P5.154SD0 · Section P5: Solids, liquids and gases

Ideal gas molecules

Key learnings

Gas pressure — collisions exert a force
Gas molecules move rapidly and in random directions. Gas pressure arises when these molecules collide with the container walls: each collision exerts a small force on the wall, and that force spread over the wall's area is the pressure ( $\text{pressure} = \text{force} \div \text{area}$ ). The chain examiners want is collisions $\rightarrow$ force $\rightarrow$ pressure; stopping at "molecules hit the walls" is only half the answer.
Absolute zero and the kelvin scale
Absolute zero is $-273\,^{\circ}\text{C}$ ( $0$ K) — the lowest possible temperature, where molecules have the least kinetic energy. The kelvin scale starts there, so gas-law temperatures must be in kelvin. Convert by adding or subtracting 273, never multiplying: $T(\text{K}) = \theta(^{\circ}\text{C}) + 273$ . Worked values: $0\,^{\circ}\text{C} = 273$ K (melting ice), $27\,^{\circ}\text{C} = 300$ K, $-273\,^{\circ}\text{C} = 0$ K (absolute zero).
Temperature = average KE; the two gas laws
Temperature (in kelvin) measures the average kinetic energy of a gas's molecules — hotter means faster on average. Two laws for a fixed mass of gas: at constant volume, $\frac{p_1}{T_1} = \frac{p_2}{T_2}$ (T in kelvin); at constant temperature (Boyle's law), $p_1V_1 = p_2V_2$ , with pressure and volume inversely related. Match the law to whatever is held constant.

Common mistakes

Drawn from real examiner reports.

"Hitting the walls" isn't the full answer

Explaining gas pressure as just molecules colliding with the walls earns only part of the marks. The mark scheme needs the full chain: molecules collide with the walls, each collision exerts a force, and that force over the wall's area is the pressure. Missing the force step or the pressure step is the lost mark; do not write vaguely that "the air pushes".

Nov 2024 1P Q7c(i); Jun 2024 1P Q9c: collisions stated but the force-to-pressure link not developed → partial marks.

Gas laws need kelvin, not Celsius

In $p_1/T_1 = p_2/T_2$ the temperatures must be in kelvin. Two slips: substituting a Celsius value straight in (the Celsius scale does not start at absolute zero, so the ratio comes out wrong), and trying to multiply by 273 or subtracting instead of adding. Always convert first with $T(\text{K}) = \theta(^{\circ}\text{C}) + 273$ , for every temperature in the question.

Nov 2024 1P Q7c(ii): only about a third of candidates converted °C to kelvin.

Absolute zero ≠ 0 °C

Absolute zero is $-273\,^{\circ}\text{C}$ ( $0$ K), the lowest possible temperature, where molecular kinetic energy is least. $0\,^{\circ}\text{C}$ is only the melting point of ice ( $273$ K), not absolute zero. Confusing the two wrecks any kelvin conversion and any $p$ - $T$ extrapolation that should pass through $-273\,^{\circ}\text{C}$ .

Pick the law by what is constant

Use $p_1V_1 = p_2V_2$ only when the temperature is constant, and $p_1/T_1 = p_2/T_2$ only when the volume is constant. Read the stem for which quantity is fixed before choosing a formula — picking the wrong law is a common zero-scorer even when the arithmetic itself is correct.

Smaller volume gives a BIGGER pressure

At constant temperature, pressure and volume are inversely related. Squeezing a gas into a smaller volume raises its pressure (the molecules are packed closer and hit the walls more often). Do not assume a smaller volume means a smaller pressure — that reverses Boyle's law, $p_1V_1 = p_2V_2$ .

Develop "faster" into more collisions

Heating a gas at constant volume makes molecules move faster, so they collide with the walls more often and harder — the rate-of-collision idea is what earns the top marks. Answers that say only "the molecules move faster", without linking to more frequent, harder collisions and hence a higher pressure, are underdeveloped.

Jun 2023 1P Q9c: the rate-of-collision idea marked the top responses when a gas was heated at fixed volume.

Motion is random — every wall equally

Gas molecules move in random directions, so on average they are equally likely to hit any wall. Candidates often miss this random, equal-probability point when explaining why a gas exerts the same pressure on all the container walls, or why the pressure is steady rather than fluctuating.

Jun 2024 1PR Q11a: the random motion / equal probability of hitting each wall was often missed.

Exam tips

Convert to kelvin first
Before touching any gas-law formula, convert every temperature with $T(\text{K}) = \theta(^{\circ}\text{C}) + 273$ . This is the mark most candidates miss, and skipping it makes the whole ratio wrong.
Check what is held constant
Decide which quantity is fixed before choosing a formula: temperature constant $\rightarrow p_1V_1 = p_2V_2$ ; volume constant $\rightarrow p_1/T_1 = p_2/T_2$ . The wrong choice scores zero however neat the algebra that follows.
Substitute first, then rearrange
Put the data into the formula before rearranging — this secures the substitution mark even if the algebra later slips. Keep the two pressures in the same unit (both Pa or both kPa); they need not be SI as long as they match.
Explaining pressure? Spell out the chain
For "explain how a gas exerts pressure", always give collisions $\rightarrow$ force $\rightarrow$ pressure, not just "molecules hit the walls". State the unit on every numeric answer (Pa or kPa for pressure, K for temperature).

Revision notes