What is the Kinetic Particle Model of Matter in the context of Cambridge IGCSE Physics?

The Kinetic Particle Model of Matter is a fundamental concept in Cambridge IGCSE Physics that explains the behavior of matter in different states—solid, liquid, and gas. It describes how particles are in constant motion, with the speed and arrangement of these particles determining the state of the matter. In solids, particles are closely packed and vibrate in place, in liquids, they are less tightly packed and move more freely, and in gases, they are far apart and move rapidly.

How does the Kinetic Particle Model explain changes of state in matter?

The Kinetic Particle Model explains changes of state by considering the energy and movement of particles. When a substance is heated, its particles gain energy and move more vigorously. For example, when a solid is heated, its particles vibrate more until they have enough energy to overcome their fixed positions, leading to melting. Similarly, when a liquid is heated, its particles gain enough energy to break free from the liquid state and become a gas, a process known as evaporation or boiling.

Why is the Kinetic Particle Model important for understanding thermal physics?

The Kinetic Particle Model is crucial for understanding thermal physics because it provides a microscopic explanation for macroscopic phenomena such as temperature, pressure, and volume. It helps explain how heat energy affects the movement of particles, leading to changes in temperature and state. This model is essential for understanding concepts like thermal expansion, gas laws, and heat transfer, which are key topics in the Cambridge IGCSE Physics curriculum.

How does temperature affect particle motion according to the Kinetic Particle Model?

According to the Kinetic Particle Model, temperature is a measure of the average kinetic energy of particles in a substance. As temperature increases, the kinetic energy of the particles also increases, causing them to move more rapidly. In solids, this means increased vibration, in liquids, it results in faster movement and more collisions, and in gases, it leads to increased speed and pressure. Conversely, a decrease in temperature reduces particle motion.

What are the limitations of the Kinetic Particle Model in explaining the behavior of matter?

While the Kinetic Particle Model is a powerful tool for understanding the behavior of matter, it has limitations. It assumes that particles are point-like and do not interact with each other except during collisions, which is not always accurate. The model also does not account for quantum effects that become significant at very low temperatures or very small scales. Additionally, it simplifies complex interactions in liquids and solids, which can lead to inaccuracies in certain conditions.

Why is "Saying particles 'expand' when heated" a common mistake in Kinetic Particle Model of Matter?

Why it happens: Confusing thermal expansion of bulk material with particle behaviour. How to avoid it: Particles do NOT change size. They move FASTER and (for solids/liquids) further apart on average. Source: 0625/42 — recurring

Why is "Using Celsius in a $p/T$ or combined gas-law ratio (note: those laws are beyond the 0625 syllabus)" a common mistake in Kinetic Particle Model of Matter?

Why it happens: Forgetting to convert — and assuming gas-law problems always involve temperature. How to avoid it: On 0625 the only gas-law calculation is Boyle's law pV = constant at constant temperature — no temperature conversion is needed there. The p/T = constant and combined gas laws (which DO require kelvin, T(K) = + 273) are not in the 0625 syllabus — they are A-level / other-board enrichment.

Why is "Saying 'particles hit walls more' without saying 'frequency'" a common mistake in Kinetic Particle Model of Matter?

Why it happens: Imprecise language. How to avoid it: Use 'frequency of collisions' AND 'change in momentum at each collision'.

Why is "Saying evaporation only happens at boiling point" a common mistake in Kinetic Particle Model of Matter?

Why it happens: Confusing evaporation with boiling. How to avoid it: Evaporation happens at the surface at ANY temperature; boiling involves the whole liquid at one specific temperature.

Cambridge IGCSE Physics (0625)

Kinetic Particle Model of Matter

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Short Notes - Kinetic Particle Model of Matter

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Kinetic Particle Model of Matter — Cambridge IGCSE 0625 Physics Extended (2026)

Particles in solids, liquids and gases — what they do, what holds them, and why pressure rises with temperature. The model that underpins thermal physics.

At a glance

Solid: particles tightly packed, vibrate about fixed positions. Strong forces.
Liquid: particles still close, but free to flow past each other.
Gas: particles far apart, fast, random motion. Weak forces.
Heating → particles gain KE → faster motion → expansion / change of state.
Brownian motion: random jiggling of small particles, evidence for molecular motion.
Gas pressure comes from particle collisions with container walls.
$pV =$ constant at constant temperature (Boyle's law) — the only gas-law equation on 0625.
Heating a fixed mass of gas at constant volume raises its pressure (described qualitatively on 0625).
Kelvin scale: $T(\text{K}) = T(\degree\text{C}) + 273$ .

What you’ll learn

Mapped to the Cambridge IGCSE 0625 syllabus (2026-2028).

2.1 — Describe the structure of solids, liquids and gases in terms of particles.
2.1 — Describe Brownian motion as evidence for the kinetic particle model.
2.1 — Describe how a gas exerts pressure on container walls.
2.1 — Describe qualitatively the effect on gas pressure of a change of temperature (constant volume) or volume (constant temperature).
2.1 — Recall and use $pV =$ constant for a fixed mass of gas at constant temperature.
2.1 — Convert temperatures between kelvin and degrees Celsius.

Particle arrangement in solids, liquids and gases

Solid: lattice. Liquid: close + flowing. Gas: spread out + fast.

Solid.

Particles in fixed positions in a regular pattern.
Vibrate about those positions.
Strong intermolecular forces.
Fixed shape, fixed volume.

Liquid.

Particles still close together but in NO regular pattern.
Slide past each other.
Weaker forces than solid.
Takes the shape of its container; fixed volume.

Gas.

Particles FAR APART (about 10× the spacing of liquids).
Fast, random motion in straight lines until they collide.
Almost no forces between them.
Fills its container; volume changes with pressure.

Heating effect. Adding heat → particles gain KE → vibrate / move faster → expansion → eventually change of state.

From solid to gas: particle spacing increases and the forces holding them together weaken.

Solid: lattice, vibrating in place.
Liquid: close but flowing.
Gas: far apart, fast, random.
Heating gives particles KE → expansion.

Brownian motion

Smoke particles seen jiggling under a microscope — evidence of invisible air-molecule collisions.

Observation. Smoke particles in air, seen through a microscope, move in random, jerky paths — never settling, never going in a straight line.

Explanation. The smoke particle is being struck constantly from all sides by invisible AIR molecules. The molecules are too small to see directly, but their effect on the visible smoke particles is the random jiggling.

Why this matters. Brownian motion was the first direct experimental evidence that gases (and liquids) consist of particles in constant random motion — i.e. that the kinetic particle model is real, not just a useful picture.

Unseen air molecules collide with the smoke particle from all directions, producing its random motion.

Worked qualitative. A bigger smoke particle moves LESS than a small one for the same gas. Why? More mass, harder to push. Same impulse from molecule collisions → smaller acceleration.

Smoke particles jiggle randomly.
Caused by invisible air molecules colliding.
Evidence for the particle model.
Bigger smoke particle → smaller jiggling for same gas.

Gas pressure and Boyle's law

Pressure comes from molecular collisions. $pV =$ constant at fixed $T$ (Boyle's law).

Why does a gas exert pressure? Gas molecules constantly collide with the container walls. Each collision transfers momentum to the wall. The total force per area = pressure.

