Chemistry: Atoms First 2e · Gases

Effusion and Diffusion of Gases

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In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

At the same temperature, all gas molecules have the same , 12mv². Since heavier molecules have more mass, they must move slower to have the same energy — and lighter molecules move faster. That single idea explains two closely related phenomena:

— a gas escapes through a tiny opening (a pinhole) into a vacuum or lower-pressure region.
— gas molecules spread through a space or mix with other gases as a result of random molecular motion.

Thomas Graham quantified the relationship in the mid-1800s: the rate at which a gas effuses (or diffuses) is inversely proportional to the square root of its molar mass:

$rate A rate B = \sqrt M B M A$

where rate is the amount of gas passing per unit time and M is molar mass. Lighter gases move faster; heavier gases lag behind. This topic explains , its molecular basis, and its surprising real-world consequences — from the smell of perfume to the industrial enrichment of uranium.

Why this matters

Uranium enrichment: gaseous diffusion of was the method used to separate fissile ²³⁵U from the more abundant ²³⁸U — Graham's law in action at industrial scale. The tiny mass difference produces only a small separation per stage, so thousands of stages were required.
Gas leaks and safety: helium and hydrogen leak through tiny openings faster than heavier gases, which is why helium is used to find leaks in vacuum systems and why hydrogen-filled balloons deflate quickly.
Respiration: oxygen and CO₂ move across the alveolar membrane partly by diffusion; rate differences matter for gas exchange and for breathing-gas mixtures (e.g., helium–oxygen mixes for divers).
Everyday chemistry: the time it takes to smell a fragrance across a room, the behavior of gas stove pilot lights, and the design of gas masks and membranes all depend on diffusion and effusion rates.
Exams: Graham's law problems are formula-driven but trap-prone — the ratio is inverted (heavier mass on top), and rates compare inversely with mass, not directly.

The college version

Core Concepts

Effusion vs. diffusion

Effusion is escape through a small hole — like air leaking from a pinhole in a balloon. Diffusion is the gradual mixing of gases due to random motion — like perfume spreading across a room. Both are driven by molecular motion, and Graham's law describes the rates of both, although effusion is the cleaner case to model: molecules pass through the hole one at a time, so the rate depends directly on how fast they travel.

Why lighter gases move faster: the kinetic-energy link

At a given temperature, the average kinetic energy of any gas is the same:

$KE = 12 m v 2$

Setting the average kinetic energies of two gases equal and solving for the speed ratio gives:

$v A v B = \sqrt M B M A$

Because speed determines how often molecules find the hole (effusion) or how quickly they spread (diffusion), the rate ratio follows the same square-root-of-inverse-mass form — Graham's law. Lighter molecules are not "more energetic"; they simply move faster for the same energy.

Graham's law in practice

For two gases A and B at the same temperature and pressure:

$rate A rate B = \sqrt M B M A$

Three equivalent operational forms appear in problems:

Rates: if A is lighter than B, A effuses faster and the ratio is greater than 1.
Times: since time is inversely proportional to rate, the time to effuse a fixed amount is proportional to the square root of molar mass:

$t A t B = \sqrt M A M B$

Distances: for diffusion in a given time, distance traveled is proportional to rate, so distances follow the same square-root ratio as rates.

The standard procedure: write the formula, decide which gas is A and which is B, substitute molar masses, and check that the lighter gas gets the faster rate (larger rate, shorter time, longer distance).

Real gases and the limits of the model

Graham's law assumes ideal behavior — no intermolecular attractions and molecules much smaller than the opening. It holds very well for gases at ordinary temperatures and pressures. Under extreme conditions (very high pressure, very low temperature, or openings comparable to molecular size) real gases deviate; the law remains an excellent approximation rather than an exact statement.

Common Confusions

Do Not Confuse	With	Difference
Effusion	Diffusion	Effusion is escape through a small hole; diffusion is spreading/mixing by random motion
Rate ratio M_B/M_A	M_A/M_B	The heavier mass goes on top (or under the lighter gas's rate). If your answer says the heavy gas is faster, the ratio is inverted
Rate	Time	Faster rate = shorter time. Time ratio is the inverse of the rate ratio: t_A/t_B = M_A/M_B
Molar mass	Density	Graham's law uses molar mass (g/mol), not density (g/L); densities of gases at the same T and P are proportional to molar mass, but the law is stated in molar mass
Lighter gas slower	Lighter gas faster	Lighter molecules have the same average kinetic energy but higher speed ⇒ faster effusion/diffusion
Temperature dependence	Mass dependence	The ratio depends only on molar mass at fixed temperature; absolute rates increase with temperature, but the ratio form cancels temperature

Eli explains

The same idea, in plain words

Explain it like I’m 10

Imagine two runners in a race: a tiny, light runner and a big, heavy runner. If both get the same amount of energy, the light one runs faster. Gas molecules are the same — light ones zip around quickly and slip through small holes sooner, while heavy ones lumber along. So a light gas like helium escapes a balloon faster than a heavy gas like the air around it. Graham's law is just the math that says "lighter means quicker."

Worked example

Example 1: Comparing effusion rates of hydrogen and oxygen

At the same temperature and pressure, how much faster does hydrogen gas (H₂, M = 2.016 g/mol) effuse than oxygen gas (O₂, M = 32.00 g/mol)?

