4.3 Particle model of matter — revision question pack

8 specification points · notes, questions, answers and worked methods

Checked against AQA 8463 section 4.3. Review basis: the qualification registry sourced from the AQA GCSE Physics (8463) specification; registry verification recorded 17 July 2026.

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4.3.1.1 · Density of materials

Explanation

Density is mass per unit volume: $\rho=m/V$ , with SI unit $\text{kg/m}^3$ .
Measure a regular solid's dimensions to calculate its volume, find an irregular solid's volume by displacement, and find a liquid's mass by subtracting the empty container's mass.
In the particle model, solids and liquids are usually denser than gases because their particles are much closer together; density also depends on particle mass and arrangement.
A common error is to mix units such as grams and cubic metres; convert both mass and volume into a compatible unit system before dividing.

Worked example

A sample has mass $0.54\,\text{kg}$ and volume $2.0\times10^{-4}\,\text{m}^3$ . Calculate its density.

1.Use $\rho=m/V$ . Thus $\rho=0.54/(2.0\times10^{-4})=2700\,\text{kg/m}^3=2.7\times10^3\,\text{kg/m}^3$ .

Answer: $2.7\times10^3\,\text{kg/m}^3$

Common mistakes

Don't mix units such as grams and cubic metres; convert both mass and volume into a compatible unit system before dividing.
Don't fall into the trap of using the dimensions of a regular object without first calculating its volume.

Exam tip

For density, show how volume was found before using $\rho=m/V$ .

Tier 1 · Easy

A fixed mass of a substance changes from a solid to a gas. Explain why the density of the substance decreases.
[2 marks]
Total for this question: 2
Three samples have the same volume. Their masses are $42\,\text{g}$ , $57\,\text{g}$ and $51\,\text{g}$ . Which sample has the greatest density? Explain your choice without calculating all three densities.
[2 marks]
Total for this question: 2

Tier 2 · Standard

An irregular mineral has mass $0.390\,\text{kg}$ . When fully submerged, it displaces $1.50\times10^2\,\text{cm}^3$ of water. Calculate its density in $\text{kg/m}^3$ .
[4 marks]
Total for this question: 4
A student measures the volume of the same irregular pebble four times by water displacement: $18.4$ , $18.6$ , $23.1$ and $18.5\,\text{cm}^3$ . The pebble's mass is $49.95\,\text{g}$ . Identify the anomalous volume, calculate the mean of the remaining results, then determine the pebble's density in $\text{g/cm}^3$ .
[4 marks]
Total for this question: 4
A rectangular block of an unknown material has dimensions $6.0\,\text{cm}$ by $4.0\,\text{cm}$ by $2.5\,\text{cm}$ and mass $162\,\text{g}$ . Candidate material X has density $2.70\,\text{g/cm}^3$ and candidate material Y has density $4.50\,\text{g/cm}^3$ . Calculate the block's density and identify its material.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Describe how to determine the densities of a rectangular metal block, a small irregular stone and a liquid. Name suitable apparatus and state the calculation used in each case.
[6 marks]
Total for this question: 6
A component has density $7.80\times10^3\,\text{kg/m}^3$ and volume $4.50\,\text{cm}^3$ . Its packaging allows a maximum mass of $40.0\,\text{g}$ . Calculate the component's mass in grams and decide whether it meets the limit.
[5 marks]
Total for this question: 5
A sealed component consists only of an insert and a casing. Its total volume is $80.0\,\text{cm}^3$ and its total mass is $440\,\text{g}$ . The insert has mass $90.0\,\text{g}$ and density $3.00\,\text{g/cm}^3$ . Calculate the density of the casing.
[5 marks]
Total for this question: 5
A rectangular casting has external dimensions $12.0\,\text{cm}$ by $8.0\,\text{cm}$ by $5.0\,\text{cm}$ . It is made from a material of density $7.50\,\text{g/cm}^3$ but contains a sealed cavity. The casting's mass is $3.24\,\text{kg}$ . Calculate the cavity's volume and the percentage of the casting's external volume occupied by the cavity.
[5 marks]
Total for this question: 5
A container is filled with different volumes of the same liquid. The total masses are $74.0\,\text{g}$ at $20.0\,\text{cm}^3$ , $101.0\,\text{g}$ at $50.0\,\text{cm}^3$ and $128.0\,\text{g}$ at $80.0\,\text{cm}^3$ . Use changes between the readings to determine the liquid's density. Then calculate the empty container's mass and the total mass when it holds $65.0\,\text{cm}^3$ of liquid.
[5 marks]
Total for this question: 5

4.3.1.2 · Changes of state

Explanation

The state changes are melting, freezing, boiling, evaporation, condensation and sublimation.
Use particle arrangements and motion to describe the change, while keeping the number and type of particles unchanged.
Mass is conserved during a change of state in a closed system because no particles are created or destroyed.
The particles gain or lose energy as their arrangement changes.
A common error is to describe a state change as a chemical reaction; it is physical because reversing it restores the material's original properties.

Worked example

Name the changes of state from gas to liquid and from solid directly to gas.

1.Follow the direction of each change: condensation brings gas particles into the liquid state, while sublimation bypasses the liquid state and takes a solid directly to a gas.

Answer: Gas to liquid: condensation. Solid to gas: sublimation.

Common mistakes

Don't describe a state change as a chemical reaction; it is physical because reversing it restores the material's original properties.
Don't fall into the trap of saying particles themselves expand when a substance changes state.

Exam tip

For particle-model explanations, compare arrangement, motion and separation before and after the change.

Tier 1 · Easy

Name the change of state when a liquid becomes a solid.
[1 mark]
Total for this question: 1
A pure substance is fully liquid at $60\,{}^\circ\text{C}$ . On further heating its temperature rises to $78\,{}^\circ\text{C}$ and stays there while some liquid remains. Name the change of state at $78\,{}^\circ\text{C}$ and state the substance's state once this stage is complete.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A sealed container holds $75.0\,\text{g}$ of ice. The ice melts completely. State the mass of water formed and explain why it is unchanged.
[3 marks]
Total for this question: 3
A pure substance is heated at a steady rate. Its melting point is $55\,{}^\circ\text{C}$ . The time-temperature readings are: $0\,\text{min}$ , $43\,{}^\circ\text{C}$ ; $1\,\text{min}$ , $47\,{}^\circ\text{C}$ ; $2\,\text{min}$ , $51\,{}^\circ\text{C}$ ; $3\,\text{min}$ , $55\,{}^\circ\text{C}$ ; $4\,\text{min}$ , $55\,{}^\circ\text{C}$ ; $5\,\text{min}$ , $55\,{}^\circ\text{C}$ ; $6\,\text{min}$ , $55\,{}^\circ\text{C}$ ; $7\,\text{min}$ , $55\,{}^\circ\text{C}$ ; $8\,\text{min}$ , $57\,{}^\circ\text{C}$ ; $9\,\text{min}$ , $59\,{}^\circ\text{C}$ ; $10\,\text{min}$ , $61\,{}^\circ\text{C}$ ; $11\,\text{min}$ , $63\,{}^\circ\text{C}$ . Identify the melting interval, state what happens to internal energy during this interval, and explain why the temperature stays constant.
[4 marks]
Total for this question: 4
Liquid in vessel A forms bubbles throughout the liquid at a fixed temperature. Liquid in vessel B slowly changes to gas only at its surface while below its boiling point. Name each process and state one difference between them.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A solid air freshener gradually forms a gas and later deposits as solid crystals on a cold surface. Explain why these are physical changes and describe the particle arrangement before and after each change.
[4 marks]
Total for this question: 4
Two equal water samples are boiled for five minutes. The measured mass of an uncovered beaker and its contents falls, whereas the total mass of a flask fitted with a condenser that returns the water to the flask is unchanged. Explain both observations, why mass has not been destroyed, and why the changes are physical.
[5 marks]
Total for this question: 5
A powder becomes a liquid when heated and forms a differently shaped solid when cooled. A student claims that the new shape proves a chemical change occurred. Describe the evidence needed to decide whether the changes were physical. Explain why an unchanged mass in a sealed container would not, by itself, prove that the changes were physical.
[4 marks]
Total for this question: 4
A sealed vessel initially contains only $90\,\text{g}$ of ice and $30\,\text{g}$ of liquid water. The ice melts completely. The vessel is then warmed, staying below the boiling point of water, until $12\,\text{g}$ of the water has become water vapour, but no water leaves the vessel. Calculate the final masses of liquid water and water vapour. Name the two changes of state and explain why the total mass of water is unchanged.
[5 marks]
Total for this question: 5
Describe an experiment to show that mass is conserved when ice melts. Include the apparatus and measurements, how to prevent water entering or leaving the measured system, the expected result and why melting is a physical change.
[6 marks]
Total for this question: 6

4.3.2.1 · Internal energy

Explanation

Internal energy is the total kinetic energy and potential energy of all the particles in a system.
When a system is heated, track whether the supplied energy raises particle kinetic energy and temperature or changes particle potential energy during a state change.
Within one state, a higher temperature means a greater average particle kinetic energy and therefore usually a greater internal energy for the same sample.
A common error is to say that temperature always rises when internal energy increases; during a change of state, internal energy changes while temperature stays constant.

Worked example

Define the internal energy of a system.

1.Include both microscopic stores in the definition: energy from particle motion is kinetic, and energy from particle positions or interactions is potential.

Answer: The total kinetic energy and potential energy of all the particles in the system.

Common mistakes

Don't say that temperature always rises when internal energy increases; during a change of state, internal energy changes while temperature stays constant.
Don't fall into the trap of confusing internal energy with temperature alone.

Exam tip

Distinguish temperature, which tracks average kinetic energy, from total internal energy.

