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

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow the OCR GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026), statements P1.3a to P1.3g. Questions 1 to 3 cover Foundation and Higher Tier content. Questions 4 to 11 test statements that are bold in the specification, so they are marked (Higher Tier only). Take g = 10 N/kg where needed and use atmospheric pressure as given in each question.

Links: study guide, revision notes, OCR GCSE Physics hub, printable checklist and the free diagnostics.

Questions

1. A sealed, rigid metal bottle of air is left in direct sunshine and its temperature rises. Explain why the pressure of the air inside rises. [3]

2. A student warms a sealed flask of air in a water bath and records the gauge pressure at four temperatures. The volume does not change.

Temperature (°C)	10	30	50	70
Pressure (kPa)	98	105	112	118

(a) Describe the pattern in the results. [2] (b) Explain the pattern using ideas about particles. [3] (c) State one quantity that must stay constant during the experiment, other than the volume. [1]

3. A student says: “When a balloon of gas is heated, the gas particles expand. They take up more room, so they hit each other more and the pressure goes up.” Identify the error and write a correct version of the explanation. [3]

4. (Higher Tier only) A syringe containing air has its nozzle sealed. (a) Explain why the air can be compressed a long way but a syringe full of sand cannot. [2] (b) Explain why the gas pressure on the inside wall of the syringe produces a force at right angles to the wall. [2]

5. (Higher Tier only) A sealed container has a flat lid of area 0.12 m². The pressure inside is 60 kPa and the pressure outside is 100 kPa. Calculate the net force on the lid and state its direction. [4]

6. (Higher Tier only) Assume the temperature is constant in each part. (a) A gas of volume 150 cm³ is at a pressure of 1.0 × 10⁵ Pa. It is compressed to 40 cm³. Calculate the new pressure. [3] (b) A tank of volume 0.30 m³ holds gas at 4.0 × 10⁵ Pa. The gas then spreads into a volume of 1.5 m³. Calculate the new pressure. [2]

7. (Higher Tier only) A marshmallow contains many tiny gas bubbles. It is placed in a bell jar and air is pumped out. The marshmallow swells. Explain why. [4]

8. (Higher Tier only) A hand pump is used to inflate a football quickly, and the metal barrel of the pump becomes warm. (a) Explain why the air in the pump becomes warmer. [3] (b) In one stroke the average force on the piston is 60 N and it moves through 0.25 m. Calculate the work done on the air. [2]

9. (Higher Tier only) (a) Describe a simple model of the Earth’s atmosphere. [2] (b) At ground level the atmospheric pressure is 1.0 × 10⁵ Pa. A flat roof has area 2.5 m². Calculate the force of the air on the roof. [2] (c) Use your answer to (b) to find the mass of air above the roof. [1]

10. (Higher Tier only) The table gives atmospheric pressure at four heights.

Height (km)	0	2	4	6
Pressure (kPa)	101	79	62	47

(a) Describe how the pressure changes with height. [2] (b) Explain why atmospheric pressure falls with height. [3]

11. (Higher Tier only) A weather balloon has volume 5.0 m³ at the ground, where the pressure is 1.0 × 10⁵ Pa. It rises to a height where the pressure is 20 kPa. (a) Assume the temperature stays constant. Calculate the volume of the balloon at that height. [3] (b) Explain, in terms of particles and forces, why the balloon expands as it rises. [3] (c) The air at that height is colder than at the ground. Suggest the effect of this on the volume you calculated. [2]

Answers

1. The air particles gain energy when heated and their average speed rises [1]. Wall collisions then happen more frequently [1], and every collision pushes harder, which raises the force on each unit of wall area [1]. [3] Examiner insight: Give both the frequency and the strength of the collisions. An answer that says only “faster particles” leaves the link to pressure unfinished.

2. (a) Pressure increases as temperature increases [1]. The increase is about the same for each 20 °C step, roughly 6 to 7 kPa [1]. (b) Heating makes the particles move faster [1]. They collide with the walls more often and with more force [1]. The volume is fixed, so the force per unit area, the pressure, rises [1]. (c) The amount (mass) of gas, so no leaks [1]. [6] Examiner insight: Support a description with a figure from the table. A particle explanation scores better when it says why the volume being fixed matters.

3. The particles do not expand when heated; they move faster [1]. Pressure is caused by collisions with the walls, not by collisions between particles [1]. Faster particles hit the walls more often and harder, so the force on each unit area increases [1]. [3] Examiner insight: A “what is wrong” question rewards naming the error and then replacing it. Repeating the student’s wording does not correct anything.

