above the critical point the vapour becomes a gas. In the case of water, the critical temperature is 364.3°, and the critical pressure 194.6 atm.; at the point representing these conditions the vapour-pressure curve of water must cease.
Sublimation Curve of Ice.—Vapour is given off not only by liquid water, but also by solid water, or ice. That this is so is familiar to every one through the fact that ice or snow, even at temperatures below the melting point, gradually disappears in the form of vapour. Even at temperatures considerably lower than 0°, the vapour pressure of ice, although small, is quite appreciable; and it is possible, therefore, to have ice and vapour coexisting in equilibrium. When we inquire into the conditions under which such a system can exist, we see again that we are dealing with a univariant system—one component existing in two phases—and that, therefore, just as in the case of the system water and vapour, there will be for each temperature a certain definite pressure of the vapour, and this pressure will be independent of the relative or absolute amounts of the solid or vapour present, and will depend solely on the temperature. Further, just as in the case of the vapour pressure of water, the condition of equilibrium between ice and water vapour will be represented by a line or curve showing the change of pressure with the temperature. Such a curve, representing the conditions of equilibrium between a solid and its vapour, is called a sublimation curve. At temperatures represented by any point on this curve, the solid (ice) will sublime or pass into vapour without previously fusing. Since ice melts at 0° (vide infra), the sublimation curve must end at that temperature.
The following are the values of the vapour pressure of ice between 0° and -50°.[30]
Vapour Pressure of Ice.
| Temperature. | Pressure in mm. mercury. | Temperature. | Pressure in mm. mercury. |
| -50° | 0.050 | -8° | 2.379 |
| -40° | 0.121 | -6° | 2.821 |
| -30° | 0.312 | -4° | 3.334 |
| -20° | 0.806 | -2° | 3.925 |
| -15° | 1.279 | 0° | 4.602 |
| -10° | 1.999 |
Equilibrium between Ice and Water. Curve of Fusion.—There is still another univariant system of the one component water, the existence of which, at definite values of temperature and pressure, the Phase Rule allows us to predict. This is the system solid—liquid. Ice on being heated to a certain temperature melts and passes into the liquid state; and since this system solid—liquid is univariant, there will be for each temperature a certain definite pressure at which ice and water can coexist or be in equilibrium, independently of the amounts of the two phases present. Since now the temperature at which the solid phase is in equilibrium with the liquid phase is known as the melting point or point of fusion of the solid, the curve representing the temperatures and pressures at which the solid and liquid are in equilibrium will represent the change of the melting point with the pressure. Such a curve is called the curve of fusion, or the melting-point curve.
It was not until the middle of the nineteenth century that this connection between the pressure and the melting point, or the change of the melting point with the pressure, was observed. The first to recognize the existence of such a relationship was James Thomson,[31] who in 1849 showed that from theoretical considerations such a relationship must exist, and predicted that in the case of ice the melting point would be lowered by pressure. This prediction was fully confirmed by his brother, W. Thomson[32] (Lord Kelvin), who found that under a pressure of 8.1 atm. the melting point of ice was -0.059°; under a pressure of 16.8 atm. the melting point was -0.129°.
The experiments which were first made in this connection were more of a qualitative nature, but in recent years careful measurements of the influence of pressure on the melting point of ice have been made more especially by Tammann,[33] and the results obtained by him are given in the following table and represented graphically in Fig. 2.
Fusion Pressure of Ice.
| Temperature. | Pressure in kilogms. per sq. cm.[34] | Change of melting point for an increase of pressure of 1 kilogm. per sq. cm. |
| -0° -2.5° -5° -7.5° -10.0° -12.5° -15.0° -17.5° -20.0° -22.1° | 1 336 615 890 1155 1410 1625 1835 2042 2200 | 0.0074° 0.0090° 0.0091° 0.0094° 0.0100° 0.0116° 0.0119° 0.0121° 0.0133° |
From the numbers in the table and from the figure we see that as the pressure is increased the melting point of ice is lowered; but we also observe that a very large change of pressure is required in order to produce a very small change in the melting point. The curve, therefore, is very steep. Increase of pressure by one atmosphere lowers the melting point by only 0.0076°,[35] or an increase of pressure of 135 atm. is required to produce a lowering of the melting point of 1°. We see further that the fusion curve bends slightly as the pressure is increased, which signifies that the variation of the melting point with the pressure changes; at -15°, when the pressure is 1625 kilogm. per sq. cm., increase of pressure by 1 kilogm. per sq. cm. lowers the melting point by 0.012°. This curvature of the fusion curve we shall later (Chap. IV.) see to be an almost universal phenomenon.
Equilibrium between Ice, Water, and Vapour. The Triple Point.—On examining the vapour-pressure curves of ice and water (Fig. 3), we see that at a temperature of about 0° and under a pressure of about 4.6 mm. mercury, the two curves cut. At this point liquid water and solid ice are each in equilibrium with vapour at the same pressure. Since this is so, they must, of course, be in equilibrium with one another, as experiment also shows. At this point, therefore, ice, water, and vapour can be in equilibrium, and as there are three phases present, the point is called a triple point.[36]
The triple point, however, does not lie exactly at 0° C., for this temperature is defined as the melting point of ice under atmospheric pressure. At the triple point, however, the pressure is equal to the vapour pressure of ice and water, and this pressure, as we see from the tables on pp. 21 and 23, is very nearly 4.6 mm., or almost 1 atm. less than in the previous case. Now, we have just seen that a change of pressure of 1 atm. corresponds to a change of the melting point of 0.0076°; the melting point of ice, therefore, when under the pressure of its own vapour, will be very nearly +0.0076°, and the pressure of the vapour will be very slightly greater than 4.579 mm., which is the pressure at 0° (p. 21). The difference is, however, slight, and may be neglected here. At the temperature, then, of +0.0076°, and under a pressure of 4.6 mm. of mercury, ice, water, and vapour will be in equilibrium; the point in our diagram representing this particular temperature and pressure is, therefore, the triple point of the system ice—water—vapour.
Since at the triple point we have three phases of one component, the