Tier 1 · Easy

Two sealed bottles contain equal masses of water. The water in bottle A is at $20\,{}^\circ\text{C}$ and the water in bottle B is at $60\,{}^\circ\text{C}$ . State which bottle of water has the greater internal energy. Give a reason for your answer.
[2 marks]
Total for this question: 2
A student says that a small cup at $80\,{}^\circ\text{C}$ must have more internal energy than a bath at $35\,{}^\circ\text{C}$ . Explain why temperature alone is not enough to support this claim.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A solid is heated but does not melt. Explain how its particles and internal energy change.
[3 marks]
Total for this question: 3
Samples A and B receive equal amounts of energy and transfer no energy to the surroundings. The observations are: A melts while its temperature remains constant; B warms with no change of state. Determine which sample has the greater increase in the potential-energy part of its internal energy. Explain your answer.
[3 marks]
Total for this question: 3
Samples A and B are two different substances, each containing the same number of particles, in the same state. In sample A, the particles have total kinetic energy $420\,\text{J}$ and total potential energy $680\,\text{J}$ . In sample B, the values are $500\,\text{J}$ and $570\,\text{J}$ . Calculate the internal energy of each sample. State which sample has the higher temperature and explain why.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A pure solid is heated at a steady rate. Its temperature rises, remains constant while it melts, then rises again. Explain the changes in kinetic energy, potential energy and internal energy during all three stages.
[5 marks]
Total for this question: 5
A substance cools through two intervals, with no energy transfers other than those stated. In interval X, $6\,\text{kJ}$ leaves while its temperature falls from $90\,{}^\circ\text{C}$ to $70\,{}^\circ\text{C}$ with no state change. In interval Y, $18\,\text{kJ}$ leaves while gas and liquid coexist at a constant $70\,{}^\circ\text{C}$ . Compare the changes in average kinetic energy, particle potential energy and total internal energy during X and Y.
[5 marks]
Total for this question: 5
A sealed sample of fixed mass is heated. Before heating, its particles have total kinetic energy $260\,\text{J}$ and total potential energy $410\,\text{J}$ . Afterwards, the values are $260\,\text{J}$ and $760\,\text{J}$ . During heating, $55\,\text{J}$ is transferred from the sample to the surroundings. Calculate the energy supplied to the sample and use the energy changes to describe what happened to its temperature and state.
[5 marks]
Total for this question: 5
Samples P and Q are the same substance in the same state. P contains $2.0\times10^{23}$ particles with total kinetic energy $5.0\times10^2\,\text{J}$ and total potential energy $7.0\times10^2\,\text{J}$ . Q contains $5.0\times10^{23}$ particles with total kinetic energy $1.0\times10^3\,\text{J}$ and total potential energy $1.75\times10^3\,\text{J}$ . Calculate each sample's internal energy and average kinetic energy per particle. Determine which sample has the higher temperature.
[5 marks]
Total for this question: 5
A hot metal block and a cool liquid are placed in an insulated container. No state changes occur. The metal's internal energy decreases by $7.4\,\text{kJ}$ while the container's internal energy increases by $0.9\,\text{kJ}$ . Calculate the change in the liquid's internal energy. Describe the changes in average particle kinetic energy and explain when the energy transfer stops.
[5 marks]
Total for this question: 5

4.3.2.2 · Temperature changes in a system and specific heat capacity

Explanation

For a temperature change without a change of state, $\Delta E=mc\Delta\theta$ links energy change, mass, specific heat capacity and temperature change.
Calculate $\Delta\theta=\theta_{\text{final}}-\theta_{\text{initial}}$ , convert mass to kilograms and rearrange the equation algebraically before inserting values.
For the same energy input, a larger mass or larger specific heat capacity gives a smaller temperature rise.
Quote the final energy in joules.
A common error is to confuse specific heat capacity in $\text{J/(kg}\,{}^\circ\text{C)}$ with specific latent heat in $\text{J/kg}$ ; the former applies when temperature changes.

Worked example

State the meaning of specific heat capacity and give its unit.

1.State the fixed mass and fixed temperature rise, then attach the compound unit: joules per kilogram per degree Celsius.

Answer: The energy required to raise the temperature of $1\,\text{kg}$ of a substance by $1\,{}^\circ\text{C}$ . $\text{J/(kg}\,{}^\circ\text{C)}$

Common mistakes

Don't confuse specific heat capacity in $\text{J/(kg}\,{}^\circ\text{C)}$ with specific latent heat in $\text{J/kg}$ ; the former applies when temperature changes.
Don't fall into the trap of using final temperature instead of temperature change in the heating equation.

Exam tip

For $\Delta E=mc\Delta\theta$ , calculate final minus initial temperature first.

Tier 1 · Easy

The base of a saucepan is made from a material with a low specific heat capacity. Explain why this helps the base of the saucepan heat up quickly.
[2 marks]
Total for this question: 2
Equal masses of materials P, Q and R each receive $12\,\text{kJ}$ . Their temperature rises are $10\,{}^\circ\text{C}$ , $18\,{}^\circ\text{C}$ and $7\,{}^\circ\text{C}$ respectively. Which material has the greatest specific heat capacity? Explain your choice.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A $2.0\,\text{kg}$ metal block warms by $35\,{}^\circ\text{C}$ . Its specific heat capacity is $450\,\text{J/(kg}\,{}^\circ\text{C)}$ . Calculate the increase in its thermal energy store.
[2 marks]
Total for this question: 2
A student investigating the specific heat capacity of a block repeats the specific-heat-capacity experiment on the same block. The results are: trial 1 - initial temperature $20\,{}^\circ\text{C}$ , calculated specific heat capacity $910\,\text{J/(kg}\,{}^\circ\text{C)}$ ; trial 2 - $21\,{}^\circ\text{C}$ , $895\,\text{J/(kg}\,{}^\circ\text{C)}$ ; trial 3 - $58\,{}^\circ\text{C}$ , $1300\,\text{J/(kg}\,{}^\circ\text{C)}$ ; trial 4 - $22\,{}^\circ\text{C}$ , $905\,\text{J/(kg}\,{}^\circ\text{C)}$ . The same electrical energy is supplied in each trial and all other conditions were intended to be kept the same. Identify the control variable that was not held constant and explain why this makes the calculated specific heat capacity in trial 3 too high.
[4 marks]
Total for this question: 4
A $0.400\,\text{kg}$ sample stays in one state while it is heated. The energy supplied and temperature readings are: $0\,\text{kJ}$ at $18\,{}^\circ\text{C}$ , $4.8\,\text{kJ}$ at $28\,{}^\circ\text{C}$ and $9.6\,\text{kJ}$ at $38\,{}^\circ\text{C}$ . Use the widest interval to calculate the sample's specific heat capacity.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Two insulated blocks each receive $54\,\text{kJ}$ . Block A has mass $1.5\,\text{kg}$ and specific heat capacity $900\,\text{J/(kg}\,{}^\circ\text{C)}$ . Block B has mass $0.75\,\text{kg}$ and specific heat capacity $450\,\text{J/(kg}\,{}^\circ\text{C)}$ . Calculate both temperature rises and explain the difference.
[5 marks]
Total for this question: 5
A heater transfers $60.0\,\text{kJ}$ while a block warms from $22\,{}^\circ\text{C}$ to $82\,{}^\circ\text{C}$ . During heating, $10.8\,\text{kJ}$ is transferred to the surroundings. The block's specific heat capacity is $820\,\text{J/(kg}\,{}^\circ\text{C)}$ . Calculate the block's mass in grams. Give your answer to 2 significant figures.
[5 marks]
Total for this question: 5
A joulemeter reads $6\,000\,\text{J}$ before a heater is switched on and $30\,000\,\text{J}$ after it heats a $0.600\,\text{kg}$ block. Exactly $75\%$ of the input energy reaches the block, producing a temperature rise of $24.0\,{}^\circ\text{C}$ . Use the Physics Equations Sheet to calculate the block's specific heat capacity. Candidate materials have specific heat capacities A: $900$ , B: $1250$ and C: $1600\,\text{J/(kg}\,{}^\circ\text{C)}$ . Identify the block's material.
[5 marks]
Total for this question: 5
A $0.200\,\text{kg}$ aluminium block at $90.0\,{}^\circ\text{C}$ is placed in $0.300\,\text{kg}$ of water at $20.0\,{}^\circ\text{C}$ in an insulated container. Calculate the final temperature. Use $c_{\text{aluminium}}=900\,\text{J/(kg}\,{}^\circ\text{C)}$ and $c_{\text{water}}=4200\,\text{J/(kg}\,{}^\circ\text{C)}$ .
[5 marks]
Total for this question: 5
A heat store must release at least $1.8\,\text{MJ}$ as it cools from $80.0\,{}^\circ\text{C}$ to $20.0\,{}^\circ\text{C}$ , and its mass must not exceed $30.0\,\text{kg}$ . Material R has $c=900\,\text{J/(kg}\,{}^\circ\text{C)}$ and material S has $c=2400\,\text{J/(kg}\,{}^\circ\text{C)}$ . Calculate the minimum mass of each material and state which meet the mass limit.
[6 marks]
Total for this question: 6

4.3.2.3 · Changes of state and specific latent heat

Explanation

Specific latent heat is the energy needed to change the state of $1\,\text{kg}$ of a substance with no temperature change, using $E=mL$ .
Use specific latent heat of fusion for solid-liquid changes and specific latent heat of vaporisation for liquid-vapour changes.
A flat section on a heating or cooling graph marks a state change: energy changes particle potential energy while average kinetic energy and temperature remain constant.
A common error is to apply $E=mc\Delta\theta$ across a state-change plateau; use $E=mL$ for that stage and calculate any temperature-changing stages separately.

A heating curve with constant-temperature melting and boiling plateaus.

Worked example

Define specific latent heat.

1.A complete definition must include the energy, the fixed mass of $1\,\text{kg}$ , the change of state and the absence of a temperature change.

Answer: The energy required to change the state of $1\,\text{kg}$ of a substance with no change in temperature.

Common mistakes

Don't apply $E=mc\Delta\theta$ across a state-change plateau; use $E=mL$ for that stage and calculate any temperature-changing stages separately.
Don't fall into the trap of saying temperature rises throughout a change of state.

Exam tip

On a heating graph, explain a flat section using energy increasing potential rather than kinetic energy.

