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Graham\'s Law of Diffusion Calculator

Ready to calculate
r₁/r₂ = √(M₂/M₁).
4-way solver.
Effusion + Diffusion modes.
100% Free.
No Data Stored.

How it Works

01Pick Effusion or Diffusion

Effusion = gas through small hole into vacuum (strict). Diffusion = gas through another gas (approximate).

02Enter Any 3 of 4 Values

Rate of Gas 1, rate of Gas 2, molar mass of Gas 1, molar mass of Gas 2 — leave one blank.

03Apply r₁/r₂ = √(M₂/M₁)

From kinetic theory: lighter gases move faster at the same T (½mv² = (3/2)kT). Rate inversely proportional to √M.

04Solve for the Missing Value

Calculator inverts the equation as needed. Use for isotope separation, leak prediction, MW determination.

What is a Graham's Law of Diffusion Calculator?

Graham's law of effusion (and approximately diffusion) is one of the most-cited results in classical physical chemistry — discovered empirically by Scottish chemist Thomas Graham in 1846 and later derived from kinetic-molecular theory, it states that the rate of effusion (or diffusion) of a gas is inversely proportional to the square root of its molar mass: r₁ / r₂ = √(M₂ / M₁). The deeper meaning is that at constant temperature, all gas molecules have the same average kinetic energy ½mv² = (3/2)kT, so lighter molecules necessarily move faster (v ∝ 1/√M). This single proportionality explains why helium balloons deflate faster than air-filled latex (He effuses 2.7× faster), why hydrogen leaks from any container faster than any other gas, and why uranium-isotope enrichment via gaseous diffusion (²³⁵UF₆ vs ²³⁸UF₆) requires thousands of stages — the per-stage separation factor is only √(352.04/349.03) = 1.0043.

Our calculator is a flexible 4-way solver: enter any 3 of {rate of Gas 1, rate of Gas 2, molar mass of Gas 1, molar mass of Gas 2}, and the calculator solves for the 4th using the appropriate algebraic rearrangement of Graham's law. Two operating modes are supported. Effusion — gas escaping through a small hole into vacuum, where Graham's law is mathematically exact (molecules pass independently with no inter-molecular collisions). Diffusion — gas spreading through another gas, where the formula is an approximation; real diffusion rates depend on molecular-collision dynamics in the surrounding medium and deviate from Graham's law by 5-20%. The result panel returns the solved value with full breakdown showing the rate ratio, the molar-mass ratio, and √(M₂/M₁) verification.

Smart warnings flag the four most common error modes: unphysical molar masses (< 1 g/mol or > 10⁶ g/mol — only sub-atomic entities are below 1, only very large biomolecules above 1 MDa); unrealistic rate ratios (> 100 or < 0.01, corresponding to molar-mass ratios > 10,000 — far beyond any real gas pair); diffusion-mode reminders that the formula is approximate. Designed for general-chemistry students learning kinetic-molecular theory, AP Chemistry / IB Chemistry students working through Graham's-law problem sets, physical-chemistry courses deriving the law from the Maxwell-Boltzmann distribution, instructors generating problem sets, and any researcher needing a quick effusion / molar-mass calculation — runs entirely in your browser, no account, no data stored.

Pro Tip: Pair this with our Molarity Calculator for solution chemistry, our Partial Pressure Calculator for gas mixtures, our Mole Calculator for the foundational n = m/M conversion, or our Grams to Moles Calculator for stoichiometry.

How to Use the Graham's Law Calculator?

