(a) Using the kinetic theory of matter, explain why; (i) Evaporation causes cooling (ii) Boiling water changes to steam without any change in temperature, a...

(a) Kinetic theory of matter

(i) Evaporation causes cooling: Molecules in a liquid are in continual random motion and possess different kinetic energies. The molecules with the greatest kinetic energy near the surface escape from the liquid during evaporation. Their escape reduces the average kinetic energy of the molecules left in the liquid. Since temperature is proportional to the average kinetic energy of the molecules, the temperature of the liquid falls; hence evaporation causes cooling.

(ii) Water boils without a rise in temperature: At its boiling point, the heat supplied to water is used as latent heat of vaporization. It is used to overcome the attractive intermolecular forces and to separate the water molecules to form steam. It does not increase the average kinetic energy of the molecules. Therefore, the temperature remains constant until all the water has changed to steam.

(b)(i) Boyle's law

For a fixed mass of gas at constant temperature, the volume of the gas is inversely proportional to its pressure.

Thus,

\[V\propto\frac{1}{P}\qquad\text{or}\qquad PV=\text{constant}.\]

(b)(ii) Experiment

The apparatus is arranged as shown below. A quantity of dry air is trapped in the graduated burette B by closing the tap. The burette is kept vertical. The mercury reservoir R is raised slowly to compress the trapped air. For each setting, read the volume, \(V\), of trapped air in the burette and the difference, \(h\), between the mercury levels. The barometer reading, \(H\), is also noted.

When the mercury level in the reservoir is higher than that in the burette, the pressure of the trapped air is:

\[P=H+h.\]

For example, if \(H=760\ \text{mmHg}\), the following readings may be obtained:

A plot of \(P\) against \(1/V\) is a straight line through the origin:

The gradient of the graph is \(22\,800\ \text{mmHg cm}^3\). Hence \(P=(22\,800)(1/V)\), so \(PV=22\,800\ \text{mmHg cm}^3\), a constant. This verifies Boyle's law.

(b)(iii) Precautions

1. Use dry air and ensure that the apparatus is airtight, so that the mass of trapped gas remains constant.
2. Raise or lower the reservoir slowly and wait for the gas to return to room temperature before taking each reading.
3. Take all scale readings at eye level to avoid parallax error.