Step 1 — Write Graham's law:

$rate H 2 rate O 2 = \sqrt M O 2 M H 2$

Step 2 — Substitute the molar masses:

$rate H 2 rate O 2 = \sqrt 32.00 2.016 = \sqrt 15.87 = 3.98$

Hydrogen effuses 3.98 times faster than oxygen. Sanity check: H₂ is much lighter than O₂, so the ratio must be well above 1. ✓ This is why hydrogen-filled balloons lose their gas quickly while oxygen tanks hold theirs.

Example 2: Identifying a gas from effusion time

A sample of an unknown gas effuses through a pinhole in 83.3 s. Under identical conditions, the same amount of oxygen (O₂, M = 32.00 g/mol) effuses in 47.1 s. What is the molar mass of the unknown gas?

Step 1 — Write Graham's law in time form:

$t unknown t O 2 = \sqrt M unknown M O 2$

Step 2 — Substitute the times:

$83.3 47.1 = 1.77 = \sqrt M unknown 32.00$

Step 3 — Square both sides and solve for M_unknown:

$(1.77) 2 = M unknown 32.00 \Rightarrow M unknown = 32.00 \times 3.13 = 100 g/mol$

The unknown takes longer, so it must be heavier than O₂ — consistent with a molar mass near 100 g/mol. The key check: longer time means larger molar mass.

Example 3: Uranium isotope separation by effusion

Natural uranium is mostly ²³⁸U (98.3%) with a small amount of ²³⁵U (0.7%). As the gas UF₆, the molar masses are approximately M₂₃₅ = 349.0 g/mol and M₂₃₈ = 352.0 g/mol. How much faster does ²³⁵UF₆ effuse than ²³⁸UF₆?

Step 1 — Write Graham's law with the lighter isotope as A:

$rate 235 rate 238 = \sqrt M 238 M 235 = \sqrt 352.0 349.0 = \sqrt 1.0086 = 1.0043$

Step 2 — Interpret. The lighter ²³⁵UF₆ effuses only about 0.43% faster — a tiny enrichment per pass. That is why real enrichment plants use thousands of cascaded diffusion stages to concentrate ²³⁵U. The small ratio is not a flaw in Graham's law; it is a direct consequence of how similar the two isotopes' masses are.

Example 4: Diffusion distances in the same time

Helium (He, M = 4.003 g/mol) and neon (Ne, M = 20.18 g/mol) begin diffusing from opposite ends of a long tube at the same temperature. In a fixed time, how much farther does helium travel than neon?

Step 1 — Write the distance form of Graham's law:

$d He d Ne = \sqrt M Ne M He = \sqrt 20.18 4.003 = \sqrt 5.04 = 2.24$

Helium travels 2.24 times farther than neon in the same time. Light gases simply cover more ground — the basis of the classic "two gases in a tube" diffusion demonstrations and of why helium leaks are so easy to spot.

Key takeaways

Graham's law: rate_A/rate_B = M_B/M_A — rates are inversely proportional to the square root of molar mass.
Lighter gas ⇒ faster effusion, shorter time, longer diffusion distance; heavier gas ⇒ the opposite.
Time form: t_A/t_B = M_A/M_B (times are directly proportional to the square root of molar mass).
The ratio is always set up so the lighter gas is the faster one — a built-in sanity check.
Both gases must be at the same temperature (and usually same pressure) for the ratio form to apply.
Effusion = escape through a small hole; diffusion = spreading/mixing by random motion.
Real-world uses: UF₆-based uranium isotope separation, helium leak detection, helium–oxygen breathing mixes.

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

State Graham's law in words and in equation form.
Show answer
The rate of effusion (or diffusion) of a gas is inversely proportional to the square root of its molar mass: rate_A/rate_B = M_B/M_A.
At the same temperature, which effuses faster: helium (He, 4.00 g/mol) or carbon dioxide (CO₂, 44.0 g/mol)? By what factor?
Show answer
Helium effuses faster: 44.0/4.00 = 11.0 = 3.32 times faster than CO₂.
A gas effuses in 2.00 times the time needed for O₂. What is its molar mass?
Show answer
t/t_O₂ = 2.00 = M/32.00, so M = 32.00 × 4.00 = 128 g/mol.
Why do hydrogen and helium balloons deflate noticeably faster than balloons filled with air?
Show answer
H₂ and He are far lighter than the N₂/O₂ mixture in air, so they effuse through the balloon's pores and leaks much faster — roughly 3–4× faster than air.
Why is separating ²³⁵U from ²³⁸U by effusion such a slow, multi-stage process?
Show answer
The two UF₆ isotopologues differ in molar mass by only ~1%, so each effusion stage enriches ²³⁵U by just ~0.4%; thousands of cascade stages are needed to reach useful enrichment.
What physical property of molecules explains Graham's law, and why must both gases be at the same temperature?
Show answer
At a given temperature all gases share the same average kinetic energy, so lighter molecules must move faster; the temperature cancels in the ratio, but the comparison is only valid at equal temperatures.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Effusion: Escape of gas through a tiny opening into lower pressure
Diffusion: Spreading and mixing of gases due to random molecular motion
Graham's law: Rate is inversely proportional to the square root of molar mass
Molar mass M: Mass of one mole of gas molecules (g/mol)
Average kinetic energy: 12mv², equal for all gases at a given temperature
Rate of effusion: Amount of gas passing through the opening per unit time
Uranium hexafluoride (UF₆): The volatile uranium compound used in isotope separation
molar mass, M: Mass of one mole of a substance, in g/mol; numerically equal to formula mass.

Sources & references

openstax.org — Chemistry Atoms First 2e

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