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This lesson provides exam-style practice questions and model answers covering the entire AQA GCSE Physics specification topic 4.3: Particle Model of Matter. Working through these questions will help you identify your strengths and weaknesses, practise applying your knowledge under exam conditions, and learn how to structure your answers for maximum marks.
Before attempting the questions, review these key exam strategies:
| Question Type | Marks | Strategy |
|---|---|---|
| Multiple choice | 1 | Read all options carefully; eliminate obviously wrong answers first |
| Short answer (1-2 marks) | 1-2 | Give concise, precise answers; one key point per mark |
| Calculation | 3-5 | Show all working: write the equation, substitute values, calculate, include units |
| "Explain" questions | 3-6 | Provide a chain of reasoning; each step in the chain can earn a mark |
| Required practical | 4-6 | Describe method step-by-step; name specific equipment; mention improvements |
Exam Tip: For calculation questions, even if you get the final answer wrong, you can still earn marks for correct working. ALWAYS write down the equation, show the substitution of values, and include units in your final answer. The mark scheme typically awards separate marks for each of these steps.
A block of aluminium has a mass of 1350 g and a volume of 500 cm3. Calculate the density of the aluminium in g/cm3.
Model Answer:
p = m / V [1 mark for correct equation or substitution]
p = 1350 / 500 = 2.7 g/cm3 [1 mark for correct answer with units]
A student measures the mass of a stone as 156 g. She uses a displacement method and finds the stone displaces 60 cm3 of water. Calculate the density of the stone in kg/m3.
Model Answer:
Convert mass: 156 g = 0.156 kg [1 mark]
Convert volume: 60 cm3 = 60 x 10^-6 m3 = 0.00006 m3 [1 mark]
p = m / V = 0.156 / 0.00006 = 2600 kg/m3 [1 mark]
(Alternative: calculate in g/cm3 first: 156/60 = 2.6 g/cm3, then convert: 2.6 x 1000 = 2600 kg/m3)
Describe how you would determine the density of an irregularly shaped stone. Include the equipment you would use and any precautions you would take to ensure accurate results.
Model Answer:
Measure the mass of the stone using a balance (reading to at least 0.1 g). Record the mass in grams. [1 mark]
Fill a eureka can (displacement can) with water until water flows from the spout. Wait until the water stops dripping from the spout. [1 mark]
Place a measuring cylinder under the spout to collect the displaced water. [1 mark]
Gently lower the stone into the eureka can using a piece of string (to avoid splashing). The stone displaces its own volume of water, which flows into the measuring cylinder. [1 mark]
Read the volume of displaced water from the measuring cylinder at eye level, from the bottom of the meniscus, to avoid parallax error. This is the volume of the stone. [1 mark]
Calculate the density using p = m / V. Repeat the experiment at least three times and calculate a mean to improve reliability. [1 mark]
Exam Tip: In a 6-mark required practical question, the mark scheme typically awards marks for: naming specific equipment, describing each step of the method, mentioning a precaution (e.g. reading at eye level), using the correct equation, and mentioning repeats/mean. Cover all of these for full marks.
Explain, in terms of particles, what happens when a solid is heated from below its melting point to above its melting point.
Model Answer:
Below the melting point, the particles in the solid vibrate around fixed positions. As the solid is heated, the particles gain kinetic energy and vibrate faster — the temperature rises. [1 mark]
At the melting point, the particles have enough energy to overcome some of the forces of attraction between them. The energy being supplied increases the potential energy of the particles, not their kinetic energy, so the temperature remains constant during melting. [1 mark]
Once all the solid has melted, continued heating causes the liquid particles to gain kinetic energy again — they move faster around each other and the temperature rises once more. [1 mark]
Explain why the temperature of a substance remains constant during boiling, even though energy is being supplied.
Model Answer:
During boiling, the energy being supplied is used to overcome the forces of attraction (break the bonds) between particles, rather than to increase their kinetic energy. [1 mark]
Since temperature is a measure of the average kinetic energy of the particles, and kinetic energy is not increasing, the temperature remains constant. The energy increases the potential energy of the particles instead. [1 mark]
A heating curve for a pure substance is shown below. The substance starts as a solid and is heated until it becomes a gas.
Describe what is happening at each stage of the heating curve:
| Stage | Description |
|---|---|
| A (rising) | The solid is being heated. Particles gain kinetic energy and vibrate faster. Temperature increases. |
| B (flat) | Melting is occurring. Energy is used to overcome forces of attraction (increase potential energy). Temperature is constant at the melting point. |
| C (rising) | The liquid is being heated. Particles gain kinetic energy and move faster. Temperature increases. |
| D (flat) | Boiling is occurring. Energy is used to completely overcome all forces of attraction. Temperature is constant at the boiling point. |
[1 mark for each stage correctly described]
Exam Tip: Heating curve questions are extremely common. Always identify the flat sections as changes of state and the rising sections as temperature increases. For the flat sections, use the key phrase "energy is used to overcome forces of attraction between particles" — this is what the mark scheme is looking for.
Calculate the energy needed to melt 0.8 kg of ice at 0 degrees C. The specific latent heat of fusion of water is 334,000 J/kg.
Model Answer:
E = m x L [1 mark for equation]
E = 0.8 x 334,000 [1 mark for substitution]
E = 267,200 J (or 267.2 kJ) [1 mark for correct answer with units]
A kettle supplies 452,000 J of energy to boil water that is already at 100 degrees C. The specific latent heat of vaporisation of water is 2,260,000 J/kg. Calculate the mass of water that boils away.
Model Answer:
E = m x L, so m = E / L [1 mark for rearranging]
m = 452,000 / 2,260,000 [1 mark for substitution]
m = 0.2 kg (or 200 g) [1 mark for correct answer with units]
Explain why the specific latent heat of vaporisation is much greater than the specific latent heat of fusion for the same substance.
Model Answer:
During vaporisation (boiling), all of the forces of attraction between particles must be completely overcome so that the particles can separate entirely and become a gas. [1 mark]
During fusion (melting), only some of the forces of attraction need to be overcome — the particles move from a rigid structure to a close but less ordered arrangement. Therefore, less energy per kilogram is needed for fusion than for vaporisation. [1 mark]
A sealed container of gas is heated. Explain, using the particle model, why the pressure of the gas increases.
Model Answer:
When the gas is heated, the particles gain kinetic energy and move faster. [1 mark]
The faster-moving particles collide with the walls of the container more frequently (more collisions per second). [1 mark]
Each collision is also more forceful because the particles have greater speed and therefore greater momentum. [1 mark]
Since there are more frequent and more forceful collisions, the total force on the walls increases. Since pressure = force / area and the area of the container walls is constant, the pressure increases. [1 mark]
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