Pick Effusion or Diffusion Mode: Effusion when gas escapes through a small hole or porous barrier into vacuum (or another low-pressure region) — Graham's law is exact. Diffusion when gas spreads through another gas (e.g. ammonia diffusing through air) — Graham's law is approximate; real diffusion rates can deviate 5-20% due to collision dynamics in the surrounding medium.
Enter Any 3 of 4 Values: Rate of Gas 1, rate of Gas 2, molar mass of Gas 1, molar mass of Gas 2. Leave the unknown blank — the calculator solves for it. Rates can be in any consistent unit (mol/s, mL/min, mm/s, mass/time) — only the ratio matters for Graham's law; units cancel.
Apply r₁/r₂ = √(M₂/M₁): the calculator algebraically rearranges to solve for the missing value: r₁ = r₂ × √(M₂/M₁); r₂ = r₁ × √(M₁/M₂); M₁ = M₂ × (r₂/r₁)²; M₂ = M₁ × (r₁/r₂)². Note the squared rate ratio when solving for molar mass — small errors in rate measurement amplify in the molar-mass calculation.
Verify Against Common Gas Pairs: H₂ (M = 2.016) vs O₂ (M = 32.00): rate ratio r(H₂)/r(O₂) = √(32/2) = √16 = 4. Hydrogen effuses 4× faster than oxygen at the same T,P. He (M = 4) vs N₂ (M = 28): rate ratio = √7 = 2.65. NH₃ (M = 17) vs HCl (M = 36.5): rate ratio = √(36.5/17) = 1.466. (Famous demo: NH₃ and HCl meet about 60% of the way along a tube from the HCl end, forming a white NH₄Cl ring.)
Use for Molar-Mass Determination: measuring effusion rate of an unknown gas vs a reference (e.g. nitrogen) and applying M_unknown = M_N₂ × (r_N₂ / r_unknown)² gives the unknown molar mass. Historical method for early identification of noble gases and small molecules; still useful for educational labs and qualitative checks.
Sanity-Check Effusion Times: at the same temperature, effusion time t ∝ 1/rate, so t₁/t₂ = √(M₁/M₂). A balloon filled with a heavier gas takes longer to deflate. Helium escapes 2.7× faster than air → He balloon deflates 2.7× faster. Hydrogen 3.8× faster than air → fastest leak rate of any common gas.
Account for Effusion vs Diffusion Carefully: Graham measured BOTH and found similar laws; effusion is exact (molecules cross hole independently), diffusion is approximate (molecules collide repeatedly with surrounding gas). For undergraduate chemistry purposes, both follow Graham's law to first approximation; for industrial gas-separation engineering, Fick's law of diffusion (with diffusion coefficient D₁₂) is preferred.

How is Graham's law calculated?

Graham's law is one of the cleanest applications of kinetic-molecular theory — a single algebraic relationship connecting macroscopic effusion rates to molecular masses via the Maxwell-Boltzmann distribution. The law is empirically simple but conceptually deep: it directly relates the average speed of gas molecules to their mass.

References: Thomas Graham, Phil. Trans. R. Soc. London 136 (1846) 573; Atkins' Physical Chemistry (12th ed., Chapter 1); Levine's Physical Chemistry (7th ed.); Maxwell's Theory of Heat (1871).

Core Formula

r₁ / r₂ = √(M₂ / M₁)

Where r is the rate of effusion (or approximately diffusion) and M is the molar mass. The lighter gas (smaller M) has the larger rate. Algebraic rearrangements: r₁ = r₂ × √(M₂/M₁); r₂ = r₁ × √(M₁/M₂); M₁ = M₂ × (r₂/r₁)²; M₂ = M₁ × (r₁/r₂)².

Derivation from Kinetic-Molecular Theory

The kinetic theory of gases gives the average kinetic energy of a gas molecule:

½ m v² = (3/2) k T

where m is the molecular mass, v is the average speed, k is Boltzmann's constant, and T is temperature. Solving for v: v = √(3kT/m). At constant T, v ∝ 1/√m, equivalently v ∝ 1/√M (since M = m × Nₐ).

Effusion rate is proportional to the average molecular speed (rate of molecules hitting the hole per unit area), so r ∝ v ∝ 1/√M, which gives r₁/r₂ = √(M₂/M₁).

Worked Example — Hydrogen vs Oxygen

M(H₂) = 2.016 g/mol, M(O₂) = 32.00 g/mol.

  • Rate ratio r(H₂) / r(O₂) = √(32.00 / 2.016) = √15.87 = 3.98.
  • Hydrogen effuses ~4× faster than oxygen at the same T and P.
  • Effusion time ratio: t(H₂) / t(O₂) = 1/3.98 = 0.251 — a hydrogen balloon deflates ~4× faster than an oxygen balloon of equivalent volume.

Worked Example — Uranium Isotope Separation

UF₆ is the only practical uranium gas (sublimes at 56 °C; usable in gaseous diffusion). Compare ²³⁵UF₆ vs ²³⁸UF₆.

