Page:Text-book of Electrochemistry.djvu/65

 50 BOILING AND FREEZING POINT. chap.

transformed into water. Bat since the freezing point is the point at which there is an equilibrium between ice and water, it necessarily follows that at this temperature they must have the same vapour pressure. Similarly, the solution whose vapour pressure is represented by |>y must at its freezing point have the same vapour pressure as the pure ice which freezes out. This point falls, therefore, exactly where the curves PP and p*p* cut each other.

(It follows from the above that neither water nor solution can exist below the freezing point in presence of ice; the introduction of a crystal of ice causes the solidification of the supercooled liquid.)

If now a line parallel to the abscissa-axis be drawn through P, and through M a line perpendiculeu: to this, the two cut at Q, and the perpendicular meets pp at N and the abscissa-axis at R. BA, which is equal to PQ, is denoted by (IT, and represents the depression of the freezing point — that is, the difference between the freezing point of the solvent and that of the solution. We then obtain —

QM = PQ tan MPQ, and QN = PQ t&Ji NPQ.

Further, according to the modified formula of Claperyon (see p. 48) —

tan VPO = ^= (A±^)P tan Ml Q-^j.- ^pr '

tanATC=^^f = A^,-

In this formula, (X + t*) is the heat of vaporisation of ice, ix, the sum of the heat of vaporisation X of the water at 0°, and the heat of fusion u of the ice at the same temperature.

If we denote the vapour pressure of ice at the freezing point R of the solution by jp'^, and the corresponding value of water by j9«, then —

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