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5.1.3Additional lessons

Avogadro’s Law and Molar Volume of Gas — IB Chemistry SL Revision Notes & Topical Questions

5 min read 6 parts Lesson 61 of 98

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Curriculum Equation Key Moles & Stoichiometry FormulasFast Exam Solution
n = m / M_r | N = n × N_A (6.022 × 10²³) | M = n / V | V_gas = n × 22.4 L (STP)
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Part 1 of 6Avogadro's Law

5.1.3 Avogadro's Law and Molar Volume of Gas

At the same temperature and pressure, all gases occupy the same volume per mole.

Statement
  • Equal volumes of different gases at the same temperature and pressure contain the same number of molecules.
Consequence
  • The volume of a gas is directly proportional to its amount in moles when temperature and pressure are fixed.
Gas volume ratio
  • The coefficients in an equation also give the ratio of volumes of gases, so 2H₂ + O₂ → 2H₂O needs 2 volumes of H₂ for every 1 of O₂.
Ideal behaviour
  • The law is exact for an ideal gas and a good approximation for real gases at low pressure and moderate temperature.

Part 2 of 6Molar Volume

Molar Volume
Value
  • At STP, which is 273 K and 100 kPa, one mole of any ideal gas occupies 22.7 dm³.
Units
  • The molar volume is quoted in dm³ mol⁻¹ and 1 dm³ equals 1000 cm³.
Formula
  • Volume of gas = amount in moles × 22.7 dm³ mol⁻¹, and rearranged, amount = volume ÷ 22.7.
Condition
  • The value applies only when the gas is at STP; other conditions need the ideal gas equation.
QuantitySymbolTypical unit
Amountnmol
Volume of gasVdm³
Molar volumeVm22.7 dm³ mol⁻¹ at STP

Part 3 of 6Method

1
Write the balanced equation and identify the gas whose volume is given or required.
2
If the other quantity is a mass, convert it to moles with n = m/M.
3
Use the mole ratio to find the moles of the gas, then multiply by 22.7 dm³ mol⁻¹ to get its volume at STP.
4
When only gases are involved and measured under the same conditions, volumes can be compared directly using the ratio of coefficients.
5
Convert the final volume to cm³ if the question requires it.

Part 4 of 6Applications

Applications
Gas evolved
  • The volume of hydrogen formed when a metal dissolves in acid can be found from its mass.
Combustion
  • Gas volumes in combustion of gaseous fuels are in the ratio of the coefficients.
Finding molar mass
  • The mass of a known volume of gas can be used to calculate its molar mass.
Limitations
  • If the temperature is not 273 K, 22.7 dm³ mol⁻¹ cannot be used, and the ideal gas equation PV = nRT is needed.

Part 5 of 6Worked Example

Question

  • Calculate the volume of CO₂ at STP formed when 10.0 g of CaCO₃ reacts with excess hydrochloric acid: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g).
  • (Mr CaCO₃ = 100.09)

  1. Amount of CaCO₃ = 10.0 ÷ 100.09 = 0.0999 mol.
  2. The ratio CaCO₃CO₂ is 1 : 1, so n(CO₂) = 0.0999 mol.
  3. V = 0.0999 × 22.7 = 2.27 dm³.

Answer 2.27 dm³ (2270 cm³) of CO₂

Part 6 of 6Exam Checklist & Tips

5.1.3 – Key Points for the Exam

  • Avogadro's law Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.
  • Molar volume
    • The volume occupied by one mole of gas;
      • 22.7 dm³ mol⁻¹ at STP.
  • STP Standard temperature and pressure: 273 K and 100 kPa.
  • Ideal gas A model gas whose particles have no volume and no attractions between them.

Common Mistakes

  • Using 22.4 dm³ mol⁻¹ when the course and data booklet give 22.7 dm³ mol⁻¹ at 100 kPa.
  • Using molar volume for a gas at room temperature without correction.
  • Mixing dm³ and cm³ in the same calculation.
  • Applying a mole-ratio to volumes of liquids or solids, which does not hold.

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