4. (a) A gas is mostly empty space, so the particles can be pushed closer together [1]. Sand grains are already touching, so they cannot be squeezed together [1]. (b) Particles hit the wall from many directions [1]. Components of their bounces along the wall cancel, leaving a net force at right angles to it [1]. [4] Examiner insight: Mention “net” and “right angles” together in part (b). Saying “pressure acts in all directions” does not answer the question about the wall.

5. Pressure difference = 100 − 60 = 40 kPa = 4.0 × 10⁴ Pa [1]. Force = pressure × area = 4.0 × 10⁴ × 0.12 [1] = 4800 N (4.8 × 10³ N) [1]. The force acts inwards, because the outside pressure is greater [1]. [4] Examiner insight: Convert kPa to Pa before multiplying, and write the unit. Give the direction of a net force as well as its size.

6. (a) p₁V₁ = p₂V₂, so 1.0 × 10⁵ × 150 = p₂ × 40 [1]. p₂ = 1.0 × 10⁵ × 150 ÷ 40 [1] = 3.75 × 10⁵ Pa [1]. (b) p₂ = 4.0 × 10⁵ × 0.30 ÷ 1.5 [1] = 8.0 × 10⁴ Pa [1]. [5] Examiner insight: Keep volume units the same on both sides; cm³ can stay as cm³. A final pressure larger than the starting pressure after an expansion is a clue that the rearrangement is upside down.

7. Pumping out air lowers the pressure outside the marshmallow [1]. The gas inside the bubbles is at a higher pressure than the surroundings, so there is a net outward force on the bubble walls [1]. The bubbles expand [1]. As the volume increases, the particles hit the walls less often, so the pressure inside falls until it matches the outside pressure [1]. [4] Examiner insight: A good answer compares the inside and outside pressures. Avoid saying the vacuum “pulls” the gas out.

8. (a) The piston does work on the air [1]. Particles rebound from the moving piston with greater speed [1], so their average kinetic energy increases and the temperature rises [1]. (b) Work = force × distance = 60 × 0.25 [1] = 15 J [1]. [5] Examiner insight: Name the energy transfer, “work done”, rather than “heat from friction”. Friction may be mentioned as a smaller extra effect.

9. (a) A layer of gas surrounds the Earth, held by gravity [1]. The density is assumed to be the same throughout the layer [1]. (b) Force = pressure × area = 1.0 × 10⁵ × 2.5 [1] = 2.5 × 10⁵ N [1]. (c) Mass = 2.5 × 10⁵ ÷ 10 = 2.5 × 10⁴ kg [1]. [5] Examiner insight: In (a), use the idea of a uniform-density model, because the specification says layers are not expected. In (c), say that the force is the weight of the air.

10. (a) The pressure decreases as height increases [1]. The drop per 2 km gets smaller: 22 kPa, then 17 kPa, then 15 kPa [1]. (b) There is less air above at greater height, so the weight of air pressing is smaller [1]. The air is also less dense higher up, because lower air is compressed by the air above it [1]. Fewer particles hit each unit of area, so fewer collisions and less force [1]. [5] Examiner insight: Use figures from the table when you describe. The explanation needs both the smaller weight and the thinner air.

11. (a) 20 kPa = 2.0 × 10⁴ Pa [1]. 1.0 × 10⁵ × 5.0 = 2.0 × 10⁴ × V₂ [1]. V₂ = 25 m³ [1]. (b) Higher up there is less air above, so the outside pressure is lower [1]. The gas inside hits the skin harder than the air outside, giving a net outward force [1]. The balloon expands until the inside pressure falls to match, because the particles hit the walls less often in a bigger volume [1]. (c) Colder gas has slower particles, so for a given volume its pressure is lower [1]. The skin is pushed out less, so the real volume is smaller than 25 m³ [1]. [8] Examiner insight: Show the unit conversion line in (a). In (c), argue from particle speed and collisions, and state the direction of change.

Where marks are usually lost

• Saying particles expand, get bigger or get hotter instead of moving faster.
• Describing only faster particles, with no link to collisions per second and force per collision.
• Forgetting to convert kPa to Pa before using force = pressure × area.
• Rearranging p₁V₁ = p₂V₂ upside down.
• Using pV = constant without saying the temperature is constant.
• Leaving out the direction of a net force.
• Explaining the warm pump as only friction.
• Saying there is no air at height, instead of less air above and thinner air.

Next steps

• Review: revision notes and the study guide.
• Tick off statements on the OCR GCSE Physics checklist and browse the course hub.
• Find weak areas with the free 10-minute diagnostics.
• Book a free trial class.

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.