  • M(²³⁵UF₆) = 235.04 + 6 × 18.998 = 349.03 g/mol.
  • M(²³⁸UF₆) = 238.05 + 6 × 18.998 = 352.04 g/mol.
  • Per-stage separation factor: r(²³⁵UF₆) / r(²³⁸UF₆) = √(352.04 / 349.03) = √1.00863 = 1.00430.
  • Each diffusion stage enriches the lighter ²³⁵U by only 0.43%. Going from natural 0.72% ²³⁵U to weapons-grade 90% requires > 1500 stages cascaded; commercial 3-5% reactor fuel needs ~1200 stages.
  • This is why uranium enrichment was the largest engineering project of WWII (Manhattan Project K-25 plant) and is the gating step in nuclear weapon development.

Worked Example — Ammonia and HCl Diffusion (Classic Demo)

Cotton swabs at opposite ends of a glass tube, soaked in NH₃ and HCl solutions. The two gases diffuse toward each other through air; they meet and form a white ring of NH₄Cl(s).

  • M(NH₃) = 17.03 g/mol; M(HCl) = 36.46 g/mol.
  • Rate ratio r(NH₃) / r(HCl) = √(36.46 / 17.03) = √2.142 = 1.464.
  • NH₃ diffuses 1.464× faster, so the white ring forms closer to the HCl end. Distance ratio (from HCl end): NH₃ travels 1.464 / (1 + 1.464) = 59.4% of the tube length; ring at ~60% from HCl end.
  • Real demos show the ring slightly closer to HCl than predicted (50-55%) because diffusion through air involves collisions that slow heavier molecules disproportionately.

Common Gas Effusion Rate Comparisons (Relative to Air, M ≈ 28.97)

  • H₂ (M 2.016): r/r(air) = √(28.97/2.016) = 3.79× faster.
  • He (M 4.003): 2.69× faster.
  • CH₄ (M 16.04): 1.34× faster.
  • Ne (M 20.18): 1.20× faster.
  • N₂ (M 28.01): 1.02× faster (slightly faster than air, which is mostly N₂).
  • Air (M 28.97): reference 1.00.
  • O₂ (M 32.00): 0.951× (slightly slower than air).
  • Ar (M 39.95): 0.851×.
  • CO₂ (M 44.01): 0.811×.
  • Cl₂ (M 70.91): 0.639×.
  • SF₆ (M 146.06): 0.445×.
  • UF₆ (M ~352): 0.287×.
Real-World Example

Worked Example — Determine Molar Mass from Effusion Rate

Question: An unknown gas effuses 0.500× as fast as oxygen at the same temperature. What is the molar mass of the unknown gas?

Step 1 — Identify the Knowns.

  • Gas 1: oxygen (O₂), M₁ = 32.00 g/mol, r₁ = 1.0 (reference rate).
  • Gas 2: unknown, M₂ = ?, r₂ = 0.500 × r₁ = 0.500.

Step 2 — Apply Graham's Law.

  • r₁ / r₂ = √(M₂ / M₁).
  • 1.0 / 0.500 = 2.0 = √(M₂ / 32.00).
  • Square both sides: 4.0 = M₂ / 32.00.
  • M₂ = 4.0 × 32.00 = 128 g/mol.

Step 3 — Identify the Gas.

  • M = 128 g/mol matches HI (hydrogen iodide, 127.91 g/mol) closely.
  • Other plausible matches: SO₂F₂ (sulfuryl fluoride, 102.06) — no; PF₃ (87.97) — no; PF₅ (125.97) — close but slightly low.
  • Best match: hydrogen iodide HI.

Step 4 — Verify.

  • Predicted ratio: r(O₂) / r(HI) = √(127.91 / 32.00) = √4.00 = 2.00 ✓.
  • The unknown gas effuses 0.500× as fast as O₂ matches HI within measurement precision.

Step 5 — Sources of Error.

