--------------------------------------------------------------------- | Pressure in mm. mercury. ------------------------------------------------------ Temperature. | | | | Water. | Ice. | Difference. --------------------------------------------------------------------- 0° | 4.618 | 4.602 | 0.016[40] -2° | 3.995 | 3.925 | 0.070 -4° | 3.450 | 3.334 | 0.116 -8° | 2.558 | 2.379 | 0.179 -10° | 2.197 | 1.999 | 0.198 -15° | 1.492 | 1.279 | 0.213 -20° | 1.005 | 0.806 | 0.199 --------------------------------------------------------------------- At all temperatures below 0° (more correctly +0.0076°), at which temperature water and ice have the same vapour pressure, the vapour pressure of supercooled water is _greater_ than that of ice at the same temperature. From the relative positions of the curves OB and OA (Fig. 4) we see that at all temperatures above 0°, the (metastable) sublimation curve of ice, if it could be obtained, would be higher than the vaporization curve of water. This shows, therefore, that at 0° a "break" must occur in the curve of states, and that in the neighbourhood of this break the curve above that point must ascend less rapidly than the curve below the break. Since, however, the differences in the vapour pressures of supercooled water and of ice are very small, the change in the direction of the vapour-pressure curve on passing from ice to water was at first not observed, and Regnault regarded the sublimation curve as passing continuously into {32} the vaporization curve. The existence of a break was, however, shown by James Thomson[41] and by Kirchhoff[42] to be demanded by thermo-dynamical considerations, and the prediction of theory was afterwards realized experimentally by Ramsay and Young in their determinations of the vapour pressure of water and ice, as well as in the case of other substances.[43] From what has just been said, we can readily understand why ice and water cannot exist in equilibrium below 0°. For, suppose we have ice and water in the same closed space, but not in contact with one another, then since the vapour pressure of the supercooled water is higher than that of ice, the vapour of the former must be supersaturated in contact with the latter; vapour must, therefore, condense on the ice; and in this way there will be a slow distillation from the water to the ice, until at last all the water will have disappeared, and only ice and vapour remain.[44]