AQA GCSE Chemistry

Revision Notes

3.5.1 Amount of Substance in Relation to Volumes of Gases

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Calculating Gas Volumes

Higher Tier Only

Avogadro's Law

  • Avogadro’s Law states that at the same conditions of temperature and pressure, equal amounts of gases occupy the same volume of space
  • At room temperature and pressure, the volume occupied by one mole of any gas was found to be 24 dm3 or 24,000 cm3
  • This is known as the molar gas volume at RTP
  • RTP stands for “room temperature and pressure” and the conditions are 20ºC and 1 atmosphere (atm)
  • From the molar gas volume the following formula triangle can be derived:

Molar Volume dm3 (decimetre) Formula Triangle, IGCSE & GCSE Chemistry revision notes

Formula triangle showing the relationship between moles of gas, volume in dm3 and the molar volume

  • If the volume is given in cm3 instead of dm3, then divide by 24,000 instead of 24:

Molar Volume cm3 Formula Triangle, IGCSE & GCSE Chemistry revision notes

Formula triangle showing the relationship between moles of gas, volume in cmand the molar volume

  • The formula can be used to calculate the number of moles of gases from a given volume or vice versa
  • Simply cover the one you want and the triangle tells you what to do

To find the volume

Volume = Moles x Molar Volume

Examples of Converting Moles into Volumes Table

Examples of Converting Moles into Volumes Table, downloadable IGCSE & GCSE Chemistry revision notes

To find the moles

Moles = Volume ÷ Molar Volume

Examples of Converting Volumes into Moles Table

Examples of Converting Volumes into Moles Table, downloadable IGCSE & GCSE Chemistry revision notes

Calculations Involving Reacting Gases

  • You may be asked to calculate the volume of a gas from a given amount stated in grams instead of moles
  • To answer these type of questions you must first convert grams to moles and then calculate the volume.

Worked example

Example 1

What is the volume of 154 g of nitrogen gas at RTP?

Answer 

Gas Volumes from Masses WE1, downloadable IGCSE & GCSE Chemistry revision notes

  • A second style of gas calculation involves calculating the volumes of gaseous reactants and products from a balanced equation and a given volume of a gaseous reactant or product
  • These problems are straightforward as you are applying Avogadro's Law, so the moles ( and coefficients) in equations are in the same ratio as the gas volumes

Worked example

Example 2

The complete combustion of propane gives carbon dioxide and water vapour as the products

C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g)

Determine the volume of oxygen needed to react with 150 cm3 of propane and the total volume of the gaseous products

Answer

    • The balanced equation shows that 5 moles of oxygen are needed to completely react with 1 mole of propane
    • Therefore the volume of oxygen needed would be = 5 moles x 150 cm3 = 750 cm3
    • The total number of moles of gaseous products is = 3 + 4 = 7 moles
    • The total volume of gaseous products would be = 7 moles x 150 cm3 = 1050 cm3

Exam Tip

Make sure you use the correct units as asked by the question when working through reacting gas volume questions.

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