This how it works covers How It Works: Why 1 Mole of Any Gas = 24 dm³ within Gas Volume for GCSE Chemistry. Topic 19: Gas Volume It is section 3 of 11 in this topic. Use this how it works to connect the idea to the wider topic before moving on to questions and flashcards.
⚙️ How It Works: Why 1 Mole of Any Gas = 24 dm³
In a gas, the particles are very far apart from each other — the space between particles is much larger than the particles themselves. This means the volume of a gas is determined almost entirely by how many particles there are and how energetically they are moving, not by the size or mass of the individual particles.
At a given temperature (which determines how fast particles move) and a given pressure (which determines how hard they push on the container walls), the volume depends only on the number of particles. Since one mole always contains the same number of particles (Avogadro's constant, 6.02 × 10²³), one mole of any gas at the same temperature and pressure will occupy the same volume.
At RTP (Room Temperature and Pressure, defined as 20°C and 1 atmosphere), this volume is 24 dm³ (24,000 cm³). This is different from STP (standard temperature and pressure, 0°C and 1 atm) where the molar volume is 22.4 dm³ — GCSE uses the RTP value of 24 dm³.
The practical consequence is elegant: to find the volume of a gas from a reaction, you simply: (1) find the moles of gas from the balanced equation, then (2) multiply by 24. No complicated density calculations needed.
Worked method: gas volume calculations
Turn the 24 dm³ rule into working you can actually use in the exam.
- The formula both waysmoles = volume (dm³) ÷ 24. Rearranged: volume (dm³) = moles × 24. This only works at room temperature and pressure (RTP: 20°C, 1 atmosphere) — the value the exam assumes unless told otherwise.
- Volume given, moles wantedDivide the volume in dm³ by 24. Convert cm³ to dm³ first by dividing by 1000. Example: 480 cm³ = 0.48 dm³, so moles = 0.48 ÷ 24 = 0.02 mol.
- Moles given, volume wantedMultiply moles by 24. Example: 0.1 mol of any gas occupies 0.1 × 24 = 2.4 dm³ at RTP — true whether the gas is hydrogen, oxygen or carbon dioxide.
- Starting from a mass insteadYou cannot multiply grams by 24 directly. Find moles first using n = mass ÷ Mr, then use that in the volume formula: volume = (mass ÷ Mr) × 24.
Calculate the volume, in dm³, of carbon dioxide produced at RTP when 11 g of CO2 is formed. (Mr of CO2 = 44)
moles of CO2 = mass ÷ Mr = 11 ÷ 44 = 0.25 mol. Volume = moles × 24 = 0.25 × 24 = 6 dm³.
The most commonly lost mark is skipping the mole step — multiplying the mass in grams straight by 24. Grams never go directly into the ×24 formula.
Practice questions for Gas Volume
What is the molar gas volume at RTP (room temperature and pressure)?
Explain why the molar gas volume of 24 dm³/mol is only valid at RTP.