Volumes of Gases

This lesson covers how to calculate the volume of a gas at room temperature and pressure (RTP), and how to use gas volumes in calculations involving balanced equations. This is required by the Edexcel GCSE Chemistry specification (1CH0) and is a Higher tier topic.

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Molar Gas Volume

At room temperature and pressure (RTP) — defined as 20°C (293 K) and 1 atmosphere (101.3 kPa) — one mole of any gas occupies a volume of 24 dm³ (or 24,000 cm³).

This value is called the molar gas volume.

Condition	Temperature	Pressure	Molar volume
RTP	20°C	1 atm	24 dm³/mol

This means:

• 1 mol of hydrogen gas (H₂) occupies 24 dm³ at RTP
• 1 mol of oxygen gas (O₂) occupies 24 dm³ at RTP
• 1 mol of carbon dioxide (CO₂) occupies 24 dm³ at RTP
• 1 mol of any gas occupies 24 dm³ at RTP

Exam Tip: The molar gas volume (24 dm³ at RTP) applies to all gases, regardless of their Mr. This is a key principle. You will always be told the molar gas volume in the question if you need it.

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The Gas Volume Formula

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

Or, in cm³:

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

These can be rearranged:

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

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

flowchart LR
    M["Mass (g)"] -- "÷ Mr" --> N[Moles]
    N -- "× 24" --> V1["Volume at RTP (dm³)"]
    N -- "× 24 000" --> V2["Volume at RTP (cm³)"]
    V1 -- "÷ 24" --> N
    V2 -- "÷ 24 000" --> N
    N -- "× Mr" --> M

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Worked Examples

Example 1: Moles to Volume

Question: Calculate the volume of 0.5 mol of carbon dioxide at RTP.

Solution:

• volume = moles × 24
• volume = 0.5 × 24
• volume = 12 dm³

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Example 2: Volume to Moles

Question: Calculate the number of moles in 6 dm³ of oxygen gas at RTP.

Solution:

• moles = volume ÷ 24
• moles = 6 ÷ 24
• moles = 0.25 mol

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Example 3: Volume in cm³

Question: What volume (in cm³) does 0.1 mol of hydrogen gas occupy at RTP?

Solution:

• volume = moles × 24,000
• volume = 0.1 × 24,000
• volume = 2,400 cm³

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Example 4: Mass to Volume

Question: Calculate the volume of gas produced when 11 g of carbon dioxide is released at RTP. (Mr of CO₂ = 44)

Solution:

1. Calculate moles: moles = mass ÷ Mr = 11 ÷ 44 = 0.25 mol
2. Calculate volume: volume = 0.25 × 24 = 6 dm³

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Example 5: Volume to Mass

Question: What is the mass of 4.8 dm³ of methane (CH₄) at RTP? (Mr of CH₄ = 16)

Solution:

1. Calculate moles: moles = 4.8 ÷ 24 = 0.2 mol
2. Calculate mass: mass = 0.2 × 16 = 3.2 g

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Calculating Gas Volumes from Equations

You can combine gas volume calculations with balanced equations to predict volumes of gaseous products.

Example 6: Volume from a Reaction

Question: What volume of hydrogen gas (at RTP) is produced when 2.4 g of magnesium reacts with excess hydrochloric acid?

Mg + 2HCl → MgCl₂ + H₂

Solution:

1. Mr of Mg = 24
2. Moles of Mg = 2.4 ÷ 24 = 0.1 mol
3. From the equation: 1 mol Mg produces 1 mol H₂
4. Moles of H₂ = 0.1 mol
5. Volume of H₂ = 0.1 × 24 = 2.4 dm³

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Example 7: Volume of CO₂ from Decomposition

Question: Calculate the volume of carbon dioxide produced (at RTP) when 25 g of calcium carbonate is heated.

CaCO₃ → CaO + CO₂

Solution:

1. Mr of CaCO₃ = 100
2. Moles of CaCO₃ = 25 ÷ 100 = 0.25 mol
3. From the equation: 1 mol CaCO₃ produces 1 mol CO₂
4. Moles of CO₂ = 0.25 mol
5. Volume of CO₂ = 0.25 × 24 = 6 dm³

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Example 8: Using Gas Volumes to Find Moles of a Solid

Question: 480 cm³ of hydrogen gas was collected at RTP when zinc reacted with excess sulfuric acid. Calculate the mass of zinc that reacted.

Zn + H₂SO₄ → ZnSO₄ + H₂

Solution:

1. Moles of H₂ = 480 ÷ 24,000 = 0.02 mol
2. From the equation: 1 mol Zn produces 1 mol H₂
3. Moles of Zn = 0.02 mol
4. Mr of Zn = 65
5. Mass of Zn = 0.02 × 65 = 1.3 g

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Avogadro's Law

Avogadro's law states:

Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

This means that if you have two containers, each holding 1 dm³ of gas at the same temperature and pressure, they contain the same number of molecules — even if the gases are different.

Consequences of Avogadro's Law

The molar ratios in a balanced equation are also the volume ratios for gases.

Example:

2H₂ + O₂ → 2H₂O

This means:

• 2 volumes of H₂ react with 1 volume of O₂
• If 100 cm³ of H₂ reacts, it requires 50 cm³ of O₂
• If 2 dm³ of H₂ reacts, it requires 1 dm³ of O₂

Example 9: Using Volume Ratios

Question: What volume of oxygen is needed to completely combust 60 cm³ of methane at RTP?

CH₄ + 2O₂ → CO₂ + 2H₂O

Solution:

• Volume ratio: 1 CH₄ : 2 O₂
• 60 cm³ of CH₄ requires 60 × 2 = 120 cm³ of O₂

Exam Tip: When a question involves gas volumes only (no masses), you can use the volume ratio directly from the equation without converting to moles first. This saves time.

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Summary of Formulae

What you want	Formula
Volume in dm³	moles × 24
Volume in cm³	moles × 24,000
Moles from dm³	volume (dm³) ÷ 24
Moles from cm³	volume (cm³) ÷ 24,000

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Common Mistakes