Revision NotesExam QuestionsFlashcards
IBChemistrySLRevision NotesHow Much, How Fast & How Far?How Much? The Amount of Chemical ChangeAvogadro's Law & Molar Volume of Gas

Syllabus Edition

First teaching 2023

First exams 2025

|

Avogadro's Law & Molar Volume of Gas (SL IB Chemistry)

Revision Note

Philippa Platt

Last updated

Avogadro's Law & Molar Volume of Gas

Volumes of gases

  • In 1811 the Italian scientist Amedeo Avogadro developed a theory about the volume of gases
  • Avogadro’s law (also called Avogadro’s hypothesis) enables the mole ratio of reacting gases to be determined from volumes of the gases
  • Avogadro deduced that equal volumes of gases must contain the same number of molecules
  • At standard temperature and pressure(STP) one mole of any gas has a volume of 22.7 dm3
  • The units are normally written as dm3 mol-1(since it is 'per mole')
  • The conditions of STP are
    • a temperature of 0o C (273 K)
    • a pressure of 100 kPa

Stoichiometric relationships

  • The stoichiometry of a reaction and Avogadro's Law can be used to deduce the exact volumes of gaseous reactants and products
    • Eg. in the combustion of 50 cm3 of propane, the volume of oxygen needed is (5 x 50) 250 cm3, and (3 x 50) 150 cm3 of carbon dioxide is formed, using the ratio of propane: oxygen: carbon dioxide, which is 1: 5: 3 respectively, as seen in the equation

C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (l)

  • Remember that if the gas volumes are not in the same ratio as the coefficients then the amount of product is determined by the limiting reactant so it is essential to identify it first

Worked example

What is the total volume of gases remaining when 70 cm3 of ammonia is combusted completely with 50 cm3 of oxygen according to the equation shown?

4NH3 (g) + 5O2 (g) → 4NO (g) + 6H2O (l)

Answer:

Step 1: From the equation deduce the molar ratio of the gases, which is NH3 :O2 :NO or 4:5:4 (water is not included as it is in the liquid state)

Step 2: We can see that oxygen will run out first (the limiting reactant) and so 50 cm3 of O2 requires 4/5 x 50 cm3 of NH3 to react = 40 cm3 

Step 3: Using Avogadro's Law, we can say 40 cm3 of NO will be produced

Step 4: There will be 70-40 = 30 cm3 of NH3 left over

Therefore the total remaining volume will be 40 + 30 = 70 cm3 of gases

Examiner Tip

Since gas volumes work in the same way as moles, we can use the 'lowest is limiting' technique in limiting reactant problems involving gas volumes. This can be handy if you are unable to spot which gas reactant is going to run out first. Divide the volumes of the gases by the cofficients and whichever gives the lowest number is the limiting reactant

  • E.g. in the previous problem we can see that
    • For NH3  70/4 gives 17.5
    • For O2 50/5 gives 10, so oxygen is limiting

Molar Gas Volume

  • The molar gas volume of 22.7 dm3 mol-1 can be used to find:
    • The volume of a given number of moles of gas:

volume of gas (dm3) = amount of gas (mol) x 22.7 dm3 mol-1

    • The number of moles of a given volume of gas:

               begin mathsize 20px style amount space of space gas space left parenthesis moles right parenthesis space equals space fraction numerator volumes space of space gas space in space dm cubed over denominator 22.7 space dm cubed space mol to the power of negative 1 end exponent end fraction end style

  • The relationships can be expressed using a formula triangle

Gas formula triangle  

Gas formula volume triangle

To use the gas formula triangle cover the one you want to find out about with your finger and follow the instructions

Worked example

What is the volume occupied by 3.0 moles of hydrogen at stp?

Answer:

volume of gas (dm3) = amount of gas (mol) x 22.7 dm3 mol-1

3.0 mol x 22.7 dm3 mol-1= 68 dm3

Worked example

How many moles are in the following volumes of gases?

  1. 7.2 dm3 of carbon monoxide
  2. 960 cm3 of sulfur dioxide

Answer 1:

Step 1: Use the formula

amount space of space gas space left parenthesis moles right parenthesis space equals space fraction numerator volumes space of space gas space in space dm cubed over denominator 22.7 space dm cubed space mol to the power of negative 1 end exponent end fraction

amount space of space gas space left parenthesis moles right parenthesis space equals space fraction numerator 7.2 space dm cubed over denominator 22.7 space dm cubed space mol to the power of negative 1 end exponent end fraction space equals space 0.32 space mol

Answer 2

Step 1: Convert the volume from cm3 to dm3

960 over 1000 = 0.960 dm3

Step 2: Use the formula

amount space of space gas space left parenthesis moles right parenthesis space equals space fraction numerator 0.960 space dm cubed over denominator 22.7 space dm cubed space mol to the power of negative 1 end exponent end fraction space equals space 4.22 space cross times 10 to the power of negative 2 end exponent space mol

 

You've read 0 of your 5 free revision notes this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

    Test yourselfFlashcards

    Did this page help you?

    Previous:Reacting Mass CalculationsNext:Concentration Calculations
    Philippa Platt

    Author: Philippa Platt

    Expertise: Chemistry

    Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener.