AQA GCSE Combined Science: Trilogy · 6.3

Particle model of matter

Revision questions with answers for 6.3 Particle model of matter in AQA GCSE Combined Science: Trilogy (spec 8464). Practise with flashcards, a topic quiz, a random quiz across the course, 5 crosswords, or read every question and answer below.

All questions and answers

Question 1 · Spec 6.3.1.1
Define density and give its equation.
Answer:
Density is mass per unit volume: density = mass / volume.
Question 2 · Spec 6.3.1.1
What are the standard units for density, mass and volume?
Answer:
Density: kg/m³; mass: kg; volume: m³.
Question 3 · Spec 6.3.1.1
Find density for a 0.54 kg block of volume 0.00020 m³.
Answer:
Density = 0.54 / 0.00020 = 2,700 kg/m³.
Question 4 · Spec 6.3.1.1
Find the mass of a material with density 800 kg/m³ and volume 0.0050 m³.
Answer:
Mass = density x volume = 800 × 0.0050 = 4.0 kg.
Question 5 · Spec 6.3.1.1
Find the volume of 1.2 kg of a material with density 600 kg/m³.
Answer:
Volume = mass / density = 1.2 / 600 = 0.0020 m³.
Question 6 · Spec 6.3.1.1
Convert 1.0 g/cm³ to kg/m³.
Answer:
1.0 g/cm³ = 1,000 kg/m³.
Question 7 · Spec 6.3.1.1
Describe particle arrangement and motion in a solid.
Answer:
Particles are closely packed in fixed positions and vibrate about those positions.
Question 8 · Spec 6.3.1.1
Describe particle arrangement and motion in a liquid.
Answer:
Particles remain close together but are irregularly arranged and move past one another.
Question 9 · Spec 6.3.1.1
Describe particle arrangement and motion in a gas.
Answer:
Particles are far apart and move rapidly and randomly in all directions.
Question 10 · Spec 6.3.1.1
Why are gases much less dense than solids or liquids?
Answer:
There are far fewer particles, and therefore less mass, in each unit volume.
Question 11 · Spec 6.3.1.1
Does every substance become denser when it freezes?
Answer:
No. Density depends on particle arrangement; for example, ice is less dense than liquid water.
Question 12 · Spec 6.3.1.2
Why is mass conserved during a change of state?
Answer:
The same particles remain present; their arrangement and motion change, not their total mass.
Question 13 · Spec 6.3.1.2
Why are changes of state physical changes?
Answer:
No new substance forms, and reversing the change restores the material's original properties.
Question 14 · Spec 6.3.1.2
What is melting, and what is its reverse?
Answer:
Melting is solid to liquid; freezing is liquid to solid.
Question 15 · Spec 6.3.1.2
How does evaporation differ from boiling?
Answer:
Evaporation occurs at a liquid's surface over a range of temperatures; boiling occurs throughout at its boiling temperature under the given pressure.
Question 16 · Spec 6.3.1.2
What is condensation?
Answer:
The change from gas to liquid, transferring energy to the surroundings.
Question 17 · Spec 6.3.1.2
What is sublimation?
Answer:
A change directly from solid to gas without passing through the liquid state.
Question 18 · Spec 6.3.1.1 · Required practical
How can the density of a regular solid be measured?
Answer:
Measure mass with a balance, calculate volume from measured dimensions, then divide mass by volume.
Question 19 · Spec 6.3.1.1 · Required practical
How can the density of an irregular solid be measured?
Answer:
Measure its mass and find volume from displaced water, then calculate mass / volume.
Question 20 · Spec 6.3.1.1 · Required practical
How can the density of a liquid be measured?
Answer:
Measure a known volume and its mass, subtracting the empty container's mass, then calculate mass / volume.
Question 21 · Spec 6.3.1.1 · Required practical
Why tare a balance with the empty liquid container?
Answer:
So the reading gives only the liquid's mass rather than the liquid plus container.
Question 22 · Spec 6.3.1.1 · Required practical
How should a measuring cylinder's water level be read?
Answer:
Place it upright on a level surface and read the bottom of the meniscus at eye level.
Question 23 · Spec 6.3.1.1 · Required practical
Why use callipers or a micrometer for a small solid's dimensions?
Answer:
Their finer resolution can reduce the relative uncertainty compared with a ruler.
Question 24 · Spec 6.3.1.1 · Required practical
Why must trapped air bubbles be removed during displacement measurements?
Answer:
They increase the apparent displaced volume, causing the calculated density to be too low.
Question 25 · Spec 6.3.1.1 · Required practical
