Avogadro's law relates gas volumes to particle numbers

In 1811, Italian scientist Amedeo Avogadro proposed that equal volumes of different gases at the same temperature and pressure contain an equal number of molecules. This principle, now known as Avogadro's law, has been extensively verified through experiments.

Since the amount of a substance is directly proportional to the number of particles it contains, we can state that:

• The amount of a gas is proportional to its volume under constant temperature and pressure conditions.
• Consequently, the volumes of two reacting gases measured under identical conditions are proportional to their respective amounts:

$\frac{\text{n}_1}{\text{n}_2}=\frac{\text{V}_1}{\text{V}_2}$

Furthermore, the amounts of reactants and products in a chemical reaction are proportional to their stoichiometric coefficients in a balanced equation. This allows us to determine the volumes of other gaseous substances involved in the reaction if we know the volume of any one gas consumed or produced, without the need to calculate their amounts.

Worked example 1 - Calculating volumes of gases in a combustion reaction

Calculate the volumes of oxygen (O₂) consumed and carbon dioxide (CO₂) produced when 1.22 dm³ of methane (CH₄) undergoes complete combustion. Assume all volumes are measured under the same conditions.

Step 1: Write balanced combustion equation

CH_4(g) + 2O_2(g) ➔ CO_2(g) + 2H₂O_(l)

Step 2: Equation

$\frac{\text{n}_1}{\text{n}_2}=\frac{\text{V}_1}{\text{V}_2}$

Step 3: Calculate volume of O₂ consumed

O₂ : CH₄ mole ratio = 2:1

Volume of O₂ = 2 x 1.22 = 2.44 dm³

Step 4: Calculate volume of CO₂ produced

CO₂ : CH₄ mole ratio = 1:1

Volume of CO₂ = 1.22 dm³

Note that Avogadro's law only applies to gases, so the volume of liquid water produced cannot be determined using this method.

The ideal gas model

The ideal gas model describes the constant motion of gas particles based on the following assumptions:

1. Gas particles move rapidly and randomly.
2. The volume of the gas particles themselves is negligible compared to the empty space between them.
3. There are no attractive or repulsive forces between the particles.
4. Collisions between particles are perfectly elastic, meaning no energy is lost during collisions.
5. The temperature of the gas is directly related to the average kinetic energy of its particles.

A hypothetical gas that perfectly follows these assumptions is called an ideal gas.

Limitations of the ideal gas model

Experimental observations show that real gases do not always adhere perfectly to the predictions of the ideal gas model.

This discrepancy arises from two main factors:

1. The existence of intermolecular attractions between gas particles.
2. The non-negligible volume occupied by the gas particles themselves.

These limitations become more pronounced under extreme conditions, such as high pressure and low temperature, because:

• Gas particles are forced closer together, allowing intermolecular attractions to become more significant.
• The volume of the particles themselves occupies a larger fraction of the total gas volume.

As a result, real gases tend to occupy smaller volumes and exert lower pressures than predicted by the ideal gas model under these conditions. However, noble gases like helium and neon exhibit near-ideal behaviour due to their minimal intermolecular forces.

The molar volume of an ideal gas

Avogadro's law implies that the molar volume of an ideal gas (the volume occupied by one mole of the gas) is a constant at a specified temperature and pressure. This molar gas volume is:

• Measured in dm³ mol^-1.
• Denoted as V_m.

Under standard temperature and pressure (STP):

• An ideal occupies a volume of 22.7 dm³ mol^-1.
• STP corresponds to 273 K (0°C) and 100 kPa.

To determine the number of moles in a given gas volume, use the formula:

$ \text{n }=\frac{\text{V}}{\text{V}_\text{m}} $

Where:

• n = number of moles of gas (mol)
• V = volume of gas (dm³)
• V_m = molar volume (dm³ mol^-1)

Worked example 2- Calculating molar mass of a gas

A 1.81 dm³ sample of an unknown gas at STP has a mass of 2.55 g.

Calculate the molar mass of the gas in g mol^-1. Give your answew to 3 significant figures.

Step 1: Equation

$n = \frac{V}{V_m}$

Step 2: Substitution and correct evaluation

$n = \frac{1.81}{22.7} = 0.0797 \text{ mol}$

Step 3: Calculate molar mass

$M = \frac{m}{n}$

Step 4: Substitution and correct evaluation

$M = \frac{2.55}{0.0797} = 32.0 \text{ g mol}^{-1}$

Therefore, the molar mass of the unknown gas is 32.0 g mol^-1.