752 



treated P, T diagrams for the case that F is a ternary compound). 

 After the previous considerations, the reader can easily deduce 

 the P, 1 diagr-am for the solutionpaths of F, when the curves are 

 situated as in fig. 6 (XI). 



Formerly [5 (IV)] we have deduced for a solutionpath 



ff BU-BN 



dT CM-AN 

 now, as « = 0, herein is : 

 M = x' r + 2 w (//-/?) s-\-{y-^Y t 

 N = X {x, -x) r + \_x iy,-y) + {x, -x) (//— /?)J s + {y,-y) (y-^) t 



In the point F becomes .x' = and y =^ i^, therefore M=0 and 

 jY = 0. Let us now contemplate a solution path and let us call the 

 angle, which it forms with the A'-axis, (f . If we imagine for the 

 sake of simplicity that the coordinatesystem is rectangular, then it 

 follows: cotg. ((j = x: (y—^). We then obtain: 



M X r cotg (f-\-2xs-\- (y — /?) t 



N^ix,- x) r cotg (p + [{y^ - y) cotg ip + x,—x'] s -\- {y,-y)t 

 In the point F becomes x = and ij = |? therefore : 

 M RT 



N fx 



{^^-\\hT-\{y-^){.^ttgi{) 



(35) 



Tiie question now arises, what P, Tcurve touches the meltingline 

 Fd in F. For this must, according to (34): 



DM—BN B 

 CM— AN ~ A 



therefore, M : N =^0. It is apparent from (35) that this is only the 

 case when tg (f is infinitely great, consequently for (p = 90° and 

 fp = 270°. Then the solutionpath coincides either with FF or with 

 F(J (fig. 1). Therefore, both the binary solutionpaths EF and UF 

 only touch in F the meltingline Fd; the ternary paths do not touch 

 this meltingline. 



In order that the tangent to the P, T curve of a solutionpath may 

 be vertical in F we have, according to (34) CM—AN^^O. As M.N 

 is fixed by (35), it follows that this is the case, when 



0^-i\rT + {y-^)s-RT^ 



From (31) it follows that in F this solutionpath must touch the 

 saturationcurve under its own vapourpressure going through the 



