Edexcel Separate Sciences · Physics · Paper 2

SP14 · Particle modelTopic 14 — Particle model

Density, temperature and changes of state

Revise the key ideas

Particles, states and density

  • In a solid, particles are closely packed and vibrate about fixed positions; a solid has a fixed shape and volume. In a liquid, close particles move past each other, giving fixed volume but variable shape.
  • In a gas, particles are widely separated and move rapidly in random directions; a gas fills its container and is readily compressed. Particles themselves do not swell when the gas expands.
    Particle arrangementsSolid: close regular particles; liquid: close irregular particles; gas: widely separated particles.SolidLiquidGas
    Gas particles are further apart; their individual sizes need not change.
  • Density is mass per unit volume: ρ = m/V. Use mass in kg and volume in m³ for density in kg/m³; g and cm³ instead give g/cm³.
  • Rearrange to m = ρV and V = m/ρ. Convert carefully: 1 g/cm³ = 1000 kg/m³; 1 cm³ = 10⁻⁶ m³.
  • The same material generally has lower density as a gas because particles are much further apart. Solids are often denser than liquids, but ice is less dense than liquid water because of its more open structure.
  • If mass stays constant while volume increases, density decreases. Changes of state conserve mass in a closed system even when volume changes.

Density of solids and liquids: core practical

  • Measure an object's mass using a zeroed balance. For a regular block, measure length, width and height and calculate volume = length × width × height.
  • For an irregular solid that does not absorb water (a non-porous solid), measure its volume by water displacement using a measuring cylinder or overflow can. Fully submerge it and avoid trapped bubbles. The volume of water displaced equals the solid’s volume.
    Displacement volumeCylinder reading changes from 40 to 60 cubic centimetres when the object is fully submerged.40 cm³60 cm³Object volume = 60 − 40 = 20 cm³
    Subtract the starting volume; avoid trapped bubbles.
  • Choose a method suitable for the material: an object that dissolves, absorbs water or floats needs a different liquid or an adapted procedure. Do not simply call a partly submerged volume its whole volume.
  • For a liquid, weigh the empty dry container and then container plus a measured liquid volume. Subtract the empty-container mass before calculating density.
  • Read liquid volume at eye level. For water, read the bottom of the curved surface (meniscus). Use an instrument with suitable scale divisions (resolution) and repeat readings.
  • Use the same unit system throughout. A 54 g block with volume 20 cm³ has density 2.7 g/cm³, equal to 2700 kg/m³.

Temperature, internal energy and heat capacity

  • Temperature is linked to the average kinetic energy of a substance’s particles. Internal energy is the total kinetic and potential energy of all its particles. It depends on how much substance there is, the material, its temperature and its state.
  • Heating transfers energy into a system, usually increasing particle kinetic energy and temperature, or changing particle potential energy during a state change. A temperature is not an amount of energy.
  • Specific heat capacity c is the energy needed to raise the temperature of 1 kg of a substance by 1 °C (or 1 K). Units are J/(kg °C), equivalently J/(kg K).
  • Change in thermal energy ΔQ = mcΔθ, with m in kg, c in J/(kg °C) and temperature change Δθ in °C. A temperature increase of 1 °C equals an increase of 1 K.
  • For 0.5 kg water, c = 4200 J/(kg °C) and a 10 °C rise, ΔQ = 0.5 × 4200 × 10 = 21000 J.
    Water heating calculationm = 0.5 kg; c = 4200 J/(kg °C); Δθ = 10 °C → ΔQ = mcΔθ = 0.5 × 4200 × 10 → Energy required = 21000 Jm = 0.5 kg; c = 4200 J/(kg °C); Δθ = 10 °CΔQ = mcΔθ = 0.5 × 4200 × 10Energy required = 21000 J
    Specific heat capacity includes “per kilogram” in its units.
  • A material with greater c needs more energy for the same mass and temperature rise. A larger mass also needs more energy; c is a material property, not the mass itself.
  • Insulation, a lid and reducing exposed area limit unwanted energy transfers during heating or cooling. Low thermal conductivity and trapped air can reduce cooling.

