Chemistry 2e · Gases
Effusion and Diffusion of Gases
On this page 9 sections
In 30 seconds
Open a bottle of perfume across the room and within seconds you can smell it — the scent molecules traveled through the air on their own, with no fan and no stirring. That spreading is Diffusion Spreading and mixing of gases by random molecular motion Full entry →: the gradual mixing of gases caused by the random motion of their particles. A closely related process is Effusion Escape of gas through a tiny hole or porous barrier Full entry →: the escape of gas molecules through a tiny hole (a pinhole or Porous barrier Material with many microscopic holes (clay, membranes) Full entry →) into a vacuum or lower-pressure region. Both processes depend on how fast gas molecules move, which depends on their mass: lighter molecules move faster, so they diffuse and effuse faster. This idea is quantified by Graham's law Rate of effusion/diffusion is inversely proportional to the square root of molar mass Full entry →, discovered by Thomas Graham in 1846: the rate of effusion (or diffusion) of a gas is inversely proportional to the square root of its molar mass.
Why this matters
- Isotope separation: During World War II, the Manhattan Project separated uranium-235 from uranium-238 by effusing uranium hexafluoride gas (UF₆ Uranium hexafluoride, the gaseous compound used in isotope separation Full entry →) through porous barriers in a cascade. The tiny mass difference produces a real difference in effusion rates, exploited stage after stage.
- Everyday gas behavior: Helium balloons deflate overnight because helium effuses through the rubber faster than air enters; natural gas (mostly methane, M ≈ 16) leaks and disperses far faster than heavier gases, which is why ventilation and detectors matter.
- Respiration: Oxygen (M = 32) and carbon dioxide (M = 44) diffuse across the alveolar membrane; diffusion of gases in the lungs is a core physiological process described by these concepts.
- Industrial separations and safety: Effusion membranes separate gases in refineries; knowing that light gases spread fast guides ventilation design, fumigation, and flammable-gas safety.
- Exams: Graham's law problems (rate ratios, molar-mass identification, effusion times) are classic test items, often paired with kinetic-molecular-theory reasoning.
The college version
Core Concepts
Diffusion versus effusion
Diffusion is the mixing of two or more gases by random molecular motion — molecules wander, collide, and gradually spread from high concentration to low. It happens everywhere: perfume spreading, food smells reaching you from the kitchen, CO₂ dissolving out of a soda into the air above it. Effusion is specifically the escape of gas through a small opening — a pinhole in a balloon, a porous clay cup, a membrane with microscopic pores. The distinction matters for experiments: effusion is easier to quantify because it depends on the hole's size and molecular speeds, whereas diffusion in a mixture is complicated by collisions with other molecules. In both cases, the lighter the molecule, the faster the process.
Graham's law
Graham found that the rates of effusion of two gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses:
rate1rate2 = M2M1
where rate is amount of gas escaping per unit time (mol/s, molecules/s, or volume per time), and M is molar mass. The reasoning: at a given temperature, all gas molecules have the same Average kinetic energy 12mv2, equal for all gases at the same temperature Full entry →, KE = 12mv2. A lighter molecule must therefore move faster to have the same kinetic energy, so it reaches the hole and escapes more often. Because rate is proportional to average speed, and speed scales as 1/m (thus 1/M), the rate scales as 1/M.
Effusion time and amount
Because rate = amount/time, a gas that effuses more slowly takes proportionally longer to escape. The ratio of times for equal amounts is the inverse of the rate ratio:
t1t2 = M1M2
And the ratio of amounts effused in equal times equals the rate ratio:
amount1amount2 = M2M1
A common experimental design: allow equal volumes of two gases to effuse through the same hole and time them. The unknown gas's molar mass can then be found from its effusion time relative to a known gas.
A classic demonstration: ammonia and hydrogen chloride
A classic classroom experiment places a cotton plug soaked in concentrated ammonia solution (NH₃, M = 17.03 g/mol) at one end of a glass tube and a plug soaked in concentrated HCl (M = 36.46 g/mol) at the other. The gases diffuse toward each other and react where they meet, forming a white ring of solid ammonium chloride:
NH3(g) + HCl(g) → NH4Cl(s)
Because NH₃ is lighter, it diffuses faster and travels farther; the white ring forms measurably closer to the HCl end. The ring's position depends on the ratio of diffusion distances, which follows the square root of the molar mass ratio. (Note: concentrated ammonia and HCl are hazardous — this is performed with proper ventilation and protective equipment as a general lab-safety principle, never casually.)
Uranium enrichment by effusion
Natural uranium is 99.3% uranium-238 and only 0.7% uranium-235 (the fissile isotope). Both form the same compound, UF₆, whose molar masses differ slightly: about 352.04 g/mol for ²³⁸UF₆ and 349.03 g/mol for ²³⁵UF₆. Graham's law gives the effusion rate ratio:
rate235UF6rate238UF6 = 352.04349.03 = 1.0043
Each pass through a porous barrier enriches the ²³⁵ fraction by only ~0.4%, so thousands of stages are needed — a striking example of a tiny mass difference, applied repeatedly, separating isotopes on an industrial scale. Modern enrichment largely uses gas centrifuges, but the effusion principle remains the classic explanation.