Each molecule (green) rebounds off the walls; the many outward pushes (red) add up to the gas pressure.

Effect of changing variables (qualitative — this is what 0625 examines).

Heat the gas (V constant): molecules gain kinetic energy and move faster → they collide with the walls more often and with a greater change of momentum → pressure rises.
Compress the gas (T constant): molecules collide with the walls more often (less travel between collisions) → pressure rises.

Boyle's law (constant $T$ ). This is the one gas-law equation in the 0625 syllabus: $p_{1} V_{1} = p_{2} V_{2}.$

Worked. A gas at $1 \times 10^{5}\,\text{Pa}$ occupies $200\,\text{cm}^{3}$ . Compressed to $50\,\text{cm}^{3}$ at constant $T$ . Find the new pressure.

$1 \times 10^{5} \times 200 = p_{2} \times 50 \Rightarrow p_{2} = 4 \times 10^{5}\,\text{Pa}$ .
Boyle's law is used at constant temperature, so no kelvin conversion is needed here.

Quarter the volume → four times the pressure: $pV$ stays constant ($1\times10^{5}\times200 = 4\times10^{5}\times50$).

Kelvin (absolute) temperature scale. 0625 §2.1.3 requires you to convert between scales: $T(\text{K}) = T(\degree\text{C}) + 273.$ Absolute zero ( $0\,\text{K} = -273\degree\text{C}$ ) is the temperature at which particles have minimum kinetic energy.

Both scales use the same size degree — kelvin simply starts at absolute zero, so you add 273 to convert.

Gas pressure: molecule-wall collisions.
Boyle's law: $pV$ constant at constant $T$ .
Heating at constant volume raises pressure (qualitative on 0625).
Kelvin: $T(\text{K}) = T(\degree\text{C}) + 273$ .

Going deeper for A*

Beyond the 0625 syllabus — the pressure-temperature relation can be made quantitative. At constant volume, $\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}$ (the pressure law), with $T$ in kelvin. For example, gas at $20\degree\text{C}$ ( $293\,\text{K}$ ) and $1 \times 10^{5}\,\text{Pa}$ heated to $80\degree\text{C}$ ( $353\,\text{K}$ ) reaches $p_{2} = 1 \times 10^{5} \times \tfrac{353}{293} \approx 1.20 \times 10^{5}\,\text{Pa}$ . When all three quantities vary, the combined gas law $\dfrac{p_{1}V_{1}}{T_{1}} = \dfrac{p_{2}V_{2}}{T_{2}}$ applies. These pressure-law / combined-gas-law equations are NOT in the Cambridge IGCSE 0625 syllabus — §2.1.3 examines the pressure-temperature dependence qualitatively only and lists Boyle's law as the single recall equation. They belong to A-level (and some other-board IGCSE) physics.

Quick recap

Solid / liquid / gas: differ in spacing and forces.
Brownian motion: evidence for particles.
Gas pressure: molecule-wall collisions.
Boyle's law: $pV$ constant at fixed $T$ — the one 0625 gas-law equation.
Heating at constant volume raises pressure (qualitative on 0625).
Kelvin: $T(\text{K}) = T(\degree\text{C}) + 273$ .

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Brownian motion — random motion of small particles in a fluid, caused by collisions with invisible molecules.
Boyle's law — $pV =$ constant at constant temperature.
Absolute zero — $0\,\text{K} = -273\degree\text{C}$ , where particle kinetic energy is a minimum.
Kelvin scale — absolute temperature scale; add $273$ to celsius.

How it’s examined

Particle model + Boyle's law appear every Paper 4 (5-7 marks): describe arrangement, explain pressure qualitatively, then a Boyle's-law calculation. Examiner reports flag confusing 'liquid' particles with 'solid' (they're close but NOT in a regular lattice) and giving vague collision descriptions. Note: the quantitative pressure law ( $p/T =$ constant) is beyond the 0625 syllabus — pressure-temperature dependence is examined qualitatively only.

Sources: Cambridge IGCSE Physics 0625 syllabus 2026-2028 (2.1); 0625/42 Oct/Nov 2024 — Q11 (Boyle's law); 0625 Examiner Reports 2022-2024. Last reviewed 2026-05-06.

Take this whole topic with you

Step-by-step worked examples — Kinetic Particle Model of Matter

Step-by-step solutions to past-paper-style questions on kinetic particle model of matter, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Kinetic Particle Model of Matter

Almost every kinetic particle model of matter exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A fixed mass of gas with two of $p$ , $V$ , $T$ given and one missing — find the new pressure / volume. One gas law, one substitution.

How to approach it

Identify which quantity is held constant: constant $T$ → $p_{1}V_{1}=p_{2}V_{2}$ ; constant $V$ → $\dfrac{p_{1}}{T_{1}}=\dfrac{p_{2}}{T_{2}}$ . Convert every temperature to kelvin first, then substitute and solve.

Common trap

Examiner reports flag candidates using Celsius directly in a gas-law ratio. Temperatures in $p$ – $T$ or $p$ – $V$ – $T$ ratios must be absolute: $T(\text{K})=\theta+273$ .

Show that / prove

Recognise it by

The stem says describe, explain why or explain in particle terms — Brownian motion, evaporation cooling, why pressure rises on heating. No number is wanted, a chain of reasoning is.

How to approach it

Build the explanation as a particle-level cause-and-effect chain: speed/KE change → collision frequency and momentum change → bulk effect (pressure, temperature, expansion). State each link explicitly.

Common trap

Vague phrasing such as 'particles hit the walls more'. Examiners want the precise terms 'frequency of collisions' and 'change in momentum / force', and 'random' for Brownian motion.

Graph or diagram

Recognise it by

A plotted relationship — $p$ against $1/V$ , or $p$ against $V$ — with a gradient or intercept to read off and use.

How to approach it

Read what the straight line tells you ( $p$ vs $1/V$ through the origin confirms $pV=\text{constant}$ ), take the gradient as that constant, then substitute the required value.

Common trap

Treating the gradient of a $p$ vs $V$ curve as the constant — only the linearised $p$ vs $1/V$ plot has a gradient equal to $pV$ .

Multi-step problem

Recognise it by

A real context (syringe, two containers) where you must chain a gas law to a second relation — force from pressure, or KE from temperature.

How to approach it

Do the gas-law step first, then carry the result into the second formula ( $F=p\,A$ , or $\overline{\text{KE}}\propto T$ ). Keep track of which pressure is absolute and which is the excess above atmospheric.

Common trap

Using the total internal pressure instead of the excess above atmospheric when finding the extra force on a plunger.