Why measure a block's dimensions at several positions?
Answer:
Repeated measurements reveal irregularity and reduce the effect of random reading errors on the estimated volume.
Question 26 · Spec 6.3.1.1 · Required practical
A 54 g object raises water from 40 to 60 cm³. Find density.
Answer:
Volume = 20 cm³; density = 54 / 20 = 2.7 g/cm³.
Question 27 · Spec 6.3.2.1
What is internal energy?
Answer:
The total kinetic and potential energy of all the particles in a system.
Question 28 · Spec 6.3.2.1
What particle property is linked to temperature?
Answer:
The particles' average kinetic energy.
Question 29 · Spec 6.3.2.1
What can happen when energy is transferred into a substance by heating?
Answer:
Its temperature can increase, or its state can change without a temperature increase.
Question 30 · Spec 6.3.2.1
Do two objects at the same temperature necessarily have the same internal energy?
Answer:
No. Internal energy also depends on particle number, substance and state, not just average kinetic energy.
Question 31 · Spec 6.3.2.2
Define specific heat capacity.
Answer:
Energy required to raise the temperature of 1 kg of a substance by 1 °C.
Question 32 · Spec 6.3.2.2
State the equation for heating without a change of state.
Answer:
E = mc x temperature change; c is in J/(kg °C).
Question 33 · Spec 6.3.2.2
Find energy to heat 2.0 kg by 15 °C if c = 500 J/(kg °C).
Answer:
E = 2.0 × 500 × 15 = 15,000 J.
Question 34 · Spec 6.3.2.2
A 1.0 kg material rises 20 °C after receiving 9,000 J. Find c.
Answer:
c = 9,000 / (1.0 × 20) = 450 J/(kg °C).
Question 35 · Spec 6.3.2.2
Find the temperature rise when 6,000 J heats 0.50 kg with c = 400 J/(kg °C).
Answer:
Rise = 6,000 / (0.50 × 400) = 30 °C.
Question 36 · Spec 6.3.2.2
Equal masses receive equal energy. Which warms more: the material with high or low specific heat capacity?
Answer:
The material with lower specific heat capacity has the greater temperature rise.
Question 37 · Spec 6.3.2.3
Define specific latent heat.
Answer:
Energy required to change the state of 1 kg of a substance without changing its temperature.
Question 38 · Spec 6.3.2.3
What is specific latent heat of fusion?
Answer:
Energy per kilogram for changing between solid and liquid at constant temperature.
Question 39 · Spec 6.3.2.3
What is specific latent heat of vaporisation?
Answer:
Energy per kilogram for changing between liquid and gas at constant temperature.
Question 40 · Spec 6.3.2.3
State the latent-heat equation and units.
Answer:
E = mL; energy in joules, mass in kilograms and specific latent heat in J/kg.
Question 41 · Spec 6.3.2.3
Find energy to melt 0.20 kg of ice. L = 334,000 J/kg.
Answer:
E = 0.20 × 334,000 = 66,800 J.
Question 42 · Spec 6.3.2.3
Find energy to vaporise 0.050 kg of water. L = 2,260,000 J/kg.
Answer:
E = 0.050 × 2,260,000 = 113,000 J.
Question 43 · Spec 6.3.2.3
What does a flat section on a pure substance's heating curve indicate?
Answer:
A change of state while temperature remains constant, despite continued energy transfer.
Question 44 · Spec 6.3.2.3
Why does temperature stay constant during melting?
Answer:
Supplied energy changes the particles' potential energy by overcoming attractions, rather than increasing average kinetic energy.
Question 45 · Spec 6.3.2.3
What happens to internal energy during freezing at constant temperature?
Answer:
It decreases as energy is transferred to the surroundings while the particles form the solid arrangement.
Question 46 · Spec 6.3.3.1
How do gas particles move?
Answer:
They are in constant random motion and frequently collide with each other and their container.
Question 47 · Spec 6.3.3.1
How do gas particles produce pressure?
Answer:
Collisions with container walls exert forces; the total force per unit area gives pressure.
Question 48 · Spec 6.3.3.1
Why does heating a fixed mass of gas at constant volume increase pressure?
Answer:
Particles move faster, colliding with walls more often and with greater force, increasing pressure.
Question 49 · Spec 6.3.3.1
Why does cooling gas in a rigid sealed container lower pressure?
Answer:
Particles have lower average kinetic energy, causing less frequent and weaker wall collisions.
Question 50 · Spec 6.3.2.3
How do the equations for specific heat capacity and specific latent heat differ in use?
Answer:
Use E = mc x temperature change for a temperature change; use E = mL for a state change at constant temperature.

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