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
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.
Section 1 of 6Avogadro's Law
At the same temperature and pressure, all gases occupy the same volume per mole.
Statement
- Statement Equal volumes of different gases at the same temperature and pressure contain the same number of molecules.
Consequence
- Consequence The volume of a gas is directly proportional to its amount in moles when temperature and pressure are fixed.
Gas volume ratio
- 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
- Ideal behaviour The law is exact for an ideal gas and a good approximation for real gases at low pressure and moderate temperature.
Section 2 of 6Molar Volume
Molar Volume
Value
- Value At STP, which is 273 K and 100 kPa, one mole of any ideal gas occupies 22.7 dm³.
Units
- Units The molar volume is quoted in dm³ mol⁻¹ and 1 dm³ equals 1000 cm³.
Formula
- Formula Volume of gas = amount in moles × 22.7 dm³ mol⁻¹, and rearranged, amount = volume ÷ 22.7.
Condition
- Condition The value applies only when the gas is at STP; other conditions need the ideal gas equation.
| Quantity | Symbol | Typical unit |
|---|---|---|
| Amount | n | mol |
| Volume of gas | V | dm³ |
| Molar volume | Vm | 22.7 dm³ mol⁻¹ at STP |
Section 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.
Section 4 of 6Applications
Applications
Gas evolved
- Gas evolved The volume of hydrogen formed when a metal dissolves in acid can be found from its mass.
Combustion
- Combustion Gas volumes in combustion of gaseous fuels are in the ratio of the coefficients.
Finding molar mass
- Finding molar mass The mass of a known volume of gas can be used to calculate its molar mass.
Limitations
- 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.
Section 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)
- Amount of CaCO₃ = 10.0 ÷ 100.09 = 0.0999 mol.
- The ratio CaCO₃CO₂ is 1 : 1, so n(CO₂) = 0.0999 mol.
- V = 0.0999 × 22.7 = 2.27 dm³.
Answer 2.27 dm³ (2270 cm³) of CO₂
Section 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.
- The volume occupied by one mole of gas;
- 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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