1Compare the three states
CoreShow that / prove• states
Question
Describe particle arrangement, motion, and forces in solids, liquids and gases.
Step-by-step solution
1. Step 1
  Solid: closely packed, vibrating about fixed positions, strong forces.
2. Step 2
  Liquid: closely packed but free to move past each other, moderate forces.
3. Step 3
  Gas: far apart, fast random motion, very weak forces (treated as none in ideal gas).
Answer
See descriptions above.
2Boyle's law
ExtendedDirect calculation• Adapted from 0625/42 May/Jun 2024 Q9• pV
Question
A gas occupies $200\,\text{cm}^{3}$ at $1.0 \times 10^{5}\,\text{Pa}$ . The volume is reduced to $50\,\text{cm}^{3}$ at constant temperature. Find the new pressure.
Step-by-step solution
1. Step 1
  $p_{1}V_{1} = p_{2}V_{2}$ at constant $T$ .
  $(1.0 \times 10^{5})(200) = p_{2}(50)$
2. Step 2
  Solve for $p_{2}$ .
  $p_{2} = \dfrac{(1.0 \times 10^{5})(200)}{50} = 4.0 \times 10^{5}\,\text{Pa}$
Answer
$4.0 \times 10^{5}\,\text{Pa}$
3Why heating increases gas pressure (constant volume)
ExtendedShow that / prove• explain
Question
Explain in particle terms why the pressure of a fixed mass of gas in a sealed container rises when it is heated.
Step-by-step solution
1. Step 1
  Heating increases the mean kinetic energy of the particles → they move faster.
2. Step 2
  Faster particles → collisions with the walls happen MORE OFTEN AND with MORE momentum.
3. Step 3
  More force per unit area → higher pressure.
Answer
Increased KE → faster, more frequent, more forceful collisions with walls → higher pressure.
Examiner tip
Examiners look for the words 'frequency of collisions' AND 'change in momentum / force'.
4Cooling effect of evaporation
ExtendedShow that / prove• evaporation
Question
Explain why evaporation cools the remaining liquid.
Step-by-step solution
1. Step 1
  The fastest molecules escape from the surface.
2. Step 2
  Average KE of remaining molecules drops → temperature falls.
Answer
Most energetic molecules leave; mean KE of those remaining decreases → cooling.
5Brownian motion observation
CoreShow that / prove• Adapted from 0625/42 May/Jun 2023 Q9• Brownian
Question
Smoke particles viewed through a microscope in a brightly-lit smoke cell are seen to jiggle randomly. Explain this observation in particle terms.
Step-by-step solution
1. Step 1
  The smoke particles are visible (large) but the surrounding air molecules are not.
2. Step 2
  Air molecules move at high speed in random directions and collide with the smoke particles.
3. Step 3
  Each collision gives the smoke particle a small change in momentum; many random collisions per second produce the observed random (Brownian) motion.
Answer
Random collisions of fast-moving, invisible air molecules with the larger smoke particles produce the observed jiggling.
Examiner tip
The examiner report flags candidates often forget to say the air molecules are 'invisible/much smaller' and that the motion is 'random' — both phrases are expected for full marks.
6Gas pressure from particle collisions
ExtendedShow that / prove• pressure, explain
Question
Using the kinetic model, explain how a gas exerts a pressure on the walls of its container.
Step-by-step solution
1. Step 1
  Gas particles move in random directions at high speed.
2. Step 2
  They collide with the container walls. At each collision the particle's momentum is reversed, so the wall exerts a force on the particle and (by Newton's third law) the particle exerts an equal and opposite force on the wall.
3. Step 3
  Many collisions per second over the wall area give an average force per unit area — this is the gas pressure.
  $p = \dfrac{F}{A}$
Answer
Gas pressure = total force per unit area from the many random momentum-changing collisions of particles with the wall.
7Beyond 0625 — pressure-temperature law at constant volume
Challenge• pT, gas law, beyond-0625
Question
A sealed rigid container holds gas at $27\degree\text{C}$ and pressure $1.0 \times 10^{5}\,\text{Pa}$ . It is heated to $127\degree\text{C}$ . Find the new pressure.
Step-by-step solution
1. Step 1
  Convert both temperatures to kelvin.
  $T_{1} = 27 + 273 = 300\,\text{K},\ T_{2} = 127 + 273 = 400\,\text{K}$
2. Step 2
  At constant volume, $\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}$ .
  $p_{2} = p_{1}\,\dfrac{T_{2}}{T_{1}} = 1.0 \times 10^{5} \times \dfrac{400}{300}$
3. Step 3
  Compute.
  $p_{2} = 1.33 \times 10^{5}\,\text{Pa}$
Answer
$1.33 \times 10^{5}\,\text{Pa}$
Examiner tip
Enrichment beyond the Cambridge IGCSE 0625 syllabus. The pressure-temperature relation $p/T = \text{constant}$ is not a recall equation on 0625 — §2.1.3 requires only Boyle's law $pV = \text{constant}$ and a qualitative description of how heating a fixed mass of gas at constant volume raises its pressure. This quantitative pressure law belongs to A-level (and other-board IGCSE) physics and is not examined on 0625.
8Interpret a Boyle's-law graph
ExtendedGraph or diagram• pV, graph
Question
A student plots $p$ (y-axis) against $1/V$ (x-axis) for a fixed mass of gas at constant temperature. The graph is a straight line through the origin with gradient $2.0 \times 10^{3}\,\text{Pa}\cdot\text{m}^{3}$ . Find the pressure of the gas when its volume is $0.020\,\text{m}^{3}$ .
Step-by-step solution
1. Step 1
  A straight line through the origin on a $p$ vs $1/V$ plot confirms $pV = \text{constant}$ at constant $T$ . The gradient equals that constant.
  $pV = 2.0 \times 10^{3}\,\text{Pa}\cdot\text{m}^{3}$
2. Step 2
  Rearrange for $p$ .
  $p = \dfrac{2.0 \times 10^{3}}{0.020} = 1.0 \times 10^{5}\,\text{Pa}$
Answer
$1.0 \times 10^{5}\,\text{Pa}$
9A* — Boyle's law applied to a syringe
ChallengeMulti-step problem• pV, synoptic
Question
The plunger of a sealed syringe is pushed in so the trapped air volume changes from $40\,\text{cm}^{3}$ to $25\,\text{cm}^{3}$ at constant temperature. The initial pressure equals atmospheric pressure $1.0 \times 10^{5}\,\text{Pa}$ . Find (a) the new pressure inside the syringe, (b) the extra force the user must apply if the plunger has cross-section $4.0 \times 10^{-4}\,\text{m}^{2}$ .
Step-by-step solution
1. Step 1
  Apply Boyle's law $p_{1}V_{1} = p_{2}V_{2}$ .
  $p_{2} = \dfrac{(1.0 \times 10^{5})(40)}{25} = 1.6 \times 10^{5}\,\text{Pa}$
2. Step 2
  Extra pressure above atmospheric pushes back on the plunger.
  $\Delta p = 1.6 \times 10^{5} - 1.0 \times 10^{5} = 6.0 \times 10^{4}\,\text{Pa}$
3. Step 3
  Force from this excess pressure on the plunger area.
  $F = \Delta p \times A = 6.0 \times 10^{4} \times 4.0 \times 10^{-4} = 24\,\text{N}$
Answer
(a) $1.6 \times 10^{5}\,\text{Pa}$ . (b) Extra force $\approx 24\,\text{N}$ .
10A* — Linking temperature to particle KE
ChallengeMulti-step problem• synoptic, explain
Question
Two identical sealed containers hold the same gas. Container X is at $300\,\text{K}$ and container Y is at $600\,\text{K}$ . (a) Compare the average kinetic energy of the particles. (b) Explain what happens to the pressure if container Y's volume is then halved at constant temperature.
Step-by-step solution
1. Step 1
  Average KE of particles is proportional to absolute temperature in kelvin.
  $\dfrac{\overline{\text{KE}}_{Y}}{\overline{\text{KE}}_{X}} = \dfrac{600}{300} = 2$
2. Step 2
  So particles in Y have, on average, twice the KE of those in X (faster average speed).
3. Step 3
  Halving Y's volume at constant $T$ : particles travel half the distance between wall collisions on average → frequency of collisions doubles → pressure doubles (Boyle's law: $pV = \text{const}$ ).
Answer
(a) Average KE of Y particles is twice that of X. (b) Halving the volume doubles the pressure at constant temperature.
Examiner tip
The examiner report flags candidates often say 'particles get hotter' instead of 'particles gain KE/move faster'. Temperature is a property of the bulk, not of individual particles.

