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Edexcel International GCSE Physics (4PH1) · Section 5 — Solids, liquids and gases
Mini-Lesson

Solids, Liquids & Gases

This mini-lesson covers the whole of 4PH1 Section 5: density and pressure, pressure in a liquid column, change of state and the particle model, and the kinetic theory of gases with the Kelvin scale and the gas laws.

SOLIDLIQUIDGAS
Same particles, three arrangements — the spacing and motion change, not the particles themselves.

Work through each screen, answer the questions as you go (some wordy, some calculations) and collect ⭐ stars. Press Start when you're ready.

Spec 5.1 / 5.2P · Units

The units you must use

Section 5 expects you to work confidently in these SI units:

  • °C degree Celsius and K kelvin — temperature
  • kg kilogram — mass; m, m², m³ — length, area, volume
  • kg/m³ — density; Pa pascal — pressure; N newton — force
  • J joule — energy; m/s and m/s² — speed and acceleration

Paper 2 only (5.2P): for specific heat capacity you also use J/kg °C (joules per kilogram per degree Celsius).

Spec 5.3–5.4 · Density

Density: how tightly packed

Density tells you how much mass is squeezed into each cubic metre of a material:

ρ = m / Vdensity (kg/m³) = mass (kg) ÷ volume (m³)

Measuring it (5.4 practical): find the mass on a balance. For a regular solid, measure its sides and calculate the volume; for an irregular solid or a liquid, use a displacement can or measuring cylinder. Then divide.

Worked example

A metal block has mass 240 g (= 0.24 kg) and volume 0.00003 m³.

ρ = 0.24 ÷ 0.00003 = 8000 kg/m³ (this is copper).

Calculate

Your turn — density

1An aluminium block has a mass of 5.4 kg and a volume of 0.002 m³. Calculate its density.
kg/m³
Hint: ρ = m ÷ V = 5.4 ÷ 0.002.
Spec 5.5–5.6 · Pressure

Pressure: force spread over area

Pressure is the force acting on each square metre of a surface:

p = F / Apressure (Pa) = force (N) ÷ area (m²)

The same force over a smaller area gives a bigger pressure — which is why a sharp knife or a drawing pin works.

Acts in all directions (5.6): at any point in a gas or a liquid at rest, the pressure pushes equally in every direction — up, down and sideways — not just downwards.

Worked example

A box pushes down with a force of 600 N over an area of 0.5 m².

p = 600 ÷ 0.5 = 1200 Pa

Calculate

Your turn — pressure

2A crate exerts a force of 900 N on the floor through a base of area 0.3 m². Calculate the pressure it exerts.
Pa
Hint: p = F ÷ A = 900 ÷ 0.3.
Spec 5.7 · Pressure in a liquid

Pressure grows with depth

Dive deeper and the weight of water above you grows, so the pressure difference increases:

p = h × ρ × gpressure difference (Pa) = height/depth (m) × density (kg/m³) × g (N/kg)
surface small pbigger pbiggest p depth h
Longer arrows = larger pressure. Only the depth and the liquid's density matter — not the shape or width of the container.

Watch out: pressure in a liquid depends on depth and density, not the shape of the container or how much liquid there is in total. A narrow tube and a wide tank give the same pressure at the same depth.

Worked example

Depth 2 m, water density 1000 kg/m³, g = 10 N/kg.

p = 2 × 1000 × 10 = 20 000 Pa

Calculate

Your turn — depth pressure

3A diver is 5 m below the surface of seawater of density 1030 kg/m³. Using g = 10 N/kg, calculate the pressure difference due to the water.
Pa
Hint: p = h × ρ × g = 5 × 1030 × 10.
Spec 5.8P–5.10P · Paper 2 only

Three states, three arrangements

The kinetic theory explains the states by how the particles are arranged and moving:

SOLIDLIQUIDGAS fixed lattice,vibrate in place touching but freeto slide past far apart, fast,random motion
Solid: regular, fixed, only vibrating. Liquid: close but mobile. Gas: widely spaced, fast and random.

Heating (5.8P–5.9P): heating raises the energy stored in the particles, so the temperature rises or the state changes — a solid melts, then the liquid evaporates/boils to a gas. During a change of state the temperature stays constant while bonds are broken.

Quick check · Paper 2 only

Which state?

?In which state are the particles far apart, moving fast in random directions, with almost no forces between them?
Spec 5.12P–5.14P · Paper 2 only

Specific heat capacity

The specific heat capacity c is the energy needed to change the temperature of 1 kg by 1 °C:

ΔQ = m × c × ΔTchange in thermal energy (J) = mass (kg) × s.h.c. (J/kg °C) × temperature change (°C)

Water has a high c (4200 J/kg °C), so it heats and cools slowly — useful in radiators and for the climate near the sea.

Practical (5.14P): heat a known mass with an electric heater, measure the energy supplied and the temperature rise, then find c = ΔQ ÷ (m × ΔT). A temperature–time graph (5.11P) shows the flat region while a substance changes state.