  • The unknown might be a polyatomic molecule with similar M (PF₅ 126, HI 128, SiH₄I (silyl iodide) 158 — too heavy). Effusion alone cannot distinguish; combine with mass-spec or NMR for definitive ID.
  • Real-gas deviations: at high pressure or near critical T, effusion rates deviate from Graham's law by 1-5%; for educational problems, ideal-gas behavior is assumed.
  • Molar-mass uncertainty in the rate-ratio formula scales as 2× the rate measurement uncertainty (because of the square in M = M_ref × (r_ref/r)²) — so a 5% error in rate measurement gives ~10% error in M.

Who Should Use the Graham's Law Calculator?

1
Graham's law is in every general-chemistry curriculum (gas laws unit) and a standard AP / IB Chemistry topic. The 4-way solver lets students verify problem-set answers and check their algebra.
2
Connect macroscopic effusion rates to microscopic molecular speeds via the Maxwell-Boltzmann distribution. Useful for derivations and conceptual problems.
3
Compute per-stage separation factors √(M_heavy/M_light) for gaseous diffusion plants. Determine number of cascaded stages required for target enrichment (e.g. ²³⁵U from 0.72% to 5% reactor fuel grade).
4
Determine M of an unknown gas by measuring its effusion rate vs a reference (typically N₂ or air). Useful for educational labs and qualitative gas identification.
5
Hydrogen leaks 3.8× faster than air through any container; helium 2.7× faster. The calculator quantifies relative leak rates for design of pressurized-gas systems and balloon longevity.
6
Mass spec separates ions by m/z; Graham's law is the kinetic-theory analog for neutral gases and gives intuition for why mass-dependent separation works.
7
Membrane-based gas separation (H₂ purification from natural gas, He recovery from air) follows Graham's-law-like scaling at the molecular level (modified by membrane permeability constants).

Technical Reference

Historical Origin. Thomas Graham (1805-1869), a Scottish chemist at University of Glasgow, published his law of effusion in Philosophical Transactions of the Royal Society in 1846 ("On the Motion of Gases"). He measured the time required for various gases to escape through small holes in graphite plates, observing that the times varied as the square root of the densities (which, at constant T and P, equals the square root of the molar masses). The kinetic-theory derivation came later (Clausius 1857, Maxwell 1860, Boltzmann 1872) — providing the molecular interpretation.

Effusion vs Diffusion. Effusion is gas passing through a hole or porous barrier whose diameter is small compared to the molecular mean free path; molecules cross independently with no inter-molecular collisions. Graham's law is exact for this case. Diffusion is gas spreading through another gas at comparable pressure; molecules undergo many collisions per second with the surrounding medium. Graham's law gives the limiting case (zero pressure of surrounding gas) but real diffusion rates depend on the binary diffusion coefficient D₁₂. For ideal-gas mixtures: D₁₂ = (3/16) × √(2πk_B T (1/m₁ + 1/m₂)) / (n × π × σ²₁₂), where σ₁₂ is the collision diameter. Numerically D₁₂ ≈ √(M₁ + M₂) / (M₁ × M₂)^(1/2) — close to but not identical to Graham's scaling.

Average Molecular Speeds. Three different "average" speeds arise in kinetic theory; all scale as 1/√M:

  • Most probable speed v_p: v_p = √(2RT/M).
  • Mean (average) speed v̄: v̄ = √(8RT/πM).
  • Root-mean-square speed v_rms: v_rms = √(3RT/M).
  • Ratio v_p : v̄ : v_rms = √2 : √(8/π) : √3 = 1 : 1.128 : 1.225.
  • For O₂ at 25 °C (298 K): v_p ≈ 394 m/s; v̄ ≈ 444 m/s; v_rms ≈ 482 m/s. All scale as 1/√M.

Effusion Rate (Quantitative). Number of molecules effusing through a hole of area A per unit time: dN/dt = (n × v̄ / 4) × A, where n is the number density and v̄ is the mean molecular speed. Substituting v̄ = √(8RT/πM): dN/dt = A × n × √(2RT/πM). Rate ∝ 1/√M, recovering Graham's law. The factor of 1/4 reflects the cosine-squared distribution of molecular trajectories perpendicular to the hole.

