Volumes of Gases

This lesson covers how to calculate the volume of a gas produced or consumed in a chemical reaction, using the concept of molar gas volume, as required by the Edexcel GCSE Combined Science specification (1SC0). This topic is primarily assessed at Higher tier.

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Molar Gas Volume

At room temperature and pressure (RTP) — defined as approximately 20 °C and 1 atmosphere (1 atm) — one mole of any gas occupies a volume of:

$\text{molar gas volume} = 24 \text{ dm}^3 = 24{,}000 \text{ cm}^3$molar gas volume=24 dm3=24,000 cm3

This applies to all gases, regardless of their identity, because at the same temperature and pressure, gas particles are spaced equally far apart (Avogadro's law).

Exam Tip: You will be given the value of 24 dm³ in the exam — you do not need to memorise it. But you do need to know how to use it.

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The Key Equation

$\text{volume (dm}^3) = \text{moles} \times 24$volume (dm3)=moles×24

Or equivalently:

$\text{moles} = \frac{\text{volume (dm}^3)}{24}$moles=24volume (dm3)​

If working in cm³:

$\text{volume (cm}^3) = \text{moles} \times 24{,}000$volume (cm3)=moles×24,000

$\text{moles} = \frac{\text{volume (cm}^3)}{24{,}000}$moles=24,000volume (cm3)​

The Triangle

graph TD
    A["<b>volume (dm³)</b>"] --- B["<b>moles</b>"]
    A --- C["<b>24</b>"]
    B --- C
    style A fill:#f59e0b,color:#000,stroke:#d97706
    style B fill:#8b5cf6,color:#fff,stroke:#7c3aed
    style C fill:#8b5cf6,color:#fff,stroke:#7c3aed

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Worked Examples

Example 1 — Volume from Moles

Question: What volume of hydrogen gas (at RTP) is produced when 0.25 mol of zinc reacts with excess hydrochloric acid?

$\text{Zn}(s) + 2\text{HCl}(aq) \rightarrow \text{ZnCl}_2(aq) + \text{H}_2(g)$Zn(s)+2HCl(aq)→ZnCl2​(aq)+H2​(g)

Step 1: From the equation, Zn : $\text{H}_2$H2​ = 1 : 1

$\text{moles of H}_2 = 0.25 \text{ mol}$moles of H2​=0.25 mol

Step 2: Calculate volume: $\text{volume} = 0.25 \times 24 = 6.0 \text{ dm}^3$volume=0.25×24=6.0 dm3

Example 2 — Moles from Volume

Question: A gas syringe collects 120 cm³ of carbon dioxide at RTP. How many moles is this?

$\text{moles} = \frac{120}{24{,}000} = 0.005 \text{ mol}$moles=24,000120​=0.005 mol

Example 3 — Mass to Volume

Question: What volume of carbon dioxide (at RTP) is produced when 5.0 g of calcium carbonate reacts with excess hydrochloric acid?

$\text{CaCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)$CaCO3​(s)+2HCl(aq)→CaCl2​(aq)+H2​O(l)+CO2​(g)

Step 1: $M_r$Mr​ of $\text{CaCO}_3$CaCO3​ = 100

$\text{moles of CaCO}_3 = \frac{5.0}{100} = 0.05 \text{ mol}$moles of CaCO3​=1005.0​=0.05 mol

Step 2: Ratio of $\text{CaCO}_3$CaCO3​ : $\text{CO}_2$CO2​ = 1 : 1

$\text{moles of CO}_2 = 0.05 \text{ mol}$moles of CO2​=0.05 mol

Step 3: Calculate volume: $\text{volume} = 0.05 \times 24 = 1.2 \text{ dm}^3$volume=0.05×24=1.2 dm3

Or in cm³: $\text{volume} = 0.05 \times 24{,}000 = 1{,}200 \text{ cm}^3$volume=0.05×24,000=1,200 cm3

Example 4 — Volume to Mass

Question: 480 cm³ of oxygen gas at RTP reacts completely with magnesium. What mass of magnesium oxide is formed?

$2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s)$2Mg(s)+O2​(g)→2MgO(s)

Step 1: Moles of $\text{O}_2$O2​: $\text{moles of O}_2 = \frac{480}{24{,}000} = 0.02 \text{ mol}$moles of O2​=24,000480​=0.02 mol

Step 2: Ratio of $\text{O}_2$O2​ : MgO = 1 : 2

$\text{moles of MgO} = 0.02 \times 2 = 0.04 \text{ mol}$moles of MgO=0.02×2=0.04 mol

Step 3: $M_r$Mr​ of MgO = 40 $\text{mass} = 0.04 \times 40 = 1.6 \text{ g}$mass=0.04×40=1.6 g

Exam Tip: Always check whether the question gives volume in dm³ or cm³. If cm³, divide by 24,000 (not 24). If dm³, divide by 24.