Tier 1 · Easy

Ice at its melting point is changing into water. State which equation should be used to calculate the energy transferred during the change of state: $\Delta E=mc\Delta\theta$ or $E=mL$ . Give a reason for your answer.
[2 marks]
Total for this question: 2
A heating graph for one substance has flat sections during melting and boiling. The same constant-power heater is used, and the boiling section lasts four times as long as the melting section. What does this show about the two specific latent heats? Explain.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A $0.28\,\text{kg}$ frozen material is already at its melting point. Calculate the transfer required to melt it completely, given $L_f=3.10\times10^5\,\text{J/kg}$ .
[2 marks]
Total for this question: 2
On a heating graph, the energy reading is $18\,\text{kJ}$ at the start of a melting section and $126\,\text{kJ}$ at its end. The sample mass is $0.300\,\text{kg}$ . Calculate its specific latent heat of fusion in $\text{J/kg}$ .
[4 marks]
Total for this question: 4
A solid melts above room temperature. A student starts with the solid at its melting point. The completed results table is: joulemeter reading before heating $=8.0\,\text{kJ}$ ; joulemeter reading after heating $=44.0\,\text{kJ}$ ; heater energy transferred by subtraction $=36.0\,\text{kJ}$ ; mass of solid before heating $=0.250\,\text{kg}$ ; mass of dry solid remaining afterwards $=0.130\,\text{kg}$ ; mass melted by subtraction $=0.120\,\text{kg}$ . Determine the substance's specific latent heat of fusion, then calculate the additional energy needed to melt the remaining $0.130\,\text{kg}$ of solid.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A $0.15\,\text{kg}$ sample of water is heated from $20\,{}^\circ\text{C}$ to $100\,{}^\circ\text{C}$ and then completely vaporised at $100\,{}^\circ\text{C}$ . Calculate the total energy supplied. Use $c=4200\,\text{J/(kg}\,{}^\circ\text{C)}$ and $L_v=2.26\times10^6\,\text{J/kg}$ .
[5 marks]
Total for this question: 5
A $0.200\,\text{kg}$ solid is heated from $15.0\,{}^\circ\text{C}$ to its melting point of $65.0\,{}^\circ\text{C}$ and then completely melted. The total energy supplied is $74.0\,\text{kJ}$ and the solid's specific heat capacity is $5.00\times10^2\,\text{J/(kg}\,{}^\circ\text{C)}$ . Calculate its specific latent heat of fusion.
[5 marks]
Total for this question: 5
Use the Physics Equations Sheet. A $90.0\,\text{W}$ heater operates for $4.0\,\text{minutes}$ while a solid melts at constant temperature. During this interval, $20\%$ of the heater's energy is transferred to the surroundings and $0.0960\,\text{kg}$ of solid melts. Calculate the material's specific latent heat of fusion.
[5 marks]
Total for this question: 5
Material P has specific latent heat of fusion $2.4\times10^5\,\text{J/kg}$ and a mass of $0.150\,\text{kg}$ . Material Q has specific latent heat of fusion $1.5\times10^5\,\text{J/kg}$ and a mass of $0.250\,\text{kg}$ . Both are at their melting points. Calculate the energy needed to melt each material completely. Determine which needs more energy and explain why the smaller latent heat does not determine the result by itself.
[4 marks]
Total for this question: 4
An unnamed phase-change material freezes at constant temperature. Its mass is $0.120\,\text{kg}$ and $L_f=2.24\times10^5\,\text{J/kg}$ . During freezing, $75\%$ of the released energy warms $0.400\,\text{kg}$ of water initially at $20.0\,{}^\circ\text{C}$ . Calculate the water's final temperature. Use $c_{\text{water}}=4200\,\text{J/(kg}\,{}^\circ\text{C)}$ .
[5 marks]
Total for this question: 5

4.3.3.1 · Particle motion in gases

Explanation

Gas molecules are in constant random motion, and gas temperature is related to their average kinetic energy.
Explain gas pressure through molecules colliding with container walls and changing momentum, which exerts a force on the walls.
At constant volume, heating increases average molecular speed, making collisions more frequent and harder, so pressure increases.
This increases force per unit area.
A common error is to say that heating creates more particles or makes each particle larger; the same molecules move faster unless gas enters or leaves.

Worked example

State how the average kinetic energy of gas molecules changes when the gas temperature increases.

1.Temperature is linked to the average kinetic energy of the molecules, so a higher temperature means a greater average kinetic energy.

Answer: The average kinetic energy increases.

Common mistakes

Don't say that heating creates more particles or makes each particle larger; the same molecules move faster unless gas enters or leaves.
Don't fall into the trap of saying gas pressure falls when molecules collide more frequently with the walls.

Exam tip

For gas pressure, link molecular motion to collision frequency and force on the container walls.

Tier 1 · Easy

State how the molecules in a gas move.
[1 mark]
Total for this question: 1
A detector counts collisions with the same small wall area for equal times. Gas P produces $820$ collisions and gas Q produces $1260$ collisions. The samples contain the same number of molecules of the same gas in equal fixed volumes. Which gas is likely to be hotter? Explain.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A sealed rigid flask of gas is heated. Explain why the gas pressure increases.
[4 marks]
Total for this question: 4
A rigid can contains gas at constant temperature. Gas leaks slowly through a faulty valve and the pressure falls. Explain why the pressure falls in terms of molecule number, collisions and force per unit area.
[4 marks]
Total for this question: 4
Identical pressure sensors cover equal areas on the top, bottom and sides of a sealed cube containing a gas. At a uniform temperature, the sensors record approximately the same pressure. Explain this observation using molecular motion and collisions.
[4 marks]
Total for this question: 4

Tier 3 · Hard

Two identical sealed rigid containers hold the same number of molecules of the same gas. Gas A is at a higher temperature than gas B. Compare the molecular motion and pressures, and explain why the comparison would be less certain if the containers had different volumes.
[5 marks]
Total for this question: 5
Equal amounts of the same gas begin at the same temperature and pressure in a rigid can and a flexible balloon. Both receive the same temperature increase. The can's pressure rises by $35\,\text{kPa}$ , but the balloon expands and its pressure rises by only $4\,\text{kPa}$ . Explain these observations using average kinetic energy, wall collisions and volume.
[5 marks]
Total for this question: 5
Use the Physics Equations Sheet. A sealed rigid container has a wall sensor of area $0.020\,\text{m}^2$ . Before heating, the mean force on the sensor is $60\,\text{N}$ ; after heating, it is $78\,\text{N}$ . Calculate the pressure before and after heating and the pressure increase. Explain the increase using the motion of the gas molecules.
[5 marks]
Total for this question: 5
Two identical sealed rigid tanks contain the same gas under the same initial conditions. Tank A is heated without changing the number of molecules. Extra molecules are pumped into tank B while its temperature stays constant. In both tanks the pressure increases. Compare the molecular reasons for the two pressure increases, referring to average kinetic energy, collision frequency and force on the walls.
[5 marks]
Total for this question: 5
A sealed rigid container of gas is heated until its pressure rises. While the gas is kept at the higher temperature, a valve is opened briefly until the pressure returns to its initial value, then closed. Compare the final gas with the initial gas in terms of average molecular kinetic energy, number of molecules and wall collisions. Explain why the pressure can be the same.
[5 marks]
Total for this question: 5

4.3.3.2 · Pressure in gases (physics only)

Explanation

Gas pressure produces a net force at right angles to a container wall or any other surface.
For a fixed mass of gas at constant temperature, use $pV=\text{constant}$ , so $p_1V_1=p_2V_2$ .
Increasing volume at constant temperature makes wall collisions less frequent per unit area, so pressure decreases.
Use consistent pressure and volume units on both sides of the equation, though the units need not be SI if they match.
A common error is to use $p_1/V_1=p_2/V_2$ ; pressure and volume are inversely proportional, so their product stays constant.

Worked example

A fixed mass of gas is kept at constant temperature while its volume doubles. State what happens to its pressure.

1.At constant temperature $pV$ is constant. If $V$ is multiplied by $2$ , $p$ must be multiplied by $1/2$ to keep the product unchanged.

Answer: The pressure halves.

Common mistakes

Don't use $p_1/V_1=p_2/V_2$ ; pressure and volume are inversely proportional, so their product stays constant.
Don't fall into the trap of treating atmospheric pressure as part of a sealed gas's volume.

Exam tip

State whether pressure, volume or temperature is controlled before predicting a gas change.

Tier 1 · Easy

A gas exerts pressure on the wall of its container. State the direction of the force produced by the gas pressure.
[1 mark]
Total for this question: 1
A fixed mass of gas at constant temperature occupies $120\,\text{cm}^3$ at $180\,\text{kPa}$ and $240\,\text{cm}^3$ at $90\,\text{kPa}$ . State the relationship between pressure and volume and show one numerical check using the data.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A gas occupies $4.0\times10^{-3}\,\text{m}^3$ at a pressure of $1.2\times10^5\,\text{Pa}$ . It is compressed at constant temperature to $1.5\times10^{-3}\,\text{m}^3$ . Calculate the new pressure.
[3 marks]
Total for this question: 3
A fixed mass of gas is tested at constant temperature. The volume-pressure pairs are $80\,\text{cm}^3$ and $300\,\text{kPa}$ , $100\,\text{cm}^3$ and $240\,\text{kPa}$ , $125\,\text{cm}^3$ and $182\,\text{kPa}$ , and $160\,\text{cm}^3$ and $150\,\text{kPa}$ . Identify the anomalous pair and calculate the pressure that should replace its recorded value.
[4 marks]
Total for this question: 4
A frictionless horizontal sliding piston traps a fixed mass of gas. Initially the gas pressure is greater than the external pressure, and the gas remains at constant temperature as the piston moves. Explain the piston's motion, the resulting pressure change and the condition at which the piston stops accelerating.
[4 marks]
Total for this question: 4

