Graham's Law of Effusion — Why Lighter Gases Escape Faster
Puncture a balloon filled with helium and one filled with air, and the helium balloon goes flat first. Graham's law is the arithmetic behind that observation. It is a short formula, it is guaranteed to appear in the states-of-matter chapter, and it is one of the few places where a square root shows up in Class 11 chemistry — which is precisely where students slip.
The formula
The rate of effusion of a gas is inversely proportional to the square root of its molar mass. Note the swap: gas 1's rate is on the left, but gas 1's molar mass is on the bottom on the right. That inversion is deliberate — heavier means slower.
What each symbol means
| Symbol | Meaning | Unit |
|---|---|---|
| rate₁, rate₂ | Volume or moles escaping per unit time | any, as long as both are the same |
| M₁, M₂ | Molar masses of the two gases | g mol⁻¹ |
Two conditions are assumed: both gases are at the same temperature and pressure, and the hole is small compared with the mean free path. The law follows directly from kinetic theory, where the root-mean-square speed is urms = √(3RT/M) — the same inverse square root of M.
The time version — get this the right way up
Rate and time are inversely related. If the same volume of each gas is collected:
So the heavier gas takes longer. Compare the two boxed formulas carefully: in the rate form the molar masses are crossed over, in the time form they are not. Writing down which quantity the question gives you — rate or time — before substituting prevents the most common error in this topic.
Worked example 1 — hydrogen against oxygen
How many times faster does H₂ effuse than O₂ at the same T and P?
M(H₂) = 2 × 1.008 = 2.016 g mol⁻¹; M(O₂) = 2 × 15.999 = 31.998 g mol⁻¹
rate(H₂)/rate(O₂) = √(31.998 ÷ 2.016) = √15.872
√15.872 = 3.984 (check: 3.984² = 15.872)
H₂ effuses about 3.98 times faster than O₂.
Notice that O₂ is roughly 16 times heavier, yet it is only about 4 times slower. The square root always softens the difference — a useful mental check.
Worked example 2 — identifying an unknown gas
An unknown gas effuses at 0.355 times the rate of helium under identical conditions. Find its molar mass.
rate(X)/rate(He) = √(MHe / MX) = 0.355
Square both sides: MHe / MX = 0.355² = 0.126025
MX = MHe ÷ 0.126025 = 4.0026 ÷ 0.126025
MX = 31.8 g mol⁻¹ ≈ 32 g mol⁻¹, so the gas is O₂.
This is the standard "identify the gas" question. Square first, then divide — never the other way round.
Worked example 3 — the time form
1.00 L of N₂ effuses through a pinhole in 60 s. How long will 1.00 L of CO₂ take through the same hole under the same conditions?
M(N₂) = 2 × 14.007 = 28.014; M(CO₂) = 12.011 + 2 × 15.999 = 44.009
t(CO₂) / t(N₂) = √(44.009 ÷ 28.014) = √1.5709 = 1.2534
t(CO₂) = 60 × 1.2534 = 75.2 s
CO₂ is heavier, so it takes longer. If your answer had come out below 60 s, the ratio was inverted.
Worked example 4 — the classic NH₃ / HCl tube
Cotton plugs soaked in concentrated ammonia and concentrated hydrochloric acid are placed at opposite ends of a 100 cm glass tube. Where does the white ring of NH₄Cl form?
M(NH₃) = 14.007 + 3 × 1.008 = 17.031; M(HCl) = 1.008 + 35.45 = 36.458
rate(NH₃)/rate(HCl) = √(36.458 ÷ 17.031) = √2.1407 = 1.4631
In the same time, distances travelled are in the same ratio, and together they must add to 100 cm. So the NH₃ front travels:
d(NH₃) = 100 × 1.4631 ÷ (1.4631 + 1) = 146.31 ÷ 2.4631 = 59.4 cm from the ammonia end
which puts the ring 100 − 59.4 = 40.6 cm from the HCl end. The lighter gas covers more ground, so the ring always forms closer to the heavier gas.
Common mistakes
- Forgetting the square root. Using M₂/M₁ instead of √(M₂/M₁) gives 16 instead of 3.98 for H₂ vs O₂ — an obviously wrong scale.
- Inverting the ratio. The lighter gas is faster. Decide which gas should win before you calculate, then check your answer agrees.
- Using atomic mass for a diatomic gas. Oxygen gas is O₂ = 32, not 16. Nitrogen is N₂ = 28, not 14. This single slip changes the answer by √2.
- Mixing rate and time forms. "Which effuses faster" is a rate question; "how long does it take" is a time question. They use crossed and uncrossed molar masses respectively.
- Comparing gases at different temperatures. The law assumes identical T and P for both gases.
- Treating diffusion in air as pure effusion. The law is exact for effusion through a pinhole; for diffusion through another gas it gives only an approximate ratio, which is why the tube experiment never lands exactly on 59.4 cm.
A real application worth knowing
Uranium enrichment historically used gaseous diffusion of UF₆. With M(²³⁵UF₆) = 235.044 + 6 × 18.998 = 349.03 and M(²³⁸UF₆) = 238.051 + 113.99 = 352.04, the separation factor per stage is √(352.04 ÷ 349.03) = √1.00862 = 1.0043 — a gain of well under half a percent, which is why thousands of stages were needed. Graham's law explains both the method and its enormous cost.
Where it appears in exams
| Exam | Typical question |
|---|---|
| CBSE/ICSE Class 11 | Rate ratio for two named gases; identifying an unknown gas |
| JEE / NEET | Time-based effusion, mixed-gas problems, link to urms |
| IIT-JAM / CUET-PG | Kinetic theory derivation, composition change of a gas mixture on effusion |
| GATE / CSIR-NET | Isotope separation factors, multi-stage enrichment calculations |
Verify the square root every time. Enter the two molar masses (or one rate and one molar mass) and the calculator returns the rate ratio, the time ratio and the unknown molar mass — so you can see instantly whether your ratio was the right way up.
Open the Graham's Law Calculator →States of matter is a short chapter with a high mark-to-effort ratio if the formulas are drilled properly. ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre and online classes across India, with home tuition available in Delhi-NCR — details at abcchemistry.in.