Ideal gases

At **constant temperature** (and amount), the pressure of a gas is **inversely proportional** to its volume: $P_{1}V_{1} = P_{2}V_{2}$.

Pressure is **directly proportional** to the **kelvin** temperature: $\dfrac{P_{1}}{T_{1}} = \dfrac{P_{2}}{T_{2}}$.

$\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}$ — with T in **kelvin**. It is given in the data booklet.

**T/K = θ/°C + 273** — always do this before a gas-law calculation.

The particles have **no volume** of their own and there are **no forces** between them.

At **high temperature** and **low pressure** — particles are far apart and fast-moving.

At **low temperature** and **high pressure** — particle volume and intermolecular forces become significant.

The pressure **halves** (Boyle's law: P ∝ 1/V).

Only the **kelvin** scale starts at true zero (0 K), so only it gives the correct proportionality; °C would give wrong ratios.

Because P ∝ kelvin T — at 0 K the particles would stop and the pressure would be **zero**.

The **temperature** and the **amount** of gas; only P and V change.

Use the combined gas law: $P_{2} = P_{1}\times\dfrac{V_{1}}{V_{2}}\times\dfrac{T_{2}}{T_{1}}$ (T in kelvin).

$PV = nRT$ — links pressure, volume, amount and temperature of an ideal gas.

**Standard temperature and pressure**: 273 K (0 °C) and 100 kPa.

V_{m} = **22.7 dm³ mol⁻¹** — the volume of one mole of any ideal gas at STP.

$n = \dfrac{V}{V_{m}}$ — divide the volume (in dm³) by 22.7.

$V = n\,V_{m}$ — multiply the amount (mol) by 22.7 dm³ mol⁻¹.

**Pa** (pressure), **m³** (volume) and **K** (temperature), because R = 8.31 J K⁻¹ mol⁻¹ is in SI units.

R = **8.31 J K⁻¹ mol⁻¹** (given in the data booklet).

**Multiply by 1000** — e.g. 101 kPa = 1.01 × 10⁵ Pa.

**Divide by 1000** — e.g. 24.0 dm³ = 0.0240 m³.

**Add 273** — e.g. 25 °C = 298 K.

**At STP** use V_{m} = 22.7; at **any other conditions** use PV = nRT with SI units.

Find n from PV = nRT, then $M = \dfrac{m}{n}$ using the sample mass.