✓ Revise Right course
The molar volume is the volume occupied by one mole of any gas at room temperature and pressure (RTP), which is 24 dm³. Free GCSE Chemistry notes.
Gas volume fundamentals and molar volume is part of Monitoring chemical reactions in GCSE OCR Higher Chemistry.
4 min
Reading time
28
Lessons in this topic
GCSE OCR Higher
Pathway
Reviewed by Revise Right
Content review
The molar volume is the volume occupied by one mole of any gas at room temperature and pressure (RTP), which is 24 dm³. Free GCSE Chemistry notes.
OCR A GCSE Chemistry mock draft, 26–31 Oct. Draft coverage: Chemistry Paper 3 (Higher tier). Reviewed for an original-practice sitting; check student availability before registering. Marking and results will follow later, after separate approval. Eligible new accounts receive a five-day Pro trial when signup email verification completes. Previous trials keep their original expiry. After expiry, new papers need Pro unless you have an explicitly confirmed free sitting offer for this event. That offer adds no Pro time or AI learning. Saved attempts and released results remain readable. No card or automatic charge.
Use this lesson to stay inside Monitoring chemical reactions while connecting the detail back to the wider Topic C5: Monitoring and controlling chemical reactions topic.
Read the core explanations from Gas volume fundamentals and molar volume before testing yourself from memory.
Here's something that seems impossible: a balloon filled with helium and another filled with carbon dioxide can contain exactly the same number of particles, even though helium atoms are much lighter than CO₂ molecules. The secret lies in understanding how gases behave at the molecular level.
This concept unlocks calculations that appear in every chemistry exam. You'll calculate gas volumes from masses, predict how much space reactions will occupy, and solve problems that combine moles with volume measurements.
Picture this: you have two identical boxes. One contains 1 mole of hydrogen gas (H₂), the other contains 1 mole of carbon dioxide (CO₂). Even though CO₂ molecules are 22 times heavier than H₂ molecules, both boxes contain exactly the same volume of gas.
This works because gas particles are so far apart that their individual size becomes irrelevant. What matters is how many particles you have (the moles) and how much energy they have (temperature and pressure).
At room temperature and pressure (20°C and 1 atmosphere), one mole of any gas always occupies 24 dm³. This is your golden number for gas calculations. Whether it's helium, methane, or chlorine gas - 1 mole = 24 dm³ at RTP.
**Trap**: Many students think heavier gases occupy more volume. Actually, gas volume depends only on the number of moles and the conditions (temperature and pressure), not the mass of individual particles.
Exam questions rarely give you moles directly. Instead, you'll get a mass and need to work backwards. This requires combining the mole calculation with the volume formula.
Worked example: Calculating Gas Volume from Mass
Calculate the volume of 8.8 g of carbon dioxide (CO₂) at RTP. [Mr of CO₂ = 44]
Answer: 4.8 dm³
Notice the two-step process: mass → moles → volume. This pattern appears in every gas volume calculation. Master this sequence and you'll handle any variation the exam throws at you.
Sometimes you'll know the volume and need to find the mass. This requires rearranging the formula - a skill that appears across all chemistry calculations.
Worked example: Finding Mass from Gas Volume
A gas occupies 9.6 dm³ at RTP. If the gas is methane (CH₄), calculate its mass. [Mr of CH₄ = 16]
Answer: 6.4 g
Gas volume questions follow predictable patterns. Recognise the type and apply the correct method systematically.
**Command word focus**: 'Calculate' questions need your working shown step-by-step. 'State' questions want just the molar volume (24 dm³). 'Explain' questions should refer to the fact that equal moles of gases occupy equal volumes under the same conditions.
Exam precision note: In Gas volume fundamentals and molar volume - Use of amount of substance in relation to volumes of gases (chemistry only) (HT only) - Use of amount of substance in relation to volumes of gases (chemistry only) (HT only), always identify the key chemical idea first, then link it to evidence, data, or an equation. A high-scoring GCSE answer names the particles involved, states what changes, and justifies the conclusion using correct scientific language. Where calculations are used, show each step with units and keep significant figures sensible. Finally, check that terminology matches AQA wording so your answer is unambiguous and mark-scheme friendly.
Gas-volume questions combine two ideas: stoichiometric ratio and molar volume. At room temperature and pressure, one mole of any gas occupies 24 dm3. But students must still use the balanced equation first, because the number of moles of gas depends on the reaction ratio. For example, if the equation shows two moles of gas being formed from one mole of reactant, the gas volume must reflect that ratio before multiplying by 24. Writing the ratio step explicitly prevents many of the common errors seen in higher-tier gas-volume calculations.
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-03-23.
Move to the related lessons or back to the topic page when you need the wider sequence before exam-style practice.
The molar volume is the volume occupied by one mole of any gas at room temperature and pressure (20°C and 1 atmosphere), which is always 24 dm³. Whether the gas is helium, methane or chlorine, 1 mole = 24 dm³ at RTP.
First find moles = mass ÷ Mr, then volume = moles × 24 dm³. For 8.8 g of carbon dioxide (Mr 44), moles = 8.8 ÷ 44 = 0.2 mol, so the volume at RTP = 0.2 × 24 = 4.8 dm³.
Equal amounts in moles of gases occupy the same volume under the same temperature and pressure because gas particles are so far apart that their individual size becomes irrelevant. Volume depends only on the number of moles and the conditions, not the mass of the particles.
First find moles = volume ÷ 24, then mass = moles × Mr. For 9.6 dm³ of methane (CH₄, Mr 16) at RTP, moles = 9.6 ÷ 24 = 0.4 mol, so the mass = 0.4 × 16 = 6.4 g.
Heavier gases do not take up more volume. One mole of carbon dioxide and one mole of hydrogen occupy the same volume, even though CO₂ molecules are 22 times heavier than H₂ molecules, because gas volume depends only on moles, temperature and pressure.
RTP, room temperature and pressure, is 20°C and 1 atmosphere. At RTP one mole of any gas occupies 24 dm³.
8.8 g of CO₂ occupies 4.8 dm³ at RTP. Moles = mass ÷ Mr = 8.8 ÷ 44 = 0.2 mol, then volume = 0.2 × 24 = 4.8 dm³.
9.6 dm³ of methane (CH₄) has a mass of 6.4 g at RTP. Moles = 9.6 ÷ 24 = 0.4 mol, and mass = moles × Mr = 0.4 × 16 = 6.4 g.
Particle size does not affect gas volume because gas particles are so far apart that their individual size becomes irrelevant. What matters is the number of moles and the temperature and pressure.
The volume of a gas at RTP is volume (dm³) = moles × 24. Rearranged, moles = volume ÷ 24, so the usual route from a mass is mass → moles → volume.