OCR GCSE Physics A: Gas pressure, compression and the atmosphere (J249)

This guide follows the OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026), for first assessment in 2018. It covers statements P1.3a to P1.3g in topic P1 Matter: pressure in gases and in the atmosphere. P1.3a and P1.3b are assessed in both Foundation Tier and Higher Tier. The specification prints P1.3c to P1.3g in bold type, which makes them Higher Tier only. Liquid pressure and floating and sinking (P1.3h to P1.3j) are covered elsewhere.

Related pages: revision notes, practice questions, the wider Matter study guide, the OCR GCSE Physics hub, the printable checklist and the free diagnostics.

What this unit covers

Statement	What you must be able to do	Tier
P1.3a	Explain how gas particle motion relates to temperature and pressure (closed systems only)	Both
P1.3b	Explain how temperature and pressure of a gas are related at constant volume (qualitative only)	Both
P1.3c	Recall that pressure changes can compress or expand a gas, and that the pressure acts as a net force normal to every surface	Higher Tier only
P1.3d	Explain how increasing the volume at constant temperature can lower the pressure; use pressure × volume = constant	Higher Tier only
P1.3e	Explain how doing work on a gas can raise its temperature	Higher Tier only
P1.3f	Describe a simple model of the atmosphere and of atmospheric pressure	Higher Tier only
P1.3g	Explain why atmospheric pressure varies with height	Higher Tier only

P1.3a: Particle motion, temperature and pressure

A gas is a huge number of particles moving randomly at high speed with large gaps between them. Here the gas is always in a closed system: a sealed container, so the number of particles stays fixed.

Pressure is force per unit area. Each particle that hits a wall pushes on it briefly. So many hits happen each second that they blend into a steady force on the wall. That force per unit area is the gas pressure.

Temperature reflects how much kinetic energy the particles have on average. Hotter gas means faster particles on average. The particles do not get bigger. Only their speed changes.

Two things change when the gas is hotter and the container stays the same:

• the particles reach the walls more often, so there are more collisions each second
• each collision is harder, so the average force from one collision is larger

Both effects raise the force on the wall, and so the pressure.

Worked example. A sealed steel bottle of air is moved from a fridge to a warm room. Explain what happens to the pressure inside.

The air warms, so its particles move faster on average. They hit the walls more often and each hit is harder. The force on the walls rises, so the pressure rises. The bottle is rigid and sealed, so the volume and number of particles stay the same.

Common errors: saying the particles “expand”, saying the particles hit each other more and that is the pressure, or leaving out the words “more often” or “harder”.

P1.3b: Temperature and pressure at constant volume

The specification asks for the relationship qualitatively only. At constant volume, raising the temperature of a fixed amount of gas raises its pressure. Lowering the temperature lowers the pressure. You explain it with the same two ideas as above: faster particles, more frequent and harder collisions.

Practical contexts include a balloon that is heated and then chilled, a manometer connected to a warmed or cooled flask, the exploding can demonstration and Alka-Seltzer rockets in film canisters.

Do not write that pressure is “proportional to temperature in degrees Celsius”. The specification only asks for the direction of change and the explanation.

Worked example. A student warms a sealed flask in a water bath and reads a pressure gauge. The gauge reading rises from 100 kPa to 106 kPa. Explain the change.

Heating raises the temperature of the gas. Its particles gain kinetic energy and move faster. In the rigid flask the volume is constant, so the faster particles collide with the walls more often and with greater force. The pressure increases.

P1.3c (Higher Tier only): Compression and force at right angles

Gases can be compressed because most of a gas is empty space. A syringe of air can be squeezed a long way. A syringe of water or sand hardly moves, because those particles are already close together. Reducing the pressure lets a gas expand again.

Gas pressure pushes on any surface it touches, and the net force is at right angles (normal) to that surface. Particles strike the wall from many directions. The sideways components of their bounces cancel out along the wall, leaving a push straight at it. A curved balloon skin works the same way, patch by patch.

The word net matters, because air often presses on both sides of a surface. The net force comes from the difference:

net force = (larger pressure − smaller pressure) × area

Here you use pressure (Pa) = force normal to the surface (N) ÷ area (m²), written as force = pressure × area. This is the same relationship as in the Forces study guide.

Worked example 1. Atmospheric pressure is 1.0 × 10⁵ Pa. A window measures 0.50 m by 0.80 m. Calculate the force of the air on one side of it.

Area = 0.50 × 0.80 = 0.40 m² Force = pressure × area = 1.0 × 10⁵ × 0.40 = 4.0 × 10⁴ N

The window is not crushed because air inside the room pushes back with about the same force, so the net force is close to zero.

Worked example 2: the collapsing can. A little water is boiled in a metal can until steam has pushed most of the air out. The can is sealed and cooled. The steam condenses, so the pressure inside drops to 4.0 × 10⁴ Pa. The outside pressure is 1.0 × 10⁵ Pa and one flat side has area 0.015 m². Calculate the net force on that side.

