Graham's Law of Effusion

Introduction

At the same temperature and pressure, lighter ideal-gas molecules generally cross a tiny opening faster than heavier ones. Graham's law quantifies that comparison through the inverse square root of molar mass. The law is easy to invert accidentally because rate and time for a fixed amount run in opposite directions. It also applies to an ideal effusion setup, not every gas leak or room-scale spreading process.

Core explanation

For gases A and B under matching temperature, pressure, opening and ideal effusion conditions, rA/rB = √(MB/MA), where r is amount effusing per unit time and M is molar mass. If MA is smaller, the ratio exceeds one: the lighter gas effuses faster. The square-root dependence follows from characteristic thermal speed scaling as 1/√M at equal T. The same conditions also give equal ideal number density, so the speed difference controls the flux through the identical small opening.

Suppose helium has M ≈ 4 g/mol and neon has M ≈ 20 g/mol. Their ideal effusion-rate ratio is rHe/rNe = √(20/4) = √5 ≈ 2.24. The helium rate is about 2.24 times the neon rate, not five times. A common error is to use the mass ratio directly or to place MA and MB in the wrong order. Write a sentence stating which gas should be faster before evaluating the square root.

If equal amounts of A and B are allowed to effuse in separate comparable experiments at approximately steady rates, time is inversely related to rate: tA/tB = rB/rA = √(MA/MB). For the helium-neon example, helium would need about 1/2.24 the time neon needs for the same amount in a simplified constant-rate comparison. In a real closed container, pressure and composition fall as gas escapes, so rates need not stay constant; the equal-amount time shortcut assumes comparable conditions throughout.

The law can estimate an unknown molar mass. If A effuses 1.50 times as fast as B and MB is known, (rA/rB)² = MB/MA, so MA = MB/(1.50)². Squaring the rate ratio is essential. A faster gas should yield a lower inferred molar mass, a direction check that catches many algebra mistakes.

Mixtures and isotope separation add nuance. A lighter isotopic molecule can effuse slightly faster than a heavier one, but because isotope molar masses may differ only a little, the rate ratio can be close to one. Repeated separation stages may be required for a substantial composition change. This is a conceptual consequence of the square-root relation, not an invitation to ignore practical engineering and safety requirements.

Effusion assumes an opening small enough that passage is governed by individual molecular arrivals rather than bulk viscous flow. The receiver is usually at low pressure, and the gases should be compared at equal source T and P. A leak through a long crack may involve collisions within the channel. Diffusion through air depends on collisions with air molecules. Applying Graham's ideal ratio to either setup without qualification can be inaccurate.

Density can replace molar mass in the ratio only at the same P and T under the ideal model, because ρ ∝ M then. Thus rA/rB = √(ρB/ρA) under those conditions. Density readings from different states cannot be inserted directly. Convert them or use molar masses. Similarly, temperature differences require a more general speed comparison; the simple Graham ratio presumes equal T.

Step-by-step reasoning

1. Verify same temperature, pressure and tiny-opening effusion conditions. 2. Identify which gas is lighter and predict that it should have the larger rate. 3. Write rA/rB = √(MB/MA) with labels intact. 4. Square or invert only after deciding whether the question asks for rate, time or unknown mass. 5. Check the result's direction and whether steady-rate assumptions are reasonable.

Visual explanation

Draw two identical containers with equal P and T and matching tiny openings. Make light-molecule arrows longer on average and heavy-molecule arrows shorter. Show more light particles crossing per unit time. Beneath write the labelled ratio rlight/rheavy = √(Mheavy/Mlight), avoiding an unlabeled fraction that could be inverted.

Real-world analogy

If two groups have the same number density near a narrow gate but one group moves faster, more members can cross per second. The idea suggests a rate ratio. Molecular direction distributions and collision conditions, however, are what determine the actual effusion law.

Real-world example

A laboratory might compare small-hole effusion rates of two inert gases to illustrate relative molecular speeds. The observed rate ratio can be compared with the square-root molar-mass prediction, while pressure changes during the experiment and imperfect hole geometry explain deviations.

Why?

Why is the mass ratio under a square root? Equal-temperature particles have comparable average translational kinetic energy, proportional to mv². Characteristic speed therefore scales as 1/√m, and the ideal effusion rate follows the arrival speed at the opening.

Common misconception

“If gas A effuses twice as fast, it takes twice as long to release the same amount.” A faster rate means less time for the same amount in a constant-rate comparison. Time ratios are the inverse of rate ratios, not the same ratio.

Worked example

An unknown gas A effuses 1.50 times as fast as oxygen gas B (MB = 32.0 g/mol) under matching ideal conditions. Set 1.50 = √(32.0/MA). Square: 2.25 = 32.0/MA. Thus MA = 32.0/2.25 ≈ 14.2 g/mol. The inferred molar mass is lower than oxygen's, consistent with faster effusion. This result estimates mass; it does not identify a unique substance without other evidence.

Quick check

1. At equal T and P, how does effusion rate compare for gases of 4 and 36 g/mol? Answer: The 4-g/mol gas effuses √(36/4) = 3 times as fast in the ideal pinhole model.

Exam focus

Label each gas in the ratio before substituting, square carefully for unknown M and invert for equal-amount time. State matched conditions and distinguish effusion through a tiny opening from diffusion or bulk leakage.

Advanced insight

The molecular flux to a wall is proportional to number density times mean molecular speed. At equal ideal P and T, number density is the same for both gases, leaving a 1/√M speed factor. If source pressures differ, the number-density factor also differs, so a molar-mass-only ratio is no longer sufficient.

Summary

Graham's law gives rA/rB = √(MB/MA) for comparable ideal effusion. Lighter gas is faster, and equal-amount time ratios are inverse to rate ratios under steady conditions. The square-root relation requires matched state and pinhole geometry.

Practice questions

1. Find the ideal rate ratio rHe/rO₂ for MHe = 4 and MO₂ = 32 g/mol. Answer: √(32/4) = √8 ≈ 2.83. 2. If A effuses twice as fast as B, what is the ideal time ratio tA/tB for equal amounts at steady rates? Answer: 1/2; A takes half as long. 3. Why might a leak through a wide valve not follow Graham's law? Answer: Wide openings can produce bulk flow rather than individual-molecule effusion through a tiny hole.