Uranium-Enrichment Cascade Engineering. Per-stage separation factor α = √(M(²³⁸UF₆) / M(²³⁵UF₆)) = √(352.04/349.03) = 1.00430. To enrich from natural composition x_natural = 0.00720 (²³⁵U fraction) to weapons-grade x_target = 0.90, the required number of theoretical stages N satisfies α^N = (x_target/(1-x_target)) / (x_natural/(1-x_natural)) = (0.90/0.10) / (0.0072/0.9928) = 9.0 / 0.00725 = 1241; N = ln(1241) / ln(1.00430) ≈ 1664 stages. Real cascades are imperfect (concentration jumps less than ideal); ~3000-4000 stages were used in the Oak Ridge K-25 plant. Modern centrifuge separation (per-stage factor ~1.3) is ~50× more efficient and has fully replaced gaseous diffusion.

Real-Gas Deviations. Graham's law assumes ideal-gas behavior — no intermolecular forces, no molecular volume. Real-gas deviations:

  • High pressure (P > 10 bar): non-ideality reduces effusion rate; corrections via the compressibility factor Z.
  • Near critical point: dramatic deviations; Graham's law fails as molecular interactions dominate.
  • Polar gases (HCl, NH₃, H₂O vapor): dipole-dipole interactions slow effusion 1-5% relative to non-polar gases of the same M.
  • Mean free path: Graham's law assumes hole diameter < λ (mean free path). At atmospheric P, λ ≈ 70 nm for air; for hole diameter ≫ 70 nm, viscous flow dominates and Graham's law fails.

Effusion Time vs Rate. If a fixed amount of gas effuses through a hole, the TIME required is inversely proportional to the rate: t₁/t₂ = r₂/r₁ = √(M₁/M₂). So the HEAVIER gas takes LONGER. Example: a balloon containing equal moles of H₂ and air takes 3.79× as long for the air to leak as the H₂. Practical: helium-filled latex balloons last 12-24 hours; air-filled balloons last weeks (mostly limited by O₂/N₂ permeation through latex, not effusion).

Connection to Mass Spectrometry. Mass spec ionizes gas molecules and accelerates them through electric and magnetic fields; the m/z (mass-to-charge ratio) determines the deflection radius. The kinetic-theory analog for neutrals is Graham's law: lighter molecules move faster at the same energy. Both rely on the same fundamental relation v² ∝ 1/m (or 1/M) at constant kinetic energy. References: Graham (1846) Phil. Trans. R. Soc.; Atkins' Physical Chemistry (12th ed., Chapter 1); Levine's Physical Chemistry (7th ed.); Maxwell's Theory of Heat (1871); Boltzmann's Lectures on Gas Theory (1898).

Conclusion

Graham's law of effusion (and approximately diffusion) is the cleanest application of kinetic-molecular theory — a single equation connecting macroscopic gas rates to molecular masses. r₁/r₂ = √(M₂/M₁) follows directly from the equipartition principle (½mv² = (3/2)kT), so lighter gases necessarily move faster. This explains everyday observations (helium-balloon deflation, hydrogen leaks) and underpins large-scale technology (uranium enrichment, gas-membrane separation, mass spectrometry).

Two operational reminders: (1) The law is exact for effusion (single hole into vacuum, independent molecule passage) and approximate for diffusion (gas through gas, molecular collisions cause 5-20% deviations). For high-precision diffusion engineering, use Fick's law with measured diffusion coefficients D₁₂ instead. (2) Solving for molar mass from rate ratio requires squaring the rate ratio: M_unknown = M_ref × (r_ref / r_unknown)². This squared dependence amplifies measurement errors — a 5% rate-measurement error becomes ~10% molar-mass uncertainty. For molar-mass determination at high precision, use mass spectrometry or freezing-point depression. The calculator handles the arithmetic; the physical interpretation (effusion vs diffusion, real-gas deviations, isotope-separation cascading) is what every chemistry student should master.

Frequently Asked Questions

What is the Graham's Law Calculator?
It implements Graham's law of effusion: r₁ / r₂ = √(M₂ / M₁), where r is rate and M is molar mass. 4-way solver: enter any 3 of {rate of Gas 1, rate of Gas 2, molar mass of Gas 1, molar mass of Gas 2} and the calculator solves for the 4th. Two modes: effusion (gas through small hole into vacuum — Graham's law exact) and diffusion (gas through another gas — formula approximate, 5-20% deviations).

Pro Tip: Pair this with our Molarity Calculator for solution chemistry.

