3.5.1 Amount of Substance in Relation to Volumes of Gases
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:
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
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
- 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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