Model Answers — Kinetic Particle Model of Matter

High-scoring sample answers for kinetic particle model of matter on the Cambridge IGCSE 0625 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
Paper 2/4 short-answer style1 mark
State one difference between the arrangement of the particles in a solid and in a gas.
Model answer
In a solid the particles are close together in a regular, fixed arrangement (vibrating about fixed positions); in a gas the particles are far apart and arranged randomly, moving freely at high speed.
Why this scores
One mark for any clear arrangement difference — close/regular vs far apart/random. A motion difference (fixed vibration vs free random movement) is equally acceptable.
Question 2
Paper 2/4 style2 marks
A fixed mass of gas occupies $300\,\text{cm}^{3}$ at a pressure of $100\,\text{kPa}$ . It is compressed to $100\,\text{cm}^{3}$ at constant temperature. Calculate the new pressure.
Model answer
At constant temperature, Boyle's law gives $p_{1}V_{1} = p_{2}V_{2}$ :

$p_{2} = \dfrac{p_{1}V_{1}}{V_{2}} = \dfrac{100 \times 300}{100} = \mathbf{300\,\text{kPa}}.$
Why this scores
One mark for using $p_{1}V_{1} = p_{2}V_{2}$ , one for $300\,\text{kPa}$ . Because both volumes are in the same unit and pressure in the same unit, no conversion is needed.
Question 3
Paper 4 explanation style3 marks
Smoke particles in a brightly lit smoke cell, viewed through a microscope, are seen to move in a jerky, random way. Explain this observation using the kinetic particle model.
Model answer
1. The relatively large smoke particles are visible, but the surrounding air molecules are far too small to see. 2. The air molecules move rapidly in random directions and constantly collide with the smoke particles. 3. Each collision gives a smoke particle a small change in momentum in a random direction; because the collisions are random and unequal at any instant, the smoke particle is knocked first one way then another — the observed random (Brownian) motion. This is evidence that air molecules are in continuous, fast, random motion.
Why this scores
Three marks: air molecules are small/invisible and fast (1); they collide with the larger smoke particles (1); random unequal collisions cause the random motion (1). 'Random' and 'invisible/much smaller molecules' are the phrases examiners require.
Question 4
Paper 4 explanation style4 marks
A fixed mass of gas is sealed in a rigid container of constant volume. Explain, in terms of the motion of the particles, why the gas pressure increases when the gas is heated.
Model answer
1. Heating the gas increases the average kinetic energy of the particles, so they move faster (on average). 2. Faster particles collide with the container walls more often — the frequency of collisions increases. 3. Each particle also hits the wall with a greater change of momentum, so it exerts a larger force on the wall. 4. With the volume fixed, more frequent and more forceful collisions mean a greater total force on each unit area of wall — so the pressure increases ( $p = F/A$ ).
Why this scores
Four marks: faster particles / more KE (1); greater frequency of collisions (1); greater change of momentum / force per collision (1); larger force per unit area → higher pressure (1). Both 'frequency of collisions' and 'change of momentum' are needed for full marks.
Question 5
Paper 4 structured style5 marks
(a) Explain, in terms of particles, why evaporation cools the liquid that is left behind. (b) State two changes that would increase the rate of evaporation from an open dish of water.
Model answer
(a) The particles in a liquid have a range of speeds (and kinetic energies). Only the fastest, most energetic particles at the surface have enough energy to escape as vapour. When these high-energy particles leave, the average kinetic energy of the particles remaining in the liquid falls, and since temperature is a measure of average kinetic energy, the temperature of the liquid drops — it cools.

(b) Any two of: increasing the temperature of the liquid; increasing the surface area of the liquid; blowing air across the surface (a draught) to remove escaped vapour; reducing the humidity of the surrounding air.
Why this scores
Five marks: particles have a range of energies (1); the most energetic escape (1); average KE of those left falls (1) so temperature falls (1); plus two correct factors that increase the rate (1 for both). Stating evaporation happens at any temperature below boiling, at the surface, is a useful distinction from boiling.
Question 6
Paper 4 multi-part structured style6 marks
(a) State Boyle's law. (b) A fixed mass of gas at a pressure of $1.0 \times 10^{5}\,\text{Pa}$ occupies $240\,\text{cm}^{3}$ . It is compressed to $160\,\text{cm}^{3}$ at constant temperature; calculate the new pressure. (c) Explain, in terms of the particles, why reducing the volume increases the pressure.
Model answer
(a) For a fixed mass of gas at constant temperature, the pressure is inversely proportional to the volume — i.e. $pV = \text{constant}$ .

(b) $p_{2} = \dfrac{p_{1}V_{1}}{V_{2}} = \dfrac{(1.0 \times 10^{5})(240)}{160} = \mathbf{1.5 \times 10^{5}\,\text{Pa}}$ .

(c) The same number of particles is now confined in a smaller volume, so they have a shorter average distance to travel between hitting the walls. They therefore collide with the walls more frequently, giving a greater total force per unit area — a higher pressure. (Their speed is unchanged, because the temperature is constant.)
Why this scores
Six marks: statement that $p \propto 1/V$ at constant temperature (1) + 'fixed mass'/'constant temperature' condition (1); correct calculation $1.5 \times 10^{5}\,\text{Pa}$ (1+1); smaller volume → more frequent wall collisions (1); same speed because temperature is constant (1). A common slip in (c) is saying the particles 'move faster' — at constant temperature the speed does not change.

Key Formulae — Kinetic Particle Model of Matter

The formulae you need to memorise for kinetic particle model of matter on the Cambridge IGCSE 0625 paper, with every variable defined in plain English and a note on when to use it.

Boyle's law (constant temperature)
$p_{1}V_{1} = p_{2}V_{2}$
$p$
pressure (Pa)
$V$
volume
When to use
Fixed mass of gas at constant temperature.
Kelvin / Celsius conversion
$T(\text{K}) = \theta(\degree\text{C}) + 273$
When to use
Converting between the Celsius and absolute (kelvin) temperature scales — a 0625 §2.1.3 requirement.

Key Definitions and Keywords — Kinetic Particle Model of Matter

Definitions to memorise and the exact keywords mark schemes credit for kinetic particle model of matter answers — sharpened from recent examiner reports for the 2026 0625 sitting.

