Study Guides
AQA GCSE Physics 8463: Particle model of matter – Study Guide
AQA GCSE Physics 8463 Particle model of matter from scratch: density, changes of state, internal energy, latent heat and gas pressure, worked examples.
- Subject
- Physics
- Level
- GCSE
- Topic
- Particle model of matter
- Author
- Marlbridge Academic Team
- Updated
- Reviewed by
- Iftikhar Azeemi (what this means)
Aligned to AQA GCSE Physics (8463), For first teaching 2016. Official specification .
Syllabus page (what it covers and how it is assessed): AQA GCSE Physics.
Syllabus points this page covers
8463
- 3 Particle model of matter (whole topic)
- 4.3.1 Changes of state and the particle model
- 4.3.2 Internal energy and energy transfers
- 4.3.3 Particle model and pressure
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This guide teaches section 4.3 Particle model of matter of the AQA GCSE Physics (8463) specification, for teaching from September 2016 and exams from 2018 onwards. It covers every point from 4.3.1.1 to 4.3.3.3: density, changes of state, internal energy, specific heat capacity, specific latent heat and the pressure of gases. The topic is assessed on Paper 1 (topics 1 to 4), set at Foundation and Higher Tier. Section 4.3.3.3 is marked (HT only) and is labelled Higher tier only below; everything else applies to both tiers. Required practical activity 5 (density) belongs to this topic.
For quick recall, use the Particle model of matter revision notes. To test yourself, use the Particle model of matter practice questions. The course hub is AQA GCSE Physics, and the printable checklist lists every specification point. To find gaps quickly, try a free diagnostic. The previous topic is covered in the Electricity study guide.
What this unit covers
| Spec point | What you must be able to do | Tier |
|---|---|---|
| 4.3.1.1 Density of materials | Recall and apply ρ = m/V; explain density differences between states; draw particle diagrams; required practical 5 | Both |
| 4.3.1.2 Changes of state | Describe how mass is conserved; explain why changes of state are physical changes | Both |
| 4.3.2.1 Internal energy | Define internal energy; explain what heating does to it | Both |
| 4.3.2.2 Specific heat capacity | Apply ΔE = m c Δθ (given on the equation sheet) | Both |
| 4.3.2.3 Specific latent heat | Apply E = m L (given on the equation sheet); interpret heating and cooling graphs; distinguish c from L | Both |
| 4.3.3.1 Particle motion in gases | Relate molecular motion to temperature and pressure; pressure and temperature at constant volume | Both |
| 4.3.3.2 Pressure in gases | Explain pressure as a force at right angles to a surface; apply pV = constant (given on the equation sheet) | Both |
| 4.3.3.3 Increasing the pressure of a gas | Explain why doing work on a gas raises its temperature | Higher tier only |
4.3.1 Changes of state and the particle model
The three states
| State | Arrangement | Motion | Density |
|---|---|---|---|
| Solid | Particles touching, in a regular pattern | Vibrate about fixed positions | High |
| Liquid | Particles touching, irregular | Move around each other | High, usually a little lower than the solid |
| Gas | Particles far apart, random | Move quickly in all directions | Very low |
In your diagrams, draw solid particles in neat rows, liquid particles touching but jumbled, and gas particles widely spaced. Keep the particles the same size in all three.
Why density differs. Density depends on how much mass is packed into each cubic metre. In solids and liquids the particles are closely packed, so there is a lot of mass in a small volume. In a gas the particles are far apart, so the same mass takes up far more volume and the density is much lower.
Density
density = mass ÷ volume ρ = m / V
ρ in kg/m³, m in kg, V in m³
You must recall this equation. Watch the units: 1 cm³ = 1 × 10⁻⁶ m³, and 1 g/cm³ = 1000 kg/m³.
Worked example 1 (regular solid). A block measures 4.0 cm × 5.0 cm × 10.0 cm and has a mass of 0.54 kg. Calculate its density in kg/m³.
V = 0.040 × 0.050 × 0.100 = 0.00020 m³ (2.0 × 10⁻⁴ m³)
ρ = m / V = 0.54 / 0.00020
ρ = 2700 kg/m³
Worked example 2 (irregular solid). A stone of mass 153 g is lowered into a measuring cylinder. The water level rises from 50.0 cm³ to 68.0 cm³. Calculate its density.
V = 68.0 − 50.0 = 18.0 cm³ = 1.80 × 10⁻⁵ m³
m = 153 g = 0.153 kg
ρ = 0.153 / 1.80 × 10⁻⁵ = 8500 kg/m³
Required practical 5 – density
- Regular solid: measure the sides with a ruler, Vernier callipers or a micrometer; calculate the volume; measure the mass on a balance.
- Irregular solid: find the volume by displacement. Either read the rise in a measuring cylinder, or fill a displacement (eureka) can to the spout, lower the object in and collect the water that overflows in a measuring cylinder.
- Liquid: zero a balance with an empty measuring cylinder on it (or weigh it empty), add a measured volume of liquid, then read the mass of the liquid.
- Read the measuring cylinder at eye level, from the bottom of the meniscus.
Changes of state
The changes of state are melting, freezing, boiling, evaporating, condensing and sublimating (solid straight to gas). In every change of state mass is conserved: the number of particles stays the same, and only their arrangement and energy change. Changes of state are physical changes. They differ from chemical changes because the material recovers its original properties if the change is reversed. Melt ice and refreeze the water and you have ice again.
