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 dm³mol^-1(since it is 'per mole')
The conditions of STP are
- a temperature of 0^oC (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 cm³of propane, the volume of oxygen needed is (5 x 50) 250 cm³, and (3 x 50) 150 cm³of 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 cm³of 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 O₂requires 4/5 x 50 cm³ of NH₃to react = 40 cm³

Step 3: Using Avogadro's Law, we can say 40 cm³of NO will be produced

Step 4: There will be 70-40 = 30 cm³of NH₃left over

Therefore the total remaining volume will be 40 + 30 = 70 cm³of 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 NH₃70/4 gives 17.5
- For O₂50/5 gives 10, so oxygen is limiting

Molar Gas Volume

The molar gas volume of 22.7 dm³mol^-1can 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 dm³mol^-1

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

$amount of gas (moles) = \frac{volumes of gas in {dm}^{3}}{22.7 {dm}^{3} {mol}^{- 1}}$

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 dm³mol^-1

3.0 mol x 22.7 dm³mol^-1= 68 dm³

Worked example

How many moles are in the following volumes of gases?

7.2 dm³ of carbon monoxide
960 cm³ of sulfur dioxide

Answer 1:

Step 1: Use the formula

$amount of gas (moles) = \frac{volumes of gas in {dm}^{3}}{22.7 {dm}^{3} {mol}^{- 1}}$

$amount of gas (moles) = \frac{7.2 {dm}^{3}}{22.7 {dm}^{3} {mol}^{- 1}} = 0.32 mol$

Answer 2

Step 1: Convert the volume from cm³ to dm³

$\frac{960}{1000}$ = 0.960 dm³

Step 2: Use the formula

$amount of gas (moles) = \frac{0.960 {dm}^{3}}{22.7 {dm}^{3} {mol}^{- 1}} = 4.22 \times 10^{- 2} mol$

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

Revision Note