Kinetic-particle model
Examiner keyword
Model that treats matter as small particles in constant motion; their motion explains thermal and gas behaviour.
Gas pressure (cause)
Examiner keyword
Caused by collisions of fast-moving particles with the container walls; pressure = force per unit area.
Evaporation
Examiner keyword
A change of state from liquid to gas at the surface, at any temperature below boiling point.
Absolute zero
Examiner keyword
The temperature at which particles have minimum kinetic energy. $0\,\text{K} = -273\degree\text{C}$ .

Common Mistakes and Misconceptions — Kinetic Particle Model of Matter

The traps other students keep falling into on kinetic particle model of matter questions — taken from recent Cambridge IGCSE 0625 examiner reports and mark schemes — and how to avoid them.

✕Saying particles 'expand' when heated
0625/42 — recurring
Why it happens
Confusing thermal expansion of bulk material with particle behaviour.
How to avoid it
Particles do NOT change size. They move FASTER and (for solids/liquids) further apart on average.
✕Using Celsius in a $p/T$ or combined gas-law ratio (note: those laws are beyond the 0625 syllabus)
Why it happens
Forgetting to convert — and assuming gas-law problems always involve temperature.
How to avoid it
On 0625 the only gas-law calculation is Boyle's law $pV = \text{constant}$ at constant temperature — no temperature conversion is needed there. The $p/T = \text{constant}$ and combined gas laws (which DO require kelvin, $T(\text{K}) = \theta + 273$ ) are not in the 0625 syllabus — they are A-level / other-board enrichment.
✕Saying 'particles hit walls more' without saying 'frequency'
Why it happens
Imprecise language.
How to avoid it
Use 'frequency of collisions' AND 'change in momentum at each collision'.
✕Saying evaporation only happens at boiling point
Why it happens
Confusing evaporation with boiling.
How to avoid it
Evaporation happens at the surface at ANY temperature; boiling involves the whole liquid at one specific temperature.

Kinetic Particle Model of Matter — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Cambridge IGCSE Physics (0625)

Kinetic Particle Model of Matter

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Short Notes - Kinetic Particle Model of Matter

Detailed Study Notes

Detailed notes on Thermal Physics for Cambridge IGCSE Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Kinetic Particle Model of Matter — Cambridge IGCSE 0625 Physics Extended (2026)

Particles in solids, liquids and gases — what they do, what holds them, and why pressure rises with temperature. The model that underpins thermal physics.

At a glance

Solid: particles tightly packed, vibrate about fixed positions. Strong forces.
Liquid: particles still close, but free to flow past each other.
Gas: particles far apart, fast, random motion. Weak forces.
Heating → particles gain KE → faster motion → expansion / change of state.
Brownian motion: random jiggling of small particles, evidence for molecular motion.
Gas pressure comes from particle collisions with container walls.
$pV =$ constant at constant temperature (Boyle's law) — the only gas-law equation on 0625.
Heating a fixed mass of gas at constant volume raises its pressure (described qualitatively on 0625).
Kelvin scale: $T(\text{K}) = T(\degree\text{C}) + 273$ .

What you’ll learn

Mapped to the Cambridge IGCSE 0625 syllabus (2026-2028).

2.1 — Describe the structure of solids, liquids and gases in terms of particles.
2.1 — Describe Brownian motion as evidence for the kinetic particle model.
2.1 — Describe how a gas exerts pressure on container walls.
2.1 — Describe qualitatively the effect on gas pressure of a change of temperature (constant volume) or volume (constant temperature).
2.1 — Recall and use $pV =$ constant for a fixed mass of gas at constant temperature.
2.1 — Convert temperatures between kelvin and degrees Celsius.

Particle arrangement in solids, liquids and gases

Solid: lattice. Liquid: close + flowing. Gas: spread out + fast.

Solid.

Particles in fixed positions in a regular pattern.
Vibrate about those positions.
Strong intermolecular forces.
Fixed shape, fixed volume.

Liquid.

Particles still close together but in NO regular pattern.
Slide past each other.
Weaker forces than solid.
Takes the shape of its container; fixed volume.

Gas.

Particles FAR APART (about 10× the spacing of liquids).
Fast, random motion in straight lines until they collide.
Almost no forces between them.
Fills its container; volume changes with pressure.

Heating effect. Adding heat → particles gain KE → vibrate / move faster → expansion → eventually change of state.

From solid to gas: particle spacing increases and the forces holding them together weaken.

Solid: lattice, vibrating in place.
Liquid: close but flowing.
Gas: far apart, fast, random.
Heating gives particles KE → expansion.

Brownian motion

Smoke particles seen jiggling under a microscope — evidence of invisible air-molecule collisions.

Observation. Smoke particles in air, seen through a microscope, move in random, jerky paths — never settling, never going in a straight line.

Unseen air molecules collide with the smoke particle from all directions, producing its random motion.

Worked qualitative. A bigger smoke particle moves LESS than a small one for the same gas. Why? More mass, harder to push. Same impulse from molecule collisions → smaller acceleration.

Smoke particles jiggle randomly.
Caused by invisible air molecules colliding.
Evidence for the particle model.
Bigger smoke particle → smaller jiggling for same gas.

Gas pressure and Boyle's law

Pressure comes from molecular collisions. $pV =$ constant at fixed $T$ (Boyle's law).

Why does a gas exert pressure? Gas molecules constantly collide with the container walls. Each collision transfers momentum to the wall. The total force per area = pressure.

Each molecule (green) rebounds off the walls; the many outward pushes (red) add up to the gas pressure.

Effect of changing variables (qualitative — this is what 0625 examines).

Heat the gas (V constant): molecules gain kinetic energy and move faster → they collide with the walls more often and with a greater change of momentum → pressure rises.
Compress the gas (T constant): molecules collide with the walls more often (less travel between collisions) → pressure rises.

Boyle's law (constant $T$ ). This is the one gas-law equation in the 0625 syllabus: $p_{1} V_{1} = p_{2} V_{2}.$

Worked. A gas at $1 \times 10^{5}\,\text{Pa}$ occupies $200\,\text{cm}^{3}$ . Compressed to $50\,\text{cm}^{3}$ at constant $T$ . Find the new pressure.

$1 \times 10^{5} \times 200 = p_{2} \times 50 \Rightarrow p_{2} = 4 \times 10^{5}\,\text{Pa}$ .
Boyle's law is used at constant temperature, so no kelvin conversion is needed here.

Quarter the volume → four times the pressure: $pV$ stays constant ($1\times10^{5}\times200 = 4\times10^{5}\times50$).

Both scales use the same size degree — kelvin simply starts at absolute zero, so you add 273 to convert.

Gas pressure: molecule-wall collisions.
Boyle's law: $pV$ constant at constant $T$ .
Heating at constant volume raises pressure (qualitative on 0625).
Kelvin: $T(\text{K}) = T(\degree\text{C}) + 273$ .

Going deeper for A*

Quick recap

Solid / liquid / gas: differ in spacing and forces.
Brownian motion: evidence for particles.
Gas pressure: molecule-wall collisions.
Boyle's law: $pV$ constant at fixed $T$ — the one 0625 gas-law equation.
Heating at constant volume raises pressure (qualitative on 0625).
Kelvin: $T(\text{K}) = T(\degree\text{C}) + 273$ .

