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1.5.2Additional lessons

Molar Gas Volume — IB Chemistry SL Revision Notes & Topical Questions

5 min read 5 parts Lesson 16 of 98

📚 IB Chemistry SL Hub / Topical Revision
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Curriculum Equation Key قوانين المول والحسابات الكيميائية - 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)
Word Equation & Conditions: عدد المولات = الكتلة بالجرام ÷ الكتلة المولية | عدد الجسيمات = عدد المولات × عدد أفوجادرو
Location & Specifications: كيمياء التحصيلي وثانوي وبكالوريا و Cambridge IGCSE Chemistry.
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Part 1 of 5Ideal Gas Assumptions

The kinetic theory of gases describes an ideal gas.

Random motion
  • Random motion Gas particles are in continuous, rapid and random motion in straight lines between collisions.
Negligible volume
  • Negligible volume The volume of the particles is negligible compared with the volume of the container.
No attractions
  • No attractions There are no intermolecular forces between the particles, so they do not attract or repel each other.
Elastic collisions
  • Elastic collisions Collisions between particles, and with the walls, are elastic, so no kinetic energy is lost overall.

Part 2 of 5Molar Volume

One mole of any ideal gas has the same volume under given conditions.

1
Standard temperature and pressure (STP) are 273 K and 1.00 × 10⁵ Pa.
2
At STP the molar volume of an ideal gas is 22.7 dm³ mol⁻¹.
3
The amount of gas is found by n = V ÷ 22.7 when V is in dm³ at STP.
4
The volume at STP is V = n × 22.7, so 0.50 mol occupies 11.35 dm³.

Part 3 of 5Using Molar Volume

Using Molar Volume
Gas from a reaction
  • Gas from a reaction Find the moles of the limiting reactant, use the mole ratio to get moles of gas, then multiply by 22.7.
Conditions
  • Conditions Only use 22.7 dm³ mol⁻¹ when the question says STP; otherwise use the ideal gas equation.
Volume units
  • Volume units 1 dm³ = 1000 cm³, so 22.7 dm³ is 22 700 cm³ and conversions need care.
Density of a gas
  • Density of a gas Density at STP equals M ÷ 22.7 in g dm⁻³, so heavier molecules give denser gases.

Part 4 of 5Worked Example

Question

  • Calcium carbonate reacts with excess hydrochloric acid: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g).
  • Calculate the volume of CO₂ at STP from 10.0 g CaCO₃ (M = 100.09 g mol⁻¹).

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

Answer 2.27 dm³ of CO₂

Part 5 of 5Exam Checklist & Tips

1.5.2 – Key Points for the Exam

  • STP Standard temperature and pressure: 273 K and 1.00 × 10⁵ Pa.
  • Molar volume The volume occupied by one mole of gas under stated conditions, 22.7 dm³ mol⁻¹ at STP.
  • Ideal gas A model gas whose particles have no volume and no intermolecular forces.
  • Elastic collision A collision in which the total kinetic energy is conserved.

Common Mistakes

  • Using 22.7 dm³ mol⁻¹ for a gas at room temperature and pressure.
  • Applying the molar volume to solids or liquids.
  • Mixing up dm³ and cm³ in the calculation.
  • Saying that ideal gas particles attract one another weakly.

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