AQA GCSE Physics (8463), Higher tier · Particle model of matter › Particle model and pressure
Practise Pressure in gases. 14 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
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A fixed mass of gas has a pressure of 100 kPa and a volume of 0.030 m3. The gas is compressed to a volume of 0.010 m3. The temperature of the gas does not change. Calculate the new pressure of the gas. Use the equation: pressure × volume = constant
300 kPa
State what happens to the volume of a fixed mass of gas, at constant temperature, if its pressure is increased.
The volume decreases.
Sample questions
Written for this site in the style of AQA exam questions. They are not taken from real past papers.
Question 1Easy4 marks
(a) The pressure of a gas produces a net force on the walls of its container. What is the direction of this force? Tick (✓) one box.[1]
At right angles to the wall
Parallel to the wall
Always downwards
Always upwards
(b) A fixed mass of gas has a pressure of 100 kPa and a volume of 0.030 m3. The gas is compressed to a volume of 0.010 m3. The temperature of the gas does not change. Calculate the new pressure of the gas. Use the equation: pressure × volume = constant[2]
(c) What happens to the pressure of a fixed mass of gas at constant temperature when its volume is increased?[1]
Show the answer and mark scheme
(a)Answer: At right angles to the wall
(b)Answer: 300 kPa
100 × 0.030 = p × 0.010
p = 300 (kPa)
(c)Answer: It decreases.
it decreases
Question 2Medium6 marks
A student sealed the end of a plastic syringe. The syringe contained 60 cm3 of air at a pressure of 100 kPa. The student slowly pushed the plunger in until the volume of the air was 40 cm3. The temperature of the air did not change.
(a) Calculate the pressure of the air when its volume was 40 cm3. Use the Physics Equations Sheet.[2]
(b) The student then pulled the plunger out so that the volume of the air was greater than 60 cm3. The temperature of the air stayed the same. Explain, in terms of the air molecules, why the pressure of the air decreased.[3]
(c) Explain why the student pushed the plunger in slowly.[1]
Show the answer and mark scheme
(a)Answer: 150 kPa
100 × 60 = p × 40
p = 150 (kPa)
(b)Answer: The molecules move at the same average speed but are more spread out, so they hit each unit area of the walls less often and the pressure falls.
the (average) speed / kinetic energy of the molecules does not change (because the temperature is constant)
the molecules are more spread out / travel further between collisions with the walls
so there are fewer collisions with each unit area of the walls each second, so the pressure decreases
(c)Answer: Pushing quickly would do work on the air and warm it; pushing slowly keeps its temperature constant.
so that the temperature of the air stayed constant / pushing quickly would do work on the air and raise its temperature
Question 3Hard8 marks
A diver is swimming at a depth where the total pressure is 250 kPa. The diver breathes out a bubble of air with a volume of 2.0 cm3. The bubble rises to the surface, where the pressure is 100 kPa.
(a) Calculate the volume of the bubble when it reaches the surface. Assume that the temperature of the air in the bubble does not change. Use the Physics Equations Sheet.[2]
(b) Explain, in terms of particles, why the volume of the bubble increases as it rises.[3]
(c) The water near the surface is warmer than the water deeper down. Explain how this would affect the volume of the bubble at the surface compared with your calculated value.[2]
(d) Divers are trained never to hold their breath while swimming up to the surface. Suggest why.[1]
Show the answer and mark scheme
(a)Answer: 5.0 cm3
250 × 2.0 = 100 × V
V = 5.0 (cm3)
(b)Answer: The water pressure outside falls, so the air inside pushes the bubble outwards. As it expands the particles hit the bubble surface less often, until the pressure inside equals the pressure outside.
the pressure of the water outside the bubble decreases as the bubble rises
the air particles hitting the inside of the bubble now exert a greater pressure than the water outside, so the bubble expands
as the volume increases, the particles hit the surface of the bubble less often, so the pressure inside falls until it equals the pressure outside
(c)Answer: The warmer air particles move faster, so the bubble would be larger than calculated.
the air in the bubble would be warmer, so its particles would move faster (on average)
so the volume at the surface would be greater than the calculated value (5.0 cm3)
(d)Answer: The air in the lungs would expand as the pressure falls and could damage them.
the air in the lungs would expand as the pressure decreases, which could damage the lungs