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Brownian motion — random motion of small particles in a fluid, caused by collisions with invisible molecules.
Boyle's law — $pV =$ constant at constant temperature.
Absolute zero — $0\,\text{K} = -273\degree\text{C}$ , where particle kinetic energy is a minimum.
Kelvin scale — absolute temperature scale; add $273$ to celsius.

How it’s examined

Sources: Cambridge IGCSE Physics 0625 syllabus 2026-2028 (2.1); 0625/42 Oct/Nov 2024 — Q11 (Boyle's law); 0625 Examiner Reports 2022-2024. Last reviewed 2026-05-06.

Take this whole topic with you

Step-by-step worked examples — Kinetic Particle Model of Matter

Step-by-step solutions to past-paper-style questions on kinetic particle model of matter, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Kinetic Particle Model of Matter

Almost every kinetic particle model of matter exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A fixed mass of gas with two of $p$ , $V$ , $T$ given and one missing — find the new pressure / volume. One gas law, one substitution.

How to approach it

Common trap

Examiner reports flag candidates using Celsius directly in a gas-law ratio. Temperatures in $p$ – $T$ or $p$ – $V$ – $T$ ratios must be absolute: $T(\text{K})=\theta+273$ .

Show that / prove

Recognise it by

The stem says describe, explain why or explain in particle terms — Brownian motion, evaporation cooling, why pressure rises on heating. No number is wanted, a chain of reasoning is.

How to approach it

Common trap

Vague phrasing such as 'particles hit the walls more'. Examiners want the precise terms 'frequency of collisions' and 'change in momentum / force', and 'random' for Brownian motion.

Graph or diagram

Recognise it by

A plotted relationship — $p$ against $1/V$ , or $p$ against $V$ — with a gradient or intercept to read off and use.

How to approach it

Read what the straight line tells you ( $p$ vs $1/V$ through the origin confirms $pV=\text{constant}$ ), take the gradient as that constant, then substitute the required value.

Common trap

Treating the gradient of a $p$ vs $V$ curve as the constant — only the linearised $p$ vs $1/V$ plot has a gradient equal to $pV$ .

Multi-step problem

Recognise it by

A real context (syringe, two containers) where you must chain a gas law to a second relation — force from pressure, or KE from temperature.

How to approach it

Common trap

Using the total internal pressure instead of the excess above atmospheric when finding the extra force on a plunger.

1Compare the three states
CoreShow that / prove• states
Question
Describe particle arrangement, motion, and forces in solids, liquids and gases.
Step-by-step solution
1. Step 1
  Solid: closely packed, vibrating about fixed positions, strong forces.
2. Step 2
  Liquid: closely packed but free to move past each other, moderate forces.
3. Step 3
  Gas: far apart, fast random motion, very weak forces (treated as none in ideal gas).
Answer
See descriptions above.
2Boyle's law
ExtendedDirect calculation• Adapted from 0625/42 May/Jun 2024 Q9• pV
Question
A gas occupies $200\,\text{cm}^{3}$ at $1.0 \times 10^{5}\,\text{Pa}$ . The volume is reduced to $50\,\text{cm}^{3}$ at constant temperature. Find the new pressure.
Step-by-step solution
1. Step 1
  $p_{1}V_{1} = p_{2}V_{2}$ at constant $T$ .
  $(1.0 \times 10^{5})(200) = p_{2}(50)$
2. Step 2
  Solve for $p_{2}$ .
  $p_{2} = \dfrac{(1.0 \times 10^{5})(200)}{50} = 4.0 \times 10^{5}\,\text{Pa}$
Answer
$4.0 \times 10^{5}\,\text{Pa}$
3Why heating increases gas pressure (constant volume)
ExtendedShow that / prove• explain
Question
Explain in particle terms why the pressure of a fixed mass of gas in a sealed container rises when it is heated.
Step-by-step solution
1. Step 1
  Heating increases the mean kinetic energy of the particles → they move faster.
2. Step 2
  Faster particles → collisions with the walls happen MORE OFTEN AND with MORE momentum.
3. Step 3
  More force per unit area → higher pressure.
Answer
Increased KE → faster, more frequent, more forceful collisions with walls → higher pressure.
Examiner tip
Examiners look for the words 'frequency of collisions' AND 'change in momentum / force'.
4Cooling effect of evaporation
ExtendedShow that / prove• evaporation
Question
Explain why evaporation cools the remaining liquid.
Step-by-step solution
1. Step 1
  The fastest molecules escape from the surface.
2. Step 2
  Average KE of remaining molecules drops → temperature falls.
Answer
Most energetic molecules leave; mean KE of those remaining decreases → cooling.
5Brownian motion observation
CoreShow that / prove• Adapted from 0625/42 May/Jun 2023 Q9• Brownian
Question
Smoke particles viewed through a microscope in a brightly-lit smoke cell are seen to jiggle randomly. Explain this observation in particle terms.
Step-by-step solution
1. Step 1
  The smoke particles are visible (large) but the surrounding air molecules are not.
2. Step 2
  Air molecules move at high speed in random directions and collide with the smoke particles.
3. Step 3
  Each collision gives the smoke particle a small change in momentum; many random collisions per second produce the observed random (Brownian) motion.
Answer
Random collisions of fast-moving, invisible air molecules with the larger smoke particles produce the observed jiggling.
Examiner tip
The examiner report flags candidates often forget to say the air molecules are 'invisible/much smaller' and that the motion is 'random' — both phrases are expected for full marks.
6Gas pressure from particle collisions
ExtendedShow that / prove• pressure, explain
Question
Using the kinetic model, explain how a gas exerts a pressure on the walls of its container.
Step-by-step solution
1. Step 1
  Gas particles move in random directions at high speed.
2. Step 2
  They collide with the container walls. At each collision the particle's momentum is reversed, so the wall exerts a force on the particle and (by Newton's third law) the particle exerts an equal and opposite force on the wall.
3. Step 3
  Many collisions per second over the wall area give an average force per unit area — this is the gas pressure.
  $p = \dfrac{F}{A}$
Answer
Gas pressure = total force per unit area from the many random momentum-changing collisions of particles with the wall.
7Beyond 0625 — pressure-temperature law at constant volume
Challenge• pT, gas law, beyond-0625
Question
A sealed rigid container holds gas at $27\degree\text{C}$ and pressure $1.0 \times 10^{5}\,\text{Pa}$ . It is heated to $127\degree\text{C}$ . Find the new pressure.
Step-by-step solution
1. Step 1
  Convert both temperatures to kelvin.
  $T_{1} = 27 + 273 = 300\,\text{K},\ T_{2} = 127 + 273 = 400\,\text{K}$
2. Step 2
  At constant volume, $\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}$ .
  $p_{2} = p_{1}\,\dfrac{T_{2}}{T_{1}} = 1.0 \times 10^{5} \times \dfrac{400}{300}$
3. Step 3
  Compute.
  $p_{2} = 1.33 \times 10^{5}\,\text{Pa}$
Answer
$1.33 \times 10^{5}\,\text{Pa}$
Examiner tip
Enrichment beyond the Cambridge IGCSE 0625 syllabus. The pressure-temperature relation $p/T = \text{constant}$ is not a recall equation on 0625 — §2.1.3 requires only Boyle's law $pV = \text{constant}$ and a qualitative description of how heating a fixed mass of gas at constant volume raises its pressure. This quantitative pressure law belongs to A-level (and other-board IGCSE) physics and is not examined on 0625.
8Interpret a Boyle's-law graph
ExtendedGraph or diagram• pV, graph
Question
A student plots $p$ (y-axis) against $1/V$ (x-axis) for a fixed mass of gas at constant temperature. The graph is a straight line through the origin with gradient $2.0 \times 10^{3}\,\text{Pa}\cdot\text{m}^{3}$ . Find the pressure of the gas when its volume is $0.020\,\text{m}^{3}$ .
Step-by-step solution
1. Step 1
  A straight line through the origin on a $p$ vs $1/V$ plot confirms $pV = \text{constant}$ at constant $T$ . The gradient equals that constant.
  $pV = 2.0 \times 10^{3}\,\text{Pa}\cdot\text{m}^{3}$
2. Step 2
  Rearrange for $p$ .
  $p = \dfrac{2.0 \times 10^{3}}{0.020} = 1.0 \times 10^{5}\,\text{Pa}$
Answer
$1.0 \times 10^{5}\,\text{Pa}$
9A* — Boyle's law applied to a syringe
ChallengeMulti-step problem• pV, synoptic
Question
The plunger of a sealed syringe is pushed in so the trapped air volume changes from $40\,\text{cm}^{3}$ to $25\,\text{cm}^{3}$ at constant temperature. The initial pressure equals atmospheric pressure $1.0 \times 10^{5}\,\text{Pa}$ . Find (a) the new pressure inside the syringe, (b) the extra force the user must apply if the plunger has cross-section $4.0 \times 10^{-4}\,\text{m}^{2}$ .
Step-by-step solution
1. Step 1
  Apply Boyle's law $p_{1}V_{1} = p_{2}V_{2}$ .
  $p_{2} = \dfrac{(1.0 \times 10^{5})(40)}{25} = 1.6 \times 10^{5}\,\text{Pa}$
2. Step 2
  Extra pressure above atmospheric pushes back on the plunger.
  $\Delta p = 1.6 \times 10^{5} - 1.0 \times 10^{5} = 6.0 \times 10^{4}\,\text{Pa}$
3. Step 3
  Force from this excess pressure on the plunger area.
  $F = \Delta p \times A = 6.0 \times 10^{4} \times 4.0 \times 10^{-4} = 24\,\text{N}$
Answer
(a) $1.6 \times 10^{5}\,\text{Pa}$ . (b) Extra force $\approx 24\,\text{N}$ .
10A* — Linking temperature to particle KE
ChallengeMulti-step problem• synoptic, explain
Question
Two identical sealed containers hold the same gas. Container X is at $300\,\text{K}$ and container Y is at $600\,\text{K}$ . (a) Compare the average kinetic energy of the particles. (b) Explain what happens to the pressure if container Y's volume is then halved at constant temperature.
Step-by-step solution
1. Step 1
  Average KE of particles is proportional to absolute temperature in kelvin.
  $\dfrac{\overline{\text{KE}}_{Y}}{\overline{\text{KE}}_{X}} = \dfrac{600}{300} = 2$
2. Step 2
  So particles in Y have, on average, twice the KE of those in X (faster average speed).
3. Step 3
  Halving Y's volume at constant $T$ : particles travel half the distance between wall collisions on average → frequency of collisions doubles → pressure doubles (Boyle's law: $pV = \text{const}$ ).
Answer
(a) Average KE of Y particles is twice that of X. (b) Halving the volume doubles the pressure at constant temperature.
Examiner tip
The examiner report flags candidates often say 'particles get hotter' instead of 'particles gain KE/move faster'. Temperature is a property of the bulk, not of individual particles.

