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How does the particle model explain density, and how do we measure it?

Density of materials: the density equation, how the particle model explains the densities of solids, liquids and gases, and the required practical to measure density.

A focused answer to AQA GCSE Physics 4.3.1, covering the density equation, how the arrangement of particles in solids, liquids and gases explains their relative densities, and the required practical for measuring density.

Generated by Claude Opus 4.88 min answer

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  1. What this dot point is asking
  2. The density equation
  3. The particle model and density
  4. Required practical: measuring density
  5. Try this

What this dot point is asking

AQA wants you to use the density equation, explain the differences in density between solids, liquids and gases using the particle model, and describe the required practical for measuring the density of a material. This is part of topic 4.3.1 of the AQA GCSE Physics (8463) specification, and the density measurement is a named required practical.

The density equation

The particle model and density

The same material has the same number of particles whether solid, liquid or gas, but spreading them further apart increases the volume and lowers the density. This is why a substance is usually densest as a solid and least dense as a gas. Water is a famous exception: ice is slightly less dense than liquid water because the particles arrange themselves in an open structure on freezing, which is why ice floats. Density also explains everyday floating and sinking: an object floats on a liquid if its density is less than that of the liquid, and sinks if it is greater. A consistent set of units is essential, so kgkg with m3m^3 gives kg/m3kg/m^3, while gg with cm3cm^3 gives g/cm3g/cm^3; the two systems differ by a factor of 10001000.

Required practical: measuring density

You find the mass with a balance, and the volume in one of two ways:

  • For a regular solid, measure its dimensions with a ruler or vernier callipers and calculate the volume (for example length by width by height for a cuboid).
  • For an irregular solid, lower it into a displacement (eureka) can filled to the spout with water, and collect and measure the volume of water that overflows; this displaced volume equals the volume of the solid.

To measure the density of a liquid, you find the mass of an empty measuring cylinder, pour in a known volume of the liquid, find the new mass, and subtract to get the mass of the liquid alone. Then apply ρ=mV\rho = \dfrac{m}{V} in every case. The main sources of error are reading the measuring scale (reduced by reading at eye level to avoid parallax) and, for the displacement method, water clinging to the object or splashing out, both of which give a slightly wrong volume and hence a slightly wrong density.

A worked appreciation of scale helps: lead has a density of about 11,300 kg/m311{,}300\,kg/m^3 while air has a density of only about 1.2 kg/m31.2\,kg/m^3, a difference of nearly ten thousand times, which is why a small lead weight feels so heavy and a large balloon of air feels almost weightless. Comparing a calculated density against known values like these is a quick sanity check in the exam.

Try this

Q1. State the density equation and the units of each quantity. [2 marks]

  • Cue. ρ=mV\rho = \dfrac{m}{V}, with ρ\rho in kg/m3kg/m^3, mm in kgkg and VV in m3m^3.

Q2. A block has a mass of 240 g240\,g and a volume of 30 cm330\,cm^3. Calculate its density. [2 marks]

  • Cue. ρ=24030=8 g/cm3\rho = \dfrac{240}{30} = 8\,g/cm^3.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20194 marksA student measures an irregular stone. Its mass is 150 g150\,\text{g}. When lowered into a measuring cylinder containing 80 cm380\,\text{cm}^3 of water, the water level rises to 140 cm3140\,\text{cm}^3. Calculate the density of the stone in g/cm3\text{g/cm}^3.
Show worked answer →

The volume of the stone equals the volume of water it displaces, which is the rise in the water level: 140−80=60 cm3140 - 80 = 60\,\text{cm}^3 (2 marks). Then apply the density equation ρ=m/V=150/60=2.5 g/cm3\rho = m / V = 150 / 60 = 2.5\,\text{g/cm}^3 (2 marks). Markers reward finding the displaced volume as the difference in levels (not the final level) and the correct density with units. A common error is to use 140 cm3140\,\text{cm}^3 as the volume rather than the rise of 60 cm360\,\text{cm}^3. This displacement method is the standard required-practical technique for an irregular solid.

AQA 20214 marksUsing the particle model, explain why a gas has a much lower density than the same substance in its solid state, even though the number of particles is unchanged.
Show worked answer →

In the solid state the particles are packed closely together in a regular arrangement, so a large number of particles occupies a small volume, giving a high density (1 mark). When the substance becomes a gas, the same particles spread out and are very far apart from one another (1 mark), so the same number (and therefore the same mass) of particles now occupies a much larger volume (1 mark). Since density is mass divided by volume, the much larger volume for the same mass means the density is much lower (1 mark). Markers reward stating the particles are unchanged in number, the much greater spacing and volume in the gas, and the link to density as mass per unit volume.

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