What is Graham's law?
The rate of effusion (or approximately diffusion) of a gas is inversely proportional to the square root of its molar mass: r₁ / r₂ = √(M₂ / M₁). Discovered empirically by Thomas Graham in 1846 and later derived from kinetic-molecular theory (½mv² = (3/2)kT, so v ∝ 1/√M). Means lighter gases effuse and diffuse faster than heavier gases at the same temperature.
What's the formula for Graham's law?
r₁ / r₂ = √(M₂ / M₁). Algebraic rearrangements: r₁ = r₂ × √(M₂/M₁); r₂ = r₁ × √(M₁/M₂); M₁ = M₂ × (r₂/r₁)²; M₂ = M₁ × (r₁/r₂)². Effusion time ratio: t₁/t₂ = √(M₁/M₂) (heavier gas takes longer). Average molecular speed: v̄ = √(8RT/πM); v_rms = √(3RT/M); v_p = √(2RT/M) — all scale as 1/√M.
What is the difference between effusion and diffusion?
Effusion: gas escaping through a tiny hole into vacuum or another low-pressure region. The hole is smaller than the mean free path, so molecules pass independently without collisions. Graham's law is mathematically exact for effusion. Diffusion: gas spreading through another gas at comparable pressure. Molecules undergo many collisions per second with the surrounding medium; the rate depends on the binary diffusion coefficient D₁₂. Graham's law gives the rate ratio approximately (within 5-20%) for diffusion.
Which gas effuses faster, hydrogen or oxygen?
Hydrogen, by a factor of ~4. Math: r(H₂) / r(O₂) = √(M(O₂) / M(H₂)) = √(32.00 / 2.016) = √15.87 = 3.98. So at the same T and P, hydrogen effuses 3.98× faster than oxygen. This is why hydrogen leaks from pressurized vessels faster than any other gas, and why hydrogen-filled balloons (Hindenburg-era) deflated rapidly. Hydrogen is also why fuel-cell-vehicle engineering requires meticulous attention to leak detection (lower flammability limit 4 vol%, MIE 0.02 mJ — extremely flammable).
Why does helium escape from a balloon faster than air?
Because helium has lower molar mass (4.003) than air (~28.97). Math: r(He) / r(air) = √(M(air) / M(He)) = √(28.97 / 4.003) = √7.24 = 2.69. Helium escapes 2.69× faster than air. Latex balloons typically have additional permeability effects, so a He balloon deflates 12-24 hours vs air balloons lasting weeks. Mylar balloons have much lower permeability and slow He escape to days/weeks. Note: Graham's law gives the diffusion-rate ratio; actual deflation also depends on balloon material permeability, surface area, and internal pressure.
How is Graham's law used in uranium enrichment?
Per-stage separation factor in gaseous diffusion. Convert U metal to UF₆ (sublimation point 56 °C → gaseous). Compare ²³⁵UF₆ (M = 349.03) vs ²³⁸UF₆ (M = 352.04). Per-stage rate ratio = √(352.04/349.03) = 1.00430 — only 0.43% enrichment per stage. Going from natural 0.72% to weapons-grade 90% requires >1500 stages cascaded; commercial 3-5% reactor fuel needs ~1200 stages. Manhattan Project K-25 plant at Oak Ridge had thousands of stages over 1.6 km of buildings. Modern gas centrifuges (per-stage factor ~1.3) replace gaseous diffusion — ~50× more efficient.
How do I find the molar mass from effusion rate?
M_unknown = M_reference × (r_reference / r_unknown)². Setup: measure the effusion rate (or time) of an unknown gas vs a known reference (typically N₂ or air). If unknown effuses 0.5× as fast as O₂ (M 32.00): M_unknown = 32.00 × (1.0/0.5)² = 32.00 × 4 = 128 g/mol — likely HI (127.91). Caution: the squared rate ratio amplifies measurement errors — 5% rate error → 10% M error. For high-precision M determination, prefer mass spectrometry.
Why is rate inversely proportional to √M (not just M)?
From the equipartition principle of kinetic-molecular theory. All gas molecules at the same temperature have the same average kinetic energy: ½mv² = (3/2)kT. Solving for v: v = √(3kT/m), so v ∝ 1/√m, equivalently 1/√M. Effusion rate is proportional to molecular speed, hence rate ∝ 1/√M, giving r₁/r₂ = √(M₂/M₁). Why √M and not M: kinetic energy goes as v², so equating energies between different molecular masses gives v ∝ 1/√m. If energy scaled linearly with v (it doesn't), rate would scale as 1/M.
Does Graham's law work for liquids and solids?
No. Graham's law is derived for ideal gases where molecules move independently. In liquids, molecules are tightly packed with strong intermolecular forces; speeds are determined by viscosity and density, not √M. In solids, molecules are fixed in lattices with thermal motion only at lattice sites. The kinetic-theory derivation specifically requires low density (independent molecules), no significant intermolecular forces, and ballistic trajectories between collisions — only gases at moderate pressure satisfy these. For liquid diffusion, use Stokes-Einstein D = kT/(6πηr) where η is viscosity and r is molecular radius. For solid diffusion, use Arrhenius-form D = D₀ exp(−E_a/RT) with activation-energy barriers.
What's the relationship between effusion time and Graham's law?
Time is inversely proportional to rate, so heavier gases take longer. t₁ / t₂ = r₂ / r₁ = √(M₁ / M₂). Example: If equal volumes of H₂ and SF₆ effuse through the same hole at the same T,P, the SF₆ takes √(146.06 / 2.016) = √72.5 = 8.51× longer than H₂. Practical: in a leaking gas cylinder, the heavier components remain longer than lighter ones; over time the cylinder becomes enriched in heavier species. This is the basis of one method for gas-mixture analysis: measure cylinder mass change over time and infer composition from the differential effusion rates.

