(42 ) 



^' = 0,1 0,2 0,3 0,4 

 T=:0 0,36 0,64 0,84 0,96 

 Finally we get for a> ^ Vs = 



Va 



.t-(l 



yielding : 



x= 0,1 0,2 0,3 0,4 0,5 0,6 0,7 0,8 0,9 1 

 T = 0,593 0,837 1,07 1,28 1,48 1,67 1,84 1,99 2,13 2,26 2,37. 

 It appears from the diagram (see also above for .» = VJ» that the 

 temperature from Cg to C„^ is not continually ascending, but that it 

 shows a minimum very near Cq. This causes the spinodal line r = 1 

 7iot to pass through C\, but to remain under it. The point Cq, where 

 T is also := 1, is an isolated point belonging to that line. Just beyond 

 Co the two branches of one and the same spinodal line intersect in 

 a double point ; beyond that place the course is normal ; between 

 Co and this intersection the spinodal line has two separate branches, 

 one of which encloses the point Co- Now the question arises, 

 whether this will be the case for every value of (p. If we solve x 

 from (lb), we get : 



x' (2w — o}') — .r [ 1 -}- 2y (1 — (o)M -h ( -^ — (f' (1 - to)' j =: . 



This gives for x two roots of the same value for given values of 

 T and CD, when 



4 ^ ((^ + 1) (1 — to)^ -h 1 — T (2 — to) = 0. 

 The value of .i' is then: 



^.^ V.+y(i-a))' 



to(2— to) 

 Now it follows from the value of the above given discriminant, 

 that it becomes = for hoo values of o). So two branches of a 

 spinodal line intersect, when those values of to become the same. From 



4<p (y+1) to'' - to ^8^ (y + 1) _ rV ^4^ (y + 1) + l-2r^ 

 follows, that CD has two roots of the same value, when 



= 



= 16r/)(t/ + l) 



to = 1-7, 



And T being 1 at Co, the minimum disappears only, when t 

 becomes = 1 in the above expression. And this is evidently only 

 the case for ^ = go, i. e. when 2^ and 1\ should have the same 

 value. Hence in general there will always be found a minimum in 

 the neighbourhood of CV For (p = 1 we find t = 0,970, to =r 0,94, 

 a; = 0,506; for (p = 2 we find r = 0,990, to = 0,98, .^; = 0,501 ; 

 etc. etc. 



