Density and the Particle Model of Matter - Worksheets, Questions and Revision

21 original exam-style questions - 10 pages of questions with a full mark scheme - free printable PDF.

Download PDF Jump to mark scheme (page 11)Read the revision guide

Next in this strand

Revision Library

revisionlibrary.co.uk

GCSE · Physics

P3a Density and the Particle Model of Matter

AQA 8464 · Calculator allowed · about 85 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

State what is meant by the density of a material.

(Total for Question 1 is 1 mark)

Complete the sentences below about laboratory equipment.

(a)State the piece of equipment used to measure the mass of an object in the laboratory.(1)

(b)State the piece of equipment used to measure the volume of a liquid in the laboratory.(1)

(Total for Question 2 is 2 marks)

The table below describes the arrangement, movement and relative energy of particles in the three states of matter. The row for a solid has been completed for you.

State	Arrangement of particles	Movement of particles	Relative energy of particles
Solid	Regularly arranged, closely packed together	Vibrate about fixed positions	Lowest
Liquid	.....	.....	.....
Gas	.....	.....	.....

(a)Complete the Liquid row of the table by describing the arrangement, movement and relative energy of particles in a liquid.(3)

(b)Complete the Gas row of the table by describing the arrangement, movement and relative energy of particles in a gas.(3)

(Total for Question 3 is 6 marks)

Name the change of state described in each case below.

(a)A solid turns into a liquid.(1)

(b)A liquid turns into a solid.(1)

(c)A liquid turns into a gas.(1)

(d)A gas turns into a liquid.(1)

(e)A solid turns directly into a gas, without becoming a liquid.(1)

(f)A gas turns directly into a solid, without becoming a liquid.(1)

(Total for Question 4 is 6 marks)

A rectangular block of metal alloy has a mass of 540 g and a volume of 60 cm³.
Calculate the density of the metal alloy. Use the equation: density = mass / volume

(Total for Question 5 is 2 marks)

Match each key term to its correct definition by writing the correct letter next to each term.
Definitions:
A - the change of state in which a solid turns directly into a gas, without becoming a liquid
B - the mass per unit volume of a material

(a)density(1)

(b)sublimation(1)

(Total for Question 6 is 2 marks)

A concrete paving slab has a mass of 3.6 kg and a volume of 1500 cm³.
Calculate the density of the slab in kg/m³. Use the equation: density = mass / volume (1 m³ = 1000000 cm³)

(Total for Question 7 is 3 marks)

Aisha carries out the required practical to determine the density of a small aluminium cube.

(a)State the piece of equipment Aisha should use to measure the mass of the cube.(1)

(b)Aisha measures the length of one side of the cube with a ruler. Suggest one way she could improve the accuracy of this length measurement.(1)

(c)The cube has a side length of 2.0 cm. Calculate the volume of the cube.(2)

(d)The mass of the cube is 21.6 g. Calculate the density of the cube. Use the equation: density = mass / volume(2)

(e)State two safety precautions Aisha should take when carrying out this experiment.(2)

(f)A data sheet gives the accepted density of aluminium as 2.70 g/cm³. Suggest one reason why Aisha's calculated density might be slightly different from the accepted value, even if all her measurements were taken carefully.(1)

(Total for Question 8 is 9 marks)

Tomasz determines the density of a small irregular rock using a displacement method.

(a)Describe how Tomasz uses a measuring cylinder to find the volume of the irregular rock.(2)

(b)The measuring cylinder reads 50.0 cm³ before the rock is added and 62.0 cm³ after. Calculate the volume of the rock.(1)

(c)The mass of the rock is 45.6 g. Calculate the density of the rock.(2)

(d)Explain why this displacement method gives a less precise volume for a regular cube than measuring its side lengths directly with a ruler.(1)

(Total for Question 9 is 6 marks)

Priya determines the density of a sample of cooking oil.

(a)The mass of an empty measuring cylinder is 38.2 g. The mass of the same cylinder containing 40 cm³ of oil is 74.2 g. Calculate the mass of the oil.(1)

(b)Calculate the density of the oil in g/cm³.(2)

(c)Priya repeats the experiment twice more and calculates densities of 0.91 g/cm³ and 0.89 g/cm³. Calculate the mean density of the oil from all three results.(1)

(d)Explain why Priya measures the mass of the empty measuring cylinder first, rather than just measuring the mass of the cylinder and oil together.(1)

(Total for Question 10 is 5 marks)

The table shows the density of three materials.

