Calculate Volumes of Gases (Edexcel IGCSE Chemistry)

Spec point — understand how to carry out calculations involving gas volumes and the molar volume of a gas (24 dm3 and 24 000 cm3 at room temperature and pressure (rtp))

Calculate Volumes of Gases

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 20 ºC 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 dm³and 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 cm³and 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

Using mass to calculate the volume of a gas

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

Answer:

• Step 1: Calculate the moles of nitrogen:

  • M_r (N₂) = 2 x 14 = 28
  • Moles of N₂ = $\frac{154}{28}$ = 5.5 mol

• Step 2: Calculate the molar volume of nitrogen:

  • Volume = moles x 24
  • Volume = 5.5 x 24 = 132 dm³

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