Worked example

Heating 0.5 kg of water (c = 4200) by 30 °C:

ΔQ = 0.5 × 4200 × 30 = 63 000 J (63 kJ)

Calculate · Paper 2 only

Your turn — thermal energy

4How much thermal energy is needed to raise the temperature of 3 kg of water (c = 4200 J/kg °C) by 20 °C?
J
Hint: ΔQ = m × c × ΔT = 3 × 4200 × 20.
Spec 5.15 · Gas pressure

Where gas pressure comes from

Gas molecules have random motion. Each time one hits a wall it bounces off, exerting a tiny force. Billions of collisions every second add up to a steady pressure:

each collision pushes on the wall → pressure
The red marks show molecules striking and rebounding off the walls — the net effect of all these impacts is gas pressure.

Watch out: gas pressure is not the gas "pressing because it is heavy" — it is the result of countless molecular collisions with the container walls.

Sort it

What raises the pressure?

For a fixed amount of gas, tap how the molecular collisions explain each change.

Spec 5.16–5.17 · The Kelvin scale

Absolute zero and kelvin

Cool a gas and its molecules slow down. At −273 °C they have the least possible energy — you cannot go colder. This is absolute zero, the start of the Kelvin scale:

T (K) = θ (°C) + 273θ (°C) = T (K) − 273
−273 °C0 °C100 °C 0 K273 K373 K add 273 to go °C → K · subtract 273 to go K → °C
The two scales have the same size of degree — they are just shifted by 273.

Absolute zero = 0 K = −273 °C. There is no such thing as a negative kelvin temperature.

Calculate

Your turn — convert to kelvin

5A gas is at a temperature of 27 °C. Convert this temperature into kelvin.
K
Hint: T (K) = θ (°C) + 273 = 27 + 273.
Spec 5.18–5.19 · Temperature & energy

Hotter means faster

Raising the temperature gives molecules more energy, so on average they move faster. In fact:

Kelvin temperature ∝ average kinetic energydouble the Kelvin temperature → double the average kinetic energy of the molecules

This is why temperature is measured from absolute zero: at 0 K the average kinetic energy is (as near as possible) zero, so it makes sense to say the energy is proportional to the Kelvin temperature — that statement would fail with the Celsius scale.

Link to pressure: faster molecules hit the walls harder and more often, so for a fixed volume a higher temperature gives a higher pressure.

Quick check

Temperature and motion

?The Kelvin temperature of a fixed mass of gas is doubled. What happens to the average kinetic energy of its molecules?
Spec 5.20 / 5.22 · p–V at constant T

Squeeze it: pressure ∝ 1/volume

For a fixed mass of gas at constant temperature, squeezing it into a smaller volume packs the same molecules into less space, so they hit the walls more often and the pressure rises:

p₁V₁ = p₂V₂pressure × volume stays constant when temperature is fixed
p V small V, big p big V, small p constant temperature
The curve falls away: halve the volume and the pressure doubles. p and V are inversely proportional.
Worked example

Gas at 100 000 Pa in 0.6 m³ is compressed to 0.2 m³ at constant temperature.

p₂ = p₁V₁ ÷ V₂ = (100 000 × 0.6) ÷ 0.2 = 300 000 Pa

Calculate

Your turn — Boyle's law

6A fixed mass of gas at 200 000 Pa occupies 0.4 m³. At constant temperature it is compressed to 0.1 m³. Calculate the new pressure.
Pa
Hint: p₂ = (p₁ × V₁) ÷ V₂ = (200 000 × 0.4) ÷ 0.1.
Spec 5.20 / 5.21 · p–T at constant V

Heat it: pressure ∝ Kelvin temperature

For a fixed mass of gas at constant volume, raising the Kelvin temperature makes the molecules hit the walls harder and more often, so the pressure rises in proportion:

p₁ / T₁ = p₂ / T₂T must be the temperature in KELVIN, not °C

Watch out — use kelvin! The gas laws only work with Kelvin temperatures. Putting °C into p₁/T₁ = p₂/T₂ gives the wrong answer (and dividing by 0 °C is meaningless). Always convert first.

Worked example

Gas at 100 000 Pa and 300 K is heated to 450 K at constant volume.

p₂ = p₁ × T₂ ÷ T₁ = 100 000 × 450 ÷ 300 = 150 000 Pa

Calculate

Your turn — the pressure law

7A sealed rigid can of gas is at 120 000 Pa and 250 K. It is warmed to 400 K with the volume unchanged. Calculate the new pressure.
Pa
Hint: p₂ = p₁ × T₂ ÷ T₁ = 120 000 × 400 ÷ 250.
Quick check

The classic trap

?A student uses p₁/T₁ = p₂/T₂ but puts the temperatures in °C instead of kelvin. What is wrong?
Match them

Equation to job

Tap an equation, then tap what it is used for.

Recap

The equations to know

Density: ρ = m / V

Pressure: p = F / A

Pressure in a liquid: p = h × ρ × g

Thermal energy (P): ΔQ = m × c × ΔT

°C → K: T = θ + 273 (absolute zero = 0 K = −273 °C)

Boyle's law: p₁V₁ = p₂V₂ (constant T)

Pressure law: p₁/T₁ = p₂/T₂ (constant V, use kelvin!)

You've covered all of 4PH1 Section 5 — units, density & pressure, change of state, and ideal gas molecules. Press Finish to see your score.

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