Tier 3 · Hard

A syringe contains $3.20\times10^2\,\text{cm}^3$ of gas at $2.40\times10^2\,\text{kPa}$ . The outlet is sealed and the gas remains at constant temperature while the pressure rises to $3.60\times10^2\,\text{kPa}$ . Calculate the final volume and the decrease in volume. Explain the pressure rise using particles.
[5 marks]
Total for this question: 5
A sealed cylinder contains gas at $96.0\,\text{kPa}$ and $750\,\text{cm}^3$ . The gas stays at constant temperature and must not exceed $250\,\text{kPa}$ . Calculate the minimum permitted volume in $\text{cm}^3$ and $\text{m}^3$ . Decide whether compressing it to $2.80\times10^{-4}\,\text{m}^3$ is safe.
[5 marks]
Total for this question: 5
Use the Physics Equations Sheet. A fixed mass of gas is held at constant temperature. In its final state, it exerts a force of $360\,\text{N}$ on a piston of area $15.0\,\text{cm}^2$ and occupies $300\,\text{cm}^3$ . Its initial pressure was $90.0\,\text{kPa}$ . Calculate the initial volume of the gas in $\text{cm}^3$ .
[5 marks]
Total for this question: 5
Describe an experiment to investigate how pressure depends on volume for gas sealed in a syringe while its temperature is kept constant. Include the apparatus, the measurements, how temperature and gas mass are controlled, and how the results should be analysed.
[6 marks]
Total for this question: 6
A gas occupies a syringe and an attached pressure-sensor tube. The tube has an unknown fixed volume $x$ . At constant temperature, the pressure is $120\,\text{kPa}$ when the syringe reads $60\,\text{cm}^3$ and $200\,\text{kPa}$ when it reads $30\,\text{cm}^3$ . Calculate $x$ . Then calculate the pressure when the syringe reads $15\,\text{cm}^3$ .
[5 marks]
Total for this question: 5

4.3.3.3 · Increasing the pressure of a gas (physics only) (HT only)

Explanation

Work is an energy transfer by a force; doing work on an enclosed gas transfers energy to its internal energy store.
Identify the force and displacement during compression, calculate work with $W=Fs$ when appropriate, and account for any energy transferred to the surroundings.
Rapid compression can raise gas temperature because increased internal energy gives the molecules greater average kinetic energy.
A common error is to attribute warming only to friction in the pump; compression itself involves work being done on the gas.

Worked example

Explain why the air in a sealed bicycle pump can become warmer when the handle is pushed in quickly.

1.Track the energy transfer: the applied force moves the handle, so mechanical work is done on the gas. The transferred energy raises the gas's internal energy; rapid compression leaves little time for transfer to the surroundings, so its temperature rises.

Answer: The handle does work on the enclosed gas. This increases the gas's internal energy and can increase its temperature.

Common mistakes

Don't attribute warming only to friction in the pump; compression itself involves work being done on the gas.
Don't fall into the trap of applying $pV=\text{constant}$ while temperature is changing.

Exam tip

Higher tier: use matching pressure and volume units on both sides of $p_1V_1=p_2V_2$ .

Tier 1 · Easy

A piston compresses an enclosed gas. State what is meant by work and state what happens to the internal energy of the gas when work is done on it.
[2 marks]
Total for this question: 2
During one compression, a gas receives $64\,\text{J}$ by work and transfers $19\,\text{J}$ to the surroundings. Determine the change in the gas's internal energy.
[2 marks]
Total for this question: 2

Tier 2 · Standard

$180\,\text{J}$ of work is done on an enclosed gas, with negligible energy transfer to the surroundings. State the change in internal energy and explain the likely temperature change.
[3 marks]
Total for this question: 3
Four gas samples are compressed. In every row the gas is reported to have increased internal energy, and no energy transfers occur except the work done on the gas and energy transferred from it. The table gives: $\begin{array}{c|cc}\text{row}&\text{work done on gas / J}&\text{energy transferred out / J}\\ \hline A&84&31\\ B&72&50\\ C&65&69\\ D&91&44\end{array}$ . Determine which row is impossible and justify your answer using conservation of energy.
[4 marks]
Total for this question: 4
The same amount of work is done when identical enclosed gas samples are compressed from the same initial conditions. One compression is rapid and the other is slow. The rapidly compressed gas reaches a higher temperature. Explain why.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A piston exerts a constant force of $250\,\text{N}$ while moving $0.18\,\text{m}$ into a sealed cylinder. During compression, $12\,\text{J}$ is transferred from the gas to the surroundings. Calculate the increase in the gas's internal energy and explain the effect on its temperature.
[4 marks]
Total for this question: 4
After compression, a gas's internal energy has increased by $72\,\text{J}$ and $18\,\text{J}$ has been transferred to the surroundings. A piston moved $0.150\,\text{m}$ under a constant force. Calculate the force on the piston and explain why the gas may become warmer.
[5 marks]
Total for this question: 5
Use the Physics Equations Sheet. A $5.0\,\text{kg}$ mass descends $0.60\,\text{m}$ and drives a mechanism that compresses an enclosed gas. The mass is at rest before and after the descent, and the mechanism transfers all of this energy. Take $g=9.8\,\text{N/kg}$ . Of the energy transferred by the mechanism, $5.4\,\text{J}$ reaches the surroundings and all the rest increases the gas's internal energy. Calculate the internal-energy increase, then link it to a possible temperature rise.
[4 marks]
Total for this question: 4
Use the Physics Equations Sheet. During compression, the average pressure exerted by a piston on the gas is $180\,\text{kPa}$ . The piston area is $8.0\,\text{cm}^2$ and it moves $0.120\,\text{m}$ . Calculate the work done on the gas. The gas transfers $3.3\,\text{J}$ to the surroundings; calculate its increase in internal energy and explain a possible temperature change.
[6 marks]
Total for this question: 6
Describe an experiment that provides evidence that doing work by compressing a gas can raise the gas temperature. Include the measurements, a comparison that helps distinguish compression heating from frictional heating, the variables held constant and the particle explanation for the result.
[6 marks]
Total for this question: 6

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

4.3.1.1 · Density of materials

Tier 1 · Easy

Mark scheme for 4.3.1.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	The particles become much farther apart, so the same mass occupies a greater volume.	The mass is unchanged, but gas particles are much farther apart than solid particles. The volume therefore increases, so $\rho=m/V$ decreases.	2
Total Question 1			2
02.1	The $57\,\text{g}$ sample has the greatest density because it has the greatest mass in the same volume.	Density is mass divided by volume. Since all three volumes are equal, the sample with the greatest mass has the greatest density, so choose the $57\,\text{g}$ sample.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.3.1.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$2.60\times10^3\,\text{kg/m}^3$	The mineral's volume equals the displaced volume. Since $1\,\text{cm}^3=10^{-6}\,\text{m}^3$ , $1.50\times10^2\,\text{cm}^3=1.50\times10^{-4}\,\text{m}^3$ . Therefore $\rho=0.390/(1.50\times10^{-4})=2.60\times10^3\,\text{kg/m}^3$ .	4
Total Question 1			4
02.1	Anomalous volume: $23.1\,\text{cm}^3$ Mean volume: $18.5\,\text{cm}^3$ Density: $2.70\,\text{g/cm}^3$	The $23.1\,\text{cm}^3$ result is far from the other three. Excluding it, the mean volume is $(18.4+18.6+18.5)/3=18.5\,\text{cm}^3$ . Then $\rho=m/V=49.95/18.5=2.70\,\text{g/cm}^3$ .	4
Total Question 2			4
03.1	Volume $=60\,\text{cm}^3$ Density $=2.70\,\text{g/cm}^3$ The block is made from material X.	The volume is $V=6.0\times4.0\times2.5=60\,\text{cm}^3$ . Using $\rho=m/V$ , $\rho=162/60=2.70\,\text{g/cm}^3$ . This matches the stated density of material X, not material Y.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.3.1.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Measure each mass with a balance. For the block, measure its three dimensions with suitable length instruments and calculate $V=lwh$ . For the stone, measure the volume of water displaced using a measuring cylinder or displacement can. For the liquid, measure a known volume and subtract the empty container's mass from the filled container's mass. Calculate each density using $\rho=m/V$ with compatible units.	Zero a balance and measure the block's mass. Measure its length, width and height with a ruler, Vernier callipers or micrometer as appropriate, then use $V=lwh$ . Measure the stone's mass, fully submerge it without trapped air and take the rise in measuring-cylinder volume as its volume. For the liquid, record the masses of an empty measuring vessel and of the vessel holding a measured volume; subtract to obtain liquid mass. For all three, divide mass by volume after converting units consistently.	6
Total Question 1			6
02.1	Mass $=35.1\,\text{g}$ It meets the limit because $35.1\,\text{g}<40.0\,\text{g}$ .	Convert the volume explicitly: $4.50\,\text{cm}^3=4.50\times10^{-6}\,\text{m}^3$ . Rearrange $\rho=m/V$ to $m=\rho V$ , giving $m=(7.80\times10^3)(4.50\times10^{-6})=0.0351\,\text{kg}$ . Convert to grams: $0.0351\,\text{kg}=35.1\,\text{g}$ , which is $4.9\,\text{g}$ below the limit.	5
Total Question 2			5
03.1	Insert volume $=30.0\,\text{cm}^3$ Casing volume $=50.0\,\text{cm}^3$ Casing mass $=350\,\text{g}$ Casing density $=7.00\,\text{g/cm}^3$	Rearrange $\rho=m/V$ for the insert: $V=m/\rho=90.0/3.00=30.0\,\text{cm}^3$ . The casing occupies $80.0-30.0=50.0\,\text{cm}^3$ and has mass $440-90.0=350\,\text{g}$ . Therefore its density is $350/50.0=7.00\,\text{g/cm}^3$ .	5
Total Question 3			5
04.1	External volume $=480\,\text{cm}^3$ Volume of material $=432\,\text{cm}^3$ Cavity volume $=48.0\,\text{cm}^3$ The cavity occupies $10.0\%$ of the external volume.	The external volume is $12.0\times8.0\times5.0=480\,\text{cm}^3$ . Convert the mass: $3.24\,\text{kg}=3240\,\text{g}$ . The material occupies $V=m/\rho=3240/7.50=432\,\text{cm}^3$ , so the cavity volume is $480-432=48.0\,\text{cm}^3$ . Its share of the external volume is $(48.0/480)\times100=10.0\%$ .	5
Total Question 4			5
05.1	Density $=0.900\,\text{g/cm}^3$ Empty container mass $=56.0\,\text{g}$ Total mass at $65.0\,\text{cm}^3=114.5\,\text{g}$	Use two widely separated readings so the unchanged container mass cancels. The liquid mass added from $20.0$ to $80.0\,\text{cm}^3$ is $128.0-74.0=54.0\,\text{g}$ for a volume increase of $60.0\,\text{cm}^3$ , so $\rho=54.0/60.0=0.900\,\text{g/cm}^3$ . At $20.0\,\text{cm}^3$ the liquid mass is $0.900\times20.0=18.0\,\text{g}$ , giving an empty-container mass of $74.0-18.0=56.0\,\text{g}$ . At $65.0\,\text{cm}^3$ , the total mass is $56.0+(0.900\times65.0)=114.5\,\text{g}$ .	5
Total Question 5			5

