Amount of Substance in Relation to Volumes of Gases (AQA GCSE Chemistry)

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:

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:

Formula triangle showing the relationship between moles of gas, volume in cm³ and 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

To find the moles

Moles = Volume ÷ Molar Volume

Examples of Converting Volumes into Moles Table

• 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.

Answer 

• 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

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