861 
From this it follows at once that Q— ÀS is also negative and that 
the saturation line under its own vapour pressure, namely the curve 
Fab, must possess a curvature stronger than that of the liquidum 
line. It further follows from our previous considerations that they 
must intersect also the line O/' somewhere between 0 and / so 
that they must exhibit a form as indicated schematically in fig. 1. 
The change in pressure along this curve is determined in F' by (22), 
from which it follows, that, starting from /’, d P is positive whether 
towards a or towards 5. The pressure in is, therefore a mini- 
mum one and increases in the direction of the arrows. The solution 
with maximum vapour pressure is, of course, in this case situated 
on the intersecting point of this curve with the line: OF. 
We will now disregard the liquidum line de of fig. 1 and sup- 
pose it to be replaced by d’/e’ which is curved in another direc- 
tion: Q is, therefore, positive so that Q@—AS can be positive as 
well as negative. If the liquidum line is not curved too strongly 
(J— AS will be negative and the saturation line under its own 
vapour pressure again exhibits a form like the curve af°b of Fig. 1. 
If however the liquidum line is curved very strongly and À is not 
too large, then Q—2AS can also become positive, so that both 
curves in /#’ are bent in the same direction. This has been assumed 
in Fig. 2 wherein de represents the liquidum line and afb the 
saturation line under its own vapour pressure. As in this case, Q is 
larger than Q— AS it follows, as assumed in Fig. 2, that in the vicinity 
of F the curve de must be bent more strongly than the curve afb. 
V<v. À has, therefore, generally a large negative value. In the 
same way as above we find that Figs. 3 and 4 can now appear. 
The saturation ling under its own vapour pressure is again represented 
by alb, the liquidum line by de. In Fig. 3 are united two cases, 
/ 
q 
d 
i F 
1a € e 
Fig. 3 Fig. 4. 
namely a liquidum line dFe curved towards O and another d' Fe 
curved in the opposite direction. We must remember also that the 
line OF must intersect the saturation line somewhere in a point 
situated at the other side of # than the point O. A now being 
negative, it further follows from (22) that the pressure of A must 
now decrease towards a as well as towards 0; hence, the arrows 