4.3.1.2 · Changes of state

Tier 1 · Easy

Mark scheme for 4.3.1.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Freezing.	The change from the liquid state to the solid state is freezing.	1
Total Question 1			1
02.1	The change is boiling, and the substance is a gas once the stage is complete.	The sample is fully liquid before the constant-temperature stage, so the plateau cannot be melting. Heating at constant temperature while some liquid remains identifies boiling. Once all the liquid has boiled, the sample is gaseous.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.3.1.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$75.0\,\text{g}$ Melting only rearranges the particles; the sealed container loses no particles, so mass is conserved.	Treat the container and contents as a closed system. The solid becomes liquid, but the number and type of water particles remain the same and none can escape. The total mass is therefore still $75.0\,\text{g}$ .	3
Total Question 1			3
02.1	Melting occurs from $3$ to $7\,\text{minutes}$ . The internal energy increases because energy is still being supplied. The supplied energy increases particle potential energy as attractions between particles are overcome. The particles' average kinetic energy does not increase, so the temperature stays constant.	The readings remain at the melting point of $55\,{}^\circ\text{C}$ from $3$ to $7\,\text{minutes}$ , so this is the melting interval. Energy is still supplied, increasing the particles' potential energy and therefore the sample's internal energy as attractions are overcome. Their average kinetic energy does not increase, so the temperature remains constant.	4
Total Question 2			4
03.1	Vessel A shows boiling. Vessel B shows evaporation. Boiling occurs throughout a liquid at its boiling point, whereas evaporation occurs only at the surface and can occur below the boiling point.	Bubbles forming throughout at a fixed temperature identify boiling. A surface process below the boiling point identifies evaporation. Both are liquid-to-gas changes, but they occur in different parts of the liquid and under different temperature conditions.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.3.1.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Sublimation changes the closely packed, ordered solid particles into widely spaced, randomly moving gas particles. Deposition changes the gas particles back into a closely packed, ordered solid arrangement. The substance keeps the same particles and recovers its original properties, so the changes are physical.	Identify the outward change as sublimation and the reverse as deposition. Compare arrangement and motion rather than inventing new particles: the particles separate and move randomly as a gas, then become fixed in an ordered arrangement as a solid. Since the same substance is recovered, no chemical change has occurred.	4
Total Question 1			4
02.1	Water vapour escapes from the uncovered beaker, so the measured mass of that beaker and its contents decreases. In the flask, the condenser cools the vapour and returns the liquid water, so the total measured mass remains constant. No particles are created or destroyed; they only change arrangement and motion. Boiling and condensation are reversible and recover water with the same properties, so they are physical changes.	Define the measured system in each case. The open beaker loses water particles as vapour, so its measured mass falls, but those particles still exist in the surroundings. The condenser cools the vapour and returns it to the flask, so the flask and its contents retain the water and their total mass is unchanged. The water particles remain water particles and the state changes can be reversed by condensation, which makes the changes physical.	5
Total Question 2			5
03.1	Cool the liquid to reverse the change and compare the recovered solid's characteristic properties with those of the original powder. If the original properties are recovered, the melting and freezing were physical changes. A change of shape alone does not show that a new substance formed. Mass is conserved in chemical changes as well as physical changes in a closed system, so unchanged mass alone is not decisive.	Reverse the change by cooling, then compare properties that characterise the material, such as melting point or density, rather than its shape. Recovery of the original properties supports a physical state change. A sealed system should conserve mass during either a physical or a chemical change, so mass conservation cannot distinguish the two on its own.	4
Total Question 3			4
04.1	Final liquid-water mass $=108\,\text{g}$ Final water-vapour mass $=12\,\text{g}$ The changes are melting and evaporation. The total remains $120\,\text{g}$ because the same water particles remain inside the sealed vessel.	Melting changes all $90\,\text{g}$ of ice into liquid, so there is initially $90+30=120\,\text{g}$ of liquid water. Evaporation then moves $12\,\text{g}$ into the gas state, leaving $120-12=108\,\text{g}$ as liquid and $12\,\text{g}$ as vapour. The particles change arrangement and motion but none enter or leave, so the total water mass is still $108+12=120\,\text{g}$ .	5
Total Question 4			5
05.1	Dry the outside of a piece of ice, place it in a dry container with a tight lid and measure the mass of the sealed container and contents using a balance. Allow the ice to melt without opening the container, then dry the outside and measure the mass again using the same balance. Keep the whole sealed container as the measured system so neither liquid water nor water vapour can leave and external water cannot enter. The final mass should equal the initial mass within the resolution of the balance. Repeat the measurements to check that the result is repeatable. Melting is physical because the particles remain water particles and freezing can reverse the change.	Use a balance to measure a closed system before and after the state change. A dry, tightly sealed container prevents loss of water vapour or liquid, and drying its outside prevents condensation from adding mass. Do not open it between readings; use the same balance and repeat. Equal masses within the balance resolution show that no particles have entered or left. The material remains water and can be frozen again, so no new substance is formed.	6
Total Question 5			6

4.3.2.1 · Internal energy

Tier 1 · Easy

Mark scheme for 4.3.2.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Bottle B has the greater internal energy because its particles have greater average kinetic energy.	The samples have the same mass and are the same substance in the same state. The higher temperature in bottle B means its particles have greater average kinetic energy, so the total internal energy is greater.	2
Total Question 1			2
02.1	Temperature indicates average particle kinetic energy, whereas internal energy is the total kinetic and potential energy of all the particles. The bath contains many more particles, so the temperatures alone do not determine which total internal energy is greater.	Separate an average from a total. The cup's higher temperature indicates greater average kinetic energy per particle, but internal energy totals kinetic and potential energies over all particles. The very different amounts of water mean the claim cannot be decided from temperature alone.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.3.2.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The particles vibrate more rapidly about their fixed positions. Their average kinetic energy and the solid's temperature increase. The solid's internal energy increases.	Because there is no state change, focus on particle kinetic energy. Heating makes the particles vibrate faster, raising their average kinetic energy and temperature. Since kinetic energy is part of internal energy, the system's internal energy increases.	3
Total Question 1			3
02.1	Sample A has the greater increase in particle potential energy. During melting, the energy overcomes attractions between particles and increases the potential-energy part of internal energy while average kinetic energy remains constant. For B, the temperature rise shows that average particle kinetic energy increases; with no change of state, its potential-energy part does not increase in this model.	The samples receive equal total energy, but it enters different parts of internal energy. A's constant temperature means its average particle kinetic energy is unchanged, so melting increases particle potential energy as attractions are overcome. B warms without changing state, so the transferred energy increases average kinetic energy rather than particle potential energy. Therefore A has the greater potential-energy increase.	3
Total Question 2			3
03.1	Sample A internal energy $=1100\,\text{J}$ Sample B internal energy $=1070\,\text{J}$ Sample B has the higher temperature because its equal number of particles has the greater total, and therefore greater average, kinetic energy.	Internal energy is the sum of the total particle kinetic and potential energies. For A, $U=420+680=1100\,\text{J}$ ; for B, $U=500+570=1070\,\text{J}$ . Temperature depends on average kinetic energy, not on total internal energy. The equal particle numbers mean B's greater total kinetic energy gives it the greater average kinetic energy and higher temperature.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.3.2.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Before melting, average kinetic energy, temperature and internal energy increase. During melting, average kinetic energy and temperature remain constant while particle potential energy and internal energy increase. After melting, average kinetic energy, temperature and internal energy increase again.	On each sloping section, the temperature rise shows that average particle kinetic energy is increasing, so internal energy rises. On the melting plateau, constant temperature means average kinetic energy is unchanged. The supplied energy instead separates particles against their attractions, increasing potential energy; internal energy therefore continues to rise.	5
Total Question 1			5
02.1	In X, average kinetic energy decreases as temperature falls, particle potential energy is unchanged in this model, and total internal energy decreases by $6\,\text{kJ}$ . In Y, average kinetic energy remains constant because temperature is constant. In Y, particle potential energy decreases as gas condenses to liquid, and total internal energy decreases by $18\,\text{kJ}$ .	For X, the temperature fall identifies a decrease in average kinetic energy. With no state change, particle potential energy is unchanged in the GCSE particle model, and the energy leaving reduces total internal energy. For Y, the constant temperature fixes average kinetic energy while gas and liquid coexist. Condensation brings particles closer together, reducing potential energy; the larger $18\,\text{kJ}$ transfer is therefore a reduction in total internal energy without a temperature fall.	5
Total Question 2			5
03.1	Initial internal energy $=670\,\text{J}$ and final internal energy $=1020\,\text{J}$ . The internal energy increases by $350\,\text{J}$ . Energy supplied $=405\,\text{J}$ . The unchanged kinetic energy means the temperature stays constant, while the increased potential energy shows that a change of state occurs.	Initially $U=260+410=670\,\text{J}$ and finally $U=260+760=1020\,\text{J}$ , so $\Delta U=350\,\text{J}$ . The supply must provide this increase and the $55\,\text{J}$ transferred out: $E_{\text{supplied}}=350+55=405\,\text{J}$ . Because total kinetic energy is unchanged for the fixed sample, average kinetic energy and temperature stay constant. The rise in particle potential energy is evidence of a state change.	5
Total Question 3			5
04.1	Internal energy of P $=1.2\times10^3\,\text{J}$ Internal energy of Q $=2.75\times10^3\,\text{J}$ Average kinetic energy per particle in P $=2.5\times10^{-21}\,\text{J}$ Average kinetic energy per particle in Q $=2.0\times10^{-21}\,\text{J}$ P has the higher temperature.	Internal energy is total kinetic plus total potential energy: $U_P=500+700=1200\,\text{J}$ and $U_Q=1000+1750=2750\,\text{J}$ . The potential energy per particle is $700/(2.0\times10^{23})=3.5\times10^{-21}\,\text{J}$ for P and $1750/(5.0\times10^{23})=3.5\times10^{-21}\,\text{J}$ for Q, consistent with the samples being the same substance in the same state. Average kinetic energy is total kinetic energy divided by particle number. For P it is $500/(2.0\times10^{23})=2.5\times10^{-21}\,\text{J}$ ; for Q it is $1000/(5.0\times10^{23})=2.0\times10^{-21}\,\text{J}$ . Temperature depends on average kinetic energy, so P is hotter even though Q has the greater internal energy.	5
Total Question 4			5
05.1	The liquid's internal energy increases by $6.5\,\text{kJ}$ . The metal particles' average kinetic energy decreases, so the metal cools. The liquid particles' average kinetic energy increases, so the liquid warms. Net energy transfer stops when the metal, liquid and container reach the same temperature.	The insulated three-part system does not transfer energy to its surroundings, so the $7.4\,\text{kJ}$ lost by the metal equals the gains of the liquid and container. Hence $\Delta U_{\text{liquid}}=7.4-0.9=6.5\,\text{kJ}$ . With no state changes, the metal's falling temperature corresponds to lower average particle kinetic energy and the liquid's rising temperature to higher average kinetic energy. At a common temperature there is no net energy transfer between them.	5
Total Question 5			5