Material	Density (g/cm³)
Ice	0.92
Water	1.00
Polystyrene foam	0.05

(a)State whether ice or polystyrene foam would float lower in water (that is, have a greater fraction of its volume below the surface).(1)

(b)Explain your answer to part (a), using the densities given in the table.(2)

(Total for Question 11 is 3 marks)

Explain, in terms of the arrangement and spacing of particles, why gases have a much lower density than solids.

(Total for Question 12 is 2 marks)

Internal energy is the energy stored inside a system by the particles that make it up.

(a)State what is meant by the internal energy of a system.(1)

(b)State what is meant by the specific latent heat of a substance.(2)

(c)When a solid is heated at its melting point, energy continues to be transferred to it, but its temperature does not increase until it has completely melted. Explain, in terms of the energy of the particles, why the temperature does not increase during melting.(3)

(Total for Question 13 is 6 marks)

Calculate the energy needed to melt 0.50 kg of ice at 0 degC. The specific latent heat of fusion of ice is 334000 J/kg. Use the equation: energy for a change of state = mass * specific latent heat (E = mL)

(Total for Question 14 is 3 marks)

A camping stove transfers 668000 J of energy to a block of ice at 0 degC, melting it completely. The specific latent heat of fusion of ice is 334000 J/kg.
Calculate the mass of ice that melts. Use the equation E = mL

(Total for Question 15 is 3 marks)

A kettle boils dry and 15 g of the remaining water turns to steam. The specific latent heat of vaporisation of water is 2260000 J/kg.

(a)Calculate the energy needed to vaporise this water. Use the equation E = mL(3)

(b)The kettle's heating element transfers energy to the water at a power of 2200 W. Calculate how long it takes to vaporise this water. Use the equation: power = energy transferred / time(3)

(Total for Question 16 is 6 marks)

Melting 1 kg of water at 0 degC requires 3.34 * 10⁵ J of energy (the specific latent heat of fusion). Vaporising 1 kg of water at 100 degC requires 2.26 * 10⁶ J of energy (the specific latent heat of vaporisation) - almost seven times as much energy per kilogram.
Explain, in terms of the particle model, why so much more energy is needed to vaporise a mass of water than to melt the same mass of water.

(Total for Question 17 is 6 marks)

An ice cube has a mass of 36 g and a density of 0.92 g/cm³.

(a)Calculate the volume of the ice cube. Give your answer to 3 significant figures.(2)

(b)The ice cube melts completely into water, with no loss of mass. The density of water is 1.00 g/cm³. Calculate the volume of the water produced.(2)

(c)Calculate the change in volume when the ice melts, and state whether the volume increases or decreases. Give your answer to 2 significant figures.(2)

(Total for Question 18 is 6 marks)

Chidi heats a pure solid wax at a constant rate from below its melting point until it is completely liquid. A graph of temperature against time shows the temperature rising, then a flat section while the wax melts, then rising again.

(a)State what is happening to the particles of wax during the flat section of the graph.(2)

(b)The wax is heated using a heater with a constant power output of 60 W. The flat section of the graph lasts for 280 seconds. Calculate the energy transferred to the wax during this time. Use the equation: energy transferred = power * time(2)

(c)The mass of the wax is 40 g. Calculate the specific latent heat of fusion of the wax, giving your answer in J/kg. Use the equation E = mL(2)

(Total for Question 19 is 6 marks)

A sample of glass has a density of 2.50 g/cm³.
Convert this density into kg/m³. Show your working. (1 g/cm³ = 1000 kg/m³)

(Total for Question 20 is 3 marks)

Freya, a jeweller, wants to check whether a gold bar is genuine. Pure gold has a density of 19.3 g/cm³. The bar has a mass of 965 g and a volume of 50 cm³.

(a)Calculate the density of the bar.(2)

(b)State whether the bar could be made of pure gold, giving a reason for your answer.(1)

(Total for Question 21 is 3 marks)

Mark scheme · P3a Density and the Particle Model of Matter

Question 1

B1 the mass per unit volume of a material (density = mass / volume) oe
Answer: Density is the mass per unit volume of a material.

Question 2

(a) B1 a (top-pan/electronic) balance
(a) Answer: A balance (top-pan balance).
(b) B1 a measuring cylinder
(b) Answer: A measuring cylinder.

Question 3

(a) B1 randomly arranged, close together (touching) oe
(a) B1 move around/past each other, able to flow oe
(a) B1 more energy than a solid (but less than a gas)
(a) Answer: Liquid: randomly arranged and close together; particles move around each other and can flow; more energy than a solid.
(b) B1 randomly arranged, far apart oe
(b) B1 move quickly, in random directions, throughout their container
(b) B1 greatest/most energy of the three states
(b) Answer: Gas: randomly arranged and far apart; particles move quickly in random directions; greatest energy of the three states.

