ratio provides a quick prediction, a way to identify an unknown gas, or a useful check on experimental observations.

Rate of effusion examples

These comparisons show how increasing molar mass lowers the predicted rate.

Known valuesCalculated rateInterpretation
Rate 1 = 1, M1 = 2, M2 = 32Rate 2 = 0.25Hydrogen effuses four times faster than oxygen under equal conditions.
Rate 1 = 12, M1 = 28, M2 = 44Rate 2 = 9.5737Carbon dioxide is heavier than nitrogen, so its calculated rate is lower.
Rate 1 = 5, M1 = 4, M2 = 36Rate 2 = 1.6667The ninefold molar-mass ratio produces a threefold rate difference.

How to calculate an effusion rate

  1. Enter the measured or known effusion rate for gas 1.
  2. Enter the molar mass of gas 1 in grams per mole.
  3. Enter the molar mass of gas 2 using the same mass unit.
  4. Select Calculate effusion rate to apply Graham's law.
  5. Read the result in the same rate unit used for the known rate.

Rate of effusion FAQ

What is Graham's law of effusion?

Graham's law states that gas effusion rate is inversely proportional to the square root of molar mass. It lets you compare two gases at the same temperature and under equivalent conditions.

Which units should I use for the known rate?

You can use any rate unit as long as you interpret the result in that same unit. The calculator multiplies the known rate by a dimensionless molar-mass factor.

Why do lighter gases effuse faster?

At the same temperature, lighter particles have higher characteristic speeds for the same average kinetic energy. More of those faster particles reach and pass through a small opening per unit time.

Is effusion the same as diffusion?

No. Effusion describes escape through a tiny opening, while diffusion describes gases mixing through random particle motion and collisions.

When does Graham's law become inaccurate?

The ideal relationship can lose accuracy at high pressures, low temperatures, or when intermolecular forces are strong. Experimental comparisons also require equal temperatures and equivalent openings.