Gas laws

A real gas is complicated, so physicists use an idealised model — the ideal gas — that keeps the important physics and throws away the awkward details. The model rests on a short list of assumptions you must be able to state:

• The gas contains a very large number of identical particles (molecules/atoms).
• The particles are treated as point masses — their own volume is negligible compared with the volume of the container.
• There are no intermolecular forces except during the brief moments of collision (so no potential energy is stored between particles).
• All collisions (particle–particle and particle–wall) are perfectly elastic — no kinetic energy is lost.
• The particles are in continuous, random motion and obey Newton's laws of mechanics.
• The time of a collision is negligible compared with the time between collisions.

Where does pressure come from? (a favourite "explain" question)

Pressure is not something the gas "contains" — it is produced by collisions:

1. A molecule moving towards a wall collides with it and bounces back, so its momentum changes (its velocity reverses).
2. A change of momentum means the wall exerts a force on the molecule (Newton's second law), and by Newton's third law the molecule exerts an equal and opposite force on the wall.
3. Countless molecules strike the walls every second, and the total force per unit area of wall is the gas pressure ($p = F/A$).

Anything that makes the molecules hit the walls harder or more often raises the pressure — for example squeezing the gas into a smaller volume, or heating it so the particles move faster.

Gases contain colossal numbers of particles, so we count them in moles. One mole is the amount of substance that contains exactly $N_A = 6.02\times10^{23}$ particles — the Avogadro constant (from the data booklet).

To convert between the amount in moles, $n$, and the actual number of particles, $N$:

$N = nN_A$

• $n$ = amount of substance in moles (mol)
• $N$ = number of particles (molecules or atoms) — just a count, no unit
• $N_A = 6.02\times10^{23}\,\text{mol}^{-1}$

Worked mini-example: how many molecules are in 0.25 mol of a gas? $N = nN_A = 0.25 \times 6.02\times10^{23} = 1.5\times10^{23}\ \text{molecules}$

This matters because the ideal gas law comes in two flavours that must agree:

• $pV = nRT$ uses moles $n$ and the molar gas constant $R = 8.31\,\text{J mol}^{-1}\,\text{K}^{-1}$;
• $pV = Nk_BT$ uses the number of particles $N$ and the Boltzmann constant $k_B = 1.38\times10^{-23}\,\text{J K}^{-1}$.

Because $N = nN_A$, the two forms are identical provided $R = N_A k_B \quad\Rightarrow\quad 6.02\times10^{23} \times 1.38\times10^{-23} = 8.31\,\text{J mol}^{-1}\,\text{K}^{-1}\ \checkmark$

Keeping $n$/$R$ together and $N$/$k_B$ together is the single best way to avoid mixing the two forms.

If you keep the amount of gas fixed and hold one of $p$, $V$ or $T$ constant, the ideal gas law collapses into a simple relationship between the other two. These are the historical gas laws.

Law	Held constant	Relationship	Shape
Boyle's law	temperature $T$	$pV = \text{constant}$ (so $p \propto 1/V$)	isothermal
Charles's law	pressure $p$	$\dfrac{V}{T} = \text{constant}$ (so $V \propto T$)	isobaric
Gay-Lussac's / pressure law	volume $V$	$\dfrac{p}{T} = \text{constant}$ (so $p \propto T$)	isovolumetric

Boyle's law ($T$ constant): squeeze a gas into half the volume and the pressure doubles. A graph of $p$ against $V$ is a curve (a rectangular hyperbola) sloping down — each such curve is called an isotherm. A graph of $p$ against $1/V$ is a straight line through the origin.

Charles's law ($p$ constant): heat a gas and it expands so that $V/T$ stays constant. A graph of $V$ against $T$ (in kelvin) is a straight line through the origin.

Gay-Lussac's / pressure law ($V$ constant): heat a gas in a rigid container and its pressure rises so that $p/T$ stays constant. A graph of $p$ against $T$ (in kelvin) is a straight line through the origin.

For any change between two states of a fixed mass of gas you can combine all three into the combined gas law: $\frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2}$ Set whichever quantity is unchanged equal on both sides and it cancels, recovering the individual law you need.

You are not asked to reproduce a full derivation under exam conditions, but seeing how the three gas laws fit together makes the ideal gas law impossible to misremember — and IB rewards reasoning from first principles.

