The ideal gas equationAQA A-Level Chemistry: Revision notes

The behaviour of a gas is described by the ideal gas equation:

$pV = nRT$

• $p$ = pressure
• $V$ = volume
• $n$ = amount of gas in mol
• $R$ = the gas constant, 8.31 J K⁻¹ mol⁻¹ (given in the exam, so you do not need to recall it)
• $T$ = temperature

It links four properties of a gas. At constant $n$ and $V$, $p$ is proportional to $T$; at constant $n$ and $T$, $p$ is inversely proportional to $V$.

The equation only works when every variable is in SI units:

• pressure in pascals, Pa (1 kPa = 10³ Pa; 1 MPa = 10⁶ Pa)
• volume in cubic metres, m³ (1 dm³ = 10⁻³ m³; 1 cm³ = 10⁻⁶ m³)
• temperature in kelvin, K (T / K = temperature in °C + 273)
• amount in mol

With these units, $R$ = 8.31 J K⁻¹ mol⁻¹. Convert first, then substitute.

Rearrange to find whichever quantity is missing:

• amount: $n = \frac{pV}{RT}$
• pressure: $p = \frac{nRT}{V}$
• volume: $V = \frac{nRT}{p}$
• temperature: $T = \frac{pV}{nR}$

Because $n = \frac{m}{M_r}$, the equation also gives the relative molecular mass of a gas or volatile liquid: $M_r = \frac{mRT}{pV}$.

Question: Calculate the amount of nitrogen in a 250 cm³ flask at 27 °C and 120 kPa.

1. $V$ = 250 × 10⁻⁶ = 2.50 × 10⁻⁴ m³
2. $p$ = 120 × 10³ = 1.20 × 10⁵ Pa
3. $T$ = 27 + 273 = 300 K
4. $n = \frac{pV}{RT} = \frac{1.20 \times 10^5 \times 2.50 \times 10^{-4}}{8.31 \times 300}$ = 0.0120 mol

Mr of a volatile liquid: 0.223 g vaporises to give 95.0 cm³ at 373 K and 101 kPa.

• $V$ = 9.50 × 10⁻⁵ m³; $p$ = 101000 Pa
• $n = \frac{101000 \times 9.50 \times 10^{-5}}{8.31 \times 373}$ = 3.10 × 10⁻³ mol
• $M_r = \frac{0.223}{3.10 \times 10^{-3}}$ = 72.0

Mass of reactant for a given gas volume: find $n$ of gas with the ideal gas equation, then use the balanced equation's mole ratio, then multiply by $M$. For 2NaN₃ → 2Na + 3N₂ producing 1.685 mol N₂: $n$(NaN₃) = 1.685 × 2/3 = 1.123 mol, so the mass is 1.123 × 65.0 = 73.0 g.