Question 4

(a) B1 melting
(a) Answer: Melting
(b) B1 freezing (solidification) oe
(b) Answer: Freezing
(c) B1 evaporating/boiling oe
(c) Answer: Evaporating (or boiling)
(d) B1 condensing
(d) Answer: Condensing
(e) B1 sublimation
(e) Answer: Sublimation
(f) B1 deposition
(f) Answer: Deposition

Question 5

M1 540 / 60 substituted
A1 9 g/cm³ (9.0) cao
Answer: 9 g/cm³

Question 6

(a) B1 B
(a) Answer: B
(b) B1 A
(b) Answer: A

Question 7

M1 1500 / 1000000 = 0.0015 m³ (volume correctly converted)
M1 3.6 / 0.0015 substituted
A1 2400 kg/m³ cao
Answer: 2400 kg/m³

Question 8

(a) B1 a (top-pan/electronic) balance
(a) Answer: A balance (top-pan balance).
(b) B1 use digital calipers instead of a ruler (more precise) oe; or repeat the measurement at different points on the cube and calculate a mean; or take the reading at eye level to avoid a parallax error
(b) Answer: Use digital calipers instead of a ruler, or repeat the measurement and take a mean.
(c) M1 2.0 * 2.0 * 2.0 substituted
(c) A1 8.0 cm³ cao
(c) Answer: 8.0 cm³
(d) M1 21.6 / 8.0 substituted ft from part (c)
(d) A1 2.7 g/cm³ cao
(d) Answer: 2.7 g/cm³
(e) B1 handle the metal cube carefully to avoid dropping it on fingers/feet oe
(e) B1 wash hands after handling the metal sample oe
(e) Answer: Handle the cube carefully to avoid dropping it; wash hands after handling the metal.
(f) B1 the cube may not be pure aluminium (it could be an alloy) oe; or small errors in measuring the side length have a large effect on the calculated volume and density because the length is cubed
(f) Answer: The cube might not be pure aluminium (could be an alloy), or a small error in the side length has a large effect once cubed.

Question 9

(a) B1 partially fill the measuring cylinder with water and record the initial volume
(a) B1 lower the rock into the water (fully submerged, e.g. on a thread), record the new volume, and find the volume of the rock as the difference between the two readings
(a) Answer: Record the initial water volume, fully submerge the rock, record the new volume; volume of rock = difference between the two readings.
(b) A1 12.0 cm³ cao
(b) Answer: 12.0 cm³
(c) M1 45.6 / 12.0 substituted ft from part (b)
(c) A1 3.8 g/cm³ cao
(c) Answer: 3.8 g/cm³
(d) B1 reading the water level (meniscus) on a measuring cylinder scale is less precise than measuring a straight edge, so the displacement method gives a more uncertain volume for a shape that could instead be measured directly
(d) Answer: Reading the meniscus on a measuring cylinder is less precise than measuring straight edges directly, so displacement gives a more uncertain volume for a regular shape.

Question 10

(a) A1 36.0 g cao
(a) Answer: 36.0 g
(b) M1 36.0 / 40 substituted ft from part (a)
(b) A1 0.90 g/cm³ cao
(b) Answer: 0.90 g/cm³
(c) A1 0.90 g/cm³ cao
(c) Answer: 0.90 g/cm³
(d) B1 so the mass of the cylinder itself can be subtracted, leaving only the mass of the oil oe (otherwise the result would include the mass of the container as well as the liquid)
(d) Answer: So the mass of the empty cylinder can be subtracted, leaving only the mass of the oil.

Question 11

(a) B1 ice
(a) Answer: Ice
(b) B1 the closer an object's density is to the density of the liquid it floats in, the greater the fraction of its volume that is submerged
(b) B1 ice's density (0.92 g/cm³) is much closer to that of water (1.00 g/cm³) than polystyrene's density (0.05 g/cm³), so ice floats with more of its volume underwater than the polystyrene foam does
(b) Answer: Ice's density is much closer to water's density than polystyrene's is, so ice sits lower (more submerged) than polystyrene when floating.

Question 12

B1 particles in a gas are spread much further apart, with large gaps between them, compared to a solid where particles are closely packed
B1 a given volume of gas therefore contains far fewer particles (much less mass) than the same volume of solid, so the density of a gas is much lower
Answer: Gas particles are far apart with large gaps between them, so a given volume of gas has much less mass (and lower density) than the same volume of a solid.