4.3.2.2 · Temperature changes in a system and specific heat capacity

Tier 1 · Easy

Mark scheme for 4.3.2.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Less energy is needed to raise the temperature of each kilogram by $1\,{}^\circ\text{C}$ , so a given energy transfer produces a greater temperature rise.	From $\Delta E=mc\Delta\theta$ , a smaller $c$ gives a larger temperature rise for the same mass and energy transfer. The base therefore reaches a high temperature more quickly.	2
Total Question 1			2
02.1	Material R has the greatest specific heat capacity because the same energy transfer to the same mass produces the smallest temperature rise.	From $\Delta E=mc\Delta\theta$ , with equal $\Delta E$ and $m$ , specific heat capacity is inversely related to temperature rise. R has the smallest rise, $7\,{}^\circ\text{C}$ , so it has the greatest $c$ .	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.3.2.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$3.2\times10^4\,\text{J}$	Use $\Delta E=mc\Delta\theta$ . Therefore $\Delta E=2.0\times450\times35=31\,500\,\text{J}$ , which to two significant figures is $3.2\times10^4\,\text{J}$ .	2
Total Question 1			2
02.1	The initial temperature was not kept constant; trial 3 began at $58\,{}^\circ\text{C}$ rather than near $20$ to $22\,{}^\circ\text{C}$ . The hotter block has a greater temperature difference from the surroundings, so it transfers energy to the surroundings at a greater rate. Less of the supplied electrical energy raises the block's temperature, making the measured temperature rise smaller than it would be without this loss. Using all the supplied energy in $c=E/(m\Delta\theta)$ therefore makes the calculated specific heat capacity too high.	Compare the initial-temperature column: only trial 3 starts far above room temperature. Its greater temperature difference from the surroundings increases unwanted energy transfer away from the block. The calculation treats the full electrical input as energy gained by the block, even though the measured temperature rise is reduced by this loss. Dividing that overstated energy gain by $m\Delta\theta$ therefore overestimates $c$ .	4
Total Question 2			4
03.1	$1.20\times10^3\,\text{J/(kg}\,{}^\circ\text{C)}$	Across the widest interval, $\Delta E=9.6\,\text{kJ}=9600\,\text{J}$ and $\Delta\theta=38-18=20\,{}^\circ\text{C}$ . Rearrange $\Delta E=mc\Delta\theta$ to $c=\Delta E/(m\Delta\theta)$ . Thus $c=9600/(0.400\times20)=1200\,\text{J/(kg}\,{}^\circ\text{C)}=1.20\times10^3\,\text{J/(kg}\,{}^\circ\text{C)}$ .	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.3.2.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Block A: $40\,{}^\circ\text{C}$ Block B: $160\,{}^\circ\text{C}$ B's smaller product $mc$ means the same energy produces a larger temperature rise.	Convert $54\,\text{kJ}$ to $54\,000\,\text{J}$ and use $\Delta\theta=\Delta E/(mc)$ . For A, $\Delta\theta=54\,000/(1.5\times900)=40\,{}^\circ\text{C}$ . For B, $\Delta\theta=54\,000/(0.75\times450)=160\,{}^\circ\text{C}$ . Block B has both lower mass and lower specific heat capacity, so its thermal capacity $mc$ is smaller.	5
Total Question 1			5
02.1	$1.0\times10^3\,\text{g}$ or $1000\,\text{g}$	The energy transferred to the block is $60.0-10.8=49.2\,\text{kJ}=49\,200\,\text{J}$ . Its temperature change is $82-22=60\,{}^\circ\text{C}$ . Rearrange $\Delta E=mc\Delta\theta$ to $m=\Delta E/(c\Delta\theta)$ : $m=49\,200/(820\times60)=1.00\,\text{kg}=1000\,\text{g}$ , or $1.0\times10^3\,\text{g}$ to two significant figures.	5
Total Question 2			5
03.1	Heater energy $=24\,000\,\text{J}$ Energy reaching the block $=18\,000\,\text{J}$ Specific heat capacity $=1250\,\text{J/(kg}\,{}^\circ\text{C)}$ The block is material B.	Subtract the joulemeter readings to obtain the input energy: $30\,000-6\,000=24\,000\,\text{J}$ . The block receives $0.75\times24\,000=18\,000\,\text{J}$ . Using $c=\Delta E/(m\Delta\theta)$ gives $c=18\,000/(0.600\times24.0)=1250\,\text{J/(kg}\,{}^\circ\text{C)}$ , matching material B exactly and differing from the next candidate by at least $350\,\text{J/(kg}\,{}^\circ\text{C)}$ .	5
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04.1	Energy lost by aluminium $=0.200\times900\times(90.0-T)$ Energy gained by water $=0.300\times4200\times(T-20.0)$ Final temperature $T=28.8\,{}^\circ\text{C}$ (3 s.f.)	In the insulated system, energy lost by the aluminium equals energy gained by the water. Therefore $0.200\times900\times(90.0-T)=0.300\times4200\times(T-20.0)$ . This gives $180(90.0-T)=1260(T-20.0)$ , so $41\,400=1440T$ and $T=28.75\,{}^\circ\text{C}=28.8\,{}^\circ\text{C}$ (3 s.f.).	5
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05.1	Temperature change $\Delta\theta=60.0\,{}^\circ\text{C}$ Rearrange $E=mc\Delta\theta$ to $m=E/(c\Delta\theta)$ . Minimum mass of R $=33.3\,\text{kg}$ R fails the mass limit because $33.3\,\text{kg}$ is about $11\%$ over $30.0\,\text{kg}$ . Minimum mass of S $=12.5\,\text{kg}$ S meets the mass limit.	The temperature change is $\Delta\theta=80.0-20.0=60.0\,{}^\circ\text{C}$ and $1.8\,\text{MJ}=1.8\times10^6\,\text{J}$ . Rearranging $E=mc\Delta\theta$ gives $m=E/(c\Delta\theta)$ . For R, $m_R=(1.8\times10^6)/(900\times60.0)=33.3\,\text{kg}$ , which exceeds $30.0\,\text{kg}$ by $(33.3-30.0)/30.0\times100\approx11\%$ , so R fails. For S, $m_S=(1.8\times10^6)/(2400\times60.0)=12.5\,\text{kg}$ , so S meets the limit.	6
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4.3.2.3 · Changes of state and specific latent heat