Author Spotlight

The ToolsACE Team - ToolsACE.io Team

The ToolsACE Team

Our ToolsACE chemistry team built this calculator to handle the foundational kinetic-theory result <strong>Graham's law</strong>: at constant temperature and pressure, the rate of effusion (or diffusion) of a gas is inversely proportional to the square root of its molar mass — <strong>r₁ / r₂ = √(M₂ / M₁)</strong>. Discovered empirically by Thomas Graham (1846) and later derived from kinetic-molecular theory, the law is the basis of <strong>uranium isotope enrichment</strong> (²³⁵UF₆ vs ²³⁸UF₆ separation factor 1.0043), <strong>helium-balloon deflation</strong> (He effuses 2.7× faster than air through latex), and any application requiring molecular-mass-dependent gas separation. The calculator is a flexible <strong>4-way solver</strong>: enter any 3 of {rate of Gas 1, rate of Gas 2, molar mass of Gas 1, molar mass of Gas 2} and it solves for the 4th. Two operating modes: <strong>effusion</strong> (single hole into vacuum, where Graham's law is exact) and <strong>diffusion</strong> (gas through another gas, where the formula is an approximation due to molecular collisions in the surrounding medium). Smart warnings catch unphysical molar masses (&lt; 1 g/mol or &gt; 10⁶ g/mol) and unrealistic rate ratios.

Thomas Graham, Phil. Trans. R. Soc. London 136 (1846) 573Atkins' Physical Chemistry (12th ed.)Levine's Physical Chemistry (7th ed.); Maxwell-Boltzmann kinetic theory

Disclaimer

Graham's law applies strictly to EFFUSION (gas through small hole into vacuum, no inter-molecular collisions); for DIFFUSION through another gas, the formula is approximate and real diffusion rates can deviate 5-20% due to molecular collision dynamics in the surrounding medium. The law assumes IDEAL GAS behavior: low pressure, no intermolecular forces, identical T for both gases. Real-gas deviations near the critical point or at high pressure are significant. Solving for molar mass from rate ratio amplifies measurement errors (squared rate ratio): a 5% rate error gives ~10% molar-mass uncertainty. For high-precision molar-mass determination, prefer mass spectrometry or freezing-point depression. Modern uranium enrichment uses centrifuges (per-stage factor ~1.3) which are ~50× more efficient than gaseous diffusion (~1.0043 per stage). References: Graham, Phil. Trans. R. Soc. (1846); Atkins' Physical Chemistry; Levine's Physical Chemistry; CRC Handbook of Chemistry and Physics.