Molar volume of gases and Avogadro's law

Higher tier: one mole of any gas takes up the same volume at the same temperature and pressure. This makes gas calculations quick.

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Molar volume

At room temperature and pressure (rtp), one mole of any gas occupies 24 dm3 (24 000 cm3).
volume of gas (dm3) = moles × 24
moles = volume (dm3) ÷ 24

Avogadro's law

Equal volumes of different gases, at the same temperature and pressure, contain the same number of molecules.
So for gases, the volume ratio is the same as the mole ratio in the equation.
CH4 + 2O2 → CO2 + 2H2O
50 cm3 of methane needs 100 cm3 of oxygen and makes 50 cm3 of carbon dioxide (water is a liquid at rtp).

Mass to volume

Combine with moles = mass ÷ Mr.
Example: what volume does 4.4 g of carbon dioxide (Mr 44) occupy at rtp?
Moles = 4.4 ÷ 44 = 0.1 mol. Volume = 0.1 × 24 = 2.4 dm3.
Worked example

What volume of hydrogen at rtp is made when 0.48 g of magnesium reacts with excess acid? Mg + 2HCl → MgCl2 + H2 (Ar Mg = 24)

  1. Moles of Mg = 0.48 ÷ 24 = 0.02 mol
  2. Ratio Mg : H2 = 1 : 1, so 0.02 mol H2
  3. Volume = 0.02 × 24 = 0.48 dm3 = 480 cm3

Answer: 0.48 dm3 (480 cm3)

Key idea

At rtp one mole of any gas occupies 24 dm3. Volume = moles × 24. Equal volumes of gases contain equal numbers of molecules, so gas volumes react in the same ratio as the moles in the equation.

Key words
Molar volume
The volume of one mole of any gas at room temperature and pressure: 24 dm^{3} (24 000 cm^{3}).
Mole
The amount of substance containing 6.02 × 10^{23} particles. Its mass in grams equals the A_{r} or M_{r}.

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Extra practice

2 more questions

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  1. What volume, in dm3, does 0.5 mol of nitrogen occupy at rtp?

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    Answer: 12 dm3

    0.5 × 24 = 12 dm3

  2. How many moles are in 12 dm3 of oxygen at rtp?

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    Answer: 0.5 mol

    12 ÷ 24 = 0.5 mol

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Sources and references

Checked by the GCSE-ready editorial team, 17 October 2026. Last updated 2 October 2026. Spotted a mistake? Tell us and we will fix it.

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