4.3 Particle model of matter — printable coverage pack

4.3.1.1 · Density of materials

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.

Tier 1 · Easy

1. A sample has mass $0.54\,\text{kg}$ and volume $2.0\times10^{-4}\,\text{m}^3$ . Calculate its density.[2 marks]
Answer
- $2.7\times10^3\,\text{kg/m}^3$
Method: Use $\rho=m/V$ . Thus $\rho=0.54/(2.0\times10^{-4})=2700\,\text{kg/m}^3=2.7\times10^3\,\text{kg/m}^3$ .

Tier 2 · Standard

1. 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]
Answer
- $2.60\times10^3\,\text{kg/m}^3$
Method: 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 $\rho=0.390/(1.50\times10^{-4})=2.60\times10^3\,\text{kg/m}^3$ .

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

4.3.1.2 · Changes of state

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.
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.

Tier 1 · Easy

1. Name the changes of state from gas to liquid and from solid directly to gas.[2 marks]
Answer
- Gas to liquid: condensation.
- Solid to gas: sublimation.
Method: 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.

Tier 2 · Standard

1. 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]
Answer
- $75.0\,\text{g}$
- Melting only rearranges the particles; the sealed container loses no particles, so mass is conserved.
Method: 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 $75.0\,\text{g}$ .

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

4.3.2.1 · Internal energy

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.

Tier 1 · Easy

1. Define the internal energy of a system.[2 marks]
Answer
- The total kinetic energy and potential energy of all the particles in the system.
Method: Include both microscopic stores in the definition: energy from particle motion is kinetic, and energy from particle positions or interactions is potential.

Tier 2 · Standard

1. A solid is heated but does not melt. Explain how its particles and internal energy change.[3 marks]
Answer
- 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.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

4.3.2.2 · Temperature changes in a system and specific heat capacity

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.
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.

Tier 1 · Easy

1. State the meaning of specific heat capacity and give its unit.[2 marks]
Answer
- The energy required to raise the temperature of $1\,\text{kg}$ of a substance by $1\,{}^\circ\text{C}$ .
- $\text{J/(kg}\,{}^\circ\text{C)}$
Method: State the fixed mass and fixed temperature rise, then attach the compound unit: joules per kilogram per degree Celsius.

Tier 2 · Standard

1. 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]
Answer
- $3.2\times10^4\,\text{J}$
Method: Use $\Delta E=mc\Delta\theta$ . Therefore $\Delta E=2.0\times450\times35=31\,500\,\text{J}$ , which to two significant figures is $3.2\times10^4\,\text{J}$ .

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

4.3.2.3 · Changes of state and specific latent heat

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.

Tier 1 · Easy

1. Define specific latent heat.[2 marks]
Answer
- The energy required to change the state of $1\,\text{kg}$ of a substance with no change in temperature.
Method: 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.

Tier 2 · Standard

1. 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]
Answer
- $8.7\times10^4\,\text{J}$
Method: Use $E=mL$ . Hence $E=0.28\times3.10\times10^5=86\,800\,\text{J}$ , which to two significant figures is $8.7\times10^4\,\text{J}$ .

Tier 3 · Hard

1. 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]
Answer
- $3.9\times10^5\,\text{J}$
Method: 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.

4.3.3.1 · Particle motion in gases

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.
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.

Tier 1 · Easy

1. State how the average kinetic energy of gas molecules changes when the gas temperature increases.[1 mark]
Answer
- The average kinetic energy increases.
Method: Temperature is linked to the average kinetic energy of the molecules, so a higher temperature means a greater average kinetic energy.

Tier 2 · Standard

1. A sealed rigid flask of gas is heated. Explain why the gas pressure increases.[4 marks]
Answer
- 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.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

4.3.3.2 · Pressure in gases (physics only)

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.
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.

Tier 1 · Easy

1. A fixed mass of gas is kept at constant temperature while its volume doubles. State what happens to its pressure.[2 marks]
Answer
- The pressure halves.
Method: At constant temperature $pV$ is constant. If $V$ is multiplied by $2$ , $p$ must be multiplied by $1/2$ to keep the product unchanged.

Tier 2 · Standard

1. 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]
Answer
- $3.2\times10^5\,\text{Pa}$
Method: Use $p_1V_1=p_2V_2$ . Therefore $p_2=(1.2\times10^5\times4.0\times10^{-3})/(1.5\times10^{-3})=3.2\times10^5\,\text{Pa}$ .

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.

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

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.

Tier 1 · Easy

1. Explain why the air in a sealed bicycle pump can become warmer when the handle is pushed in quickly.[2 marks]
Answer
- The handle does work on the enclosed gas.
- This increases the gas's internal energy and can increase its temperature.
Method: 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.

Tier 2 · Standard

1. $180\,\text{J}$ 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]
Answer
- The internal energy increases by $180\,\text{J}$ .
- The temperature is likely to increase because the molecules gain average kinetic energy.
Method: 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.

Tier 3 · Hard

1. 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]
Answer
- 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.
Method: 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.