Tier 1 · Easy

Mark scheme for 4.3.2.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Use $E=mL$ because the ice changes state without a change in temperature.	Melting is a change of state at constant temperature, so the energy depends on mass and specific latent heat. Therefore use $E=mL$ , not the specific-heat-capacity equation.	2
Total Question 1			2
02.1	The specific latent heat of vaporisation is four times the specific latent heat of fusion. The same heater transfers four times as much energy during a section lasting four times as long, while the sample mass is unchanged.	A constant-power heater transfers energy in proportion to time. The boiling plateau therefore needs four times the energy of the melting plateau. Since the mass is the same and $E=mL$ , the latent heat of vaporisation is four times the latent heat of fusion.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.3.2.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$8.7\times10^4\,\text{J}$	Use $E=mL$ . Hence $E=0.28\times3.10\times10^5=86\,800\,\text{J}$ , which to two significant figures is $8.7\times10^4\,\text{J}$ .	2
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02.1	$3.60\times10^5\,\text{J/kg}$	The energy used for melting is the difference between the readings: $126-18=108\,\text{kJ}=108\,000\,\text{J}$ . Rearrange $E=mL$ to $L=E/m$ , so $L=108\,000/0.300=360\,000\,\text{J/kg}=3.60\times10^5\,\text{J/kg}$ .	4
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03.1	Substitution $L_f=36\,000/0.120$ Specific latent heat of fusion $=3.00\times10^5\,\text{J/kg}$ Substitution $E=0.130\times3.00\times10^5$ Additional energy $=39\,000\,\text{J}=39.0\,\text{kJ}$	The completed table gives $E=44.0-8.0=36.0\,\text{kJ}=36\,000\,\text{J}$ and $m=0.250-0.130=0.120\,\text{kg}$ . Therefore $L_f=E/m=36\,000/0.120=3.00\times10^5\,\text{J/kg}$ . The energy needed to melt the remaining solid is $E=mL_f=0.130\times3.00\times10^5=39\,000\,\text{J}=39.0\,\text{kJ}$ .	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.3.2.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	$3.9\times10^5\,\text{J}$	For warming, $E_1=mc\Delta\theta=0.15\times4200\times(100-20)=50\,400\,\text{J}$ . For vaporisation, $E_2=mL_v=0.15\times2.26\times10^6=339\,000\,\text{J}$ . The total is $50\,400+339\,000=389\,400\,\text{J}$ , or $3.9\times10^5\,\text{J}$ to two significant figures.	5
Total Question 1			5
02.1	$3.45\times10^5\,\text{J/kg}$	First find the energy for the temperature rise: $E_1=mc\Delta\theta=0.200\times(5.00\times10^2)\times(65.0-15.0)=5000\,\text{J}$ . Convert the total: $74.0\,\text{kJ}=74\,000\,\text{J}$ . The melting energy is $E_2=74\,000-5000=69\,000\,\text{J}$ . Therefore $L_f=E_2/m=69\,000/0.200=345\,000\,\text{J/kg}=3.45\times10^5\,\text{J/kg}$ .	5
Total Question 2			5
03.1	Heater energy $=21\,600\,\text{J}$ Energy used for melting $=17\,280\,\text{J}$ Specific latent heat of fusion $=1.80\times10^5\,\text{J/kg}$	Convert the time explicitly: $4.0\,\text{min}=240\,\text{s}$ . The heater transfers $E=Pt=90.0\times240=21\,600\,\text{J}$ . Since $20\%$ leaves, $80\%$ melts the solid: $E_{\text{melt}}=0.80\times21\,600=17\,280\,\text{J}$ . Then $L_f=E_{\text{melt}}/m=17\,280/0.0960=180\,000\,\text{J/kg}=1.80\times10^5\,\text{J/kg}$ .	5
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04.1	Energy for P $=36\,000\,\text{J}$ Energy for Q $=37\,500\,\text{J}$ Q needs $1500\,\text{J}$ more, about $4.2\%$ more than P. The required energy depends on the product $mL$ , so Q's larger mass outweighs its smaller specific latent heat.	Use $E=mL$ . For P, $E_P=0.150\times2.4\times10^5=36\,000\,\text{J}$ . For Q, $E_Q=0.250\times1.5\times10^5=37\,500\,\text{J}$ . Thus Q needs $1500\,\text{J}$ more; relative to P this is $(1500/36\,000)\times100=4.2\%$ . Comparing $L$ alone is insufficient because the mass is also a factor in $mL$ .	4
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05.1	Latent energy released $=26\,880\,\text{J}$ Energy transferred to the water $=20\,160\,\text{J}$ Water temperature rise $=12.0\,{}^\circ\text{C}$ Final water temperature $=32.0\,{}^\circ\text{C}$ (3 s.f.)	Freezing releases $E=mL=0.120\times2.24\times10^5=26\,880\,\text{J}$ . The water receives $0.75\times26\,880=20\,160\,\text{J}$ . Using $\Delta E=mc\Delta\theta$ , $\Delta\theta=20\,160/(0.400\times4200)=12.0\,{}^\circ\text{C}$ . The final temperature is $20.0+12.0=32.0\,{}^\circ\text{C}$ (3 s.f.).	5
Total Question 5			5

4.3.3.1 · Particle motion in gases

Tier 1 · Easy

Mark scheme for 4.3.3.1 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	The molecules move constantly and randomly.	Molecules in a gas are in constant random motion.	1
Total Question 1			1
02.1	Gas Q is likely to be hotter because its faster molecules collide with the wall area more frequently.	For identical gases with equal molecule numbers and fixed equal volumes, a higher temperature means greater average molecular kinetic energy and speed. The larger collision count for Q is therefore evidence that Q is hotter.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.3.3.1 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The molecules gain average kinetic energy and move faster. They collide with the flask walls more frequently. Each collision has a greater change of momentum and exerts a greater force. The greater force on the same wall area gives a higher pressure.	The flask is sealed, so molecule number is constant, and rigid, so volume is constant. Heating raises average kinetic energy and speed. Faster molecules strike each unit area of wall more often and with a larger momentum change, increasing the mean force and therefore the pressure.	4
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02.1	Molecules escape through the leak, so fewer gas molecules remain in the can. Constant temperature means their average kinetic energy and average speed remain unchanged. With fewer molecules, collisions with the walls occur less frequently. With fewer collisions each second, the total force on each unit area of the wall falls, so the pressure decreases.	The rigid can keeps a fixed volume, while the constant temperature keeps average molecular kinetic energy and speed unchanged. The leak reduces the number of molecules in the can. Fewer molecules produce fewer wall collisions each second, so the mean force on each unit area decreases. Since pressure is force per unit area, the pressure falls.	4
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03.1	Gas molecules move constantly and randomly in all directions. They collide with every wall of the cube. Each collision changes molecular momentum and exerts a force on the wall. With uniform conditions and equal sensor areas, the average force per unit area is approximately the same on every face.	Random motion sends molecules towards all six faces rather than only towards the bottom. Their wall collisions transfer momentum and produce forces normal to the surfaces. Over time, the large number of random collisions gives similar mean force per unit area on equal sensor areas, so the measured pressures are approximately equal.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.3.3.1 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Molecules in A have greater average kinetic energy and average speed. In identical volumes they collide with the walls more frequently and with greater momentum changes, so A has the higher pressure. With different volumes, collision frequency also depends on the distance between walls and number of molecules per unit volume, so temperature alone would not determine which pressure is higher.	Use the controlled conditions first: equal molecule number and equal volume isolate temperature, so A's faster molecules produce a greater force per unit area. If volume changes, molecular spacing and the rate at which molecules reach the walls also change; a larger volume can reduce pressure, so the temperature comparison alone is insufficient.	5
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02.1	In both containers, the temperature rise increases average molecular kinetic energy and speed. The rigid can keeps the same volume, so collisions become more frequent and harder, producing a substantial pressure rise. The balloon expands, increasing the distance between its walls and spreading the molecules through a larger volume. Expansion offsets much of the increase in collision rate per unit area, so its pressure rises by much less.	The equal temperature increases give the gas molecules comparable increases in average kinetic energy. The can cannot expand, so faster motion directly increases collision frequency and momentum transfer to each unit area. The flexible balloon gains volume; molecules travel farther between wall collisions and are less concentrated, limiting the pressure increase despite their greater speed.	5
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03.1	Initial pressure $=3000\,\text{Pa}$ Final pressure $=3900\,\text{Pa}$ Pressure increase $=900\,\text{Pa}$ Heating increases average molecular kinetic energy and speed, so collisions with the wall are more frequent. The collisions transfer momentum at a greater rate, increasing the force per unit area and therefore the pressure.	Use $p=F/A$ . Initially $p=60/0.020=3000\,\text{Pa}$ and finally $p=78/0.020=3900\,\text{Pa}$ , so the increase is $3900-3000=900\,\text{Pa}$ . In the fixed volume, heating makes the molecules move faster. Their more frequent, harder collisions increase the mean force on the unchanged sensor area and therefore increase the pressure.	5
Total Question 3			5
04.1	In A, heating increases the molecules' average kinetic energy and speed. The faster molecules collide more frequently and change momentum more in each collision, increasing the force on the walls. In B, constant temperature means the average kinetic energy and typical effect of each collision stay unchanged. The extra molecules produce more wall collisions per second. Both changes increase mean force per unit area and therefore pressure, but by different molecular mechanisms.	The rigid tanks keep volume and wall area fixed. In A, the fixed number of molecules gains average kinetic energy, so faster and harder wall collisions increase the rate of momentum transfer. In B, the unchanged temperature keeps average kinetic energy unchanged, but a greater number of molecules raises the collision frequency. Either change increases the mean force on the fixed wall area and hence the pressure.	5
Total Question 4			5
05.1	The final gas has greater average molecular kinetic energy because it is at a higher temperature. Some gas escapes through the valve, so fewer molecules remain. The remaining molecules move faster and individual collisions transfer momentum at a greater rate. The smaller number of molecules reduces the total collision rate with the walls. These opposing effects can give the same mean force per unit area and therefore the same pressure.	Heating raises average molecular kinetic energy and makes collisions faster and harder. Opening the valve allows molecules to leave, reducing the number available to collide with the walls. When enough molecules have escaped, the greater effect of the hotter molecules is balanced by the reduced total collision rate. The wall can then experience the original mean force per unit area even though the gas is hotter and has a smaller mass.	5
Total Question 5			5

4.3.3.2 · Pressure in gases (physics only)

Tier 1 · Easy

Mark scheme for 4.3.3.2 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	The force acts at right angles (normal) to the wall.	Pressure produces a net force normal to the surface, so the force acts at right angles to the container wall.	1
Total Question 1			1
02.1	Pressure is inversely proportional to volume. $180\times120=90\times240=21\,600\,\text{kPa cm}^3$ , so $pV$ is constant.	The volume doubles from $120$ to $240\,\text{cm}^3$ while the pressure halves from $180$ to $90\,\text{kPa}$ . Equivalently, both pressure-volume products equal $21\,600\,\text{kPa cm}^3$ , confirming the inverse relationship.	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.3.3.2 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	$3.2\times10^5\,\text{Pa}$	Use $p_1V_1=p_2V_2$ . Therefore $p_2=(1.2\times10^5\times4.0\times10^{-3})/(1.5\times10^{-3})=3.2\times10^5\,\text{Pa}$ .	3
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02.1	The anomalous pair is $125\,\text{cm}^3$ and $182\,\text{kPa}$ . The corrected pressure is $192\,\text{kPa}$ .	The consistent pairs give $pV=80\times300=100\times240=160\times150=24\,000\,\text{kPa cm}^3$ . The $125\times182=22\,750\,\text{kPa cm}^3$ product is inconsistent. At $125\,\text{cm}^3$ , the expected pressure is $p=24\,000/125=192\,\text{kPa}$ .	4
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03.1	The greater gas pressure produces a larger outward force than the external pressure produces inward, so the piston accelerates outwards. The outward motion increases the gas volume. At constant temperature, the larger volume reduces the frequency of wall collisions per unit area, so the gas pressure falls. The piston stops accelerating when the gas and external pressures produce equal opposing forces, so the resultant force is zero.	Both pressures act normally on the same piston area. The initially greater internal pressure gives a resultant outward force. Expansion increases the volume available to the fixed number of molecules, reducing collision frequency per unit area and therefore the internal pressure. Once the opposing pressure forces balance, there is no resultant force and hence no further acceleration.	4
Total Question 3			4

