Pressure in gases (physics only) - AQA GCSE Physics Exam Questions

Practise 29 exam-style questions on Pressure in gases (physics only) for AQA GCSE Physics. Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

29 questions
1–6 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

For a gas, pressure × volume = constant.

Give the two conditions that must be met for this relationship to apply.

Model answer

The gas must be a fixed mass (no gas enters or leaves), and its temperature must stay constant.

Explanation

Adding gas or raising the temperature would also change the pressure, so pV would no longer stay the same. The relationship only describes what happens when the volume alone changes.

Mark scheme

1 mark
the mass of gas is fixed
allow the amount of gas / number of molecules stays the same; allow no gas enters or leaves
1 mark
the temperature is constant

Question 2

2 marks

A student compresses the air in a sealed syringe in steps. After each step, she waits for a minute before reading the pressure gauge.

Explain why she waits before taking each reading.

Model answer

Pushing the plunger in does work on the air, which increases its temperature. Waiting allows the air to cool back to room temperature, so that all the readings are taken at the same temperature. The relationship pV = constant only applies when the temperature is constant.

Explanation

If the reading is taken while the air is still warm, the pressure is higher than it should be, and pV will not be constant. Pushing the plunger in slowly and waiting both help keep the temperature constant.

Mark scheme

1 mark
compressing the air (doing work on it) increases its temperature
allow the air warms up when it is compressed
1 mark
waiting lets the air cool back to room temperature, because pV is only constant at a constant temperature
allow otherwise the pressure reading would be too high

Question 3

3 marks

The gas in a cylinder has a pressure of 120 kPa. It pushes on a piston with a surface area of 0.020 m².

Calculate the force that the gas exerts on the piston.

Model answer

p = 120 kPa = 120 000 Pa
p = F ÷ A, so F = p × A
F = 120 000 × 0.020
F = 2400 N

Explanation

Gas pressure produces a force at right angles to any surface it touches, and pressure = force ÷ area. Convert kilopascals to pascals first, because 1 Pa = 1 N/m².

Mark scheme

1 mark
120 kPa = 120 000 Pa
an answer of 2400 (N) scores 3 marks
1 mark
120 000 = F ÷ 0.020
allow F = 120 000 × 0.020; allow ecf from an incorrect conversion
1 mark
F = 2400 (N)
an answer of 2.4 scores 2 marks

Question 4

4 marks

The nozzle of a syringe containing 20 cm³ of air at 100 kPa is sealed. The plunger is pulled out slowly until the volume of the air is 50 cm³. The temperature does not change.

Calculate the new pressure of the air in the syringe. Explain why the plunger is pushed back in when it is let go.

The air pressure outside the syringe is 100 kPa.

Use the equation:
pressure × volume = constant

Model answer

p₁ × V₁ = p₂ × V₂
100 × 20 = p₂ × 50
p₂ = 2000 ÷ 50 = 40 kPa
The air outside the syringe is at 100 kPa, which is greater than the 40 kPa inside. The force on the outside of the plunger is therefore larger than the force on the inside, so there is a resultant force pushing the plunger back in.

Explanation

Expanding the trapped air lowers its pressure. The plunger has gas pressure acting on both sides, each producing a force at right angles to its surface; when the outside pressure is bigger, the resultant force pushes the plunger inwards.

Mark scheme

1 mark
100 × 20 = p × 50
an answer of 40 (kPa) scores the first 2 marks
1 mark
p = 40 (kPa)
1 mark
the pressure outside (100 kPa) is greater than the pressure inside (40 kPa)
allow ecf from an incorrect pressure below 100 kPa
1 mark
so there is a resultant force on the plunger pushing it inwards
allow the force from the air outside is bigger than the force from the air inside

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