Changes of state and latent heat

  • Melting is solid to liquid; freezing is liquid to solid; boiling/evaporation is liquid to gas; condensation is gas to liquid; sublimation is solid to gas; deposition is gas to solid.
  • These are physical changes: the material can recover its original properties when reversed. In a closed system, mass is conserved; the volume may change.
  • During melting or boiling of a pure substance at fixed pressure, temperature stays constant while supplied energy changes the particle arrangements and potential energy rather than average kinetic energy.
  • Specific latent heat L is energy needed to change the state of 1 kg without changing temperature, in J/kg. Latent heat of fusion refers to melting/freezing; latent heat of vaporisation refers to liquid/gas change.
  • Use Q = mL, with mass in kg. Melting 0.2 kg ice with L = 334000 J/kg requires 66800 J for the state change alone.
  • If a question includes warming and a state change, calculate separate stages using mcΔθ and mL, then add energies. Do not use mL for an ordinary temperature rise.
  • During cooling, freezing and condensation release energy to surroundings. A flat region on a heating/cooling curve identifies a state-change stage under the stated conditions.
    Heating curve stagesTemperature rises in solid, stays flat during melting, rises in liquid, stays flat during boiling, then rises in gas.TemperatureEnergy suppliedMeltingBoilingSolidLiquidGas
    Flat stages show energy used for a state change, not a temperature rise.
  • Evaporation can occur at the liquid surface below the boiling point. Boiling occurs throughout the liquid at the boiling temperature for the pressure.

Water properties: core practical

  • To estimate water's specific heat capacity, measure its mass, initial temperature and the energy transferred by a low-voltage immersion heater over a timed interval.
  • Measure heater voltage and current and use E = VIt, or use a suitable joulemeter. Stir gently for uniform temperature and record the final temperature; calculate c = E/(mΔθ).
    Water heat capacity setupInsulated water container has immersed heater, temperature probe and stirring rod; heater energy and water temperature change are measured.Low-voltage heaterProbeMeasure mass, energy input and temperature rise
    Keep the heater immersed and use insulation to reduce heat loss.
  • Insulate the container and use a lid to reduce losses. Keep the heater immersed and follow electrical safety procedures; avoid contact with hot water or heaters.
  • Energy heating the container or escaping to surroundings can make calculated c too high if all supplied energy is incorrectly attributed to the water. Use appropriate corrections or discuss this limitation.
  • For a melting-ice temperature–time graph, place a temperature probe in ice/water and record at regular intervals while energy is supplied. Mix carefully and keep the probe away from the heater and vessel wall.
  • At ordinary pressure, a pure ice–water mixture stays near 0 °C while the ice melts. Once all the ice has melted, the water warms up. A real graph may not have a perfectly flat section (plateau) because heat flow and the thermometer’s response affect the readings.

Gas pressure and absolute temperature

  • Gas particles collide with container walls and exert forces on them. Pressure is force per area and is measured in pascals (Pa).
  • For a fixed mass of gas at constant volume, raising temperature increases average particle speed. More frequent and stronger wall impacts increase pressure.
    Gas pressure modelGas particles move randomly and collide with the container walls; higher temperature increases speed and pressure at fixed volume.Wall collisions transfer momentum and exert force
    Fixed volume and fixed gas mass are essential to the temperature comparison.
  • In the ideal particle model, absolute zero corresponds to no random thermal motion and zero extrapolated gas pressure at fixed volume. It is approximately −273 °C or 0 K; real gases condense before reaching it.
  • Convert approximately using T(K) = θ(°C) + 273 and θ(°C) = T(K) − 273. Kelvin temperatures are written without a degree symbol.
  • 20 °C is approximately 293 K. A 10 °C temperature increase is a 10 K increase; adding 273 is for converting absolute readings, not temperature differences.
  • The simple gas model has limits: particles have finite size, attractions matter and phase changes occur. Do not extrapolate an ordinary gas-pressure experiment as though the gas stays gaseous down to absolute zero.

Gas compression and Boyle’s law

  • A gas is compressible because there is substantial space between its particles. Pressure produces a net force perpendicular to a surface through particle collisions; it does not act only downwards.
  • For a fixed mass of gas at constant temperature, reducing volume increases collision frequency with the container walls and increases pressure. Pressure is inversely proportional to volume: pV is constant.
    Pressure–volume curveFor fixed gas mass and temperature, relative pressure is reciprocal relative volume. Halving volume doubles pressure.p/p₀V/V₀01212
    The curve follows p/p₀ = 1/(V/V₀), rather than a straight-line fall.
  • Use p₁V₁ = p₂V₂ with consistent pressure and volume units. Use absolute pressure, which includes atmospheric pressure. If a pressure gauge shows pressure above atmospheric pressure (gauge pressure), add atmospheric pressure first.
  • The fixed-mass and constant-temperature conditions matter. A leaking sample changes its particle number; rapid compression can heat a gas and depart from the constant-temperature relationship.
  • (Higher tier) Doing work on a gas can increase its internal energy and temperature (Higher tier). A bicycle pump can warm during compression, so allow gas to return to the controlled temperature when investigating Boyle’s law.
  • In a pressure-volume investigation, use suitable sealed apparatus, vary volume, allow temperature to settle and record corresponding pressure. Plot pressure against 1/volume to test the inverse relationship; follow the apparatus pressure limits.

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