746 



S dH 



1. J and have the same smn. rrom (26) it follows 



S, dP' ^ ' 



that I is positive, so that the parabola touches BC in H and is 

 further situated within the triangle [fig. 6 (XI)J. This is apparent 

 also jet from (25), as % becomes positive as well for dP positive 

 as negative. 



*S dH 



2. 1 — — and — ^ have an opposite sign. From (26) it follows 



that è is negative. Therefore the parabola touches BC in H, but is 



further situated outside the triangle. Therefore a similar parabola 



may be imagined in fig. 5 (XI). Then only its j^oint H represents a 



liquid, its other points have no meaning. 



dl dH 



From our deduction of — and in the point H, follows : 



dP dP' ^ 



dH 



Y—y = h. . dP\ 



^ ' dp' 



Now, in the point H of fig. 4 — 6 (XI) Y — y, therefore also 



dl^ 



— - becomes positive, as well on increase as on decrease of pressure. 



dP^ 



When, however, the point H is situated on the other side of F, 



dH 

 tiien 1 — ?/ and therefore also becomes negative. 



We now consider some cases. 



dH 



1. F melts with increase of volume (V^v). — ~^^- -SandASi 



positive. 



a. S'^Si- From (26j follows: the saturationcurve under its 

 own vapourpressure is a parabola, which touches BC in H, but is 

 situated further outside the triangle [fig. 5 (XI)]. 



b. S<^S^. From (26) follows: the saturationcurve under its 

 own vapourpressure is a parabola, which touches BC in H, but is 

 further situated within the triangle, [fig. 6 (XI)]. 



dH 



2. P' melts with decrease of volume ( K <[ v). --— <^ 0. 



We take again S and S^ with opposite sign. 



As sub l.a. In fig. 5 (XI) the point H must be imagined on the 

 other side of F, therefore, between F and B and B^ between F 

 and C. 



From (18) it follows that t] changes sign with dP, as in the 

 point IP the coefficient of dP is negative, ij and dP must have the 

 opposite sign. Therefore, the pressure increases in the direction in 



