Question bank › GCSE Physics › Particle model of matter
Particle motion and pressure in gases
Answer on paper first, then reveal the mark scheme one mark at a time.
1Grade 4–53 marks
A sealed container of gas is heated. The volume of the container does not change.
Explain, in terms of particles, why the pressure of the gas increases. [3]
the (average) kinetic energy of the particles increases / particles move faster(1)
the particles collide with the walls (of the container) more often / more frequently allow more collisions per second; ignore 'more collisions' with no reference to the walls(1)
the particles hit the walls with greater force (so the pressure increases) allow greater change in momentum per collision(1)
Answer: Particles move faster, so they hit the walls more often and with more force.
Examiner tip: Say 'collide with the walls more often' — collisions between particles do not cause pressure.
2Grade 6–7HigherTriple only5 marks
(a) A gas has a pressure of $1.2 \times 10^5$ Pa and a volume of 0.040 m³. It is compressed slowly at constant temperature to a volume of 0.025 m³.
Calculate the new pressure of the gas. Use the equation: pressure $\times$ volume $=$ constant [3]
(b) When a bicycle tyre is pumped up quickly, the air in the pump gets warmer.
Explain why. [2]
(a) $1.2 \times 10^5 \times 0.040 = p \times 0.025$(1)
(b) work is done on the gas (by pushing the piston) allow energy is transferred to the gas by the force on the piston; ignore friction(1)
(b) this increases the internal energy of the gas, so its temperature increases allow kinetic energy of particles increases(1)
Answer: (a) $1.92 \times 10^5$ Pa (b) work done on gas increases its internal energy, so temperature rises
Examiner tip: Smaller volume means larger pressure — if your answer is smaller than $1.2 \times 10^5$ Pa, you have divided the wrong way.
3Grade 1–3Foundation4 marks
This question is about the particles in a gas.
(a) Complete the sentences. Choose answers from the box.
chemical
kinetic
random
same
straight
different
The particles in a gas are in constant ________ motion.
The particles do not all move at the ________ speed.
The temperature of the gas is related to the average ________ energy of its particles. [3]
(b) What causes a gas to exert a pressure on the walls of its container? [1]
(a) random(1)
(a) same(1)
(a) kinetic(1)
(b) (gas) particles colliding with / hitting / bouncing off the walls (of the container); allow particles exert a force on the walls when they collide; ignore particles 'touching' or 'pushing' the walls without reference to collisions / hitting(1)
Answer: (a) random; same; kinetic (b) particles collide with the walls
Examiner tip: In (b) say the particles collide with the walls – answers like 'the particles take up space' or 'the gas pushes' score zero.
4Grade 4–55 marks
A student investigated how the pressure of the air inside a sealed metal can changed with temperature. The can was placed in a water bath and the volume of the can did not change.
The table shows the results.
Temperature in °C
Pressure in kPa
0
100
20
107
40
115
60
122
80
129
(a) Name the independent variable in this investigation. [1]
(b) Give one control variable in this investigation. [1]
(c) Describe the relationship between the temperature and the pressure of the air. [2]
(d) Predict the pressure of the air at 100 °C. [1]
(a) temperature (of the air / water bath)(1)
(b) volume of the can / air; allow mass / amount of air (in the can); allow (same) can(1)
(c) as temperature increases, pressure increases(1)
(c) pressure increases by equal amounts for equal rises in temperature / linear relationship / increases by about 7 (kPa) every 20 °C; do not accept 'directly proportional' (pressure is not doubled when °C temperature doubles)(1)
(d) any value from 135 to 138 kPa(1)
Answer: (a) temperature (b) volume of the can (c) pressure increases as temperature increases, by roughly equal amounts (≈7 kPa per 20 °C) – a linear pattern (d) about 136 kPa
Examiner tip: In (c) do not write 'directly proportional' – the line does not go through the origin on a °C scale, so examiners give no credit for it.
5Grade 4–5Triple only4 marks
The diagrams show the same sealed container of gas. Each diagram shows arrows for the forces that the gas exerts on the walls of the container.
(a) Which diagram, A, B or C, shows the correct direction of the forces? [1]
(b) Explain how the gas particles produce a force on the walls of the container. [2]
(c) The pressure of the gas is the same on every wall.
