All calculatorsRate of Effusion Calculator
Use Graham’s Law to relate gas effusion rates and molar masses: rate₁ / rate₂ = √(M₂ / M₁). Enter any three of rate₁, rate₂, M₁, M₂ and we’ll solve the missing one. See steps, a mini bar chart, and a ratio gauge.
Background
Graham’s Law states lighter gases effuse faster. If two gases pass through identical pinholes at the same conditions, their rates are inversely proportional to the square roots of their molar masses: rate ∝ 1/√M.
How this calculator works
- Graham’s Law: rate₁ / rate₂ = √(M₂ / M₁).
- Enter any three of rate₁, rate₂, M₁, M₂; we solve the missing one.
- Rates can be any matching units (e.g., mol·s⁻¹ vs mol·s⁻¹).
Formula & Equation Used
Graham’s Law: rate ∝ 1/√M
Relation: rate₁ / rate₂ = √(M₂ / M₁)
Rearrangements: rate₁ = rate₂ √(M₂/M₁); rate₂ = rate₁ √(M₁/M₂); M₁ = M₂ (rate₂/rate₁)²; M₂ = M₁ (rate₁/rate₂)²
Example Problems & Step-by-Step Solutions
Example 1 — He vs O₂
M₁ = 4.00 g·mol⁻¹ (He), M₂ = 32.00 g·mol⁻¹ (O₂), rate₂ = 1.00.
√(M₂/M₁) = √(32/4) = √8 ≈ 2.828 → rate₁ = 1.00 × 2.828 = 2.83.
Example 2 — H₂ vs N₂
M₁ = 2.016, M₂ = 28.014, rate₁ = 3.00 → rate₂ = rate₁/√(M₂/M₁) = 3.00 / √(13.90) ≈ 0.80.
Example 3 — Unknown mass
rate₁ = 1.00, rate₂ = 0.50, M₂ = 28.0 → M₁ = M₂ (rate₂/rate₁)² = 28.0 × 0.25 = 7.00 g·mol⁻¹.
Frequently Asked Questions
Q: Do rate units matter?
No, as long as both rates use the same units, their ratio is valid.
Q: When does Graham’s Law apply?
When gases effuse through identical small openings under the same conditions (T, P).
Q: Why does a lighter gas effuse faster?
Lighter gases have higher average speeds at the same temperature (kinetic molecular theory).