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Combining Gas Volume With Concentration Calculations

Sometimes you need to link gas volume calculations with solution concentration.

Example 5

Question: What volume of CO₂ at RTP is produced when 50 cm³ of 1.0 mol/dm³ HCl reacts with excess sodium carbonate?

$\text{Na}_2\text{CO}_3(s) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)$Na2​CO3​(s)+2HCl(aq)→2NaCl(aq)+H2​O(l)+CO2​(g)

Step 1: Moles of HCl: $n = c \times V = 1.0 \times 0.05 = 0.05 \text{ mol}$n=c×V=1.0×0.05=0.05 mol

Step 2: Ratio of HCl : $\text{CO}_2$CO2​ = 2 : 1

$\text{moles of CO}_2 = \frac{0.05}{2} = 0.025 \text{ mol}$moles of CO2​=20.05​=0.025 mol

Step 3: Volume: $\text{volume} = 0.025 \times 24 = 0.6 \text{ dm}^3 = 600 \text{ cm}^3$volume=0.025×24=0.6 dm3=600 cm3

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Why All Gases Have the Same Molar Volume

At the same temperature and pressure:

• Gas particles are far apart relative to their size.
• The actual size of the particles is negligible compared to the spaces between them.
• Therefore the volume occupied depends only on the number of particles, not their identity.

This is a consequence of Avogadro's law: equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

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Common Mistakes to Avoid

Mistake	Correction
Using 24 when volume is in cm³	Use 24,000 for cm³; 24 for dm³
Forgetting the mole ratio	Always check the balanced equation ratio
Using molar volume for liquids or solids	24 dm³/mol applies to gases only at RTP
Using molar volume at non-RTP conditions	24 dm³ is only valid at RTP (20 °C, 1 atm). At other conditions the value differs.

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Summary

• At RTP, one mole of any gas occupies 24 dm³ (24,000 cm³).
• $\text{volume (dm}^3) = \text{moles} \times 24$volume (dm3)=moles×24.
• $\text{moles} = \text{volume (dm}^3) / 24$moles=volume (dm3)/24.
• This relationship allows you to convert between mass, moles and gas volume.
• Always use the balanced equation to get the mole ratio.
• This is primarily a Higher tier topic.

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Extended Worked Examples

Each calculation should follow a strict pattern: mass/volume/concentration → moles → ratio → moles of gas → volume. The two most common errors are mixing cm³ with dm³ (use 24 dm³ or 24 000 cm³ consistently) and forgetting the mole ratio from the balanced equation.

Example 6 — Volume of Hydrogen From a Metal

Question: 0.600 g of magnesium reacts completely with excess sulfuric acid. What volume of hydrogen is produced at RTP?

$\text{Mg}(s) + \text{H}_2\text{SO}_4(aq) \rightarrow \text{MgSO}_4(aq) + \text{H}_2(g)$Mg(s)+H2​SO4​(aq)→MgSO4​(aq)+H2​(g)

Step 1: $n(\text{Mg}) = 0.600 / 24 = 0.0250 \text{ mol}$n(Mg)=0.600/24=0.0250 mol. Step 2: Ratio Mg : $\text{H}_2$H2​ = 1 : 1, so $n(\text{H}_2) = 0.0250 \text{ mol}$n(H2​)=0.0250 mol. Step 3: Volume = $0.0250 \times 24 = \mathbf{0.600 \text{ dm}^3 = 600 \text{ cm}^3}$0.0250×24=0.600 dm3=600 cm3.

Example 7 — Mass of Reactant Needed for a Given Gas Volume

Question: What mass of calcium carbonate is needed to produce 240 cm³ of carbon dioxide at RTP?

$\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$CaCO3​(s)→CaO(s)+CO2​(g)

Step 1: $n(\text{CO}_2) = 240 / 24\,000 = 0.0100 \text{ mol}$n(CO2​)=240/24000=0.0100 mol. Step 2: Ratio 1 : 1, so $n(\text{CaCO}_3) = 0.0100 \text{ mol}$n(CaCO3​)=0.0100 mol. Step 3: $M_r$Mr​($\text{CaCO}_3$CaCO3​) = 100; mass = $0.0100 \times 100 = \mathbf{1.00 \text{ g}}$0.0100×100=1.00 g.

Example 8 — Linking Concentration to Gas Volume