Gas laws

At **constant temperature**, the pressure and volume of a fixed mass of gas obey **P V = constant** (inversely proportional).

At **constant pressure**, the volume of a fixed mass of gas obeys **V ÷ T = constant** — volume is proportional to the absolute (kelvin) temperature.

At **constant volume**, the pressure of a fixed mass of gas obeys **P ÷ T = constant** — pressure is proportional to the absolute (kelvin) temperature.

$\dfrac{PV}{T} = \text{constant}$ — so $\dfrac{P_1 V_1}{T_1} = \dfrac{P_2 V_2}{T_2}$. **Given** in the data booklet.

**T (K) = θ (°C) + 273.** Always do this before using a gas law.

The laws count temperature from **absolute zero** (−273 °C = 0 K); only the kelvin scale makes V and P truly proportional to T.

A **curve** (hyperbola) that sweeps down to the right, because P V is constant.

A **straight line through the origin**, because P = K(1/V); its **slope is the constant K**.

Pressure × volume = **Pa × m³ = J** (the joule).

Volume is fixed, so **P ÷ T = constant**: the pressure rises in proportion to the kelvin temperature.

Leaving the temperature in **°C** — every gas-law T must be in **kelvin** (°C + 273).

The **combined gas law** P V ÷ T = constant: fix T → Boyle, fix P → Charles, fix V → Gay-Lussac.

$PV = nRT = N k_B T$ — given in the data booklet. T must be in **kelvin**.

A fixed-size 'pack' of particles: one mole = **6.02 × 10²³** particles (the **Avogadro constant** N_A).

$N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}$ — the number of particles in one mole. **Given**.

$n = \dfrac{N}{N_A}$, so $N = n\,N_A$. **Given** in the data booklet.

Use **R** (8.31) with the amount in **moles n**; use **k_B** (1.38 × 10⁻²³) with the **number of molecules N**. Never mix them.

**Kelvin** (K). Convert from Celsius by adding 273.

**Equal N** — same P, V, T means the same number of molecules, whatever the gas.

Write $PV = NkT$ for each and **divide** one by the other — any equal quantity (P, V or T) cancels, leaving a simple ratio.

$R = 8.31\ \text{J K}^{-1}\,\text{mol}^{-1}$ — used with the amount in moles. **Given**.

$k_B = 1.38 \times 10^{-23}\ \text{J K}^{-1}$ — used with the number of molecules N. **Given**.

$n = \dfrac{PV}{RT}$ — with P in Pa, V in m³, T in K.

Leaving **T in Celsius** (must be kelvin), or mixing **n with k_B** / **N with R**.

Gas particles **colliding with the walls** of the container — each collision pushes on the wall.

The **average kinetic energy** of its particles — hotter gas means faster particles.

It is **proportional to the absolute temperature**: average KE ∝ T (T in kelvin).

$\overline{E_k} = \tfrac{3}{2}k_B T$ — **given** in the data booklet (T in kelvin).

The **Boltzmann constant**, 1.38 × 10⁻²³ J K⁻¹ — it links energy to temperature for one particle.

The relation average KE ∝ T only works from **absolute zero** (0 K); convert Celsius with **+ 273**.

**Equal** — average kinetic energy depends only on the temperature, not the gas or particle mass.

The piston **does work** on the gas, raising the particles' average kinetic energy, so they move faster.

It is **zero** — the particles have the least possible motion.

Particles are tiny points with negligible volume; there are **no forces between them** except during (elastic) collisions.

**Only kinetic** energy — no intermolecular potential energy (no forces between particles).

All gases have the **same average KE** at a given temperature, so heavier particles need a **lower speed** to have that energy.