Model Answers — Kinetic Particle Model of Matter

High-scoring sample answers for kinetic particle model of matter on the Cambridge IGCSE 0625 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
Paper 2/4 short-answer style1 mark
State one difference between the arrangement of the particles in a solid and in a gas.
Model answer
In a solid the particles are close together in a regular, fixed arrangement (vibrating about fixed positions); in a gas the particles are far apart and arranged randomly, moving freely at high speed.
Why this scores
One mark for any clear arrangement difference — close/regular vs far apart/random. A motion difference (fixed vibration vs free random movement) is equally acceptable.
Question 2
Paper 2/4 style2 marks
A fixed mass of gas occupies $300\,\text{cm}^{3}$ at a pressure of $100\,\text{kPa}$ . It is compressed to $100\,\text{cm}^{3}$ at constant temperature. Calculate the new pressure.
Model answer
At constant temperature, Boyle's law gives $p_{1}V_{1} = p_{2}V_{2}$ :

$p_{2} = \dfrac{p_{1}V_{1}}{V_{2}} = \dfrac{100 \times 300}{100} = \mathbf{300\,\text{kPa}}.$
Why this scores
One mark for using $p_{1}V_{1} = p_{2}V_{2}$ , one for $300\,\text{kPa}$ . Because both volumes are in the same unit and pressure in the same unit, no conversion is needed.
Question 3
Paper 4 explanation style3 marks
Smoke particles in a brightly lit smoke cell, viewed through a microscope, are seen to move in a jerky, random way. Explain this observation using the kinetic particle model.
Model answer
1. The relatively large smoke particles are visible, but the surrounding air molecules are far too small to see. 2. The air molecules move rapidly in random directions and constantly collide with the smoke particles. 3. Each collision gives a smoke particle a small change in momentum in a random direction; because the collisions are random and unequal at any instant, the smoke particle is knocked first one way then another — the observed random (Brownian) motion. This is evidence that air molecules are in continuous, fast, random motion.
Why this scores
Three marks: air molecules are small/invisible and fast (1); they collide with the larger smoke particles (1); random unequal collisions cause the random motion (1). 'Random' and 'invisible/much smaller molecules' are the phrases examiners require.
Question 4
Paper 4 explanation style4 marks
A fixed mass of gas is sealed in a rigid container of constant volume. Explain, in terms of the motion of the particles, why the gas pressure increases when the gas is heated.
Model answer
1. Heating the gas increases the average kinetic energy of the particles, so they move faster (on average). 2. Faster particles collide with the container walls more often — the frequency of collisions increases. 3. Each particle also hits the wall with a greater change of momentum, so it exerts a larger force on the wall. 4. With the volume fixed, more frequent and more forceful collisions mean a greater total force on each unit area of wall — so the pressure increases ( $p = F/A$ ).
Why this scores
Four marks: faster particles / more KE (1); greater frequency of collisions (1); greater change of momentum / force per collision (1); larger force per unit area → higher pressure (1). Both 'frequency of collisions' and 'change of momentum' are needed for full marks.
Question 5
Paper 4 structured style5 marks
(a) Explain, in terms of particles, why evaporation cools the liquid that is left behind. (b) State two changes that would increase the rate of evaporation from an open dish of water.
Model answer
(a) The particles in a liquid have a range of speeds (and kinetic energies). Only the fastest, most energetic particles at the surface have enough energy to escape as vapour. When these high-energy particles leave, the average kinetic energy of the particles remaining in the liquid falls, and since temperature is a measure of average kinetic energy, the temperature of the liquid drops — it cools.