Pressure difference = 1.0 × 10⁵ − 4.0 × 10⁴ = 6.0 × 10⁴ Pa Net force = 6.0 × 10⁴ × 0.015 = 900 N, acting inwards, at right angles to the side.

Nothing “sucks” the can in. The outside air hits it harder than the few particles inside.

P1.3d (Higher Tier only): Volume, pressure and pV = constant

Keep the temperature constant and the amount of gas fixed. Now make the container bigger. The particles travel at the same average speed, but they have further to go between hitting the walls. Each wall is hit less often. The average force from a collision is unchanged. So the force on each unit of wall area falls, and the pressure falls. Making the container smaller does the opposite.

The mathematical learning outcome for this statement is:

pressure (Pa) × volume (m³) = constant

With a fixed mass of gas held at one temperature, p₁V₁ = p₂V₂. Doubling the volume halves the pressure. Because it is a ratio, any volume unit works if you use the same unit on both sides. Pressure units must also match.

Worked example 1. A syringe holds 60 cm³ of air at 1.0 × 10⁵ Pa. The nozzle is blocked and the plunger is pushed in until the volume is 24 cm³. The temperature is constant. Calculate the new pressure.

p₁V₁ = p₂V₂ 1.0 × 10⁵ × 60 = p₂ × 24 p₂ = (1.0 × 10⁵ × 60) ÷ 24 = 2.5 × 10⁵ Pa

In m³ the volumes are 6.0 × 10⁻⁵ and 2.4 × 10⁻⁵, and the answer is the same.

Worked example 2. A tank of volume 0.80 m³ holds gas at 150 kPa. It is connected to an empty chamber so that the gas fills a total volume of 2.0 m³ at the same temperature. Calculate the final pressure.

p₂ = (150 × 0.80) ÷ 2.0 = 60 kPa

The pressure falls to less than half because the volume rose to two and a half times its first value.

Marshmallow in a vacuum. Pumping air out of a bell jar lowers the pressure outside the marshmallow. The gas bubbles inside it are now at higher pressure than the surroundings, so they expand and the marshmallow swells.

P1.3e (Higher Tier only): Work done on a gas

When you push a piston into a gas, you do work on it. The pump example in the specification is a bicycle pump. Particles bouncing off the moving piston leave it faster than they arrived, in the same way that a ball leaves a bat faster than it arrived. The average kinetic energy of the particles rises, which is a rise in temperature. The energy has been transferred by the work done by the force on the piston.

Worked example. In one stroke a pump piston moves 0.20 m against an average force of 45 N from the gas. Calculate the work done on the gas.

Work done = force × distance moved along the force = 45 × 0.20 = 9.0 J

If little energy escapes, most of this raises the internal energy of the gas, so its temperature rises. A real pump also warms by friction.

Link with P1.3d: the rule pV = constant holds only at constant temperature. A fast compression warms the gas, so the pressure rises by more than pV = constant predicts.

P1.3f (Higher Tier only): A simple model of the atmosphere

The atmosphere is a layer of gas held around the Earth by gravity. Air has mass, so the air above any point has weight. Atmospheric pressure at a place is the force from that weight of air pressing on each unit area below it. It acts in all directions, not only downwards.

The specification asks for a simple model that assumes uniform density, and you are not expected to know the layers.

Worked example. Atmospheric pressure at the ground is 1.0 × 10⁵ Pa. Take g = 10 N/kg. Find the mass of air above each square metre.

Weight of the column on 1 m² = pressure × area = 1.0 × 10⁵ × 1 = 1.0 × 10⁵ N Mass = weight ÷ g = 1.0 × 10⁵ ÷ 10 = 1.0 × 10⁴ kg

That is about ten tonnes of air on every square metre.

Real air has no sharp top and thins gradually, so the uniform model is only a simplification.

P1.3g (Higher Tier only): Why pressure falls with height

Two reasons explain why atmospheric pressure decreases as you go higher:

1. There is less air above you, so the weight of air pressing down is smaller.
2. Air is compressed by the air above it, so low air is denser than high air. Higher up, there are fewer particles in each cubic metre, so fewer collisions on a surface.

The fall is not the same for every kilometre, because the density changes with height. At the summit of the highest mountains the pressure is only about a third of its sea-level value.

Worked example. A weather balloon has volume 2.4 m³ at the ground where the pressure is 100 kPa. It rises to a height where the pressure is 25 kPa. Assume constant temperature. Calculate its new volume and explain the change.

p₁V₁ = p₂V₂, so 100 × 2.4 = 25 × V₂ V₂ = 240 ÷ 25 = 9.6 m³

The outside pressure is lower, so the gas inside pushes the skin out until the inside pressure matches. In reality the air is also colder at height, which would reduce the volume a little.

Common errors in this unit

• Saying the particles get hotter or bigger, instead of faster.
• Explaining pressure as particles bumping each other, not hitting the walls.
• Using pV = constant when the temperature changes or gas leaks.
• Mixing cm³ with m³, or kPa with Pa.
• Saying the atmosphere pushes down only.

Official syllabus

OCR, GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026), first assessment 2018, topic P1.3 Pressure, statements P1.3a to P1.3g.