Take a fixed amount of gas. The three experimental laws say, one variable at a time:

• Boyle (constant $T$): $\;V \propto \dfrac{1}{p}$
• Charles (constant $p$): $\;V \propto T$
• Gay-Lussac (constant $V$): $\;p \propto T$

Boyle and Charles between them tell us how $V$ depends on both $p$ and $T$ at once. Combining the two proportionalities: $V \propto \frac{T}{p} \quad\Rightarrow\quad \frac{pV}{T} = \text{constant}$

This is the combined gas law: for a fixed mass of gas, $\dfrac{pV}{T}$ never changes, which is exactly why $\dfrac{p_1V_1}{T_1} = \dfrac{p_2V_2}{T_2}$.

What is the constant? Experiment shows that, for a fixed pressure, volume and temperature, the constant is proportional to the amount of gas. Writing the amount as $n$ moles and calling the constant of proportionality $R$ (the molar gas constant): $\frac{pV}{T} = nR \quad\Rightarrow\quad \boxed{pV = nRT}$ with $R = 8.31\,\text{J mol}^{-1}\,\text{K}^{-1}$ from the data booklet.

The particle form. Using $n = \dfrac{N}{N_A}$ (from $N = nN_A$): $pV = \frac{N}{N_A}RT = N\left(\frac{R}{N_A}\right)T = Nk_BT$ where $k_B = \dfrac{R}{N_A} = 1.38\times10^{-23}\,\text{J K}^{-1}$ is the Boltzmann constant. So the two forms are the same law written per-mole or per-particle: $\boxed{pV = nRT} \qquad\Longleftrightarrow\qquad \boxed{pV = Nk_BT}$

Why this matters for grade 9: if you ever forget a specific gas law, start from $pV = nRT$, hold one quantity constant, and the special case drops out. For example, at constant $T$ the right-hand side $nRT$ is a constant, so $pV =$ constant — that is Boyle's law.

Every gas-law relationship — $V \propto T$, $p \propto T$, $E_k \propto T$ — is a direct proportion, so it is only true if $T$ is measured from a scale whose zero is a true zero. That scale is the absolute (thermodynamic) temperature in kelvin (K).

Converting: $T(\text{K}) = \theta(°\text{C}) + 273$ So $0°\text{C} = 273\,\text{K}$, $27°\text{C} = 300\,\text{K}$, $100°\text{C} = 373\,\text{K}$. (A change of 1 °C equals a change of 1 K, but the two scales have different zeros.)

Absolute zero ($0\,\text{K} = -273°\text{C}$) is the temperature at which an ideal gas would exert zero pressure and shrink to zero volume — the particles would have the least possible kinetic energy. You never quite reach it, but extrapolating a $V$–$T$ (or $p$–$T$) graph backwards to where $V = 0$ (or $p = 0$) always points to the same temperature, $-273°\text{C}$. That is the experimental evidence for absolute zero, and the reason the kelvin scale is defined from it.

The habit to build: the very first line of any gas-law calculation should convert every temperature to kelvin. Doubling the kelvin temperature of a gas at constant pressure doubles its volume; doubling the Celsius temperature (say 20 °C → 40 °C) does not, because 293 K → 313 K is only a 7 % rise.

The deepest idea in this topic is that temperature is molecular kinetic energy. For an ideal gas the average translational kinetic energy of a single particle is $E_k = \frac{3}{2}k_BT$ where $k_B = 1.38\times10^{-23}\,\text{J K}^{-1}$ is the Boltzmann constant and $T$ is the absolute temperature in kelvin (this equation is in the data booklet).

Read what this equation is telling you:

• The average KE is directly proportional to the absolute temperature: $E_k \propto T$. Double the kelvin temperature and you double the average kinetic energy.
• It depends only on $T$ — not on the type of gas. At the same temperature, a helium atom and a much heavier oxygen molecule have the same average translational KE. (The heavier molecule simply moves more slowly on average, since $\tfrac{1}{2}mv^2$ is the same but $m$ is larger.)
• At absolute zero ($T = 0\,\text{K}$) the average translational KE would be zero — the particles have the least possible motion. This is another way of seeing why 0 K is the natural bottom of the temperature scale.

A quick number to anchor it. At room temperature, $T = 300\,\text{K}$: $E_k = \frac{3}{2}(1.38\times10^{-23})(300) = 6.21\times10^{-21}\,\text{J}$ Tiny per particle — but multiply by $\sim10^{23}$ particles and you get everyday amounts of energy.

This links straight back to pressure: heating a gas raises $T$, which raises the average KE, so the particles move faster and strike the walls harder and more often — which (at fixed volume) raises the pressure, exactly as Gay-Lussac's law describes.