(b) Any two of: increasing the temperature of the liquid; increasing the surface area of the liquid; blowing air across the surface (a draught) to remove escaped vapour; reducing the humidity of the surrounding air.
Why this scores
Five marks: particles have a range of energies (1); the most energetic escape (1); average KE of those left falls (1) so temperature falls (1); plus two correct factors that increase the rate (1 for both). Stating evaporation happens at any temperature below boiling, at the surface, is a useful distinction from boiling.
Question 6
Paper 4 multi-part structured style6 marks
(a) State Boyle's law. (b) A fixed mass of gas at a pressure of $1.0 \times 10^{5}\,\text{Pa}$ occupies $240\,\text{cm}^{3}$ . It is compressed to $160\,\text{cm}^{3}$ at constant temperature; calculate the new pressure. (c) Explain, in terms of the particles, why reducing the volume increases the pressure.
Model answer
(a) For a fixed mass of gas at constant temperature, the pressure is inversely proportional to the volume — i.e. $pV = \text{constant}$ .

(b) $p_{2} = \dfrac{p_{1}V_{1}}{V_{2}} = \dfrac{(1.0 \times 10^{5})(240)}{160} = \mathbf{1.5 \times 10^{5}\,\text{Pa}}$ .

(c) The same number of particles is now confined in a smaller volume, so they have a shorter average distance to travel between hitting the walls. They therefore collide with the walls more frequently, giving a greater total force per unit area — a higher pressure. (Their speed is unchanged, because the temperature is constant.)
Why this scores
Six marks: statement that $p \propto 1/V$ at constant temperature (1) + 'fixed mass'/'constant temperature' condition (1); correct calculation $1.5 \times 10^{5}\,\text{Pa}$ (1+1); smaller volume → more frequent wall collisions (1); same speed because temperature is constant (1). A common slip in (c) is saying the particles 'move faster' — at constant temperature the speed does not change.

Key Formulae — Kinetic Particle Model of Matter

The formulae you need to memorise for kinetic particle model of matter on the Cambridge IGCSE 0625 paper, with every variable defined in plain English and a note on when to use it.

Boyle's law (constant temperature)
$p_{1}V_{1} = p_{2}V_{2}$
$p$
pressure (Pa)
$V$
volume
When to use
Fixed mass of gas at constant temperature.
Kelvin / Celsius conversion
$T(\text{K}) = \theta(\degree\text{C}) + 273$
When to use
Converting between the Celsius and absolute (kelvin) temperature scales — a 0625 §2.1.3 requirement.

Key Definitions and Keywords — Kinetic Particle Model of Matter

Definitions to memorise and the exact keywords mark schemes credit for kinetic particle model of matter answers — sharpened from recent examiner reports for the 2026 0625 sitting.

Kinetic-particle model
Examiner keyword
Model that treats matter as small particles in constant motion; their motion explains thermal and gas behaviour.
Gas pressure (cause)
Examiner keyword
Caused by collisions of fast-moving particles with the container walls; pressure = force per unit area.
Evaporation
Examiner keyword
A change of state from liquid to gas at the surface, at any temperature below boiling point.
Absolute zero
Examiner keyword
The temperature at which particles have minimum kinetic energy. $0\,\text{K} = -273\degree\text{C}$ .

Common Mistakes and Misconceptions — Kinetic Particle Model of Matter

The traps other students keep falling into on kinetic particle model of matter questions — taken from recent Cambridge IGCSE 0625 examiner reports and mark schemes — and how to avoid them.

✕Saying particles 'expand' when heated
0625/42 — recurring
Why it happens
Confusing thermal expansion of bulk material with particle behaviour.
How to avoid it
Particles do NOT change size. They move FASTER and (for solids/liquids) further apart on average.
✕Using Celsius in a $p/T$ or combined gas-law ratio (note: those laws are beyond the 0625 syllabus)
Why it happens
Forgetting to convert — and assuming gas-law problems always involve temperature.
How to avoid it
On 0625 the only gas-law calculation is Boyle's law $pV = \text{constant}$ at constant temperature — no temperature conversion is needed there. The $p/T = \text{constant}$ and combined gas laws (which DO require kelvin, $T(\text{K}) = \theta + 273$ ) are not in the 0625 syllabus — they are A-level / other-board enrichment.
✕Saying 'particles hit walls more' without saying 'frequency'
Why it happens
Imprecise language.
How to avoid it
Use 'frequency of collisions' AND 'change in momentum at each collision'.
✕Saying evaporation only happens at boiling point
Why it happens
Confusing evaporation with boiling.
How to avoid it
Evaporation happens at the surface at ANY temperature; boiling involves the whole liquid at one specific temperature.

Kinetic Particle Model of Matter — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Kinetic Particle Model of Matter

Short Notes - Kinetic Particle Model of Matter

Detailed Study Notes

At a glance

What you’ll learn

Particle arrangement in solids, liquids and gases

Brownian motion

Gas pressure and Boyle's law

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Kinetic Particle Model of Matter

1Compare the three states

2Boyle's law

3Why heating increases gas pressure (constant volume)

4Cooling effect of evaporation

5Brownian motion observation

6Gas pressure from particle collisions

7Beyond 0625 — pressure-temperature law at constant volume

8Interpret a Boyle's-law graph

9A* — Boyle's law applied to a syringe

10A* — Linking temperature to particle KE

Boyle's law (constant temperature)

Kelvin / Celsius conversion

Kinetic-particle model

Gas pressure (cause)

Evaporation

Absolute zero

✕Saying particles 'expand' when heated

✕Using Celsius in a p/Tp/Tp/T or combined gas-law ratio (note: those laws are beyond the 0625 syllabus)

✕Saying 'particles hit walls more' without saying 'frequency'

✕Saying evaporation only happens at boiling point

Kinetic Particle Model of Matter — frequently asked questions

Kinetic Particle Model of Matter

Short Notes - Kinetic Particle Model of Matter

Detailed Study Notes

At a glance

What you’ll learn

Particle arrangement in solids, liquids and gases

Brownian motion

Gas pressure and Boyle's law

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Kinetic Particle Model of Matter

1Compare the three states

2Boyle's law

3Why heating increases gas pressure (constant volume)

4Cooling effect of evaporation

5Brownian motion observation

6Gas pressure from particle collisions

7Beyond 0625 — pressure-temperature law at constant volume

8Interpret a Boyle's-law graph

9A* — Boyle's law applied to a syringe

10A* — Linking temperature to particle KE

Boyle's law (constant temperature)

Kelvin / Celsius conversion

Kinetic-particle model

Gas pressure (cause)

Evaporation

Absolute zero

✕Saying particles 'expand' when heated

✕Using Celsius in a p/Tp/Tp/T or combined gas-law ratio (note: those laws are beyond the 0625 syllabus)

✕Saying 'particles hit walls more' without saying 'frequency'

✕Saying evaporation only happens at boiling point

Kinetic Particle Model of Matter — frequently asked questions

✕Using Celsius in a $p/T$ or combined gas-law ratio (note: those laws are beyond the 0625 syllabus)

✕Using Celsius in a $p/T$ or combined gas-law ratio (note: those laws are beyond the 0625 syllabus)