Physics 14.1-14.4 - Particle model and density

The kinetic particle model

Matter can be represented using a kinetic particle model. "Kinetic" refers to movement: the model describes both the arrangement of particles and how they move. A particle may represent an atom or a molecule, depending on the substance.

The circles in a particle diagram are not photographs and are not drawn to scale. They are a model that helps us reason about matter. The gaps are spaces between particles; they are not filled with air unless the substance itself is a mixture containing air.

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The particles in a solid are not motionless. They vibrate while remaining centred on fixed positions. In a liquid, particles remain close but can change neighbours. In a gas, particles are separated by much larger gaps and move freely through the available space.

This lesson's particle model, density equation and core practical are common to both Foundation and Higher tier.

Density as mass per unit volume

Two objects can have the same size but different masses. Density compares them fairly by asking how much mass there is in each unit of volume.

The symbol for density is the Greek letter $\rho$ (rho). Mass is $m$ and volume is $V$.

In a laboratory, mass is often measured in grams and volume in cubic centimetres, so density may be calculated in $\mathrm{g\,cm^{-3}}$. Keep one consistent unit set throughout a calculation:

• kg with $\mathrm{m^3}$ gives $\mathrm{kg\,m^{-3}}$;
• g with $\mathrm{cm^3}$ gives $\mathrm{g\,cm^{-3}}$;
• $1\,\mathrm{g\,cm^{-3}}=1000\,\mathrm{kg\,m^{-3}}$.

For volume conversion, cube the length conversion: $1\,\mathrm{cm}=10^{-2}\,\mathrm{m}$, so $1\,\mathrm{cm^3}=10^{-6}\,\mathrm{m^3}$. Hence $1\,\mathrm{g\,cm^{-3}}=10^{-3}/10^{-6}=10^3\,\mathrm{kg\,m^{-3}}$. Dividing a cubic-centimetre value by 100 alone is not a volume conversion.

You may need either rearranged form:

Worked example: calculating density

A block has a mass of $0.540\,\mathrm{kg}$ and a volume of $2.00 \times 10^{-4}\,\mathrm{m^3}$. Calculate its density.

Solution

1. The values already use kg and $\mathrm{m^3}$, so no conversion is needed.
2. Select the equation: $\rho = m/V$.
3. Substitute: $\rho = 0.540/(2.00 \times 10^{-4})$.
4. Calculate: $\rho = 2700\,\mathrm{kg\,m^{-3}} = 2.70 \times 10^3\,\mathrm{kg\,m^{-3}}$.

The answer has three significant figures, matching the data. A value of a few thousand $\mathrm{kg\,m^{-3}}$ is physically plausible for a dense solid.

Worked example: rearranging to find volume

A metal sample has mass $96.0\,\mathrm{g}$ and density $8.00\,\mathrm{g\,cm^{-3}}$. Find its volume.

Solution

The units are consistent, and multiplying the answer back by the density gives $12.0 \times 8.00 = 96.0\,\mathrm{g}$, so the rearrangement passes a sense-check.

Core practical: density of a solid

The purpose is to determine the density of an irregular, non-porous solid. Its mass is measured directly. Its volume is found from the volume of water it displaces.

Apparatus: a balance, measuring cylinder, water, thin thread and a dry irregular solid that does not dissolve in or absorb water.

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Method

1. Place the dry solid on a zeroed balance and record its mass $m$ in g.
2. Part-fill a measuring cylinder with enough water to cover the solid. At eye level, read the bottom of the water's meniscus and record the initial volume $V_1$ in $\mathrm{cm^3}$.
3. Lower the solid gently into the water using thin thread. Make sure it is fully submerged, with no trapped air bubbles and no water spilled.
4. Read the new volume $V_2$ at eye level using the same part of the meniscus.
5. Calculate the solid's volume: $V = V_2 - V_1$.
6. Calculate its density: $\rho = m/V$.
7. Return the water to a suitable starting level and repeat the displacement measurement at least three times. Investigate unexpected readings, repeat them where possible, and calculate a mean from valid results. Do not discard a value just because it disagrees with your prediction.