4.3.2 Internal energy and energy transfers
Internal energy
Energy is stored inside a system by its particles (atoms and molecules). Internal energy is the total kinetic energy and potential energy of all the particles that make up a system.
Heating increases the energy of the particles. This either raises the temperature of the system or produces a change of state.
Specific heat capacity
The specific heat capacity of a substance is the amount of energy needed to raise the temperature of one kilogram of the substance by one degree Celsius.
change in thermal energy = mass × specific heat capacity × temperature change
ΔE = m c Δθ
ΔE in J, m in kg, c in J/kg °C, Δθ in °C
This equation is on the Physics equation sheet: you must be able to select and apply it.
Worked example 3. Calculate the energy needed to heat 0.50 kg of water from 20 °C to 80 °C. The specific heat capacity of water is 4200 J/kg °C.
Δθ = 80 − 20 = 60 °C
ΔE = m c Δθ = 0.50 × 4200 × 60
ΔE = 126 000 J
Specific latent heat
The energy needed for a substance to change state is called latent heat. During a change of state, the energy supplied changes the internal energy but not the temperature.
The specific latent heat of a substance is the amount of energy needed to change the state of one kilogram of it with no change in temperature.
energy for a change of state = mass × specific latent heat
E = m L
E in J, m in kg, L in J/kg
This equation is also on the equation sheet.
- Specific latent heat of fusion: solid ↔ liquid.
- Specific latent heat of vaporisation: liquid ↔ vapour.
Worked example 4. Calculate the energy needed to melt 0.20 kg of ice at 0 °C. Specific latent heat of fusion of ice = 334 000 J/kg.
E = m L = 0.20 × 334 000
E = 66 800 J
Worked example 5 (two stages). 0.40 kg of ice at 0 °C melts and the water warms to 25 °C. Use L = 334 000 J/kg and c = 4200 J/kg °C.
Melting: E₁ = 0.40 × 334 000 = 133 600 J
Warming: E₂ = 0.40 × 4200 × 25 = 42 000 J
Total: 133 600 + 42 000 = 175 600 J ≈ 176 000 J (3 s.f.)
Heating and cooling graphs
On a temperature–time graph for a substance heated steadily:
- Sloping parts: one state is warming. Energy raises the particles’ kinetic energy, so the temperature rises.
- Flat parts: a change of state. Energy breaks bonds between particles (increasing potential energy), so the temperature stays constant. The first flat part is the melting point; the second is the boiling point.
- A longer flat part means more energy is needed for that change of state.
A cooling graph is the same shape in reverse: the flat parts show condensing and freezing, where energy is transferred to the surroundings with no change in temperature.
Specific heat capacity vs specific latent heat. c is about a temperature change with no change of state. L is about a change of state with no temperature change.
4.3.3 Particle model and pressure
Particle motion in gases
The molecules of a gas are in constant random motion. The temperature of a gas is related to the average kinetic energy of its molecules.
Gas molecules collide with the walls of their container. Each collision exerts a force on the wall, and the total force on each unit of area is the gas pressure.
Pressure and temperature at constant volume. If you heat a gas in a sealed container of fixed volume, the molecules gain kinetic energy and move faster. They hit the walls more often and with more force, so the pressure increases.
Pressure in gases
A gas can be compressed or expanded by pressure changes. The pressure produces a net force at right angles to the wall of the container (or any surface).
Increasing the volume at constant temperature. The molecules have more space, so they hit each unit area of wall less often. Their average speed is unchanged (same temperature), so the pressure decreases.
For a fixed mass of gas at constant temperature:
pressure × volume = constant p V = constant
so p₁ V₁ = p₂ V₂
p in pascals (Pa), V in m³
This equation is on the equation sheet.
Worked example 6. A fixed mass of gas at 120 kPa occupies 0.030 m³. It is compressed slowly to 0.020 m³ at constant temperature. Calculate the new pressure.
p₁ V₁ = p₂ V₂
120 × 0.030 = p₂ × 0.020
p₂ = 3.6 / 0.020 = 180 kPa
Keep the pressure units the same on both sides; you do not need to convert kPa to Pa for a ratio.
Increasing the pressure of a gas (Higher tier only)
Work is the transfer of energy by a force. When you do work on a gas, for example by pushing in the piston of a bicycle pump, you transfer energy to the gas. This increases its internal energy, and the molecules’ average kinetic energy rises, so the temperature of the gas increases. That is why the end of a bicycle pump gets warm when you pump up a tyre quickly.
Common errors
- Using cm³ with kg in ρ = m/V without converting. Convert to m³ (× 10⁻⁶) or give the answer in g/cm³ first.
- Saying mass decreases when a liquid boils. Mass is conserved in every change of state.
- Defining internal energy as kinetic energy only. It is kinetic plus potential energy of all the particles.
- Using the final temperature instead of the temperature change in ΔE = m c Δθ.
- Using ΔE = m c Δθ on the flat part of a heating graph. That part needs E = m L.
- Saying that particles “expand” when heated. The particles move faster; they do not get bigger.
- Explaining gas pressure with “more collisions” only. Say collisions are more frequent and, for a temperature rise, harder.
- Using pV = constant when the temperature changes. It needs a fixed mass at constant temperature.
Official syllabus
AQA GCSE Physics (8463) specification, for teaching from September 2016 onwards, for exams in 2018 onwards (Version 1.1, 30 September 2019), published by AQA. This guide covers section 4.3 Particle model of matter.
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