Gas volumes at room temperature and pressure
Use 24 dm³ per mole at r.t.p. and apply gas-volume ratios only under matching conditions.
Section C3.3 · The mole and the Avogadro constant
Mapped to the named 0654 section using teaching from the corresponding single-science course. A 0654-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade.
Before you begin
- Use mole ratios from a balanced equation.
- Convert between cm³ and dm³.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is 480 cm³ in dm³?
What you will learn
- Convert gas volume at r.t.p. into amount and mass.
- Use balanced ratios for gas volumes at the same temperature and pressure.
Read the stated conditions and volume unit
At room temperature and pressure, abbreviated r.t.p., use the syllabus molar gas volume 24 dm³/mol. Gas amount is volume in dm³ divided by 24 dm³/mol. The reverse calculation multiplies the amount by 24. Since 1 dm³=1000 cm³, the same molar volume is 24 000 cm³/mol. Choose one consistent volume scale.
The value is tied to the stated conditions. Do not substitute a value learned for a different temperature or pressure, or use the gas molar volume for a solid or liquid. If the question supplies another molar volume or different conditions, use its relevant information. At the same temperature and pressure, equal volumes of gases contain equal amounts in moles; they need not have equal masses.
Use a balanced volume ratio for gases only
Because equal gas amounts occupy equal volumes under matching conditions, balanced coefficients can give a gas-volume ratio. In N₂(g)+3H₂(g)→2NH₃(g), one volume of nitrogen reacts with three volumes of hydrogen and yields two volumes of ammonia if all volumes are compared at the same temperature and pressure and the theoretical reaction is complete.
Do not assign 24 dm³ to a mole of liquid water. For 2H₂(g)+O₂(g)→2H₂O(l), the starting gases have a volume ratio 2:1, but the liquid product does not have twice the oxygen's gas volume. To predict a product gas, check its state, compare the gas conditions and use the limiting reactant if both inputs are stated.
Sources for this lesson
Pause and explain
At the same temperature and pressure, equal volumes of CO₂ and H₂ have equal…
Worked example
A 600 cm³ sample of CO₂ is measured at r.t.p. Calculate its amount and mass using C=12 and O=16.
Show the worked solution
- Convert 600 cm³ to 0.600 dm³.
- Amount is 0.600/24=0.025 mol.
- CO₂ molar mass is 44 g/mol, so mass is 0.025×44=1.1 g.
Answer 0.025 mol and 1.1 g CO₂.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Calculate the amount and mass of 120 cm³ H₂ at r.t.p., using H=1.
Give me a hint
Convert to dm³ first and use the diatomic molar mass.
Compare my reasoning
- 120 cm³=0.120 dm³.
- n=0.120/24=0.0050 mol; M(H₂)=2 g/mol.
- Mass is 0.0050×2=0.010 g.
0.0050 mol and 0.010 g H₂.
Look for these in your work
- Volume units match the molar volume.
- The molar mass is for H₂ rather than H atoms.
Using N₂(g)+3H₂(g)→2NH₃(g), predict the ammonia volume from 200 cm³ N₂ and 900 cm³ H₂. Assume complete reaction and compare all volumes at the same temperature and pressure.
Give me a hint
Compare volume/coefficient; volumes share one unit and the same conditions.
Compare my reasoning
- Reaction portions in the volume scale are 200/1=200 and 900/3=300, so nitrogen limits.
- Ammonia volume is 2×200=400 cm³.
- Hydrogen used is 3×200=600 cm³, leaving 900−600=300 cm³.
400 cm³ NH₃; nitrogen limits and 300 cm³ H₂ remains.
Look for these in your work
- The limiting gas is justified.
- Product and excess volumes use the same conditions.
10 g CaCO₃ decomposes fully by CaCO₃→CaO+CO₂. Calculate CO₂ volume at r.t.p., using Ca=40, C=12 and O=16, and explain why the solid CaO has no volume from the 24 dm³/mol rule.
Give me a hint
Find carbonate moles, then the gaseous product amount.
Compare my reasoning
- M(CaCO₃)=100 g/mol, so n=10/100=0.10 mol.
- The 1:1 ratio gives 0.10 mol CO₂ and V=0.10×24=2.4 dm³.
- CaO is solid; 24 dm³/mol is a gas value and cannot give its solid volume.
2.4 dm³ CO₂; the r.t.p. gas rule does not give CaO's solid volume.
Look for these in your work
- Mass, amount and gas volume are connected in order.
- The state restriction is explained.
Common mistakes
- Dividing a cm³ volume by 24 without converting it to dm³.
- Using 22.4 dm³/mol for a question requiring the syllabus r.t.p. value.
- Applying a gas-volume ratio to a liquid product.
What can you explain now?
Find the r.t.p. volume of 0.15 mol of a gas.
Compare with the explanation
3.6 dm³
V=n×24=0.15×24=3.6 dm³, or 3600 cm³.
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Make it stick
Convert one gas volume in cm³ into moles and grams. Explain why equal gas volumes can have different masses and why liquid water cannot be assigned the r.t.p. gas molar volume.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.