Let the atmospheric pressure be A, let the air pressure in the tube be p and the s.v.p. of water at that temperature be s. Neglecting the weight of the small amount of water in the plug, we have

Therefore:

p = A - s

Now saturated vapours do not obey Boyle's law, because their pressure is constant at a given temperature and is independent of the volume of air. Therefore for the air in the tube we have:

(A - s)V/T = constant

where V is the volume of air in the tube and T is its temperature.

We can therefore measure the s.v.p. of water if its value at one temperature is known.

The results obtained for water are shown in Figure 2(a). Notice that the s.v.p. rises with temperature as predicted, and that its value at 100

Now when the liquid boils bubbles rise to the surface of the liquid and escape so the pressure within them must be equal to the external pressure (Figure 2(b).

We can say, therefore, that: