Molar Gas Volume (Cambridge O Level Chemistry)

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Molar Gas Volume

  • 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:

 

3-2-1-molar-gas-volume-1

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:

3-2-1-molar-gas-volume-2

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 of a gas

Volume = Moles x Molar Volume

Examples of Converting Moles into Volumes Table

extended_a table, IGCSE & GCSE Chemistry revision notes

 

To find the moles of a gas

Moles = Volume ÷ Molar Volume

Examples of Converting Volumes into Moles Table

extended_b table, IGCSE & GCSE Chemistry revision notes

Calculating Gas Volumes

  • 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

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

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 Tip

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

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Caroline

Author: Caroline

Expertise: Physics Lead

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.