Question 13

(a) B1 the sum of the kinetic energies and potential energies of all the particles that make up the system oe
(a) Answer: The sum of the kinetic and potential energies of all the particles that make up the system.
(b) B1 the energy needed to change the state of
(b) B1 1 kg of the substance, with no change in temperature
(b) Answer: The energy needed to change the state of 1 kg of a substance, with no change in temperature.
(c) B1 the energy transferred increases the potential energy of the particles, not their kinetic energy
(c) B1 this energy is used to (partially) overcome the forces of attraction between particles, allowing them to move away from their fixed positions
(c) B1 the average kinetic energy of the particles is unchanged, so the temperature stays constant until melting is complete
(c) Answer: Energy increases the particles' potential energy (overcoming forces of attraction) rather than their kinetic energy, so the temperature stays constant until melting finishes.

Question 14

M1 0.50 * 334000 substituted
A1 167000 (J) evaluated correctly
A1 correct unit, J (or 1.67 * 10⁵ J), stated cao
Answer: 167000 J (1.67 * 10⁵ J)

Question 15

dM1 rearranges to m = E / L
M1 668000 / 334000 substituted
A1 2.0 kg cao
Answer: 2.0 kg

Question 16

(a) M1 converts 15 g to 0.015 kg
(a) M1 0.015 * 2260000 substituted
(a) A1 33900 J (3.39 * 10⁴ J) cao
(a) Answer: 33900 J
(b) M1 rearranges to time = energy transferred / power
(b) M1 33900 / 2200 substituted ft from part (a)
(b) A1 awrt 15.4 s cao
(b) Answer: 15.4 s (awrt)

Question 17

Level 1 (1-2): Basic or isolated statements are made (for example, that vaporising needs more energy than melting) without a clear or correct explanation in terms of particles.
Level 2 (3-4): Some relevant physics is given, showing understanding that both changes of state involve overcoming forces of attraction between particles, but the explanation of why vaporisation needs more energy is incomplete or lacks a clear comparison between the two processes.
Level 3 (5-6): A detailed and coherent explanation is given, linking the difference in energy needed to the different changes in particle separation in each process, with a clear comparison between melting and vaporising.
Indicative content:
- When a solid melts, particles gain enough energy to partially overcome the forces of attraction between them, allowing them to move around each other while remaining close together, as in a liquid.
- When a liquid vaporises, particles must gain enough energy to completely overcome the forces of attraction between them and move far apart from each other, as in a gas.
- Because particles must move much further apart during vaporisation than during melting, much more work must be done against the forces of attraction between particles during vaporisation.
- This means much more energy is needed to increase the potential energy of the particles during vaporisation than during melting.
- In both processes the extra energy transferred increases the potential energy, not the kinetic energy, of the particles, which is why the temperature does not change during either process.

Question 18

(a) M1 36 / 0.92 substituted
(a) A1 39.1 cm³ cao (3 sf)
(a) Answer: 39.1 cm³
(b) M1 36 / 1.00 substituted
(b) A1 36.0 cm³ cao
(b) Answer: 36.0 cm³
(c) M1 39.1 - 36.0 substituted ft from parts (a) and (b)
(c) A1 3.1 cm³ decrease cao (2 sf)
(c) Answer: Decrease of 3.1 cm³

Question 19

(a) B1 the energy transferred is increasing the potential energy of the particles, used to overcome the forces of attraction between them so they can move out of their fixed positions
(a) B1 the kinetic energy of the particles (and therefore the temperature) stays the same during this time
(a) Answer: Energy is increasing the particles' potential energy (overcoming forces between them), while their kinetic energy, and so the temperature, stays the same.
(b) M1 60 * 280 substituted
(b) A1 16800 J cao
(b) Answer: 16800 J
(c) M1 converts 40 g to 0.040 kg and rearranges to L = E / m
(c) A1 420000 J/kg (4.2 * 10⁵ J/kg) cao ft from part (b)
(c) Answer: 420000 J/kg

Question 20

M1 correct conversion factor identified (1 g/cm³ = 1000 kg/m³, since 1 kg = 1000 g and 1 m³ = 1000000 cm³)
M1 2.50 * 1000 substituted
A1 2500 kg/m³ cao
Answer: 2500 kg/m³

Question 21

(a) M1 965 / 50 substituted
(a) A1 19.3 g/cm³ cao
(a) Answer: 19.3 g/cm³
(b) B1 yes ft, because the calculated density (19.3 g/cm³) matches the density of pure gold
(b) Answer: Yes, the calculated density (19.3 g/cm³) matches the density of pure gold.

Mark my answers