Suggest why the container does not move, even though the gas exerts forces on its walls. [1]
(a) A(1)
(b) particles collide with / hit / bounce off the walls(1)
(b) (each) collision exerts a (small) force on the wall; allow particles change direction / momentum at the wall so exert a force; allow many collisions (per second) produce an overall / net force at right angles to the wall(1)
(c) forces on opposite walls are equal and (act in) opposite (directions) so cancel / resultant force (on the container) is zero(1)
Answer: (a) A (b) particles collide with the walls; each collision exerts a force on the wall (net force at right angles to the surface) (c) forces on opposite walls are equal and opposite, so the resultant force is zero
Examiner tip: Gas pressure always gives a force at right angles (perpendicular) to a surface – pointing outwards from the gas, never along the wall or just downwards.
6Grade 4–5Triple only5 marks
A syringe contains 60 cm³ of air at a pressure of 100 kPa. The end of the syringe is sealed.
The plunger is pushed in slowly until the volume of the air is 25 cm³. The temperature of the air does not change.
(a) Calculate the new pressure of the air in the syringe.
Use the equation: pressure × volume = constant [3]
(b) The equation in part (a) only applies when two conditions are met.
Give the two conditions. [2]
(a) 100 × 60 = p × 25 (or constant = 6000)(1)
(a) p = 6000 ÷ 25(1)
(a) p = 240 kPa; allow 240 000 Pa or 2.4 × 105 Pa; a correct answer with no working scores 3 marks(1)
(b) fixed mass (of gas) / no gas escapes or enters; allow sealed container(1)
Answer: (a) 240 kPa (b) fixed mass of gas; constant temperature
Examiner tip: You can keep pressure in kPa and volume in cm³ as long as both readings use the same units – then the answer comes out in kPa.
7Grade 6–7HigherTriple only7 marks
A student investigated how the pressure of a fixed mass of gas depends on its volume. The temperature of the gas was kept constant.
The graph shows the results.
(a) Use the graph to find the pressure of the gas when its volume is 40 cm³. [1]
(b) Use data from the graph to show that pressure × volume is constant for this gas. [2]
(c) The volume of the gas is reduced to 12 cm³ at the same temperature.
Calculate the pressure of the gas. [2]
(d) Explain, in terms of particles, why the pressure of the gas decreases when its volume increases at constant temperature. [2]
(a) 75 kPa; allow 74–76 kPa(1)
(b) pressure × volume calculated correctly for at least two points from the graph, e.g. 150 × 20 = 3000 and 60 × 50 = 3000; allow 100 × 30, 75 × 40, 50 × 60(1)
(b) statement that the values are the same / (approximately) equal (so pV is constant)(1)
(c) p × 12 = 3000 (or use of any correct pair of values from the graph)(1)
(c) p = 250 kPa; allow 2.5 × 105 Pa; allow 245–255 kPa from a value of pV read from the graph(1)
(d) particles have further to travel (between walls) / the walls have a larger area; ignore particles are more spread out unless linked to collisions(1)
(d) so particles collide with the walls less often / fewer collisions per second (on each unit area); do not accept particles move slower / have less kinetic energy (temperature is constant)(1)
Answer: (a) 75 kPa (b) e.g. 150 × 20 = 3000 and 60 × 50 = 3000 – the same, so pV is constant (c) 250 kPa (d) particles travel further between collisions with the walls, so collide with the walls less often (same speed because temperature is constant)
Examiner tip: In (d) the temperature is constant, so the particles do NOT slow down – the pressure falls only because collisions with the walls happen less often.
8Grade 8–9HigherTriple only7 marks
In a diesel engine, a piston moves quickly to compress the air in a cylinder. The air becomes hot enough to ignite the fuel.
During one compression, the volume of the air is reduced from 500 cm³ to 25 cm³. At the start, the pressure of the air is 100 kPa.
(a) The average force exerted by the piston on the air is 3600 N. The piston moves a distance of 0.080 m.
Calculate the work done on the air by the piston. [2]
(b) Calculate the pressure of the air after compression if its temperature did not change.
Use the equation: pressure × volume = constant [2]
(c) The actual pressure of the air after compression is about 6600 kPa.
