1.5.10Principles of Chemistry
Calculate Volumes of Gases — Edexcel Linear - OL (Chemistry) Revision Notes & Topical Questions
5 min read 6 parts Lesson 20 of 25 Unit quiz
Section 1 of 6Avogadro's Law and Molar Volume
Because gas volume depends only on the amount of gas, moles and volume are directly linked.
Avogadro's law
- Equal volumes of any gases at the same temperature and pressure contain the same number of molecules.
Molar volume
- At room temperature and pressure (20 °C, 1 atm), one mole of any gas occupies 24 dm³, which equals 24 000 cm³.
Why it works
- Gas particles are so far apart that the volume depends on the number of particles, not on their size or identity.
Units
- Use 24 when the volume is in dm³ and 24 000 when it is in cm³, so units stay consistent.
Section 2 of 6The Calculations
The Calculations
Moles to volume
- Volume of gas (dm³) = moles × 24, for example 0.25 mol of CO₂ occupies 6 dm³.
Volume to moles
- Moles = volume (dm³) ÷ 24, so 7.2 dm³ of carbon monoxide is 0.30 mol.
From a mass
- Convert the mass to moles using Mr first, and only then multiply by 24 to reach a volume.
Volume ratios
- In a balanced equation involving only gases, the coefficients give the ratio of the gas volumes directly.
| Gas | Amount | Volume at RTP |
|---|---|---|
| Hydrogen | 3 mol | 72 dm³ |
| Ammonia | 0.02 mol | 0.48 dm³ |
| Oxygen | 0.05 mol | 1200 cm³ |
Section 3 of 6Step-by-Step Method
Following the same sequence each time prevents most slips.
1
Write the balanced equation and identify the gas whose volume is wanted.
2
Convert the given quantity to moles, using mass ÷ Mr for solids or volume ÷ 24 for gases.
3
Use the mole ratio from the equation to find the moles of the gas required.
4
Multiply moles by 24 to give dm³, or by 24 000 for cm³.
5
Check the units requested and give the answer to a suitable number of significant figures.
Section 4 of 6Gas Volume Ratios
Gas Volume Ratios
Combustion of propane
- In C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g), 100 cm³ of propane needs 500 cm³ of oxygen at the same conditions.
Total product volume
- The same 100 cm³ of propane gives 300 cm³ of carbon dioxide and 400 cm³ of steam, 700 cm³ in total.
Condition
- The ratios hold only when all volumes are measured at the same temperature and pressure.
Section 5 of 6Worked Example
Question
- 5.0 g of calcium carbonate reacts completely with excess hydrochloric acid.
- Calculate the volume of carbon dioxide produced at RTP.
- CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ (Ar: Ca = 40, C = 12, O = 16)
- Mr of CaCO₃ = 40 + 12 + (3 × 16) = 100.
- Moles of CaCO₃ = 5.0 ÷ 100 = 0.050 mol.
- The equation shows 1 mol CaCO₃ gives 1 mol CO₂, so moles of CO₂ = 0.050 mol.
- Volume = 0.050 × 24 = 1.2 dm³.
Answer Volume of carbon dioxide = 1.2 dm³ (1200 cm³)
Section 6 of 6Exam Checklist & Tips
1.5.10 – Key Points for the Exam
- Molar volume
- The volume occupied by one mole of any gas;
- 24 dm³ at RTP.
- The volume occupied by one mole of any gas;
- RTP Room temperature and pressure, taken as 20 °C and 1 atmosphere.
- Avogadro's law Equal volumes of gases at the same conditions hold equal numbers of molecules.
- Mole ratio The ratio of coefficients in a balanced equation.
Common Mistakes
- Using 24 for a volume given in cm³ instead of 24 000.
- Forgetting to convert grams to moles before using the molar volume.
- Applying volume ratios to solids or liquids in the equation.
- Using the molar volume for a different temperature without being given its value.
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