Common Confusions
| Do Not Confuse | With | Difference |
|---|---|---|
| Diffusion | Effusion | Diffusion is mixing within a space; effusion is escape through a hole. Both follow speed dependence on 1/M |
| Rate ratio | Time ratio | Rate is inversely proportional to time: t1/t2 = M1/M2, the opposite arrangement. Heavier gas = slower rate = longer time |
| Molar mass M | Molecular speed | Speed depends on 1/M, not 1/M — a gas 4× heavier moves only 2× slower |
| Average speed | Average kinetic energy | Kinetic energy is the same for all gases at a given T; speeds differ because masses differ |
| Lighter gas effusing faster | Lighter gas always diffusing fastest | In dense mixtures, collisions slow diffusion; Graham's law is exact for effusion, approximate for diffusion |
| Effusion (physical process) | Isotope decay | Effusion separates isotopes by mass through barriers; it does not change the atoms themselves |
| Using grams in Graham's law | Using molar mass | Rate depends on molar mass (g/mol), not the mass of a single molecule in grams |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Imagine two kids racing through a gate: one carrying a heavy backpack and one carrying nothing. With the same effort, the kid with no backpack is faster and gets through the gate more often. Gas molecules are the same: lighter molecules run faster, so they escape through tiny holes and spread through the air quicker than heavy ones. If you make the "backpack" four times heavier, the speed gets cut in half — that's what the square root in Graham's law means.
Worked example
Example 1: Rate ratio — helium versus oxygen
How much faster does helium (He, M = 4.00 g/mol) effuse than oxygen (O₂, M = 32.00 g/mol) at the same temperature?
Step 1 — write Graham's law:
rateHerateO2 = MO2MHe
Step 2 — substitute molar masses (heavier gas in the numerator gives a ratio > 1):
rateHerateO2 = 32.004.00 = 8.00 = 2.83
Helium effuses 2.83 times faster than oxygen — which is why a helium balloon loses gas much faster than an air-filled balloon of the same material.
Example 2: Effusion time — identifying an unknown gas
Equal volumes of an unknown gas and nitrogen (N₂, M = 28.02 g/mol) effuse through the same hole at the same temperature and pressure. The unknown takes 90.0 s; nitrogen takes 60.0 s. Find the unknown's molar mass.
Step 1 — write the time form of Graham's law:
tunktN2 = MunkMN2
Step 2 — substitute times and solve for Munk:
90.060.0 = Munk28.02
Step 3 — square both sides:
(1.50)2 = Munk28.02 → Munk = 2.25 × 28.02 = 63.0 g/mol
The molar mass is about 63 g/mol, consistent with sulfur dioxide (SO₂, 64.07 g/mol). Check: the unknown is roughly twice as heavy as N₂, so it effuses half as fast — and 90 s > 60 s, consistent.
Example 3: Diffusion distances — ammonia and hydrogen chloride
In the NH₃/HCl tube experiment at the same temperature, compare how far each gas diffuses before meeting. Take the diffusion distance ratio as equal to the rate ratio.
Step 1 — write Graham's law:
distanceNH3distanceHCl = MHClMNH3
Step 2 — substitute (MNH3 = 17.03, MHCl = 36.46):
distanceNH3distanceHCl = 36.4617.03 = 2.14 = 1.46
NH₃ travels about 1.46× farther than HCl, so the white ring of NH₄Cl forms closer to the HCl end of the tube — the classic result.
Key takeaways
- Diffusion = mixing of gases by random motion; effusion = escape through a small hole.
- Graham's law: rate1rate2 = M2M1 — lighter gas effuses/diffuses faster.
- At the same temperature, all gases have the same average kinetic energy; lighter molecules move faster.
- Effusion time ratio is the inverse of the rate ratio: t1/t2 = M1/M2.
- He (M = 4) effuses about 2.8× faster than O₂ (M = 32): 32/4 = 8 ≈ 2.83.
- NH₃/HCl tube experiment: the white NH₄Cl ring forms closer to the HCl end because NH₃ diffuses faster.
- Molar masses must be in the same units (g/mol) in both numerator and denominator; the ratio is unitless.
Check yourself
6 review questions from the chapter. Try each one, then open the answer.
State Graham's law in words and as an equation.
Show answer
The rate of effusion (or diffusion) of a gas is inversely proportional to the square root of its molar mass: rate1/rate2 = M2/M1.
Which effuses faster at the same temperature: H₂ (M = 2.02) or CO₂ (M = 44.01)? By what factor?
Show answer
H₂, by a factor of 44.01/2.02 = 21.8 ≈ 4.7. Hydrogen is the fastest effusing common gas.
A gas effuses in 100 s; an equal amount of O₂ (M = 32.00) effuses in 50 s under identical conditions. What is the unknown gas's molar mass?
Show answer
tunk/tO2 = 100/50 = 2 = Munk/32.00, so Munk = 4 × 32.00 = 128 g/mol (consistent with xenon, 131.29 g/mol).
Why do lighter molecules move faster at the same temperature?
Show answer
At the same temperature all gases have the same average kinetic energy 12mv2; a smaller mass must have a larger speed to give the same kinetic energy.
In the NH₃/HCl experiment, why does the white ring form closer to the HCl end?
Show answer
NH₃ (M = 17.03) diffuses faster than HCl (M = 36.46), traveling farther before the gases meet and react; the ring marks where the reaction occurs, closer to the HCl end.
Is Graham's law more accurate for diffusion or effusion? Why?
Show answer
Effusion — Graham's law is exact for effusion of an ideal gas through a small hole, while diffusion in a mixture is slowed by collisions with other molecules, so the law is an approximation there.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- Diffusion
- Spreading and mixing of gases by random molecular motion
- Effusion
- Escape of gas through a tiny hole or porous barrier
- Graham's law
- Rate of effusion/diffusion is inversely proportional to the square root of molar mass
- Molar mass (M)
- Mass of one mole of gas in g/mol
- Average kinetic energy
- 12mv2, equal for all gases at the same temperature
- Porous barrier
- Material with many microscopic holes (clay, membranes)
- UF₆
- Uranium hexafluoride, the gaseous compound used in isotope separation
Sources & references
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