Explain why the actual pressure is greater than your answer to part (b). [3]
(a) W = 3600 × 0.080(1)
(a) W = 288 J; allow 290 J(1)
(b) 100 × 500 = p × 25(1)
(b) p = 2000 kPa; allow 2.0 × 106 Pa(1)
(c) work is done on the air (by the piston) so the internal energy of the air increases; allow energy is transferred to the air (by the force)(1)
(c) (so) the temperature of the air increases / the air particles move faster / have more (average) kinetic energy(1)
(c) (so) particles collide with the walls more often / with more force (than they would at constant temperature); ignore 'more collisions' alone without comparison to rate or force(1)
Answer: (a) 288 J (b) 2000 kPa (c) work done on the air increases its internal energy, so its temperature rises; the faster particles hit the walls more often and with greater force, so the pressure is higher than pV = constant predicts
Examiner tip: pV = constant only works at constant temperature – a fast compression does work on the gas, heats it, and pushes the pressure far above the constant-temperature value.
9Grade 6–7Triple only6 marks
The diagram shows apparatus that can be used to investigate how the pressure of a fixed mass of air changes with its volume.
Describe how the apparatus could be used to obtain valid results, and how the results could be used to show the relationship between pressure and volume. [6]
Check the syringe and tubing are sealed / airtight so the mass of air is fixed(1)
Read and record the starting volume from the scale on the syringe and the pressure from the gauge(1)
Push in (or pull out) the plunger to change the volume by a set amount, e.g. 5 cm³ at a time(1)
Move the plunger slowly / wait a short time after each change so the temperature of the air returns to room temperature (temperature is a control variable)(1)
Read the new volume and pressure each time; read the scale at eye level to avoid parallax error(1)
Take at least five sets of readings over as wide a range of volume as possible(1)
Repeat the readings (e.g. with the plunger moving out as well as in) and calculate means(1)
Plot a graph of pressure (y-axis) against volume (x-axis) – a curve showing pressure decreases as volume increases(1)
Calculate pressure × volume for each pair of readings – values should be (approximately) the same, showing pV is constant(1)
Allow: plot pressure against 1/volume – a straight line through the origin shows pressure is inversely proportional to volume(1)
Level
Marks
What it looks like
Level 3
5–6
A clear, logically ordered method that could be followed by someone else to give valid results, including how the mass and temperature of the air are kept constant. Explains how the results are processed (graph or p × V) to show the relationship.
Level 2
3–4
A method with the main steps of changing the volume and measuring the pressure, but with some steps missing or not in order. Some attempt to control a variable or to process the results.
Level 1
1–2
Some relevant steps, e.g. pushing the plunger and reading the gauge, but these do not form a workable method. Little or no reference to control variables or analysis.
No relevant content
0
No relevant content.
Answer: Sealed syringe; change volume in steps, moving the plunger slowly and waiting so temperature stays constant; record p and V each time (≥5 readings, repeated); plot p against V (curve) or calculate p × V for each pair – constant values show pV = constant.
Examiner tip: Saying HOW temperature is kept constant (move the plunger slowly / wait before reading) is what separates a Level 3 method from a list of steps.
10Grade 4–55 marks
Hot jam is poured into a glass jar. A metal lid is screwed on tightly, trapping some air above the jam.
The jar is left to cool to room temperature.
(a) What happens to the average kinetic energy of the particles of the trapped air as it cools? [1]
(b) Explain, in terms of particles, why the pressure of the trapped air decreases as it cools. [3]
(c) After the jar has cooled, the lid is harder to unscrew.
Suggest why. [1]
(a) it decreases(1)
(b) particles move more slowly; allow particles have less kinetic energy(1)
(b) (so) particles collide with the lid / walls less often / fewer collisions per second(1)
(b) (and) each collision exerts a smaller force (on the lid / walls); allow particles hit the walls with less force(1)
(c) the pressure of the air outside the jar is now greater than the pressure inside, so the lid is pushed on (more firmly); allow atmospheric pressure pushes the lid down(1)
Answer: (a) decreases (b) particles move more slowly, so they hit the walls less often and with less force (c) air pressure outside is greater than inside, pushing the lid on
Examiner tip: Two separate effects earn two marks in (b): fewer collisions per second AND less force per collision – many students only give one.
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