Tier 3 · Hard

Mark scheme for 4.3.3.2 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Final volume $=213\,\text{cm}^3$ Decrease in volume $=107\,\text{cm}^3$ In the smaller volume, molecules collide with the walls more frequently, increasing the pressure.	Using the same pressure and volume units on both sides, $V_2=p_1V_1/p_2=(240\times320)/360=213.3\,\text{cm}^3$ . The decrease is $320-213.3=106.7\,\text{cm}^3$ , giving $107\,\text{cm}^3$ . At constant temperature average molecular kinetic energy is unchanged, but the shorter travel distance causes more wall collisions per second and a greater force per unit area.	5
Total Question 1			5
02.1	Minimum volume $=288\,\text{cm}^3=2.88\times10^{-4}\,\text{m}^3$ Compressing to $2.80\times10^{-4}\,\text{m}^3$ is not safe because this is below the minimum volume and would give a pressure above $250\,\text{kPa}$ .	At the pressure limit, $V_2=p_1V_1/p_2=(96.0\times750)/250=288\,\text{cm}^3$ . Since $1\,\text{cm}^3=10^{-6}\,\text{m}^3$ , this is $2.88\times10^{-4}\,\text{m}^3$ . The proposed $2.80\times10^{-4}\,\text{m}^3$ equals $280\,\text{cm}^3$ . It would give $p=(96.0\times750)/280=257.142\ldots\,\text{kPa}$ , above the limit, so it is unsafe.	5
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03.1	Piston area $=1.50\times10^{-3}\,\text{m}^2$ Substitution $p_2=360/(1.50\times10^{-3})$ $p_2=240\,\text{kPa}$ Rearranged substitution $90.0\times V_1=240\times300$ $V_1=800\,\text{cm}^3$	Convert the piston area explicitly: $15.0\,\text{cm}^2=15.0\times10^{-4}\,\text{m}^2=1.50\times10^{-3}\,\text{m}^2$ . Hence $p_2=F/A=360/(1.50\times10^{-3})=2.40\times10^5\,\text{Pa}=240\,\text{kPa}$ . Use matching pressure and volume units in $p_1V_1=p_2V_2$ : $V_1=(240\times300)/90.0=800\,\text{cm}^3$ .	5
Total Question 3			5
04.1	Trap a fixed mass of gas in a sealed syringe or cylinder connected to an absolute pressure sensor. Change the gas volume in measured steps using the syringe scale and record the pressure at each volume. Keep the apparatus sealed so the gas mass is constant. Change the volume slowly and wait for the gas to return to the same temperature before each reading. Repeat readings over a suitable range of volumes. Calculate $pV$ for each pair or plot pressure against $1/V$ ; constant $pV$ or a straight line through the origin supports the inverse relationship.	Connect a sealed gas syringe to an absolute pressure sensor and use the syringe scale to measure volume. Move the plunger through several positions, waiting after each slow change so compression or expansion heating does not alter the temperature. Record pressure and volume without opening the sealed system, and repeat. Test the model by checking whether $pV$ is approximately constant or whether a graph of $p$ against $1/V$ is a straight line through the origin.	6
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05.1	Equation $120(60+x)=200(30+x)$ Tube volume $x=15\,\text{cm}^3$ Pressure at a syringe reading of $15\,\text{cm}^3=300\,\text{kPa}$	The gas volume is the syringe reading plus the fixed tube volume. At constant temperature, $pV$ is constant, so $120(60+x)=200(30+x)$ . Expanding gives $7200+120x=6000+200x$ , hence $x=15\,\text{cm}^3$ . The first-state constant is $120\times(60+15)=9000\,\text{kPa cm}^3$ . At the final syringe reading the total volume is $15+15=30\,\text{cm}^3$ , so $p=9000/30=300\,\text{kPa}$ .	5
Total Question 5			5

4.3.3.3 · Increasing the pressure of a gas (physics only) (HT only)

Tier 1 · Easy

Mark scheme for 4.3.3.3 Tier 1 · Easy
Question	Answers	Extra information	Mark
01.1	Work is the transfer of energy by a force. The internal energy of the gas increases.	The force on the moving piston transfers energy to the enclosed gas. This work done on the gas increases its internal energy.	2
Total Question 1			2
02.1	The internal energy increases by $45\,\text{J}$ .	The net energy transferred to the gas is $64-19=45\,\text{J}$ . Its internal energy therefore increases by $45\,\text{J}$ .	2
Total Question 2			2

Tier 2 · Standard

Mark scheme for 4.3.3.3 Tier 2 · Standard
Question	Answers	Extra information	Mark
01.1	The internal energy increases by $180\,\text{J}$ . The temperature is likely to increase because the molecules gain average kinetic energy.	With no energy leaving, conservation of energy means all $180\,\text{J}$ of work increases the gas's internal energy. During compression this can raise average molecular kinetic energy, so the gas temperature increases.	3
Total Question 1			3
02.1	Row C is impossible. It shows $69\,\text{J}$ transferred out but only $65\,\text{J}$ transferred in by work. The resulting change would be $65-69=-4\,\text{J}$ , so internal energy would decrease rather than increase. In A, B and D, work input exceeds energy output, so a positive internal-energy change is possible.	For the stated transfers, conservation of energy gives $\Delta U=W_{\text{in}}-E_{\text{out}}$ . Row C gives $\Delta U=65-69=-4\,\text{J}$ , contradicting the report that internal energy increased. The other rows all have $W_{\text{in}}>E_{\text{out}}$ , so each permits a positive change. The impossible row is separated from the boundary by $4\,\text{J}$ .	4
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03.1	Work transfers the same amount of energy to each gas. During slow compression there is more time for energy to transfer from the gas to the surroundings, so its internal energy increases by less. During rapid compression less energy escapes, so more of the work increases internal energy and average molecular kinetic energy, giving a higher temperature.	Equal work inputs do not guarantee equal retained energy. The slow process allows more transfer to the surroundings while compression occurs. Rapid compression retains more of the transferred energy in the gas's internal energy store, so its molecules gain more average kinetic energy and it becomes hotter.	3
Total Question 3			3

Tier 3 · Hard

Mark scheme for 4.3.3.3 Tier 3 · Hard
Question	Answers	Extra information	Mark
01.1	Work done on the gas $=45\,\text{J}$ Increase in internal energy $=33\,\text{J}$ Its temperature increases because its particles have greater average kinetic energy.	The work done is $W=Fs=250\times0.18=45\,\text{J}$ . Of this, $12\,\text{J}$ leaves for the surroundings, so the internal energy increase is $45-12=33\,\text{J}$ . The increased internal energy can increase average molecular kinetic energy, raising the temperature.	4
Total Question 1			4
02.1	Work done on the gas $=90\,\text{J}$ Force $=600\,\text{N}$ The gas may warm because its increased internal energy can increase the molecules' average kinetic energy.	The work input must supply both the $72\,\text{J}$ internal-energy increase and the $18\,\text{J}$ transferred out, so $W=72+18=90\,\text{J}$ . Using $W=Fs$ , $F=W/s=90/0.150=600\,\text{N}$ . An increase in internal energy can increase average molecular kinetic energy, so the temperature may rise.	5
Total Question 2			5
03.1	Gravitational potential energy transferred $=29.4\,\text{J}$ Increase in gas internal energy $=24\,\text{J}$ (2 s.f.) The gas may warm because the internal-energy increase can raise the molecules' average kinetic energy.	The descending mass transfers $\Delta E_p=mgh=5.0\times9.8\times0.60=29.4\,\text{J}$ . Subtract the energy reaching the surroundings: $\Delta U=29.4-5.4=24\,\text{J}$ (2 s.f.). Work done through the mechanism transfers this energy to the enclosed gas. If the increase raises average molecular kinetic energy, the gas temperature increases.	4
Total Question 3			4
04.1	Pressure $=1.80\times10^5\,\text{Pa}$ and area $=8.0\times10^{-4}\,\text{m}^2$ Force on the piston $=144\,\text{N}$ Substitution $W=Fs=144\times0.120$ Work done on the gas $=17.28\,\text{J}$ Increase in internal energy $=14\,\text{J}$ (2 s.f.) The gas may warm because the internal-energy increase can raise average molecular kinetic energy.	Convert first: $p=180\,000\,\text{Pa}$ and $A=8.0\times10^{-4}\,\text{m}^2$ . From $p=F/A$ , the mean force is $F=pA=180\,000\times8.0\times10^{-4}=144\,\text{N}$ . The work done is $W=Fs=144\times0.120=17.28\,\text{J}$ . After $3.3\,\text{J}$ transfers out, $\Delta U=17.28-3.3=13.98\,\text{J}=14\,\text{J}$ (2 s.f.). If this raises average molecular kinetic energy, the gas temperature rises.	6
Total Question 4			6
05.1	Fit a fast-response temperature probe to the gas in a cylinder or pump with a closable outlet. Record the initial gas temperature, close the outlet and compress the gas rapidly through a measured distance, then record the temperature immediately. Repeat from the same initial temperature, using the same plunger distance and speed. For comparison, move the plunger through the same distance and speed with the outlet open so little pressure increase occurs but plunger friction remains. A greater temperature rise in the sealed compression provides evidence that compression, rather than friction alone, heats the gas. Work done on the gas increases its internal energy; greater average molecular kinetic energy means a higher temperature.	Use a gas cylinder or pump containing a fast temperature probe. Compare a rapid sealed compression with an open-outlet control, keeping the starting temperature, plunger travel and plunger speed the same and repeating both conditions. Record temperature immediately because energy soon transfers to the surroundings. Friction occurs in both motions, but substantial work compresses the gas only in the sealed trial. An extra temperature rise there supports the conclusion that work increased the gas's internal energy and average molecular kinetic energy.	6
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