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4.9 Volumes of Gases

GCSE Chemistry · AQA · Chapter 4: Chemical Calculations

A free GCSE Chemistry lesson for AQA (8462) from OBA School: (Higher tier) The molar gas volume at room temperature and pressure, and calculations involving volumes of gases. Watch the video, read the explanation and worked examples, then try the revision questions. You can work through the full lesson, with every video, practice quizzes and printable worksheets with mark schemes, in our free online course.

What you will learn

  • Recall that one mole of any gas occupies 24 dm³ at room temperature and pressure (rtp)
  • Convert between the volume of a gas, the number of moles and the mass
  • Use the ratio in a balanced equation to calculate volumes of gaseous reactants and products
  • Calculate the volume of gas made from a given mass of a reactant, and the reverse

Key words

  • Room temperature and pressure (rtp): 20 °C and 1 atmosphere pressure.
  • Molar gas volume: The volume occupied by one mole of any gas: 24 dm³ (24 000 cm³) at rtp.
  • Avogadro’s law: Equal volumes of different gases at the same temperature and pressure contain the same number of molecules.
  • Gas syringe: Apparatus used to collect and measure the volume of a gas made in a reaction.

One mole of any gas

In a gas the molecules are very far apart compared with their own size. So the space a gas takes up depends on the number of molecules, not on what they are. At room temperature and pressure (20 °C and 1 atmosphere), one mole of any gas occupies 24 dm³.

volume of gas (dm³) = moles × 24 moles = volume (dm³) ÷ 24

In cm³: volume (cm³) = moles × 24 000

Key fact: The figure 24 dm³ only applies to gases at rtp. It does not apply to liquids or solids (water at 20 °C is a liquid), or to gases at other temperatures and pressures.

Volume, moles and mass

Worked example: volume of 0.25 mol of oxygen at rtp

volume = 0.25 × 24 = 6.0 dm³

Answer: 6.0 dm³

Worked example: volume of 8.8 g of carbon dioxide at rtp

moles = 8.8 ÷ 44 = 0.20 mol
volume = 0.20 × 24 = 4.8 dm³

Answer: 4.8 dm³

Worked example: mass of 600 cm³ of hydrogen at rtp

moles = 600 ÷ 24 000 = 0.025 mol
mass = 0.025 × 2 = 0.050 g

Answer: 0.050 g

Reacting volumes of gases

Because equal volumes of gases contain equal numbers of molecules, the balancing numbers in an equation give the ratio of gas volumes directly (at the same temperature and pressure). No Mr is needed.

Worked example: burning methane

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
50 cm³ of methane needs 2 × 50 = 100 cm³ of oxygen and makes 50 cm³ of carbon dioxide.
The water is a liquid at rtp, so it has (almost) no volume as a gas.

Answer: 100 cm³ O₂; 50 cm³ CO₂

Exam tip: If one gas is in excess, use the limiting gas to work out what reacts, then subtract to find what is left over.

Gas volumes from masses

Combine the reacting-mass method with the molar volume.

Worked example: hydrogen from magnesium

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). 0.60 g of magnesium reacts with excess acid.
moles Mg = 0.60 ÷ 24 = 0.025 mol; ratio Mg : H₂ = 1 : 1, so 0.025 mol H₂
volume = 0.025 × 24 = 0.60 dm³ = 600 cm³

Answer: 600 cm³ of hydrogen

Worked example: mass of reactant needed for a volume of gas (ratio not 1 : 1)

Oxygen can be made from hydrogen peroxide: 2H₂O₂(aq) → 2H₂O(l) + O₂(g). What mass of H₂O₂ is needed to make 1.2 dm³ of oxygen at rtp?
moles O₂ = 1.2 ÷ 24 = 0.050 mol
ratio H₂O₂ : O₂ = 2 : 1, so moles H₂O₂ = 2 × 0.050 = 0.10 mol
Mr H₂O₂ = 34, so mass = 0.10 × 34 = 3.4 g

Answer: 3.4 g

Exam tip: Work in three steps: volume → moles of gas (÷ 24); use the equation ratio; moles → mass (× Mr). The molar gas volume (24 dm³) is always given in the exam.

Common mistakes

  • Using 24 for a volume in cm³. Use 24 000 cm³, or convert cm³ to dm³ first.
  • Using Mr to compare gas volumes. Volumes of gases depend on moles, not mass.
  • Including water as a gas at rtp. At 20 °C it is a liquid.
  • Forgetting the ratio: in 2H₂ + O₂, the volume of oxygen needed is half the volume of hydrogen.

Revision questions

Try these, then check your answers below each one.

Question 1. (Higher tier) What is the volume of 0.40 mol of carbon dioxide at room temperature and pressure?

Answer: 9.6 dm³

Working: 0.40 × 24 = 9.6 dm³.

Question 2. (Higher tier) How many moles of gas are in 1.2 dm³ at rtp?

Answer: 0.05 mol

Working: 1.2 ÷ 24 = 0.050 mol.

Question 3. What conditions are meant by room temperature and pressure (rtp)? (Choose from: 20 °C and 1 atmosphere, 0 °C and 1 atmosphere, 25 °C and 2 atmospheres, 100 °C and 1 atmosphere.)

Answer: 20 °C and 1 atmosphere

Working: rtp is 20 °C and 1 atmosphere pressure. At rtp, 1 mol of any gas occupies 24 dm³.

Related lessons

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This lesson is section 4.9 (AGCCH0409) of our free GCSE Chemistry course for AQA. In the full course you get every video for this section, a self-marking practice quiz, a printable worksheet with a mark scheme, chapter tests, and your progress saved. Go to the full GCSE Chemistry course.

 

 

 

 

 

 

 

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