Amount of Substance in Relation to Volumes of Gases (AQA GCSE Chemistry)
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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 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 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
Worked Example
Convert the following moles of gases at rtp into volume (dm3)
3 moles of hydrogen
0.25 moles of carbon dioxide
5.4 moles of oxygen
0.02 moles ammonia
Answer:
Use the equation:
volume = moles x 24
3 x 24 = 72 dm3
0.25 x 24 = 6 dm3
5.4 x 24 = 129. 6 dm3
0.02 x 24 = 0.48 dm3
Worked Example
Convert the following volumes of gases at rtp into moles
225.6 dm3 methane
7.2 dm3 carbon monoxide
960 cm3 sulfur dioxide
1200 cm3 of oxygen
Answer:
Use the equation :
moles = volume (dm3) ÷ 24
OR
moles = volume (cm3) ÷ 24 000
225.6 ÷ 24 = 9.4 mols
7.2 ÷ 24 = 0.3 mols
960 ÷ 24 000 = 0.04 mols
1200 ÷ 24 000 = 0.05 mols
Calculations involving reacting gases
Higher tier only
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:
Mr (N2) = 2 x 14 = 28
Moles of N2 =
= 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
Examiner Tips and Tricks
Make sure you use the correct units as asked by the question when working through reacting gas volume questions.
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