5 Solids, liquids and gases

The powers show what has been measured: length uses m, a surface area uses m² and a three-dimensional volume uses m³. Division means ‘per’: kg/m³ is kilograms per cubic metre, while m/s is metres travelled per second. In m/s², the speed changes by so many metres per second during each second.

$\rho$ is density, $m$ is mass and $V$ is volume. Useful rearrangements are $m=\rho V$ and $V=m/\rho$. With mass in kilograms and volume in cubic metres, density is in kg/m³; with grams and cubic centimetres, it is in g/cm³.

Example: a solid has mass 540 g and volume 200 cm³. $\rho=540/200=2.70$ g/cm³. The mass and volume units already match, so no conversion is needed; the density unit follows from g divided by cm³.

For displacement, lower the object gently, remove trapped air and avoid losing water by splashing. Use a cylinder with a scale suited to the volume. Repeat measurements where practical, calculate density for each repeat and check any anomalous result before finding a mean.

$p$ is pressure in pascals (Pa), $F$ is the force normal to the surface in newtons (N), and $A$ is the contact area in square metres (m²). Rearrangements are $F=pA$ and $A=F/p$. One pascal is one newton per square metre.

Example: a 600 N force acts normally over 0.030 m². $p=600/0.030=20\,000$ Pa. If the same force acted over twice the area, the pressure would halve.

A small stationary air bubble under water is pushed inward from all sides. Likewise, identical holes at the same depth on opposite sides of a container, both opening to the same outside pressure, have the same pressure difference; water begins to leave with the same speed in opposite directions.

$p$ is the pressure difference in pascals (Pa), $h$ is the vertical height difference in metres (m), $\rho$ is density in kg/m³, and $g$ is gravitational field strength in N/kg. A useful rearrangement is $h=p/(\rho g)$.

Example: for water with $\rho=1000$ kg/m³ and $g=10$ N/kg, the pressure difference over 0.60 m is $p=0.60\times1000\times10=6000$ Pa. The deeper point has the greater pressure.

Once a change of state has finished, further heating again increases average kinetic energy, so the temperature of the new state rises. Throughout, energy is conserved: it has been transferred into the system even when the thermometer does not rise.

During each change, energy transferred to the substance weakens or overcomes attractive forces between particles. During melting and boiling, the temperature stays constant until all of the sample has changed state.

Because solid particles keep fixed neighbours, a solid keeps its shape. Liquid particles remain close but can change neighbours, so a liquid flows while keeping nearly the same volume. Widely separated gas particles move through the available space, so a gas fills its container.

Method: 1. Place a suitable pure sample with a temperature probe in a heat-safe container. 2. Heat gently at steady power, starting below its melting point. 3. Start a stopwatch and record temperature at equal time intervals. 4. Stir gently once the sample is liquid, where safe, so the measured temperature is representative. 5. Continue until the state change is complete and the new state warms. 6. Plot temperature on the vertical axis against time on the horizontal axis.

Before the plateau the solid warms; at the plateau solid and liquid coexist and the temperature gives the melting point; after the plateau the liquid warms. A cooling experiment gives the corresponding constant-temperature section while the liquid freezes.

A larger specific heat capacity means more energy is required for the same mass and temperature rise. Therefore, when equal masses receive the same energy, the substance with the larger specific heat capacity has the smaller temperature increase.

A value of 900 J/(kg °C) means 900 J must be transferred to 1 kg of the substance to raise its temperature by 1 °C. Raising 1 kg by 2 °C would require twice that energy if the state does not change.

$\Delta Q$ is the change in thermal energy in joules (J), $m$ is mass in kilograms (kg), $c$ is specific heat capacity in J/(kg °C), and $\Delta T$ is the temperature change in °C. Rearrangements include $c=\Delta Q/(m\Delta T)$ and $m=\Delta Q/(c\Delta T)$.

Example: a 0.80 kg block with $c=900$ J/(kg °C) warms from 20 °C to 35 °C. $\Delta T=35-20=15$ °C, so $\Delta Q=0.80\times900\times15=10\,800$ J.

Record the initial temperature, heat for a measured time, then record the highest uniform temperature and calculate $\Delta T$. Repeat the run and compare values. For comparisons between materials, keep mass, input power, heating time, insulation and probe placement consistent where possible.

Use the full causal chain: random molecular motion → collisions with the walls → momentum changes and forces on the walls → total force per unit area, which is pressure. Because motion is random, collisions occur on every wall and a gas at rest exerts pressure in all directions.

Cooling a gas reduces its molecules' average kinetic energy and average speed. In the ideal particle model, extrapolating this trend reaches its minimum at absolute zero; going below would require a negative average kinetic energy, which is impossible.

On a pressure–Celsius-temperature graph for a fixed mass of gas at constant volume, extending the straight line to zero pressure gives an estimate near −273 °C. Experimental data may give a nearby value rather than exactly −273 °C.

Example: 15 °C = 15 + 273 = 288 K. In reverse, 358 K = 358 − 273 = 85 °C.

Gas molecules do not all travel at one speed. At any instant there is a range of speeds because collisions continually redistribute energy. Heating shifts the distribution toward higher speeds, which is why the word average matters.

At fixed volume, faster molecules reach the walls more often and undergo larger momentum changes in collisions. This connects a temperature increase to a pressure increase, although that pressure relationship is developed separately.

If the Kelvin temperature doubles, the average kinetic energy doubles. A graph of average kinetic energy against Kelvin temperature is a straight line through the origin: both reach their ideal-model minimum at 0 K.

For molecules of one gas, $E_k=\tfrac12mv^2$. A greater average kinetic energy therefore means a greater average molecular speed, but speed is not directly proportional to temperature because kinetic energy depends on speed squared.

The reverse changes follow the same mechanisms: increasing volume at constant temperature reduces collision frequency and pressure; decreasing temperature at constant volume produces slower, less frequent and less forceful wall collisions, so pressure falls.

Method: 1. Convert every Celsius temperature to kelvin. 2. Substitute matching state-1 and state-2 values. 3. Rearrange for the unknown. 4. Keep pressure units consistent and check that the direction is sensible.

Example: a gas at constant volume has pressure $9.95\times10^4$ Pa at 16 °C. At 32 °C, $T_1=289$ K and $T_2=305$ K. Therefore $p_2=(9.95\times10^4\times305)/289=1.05\times10^5$ Pa. The pressure rises because the Kelvin temperature rises.

Method: 1. Pair each pressure with its volume. 2. Substitute into $p_1V_1=p_2V_2$. 3. Rearrange for the unknown. 4. Use one pressure unit and one volume unit consistently on both sides, then check the inverse trend.

Example: gas at 101 kPa expands from 110 cm³ to 140 cm³ at constant temperature. $p_2=(101\times110)/140=79.4$ kPa. The larger volume gives a lower pressure; if the volume were halved, the pressure would double.