( 617 ) 
of the four assumptions leads to the same composition. This may be 
explained also in the following manner. 
If, for instance, we calculate the quantity A of this phase in the 
four ways described we always get the same result: According to 
1 the phase contains y Mol. AC and 1—«—y Mol. AB, therefore 
1—zx Mol. A; the same is found according to 2, 3 and 4. 
We also find in each of the four cases that that phase contains 
1—y Mol. B, y Mol. C and « Mol. D. 
If we draw from p the perpendicular lines z and w these are 
_ also significant. For we find: 
1 1 
ay (l—#—y) V2 and w= = (y—a#) V2. 
For the composition of the phase p we may write according to 1. 
«2 Mol. BD, y Mol. AC and 1—a«—y Mol. AB. 
As, however, between the four substances exists the relation 
AB + CD= AC+ BD 
we may express the composition also in the four substances, for 
instance : 
a—n Mol. BD, y—n Mol. AC, 1—«—y + n Mol. AB and n Mol. CD. 
From this it follows that 1—«#—y represents the number of Mols. 
AB minus the number of Mols. CD, while y—«a represents the 
number of Mols. AC minus the number of Mol. BD, Therefore: 
1 1 
En (Mol. AB—Mol. CD)V2 , u= 7 (Mol. AC— Mol. BD) 2. 
1 
The half diagonal of the square is now 5 V2; if, however, the 
half diagonal is taken as 1 we have: | 
z= Mol. AB — Mol. CD and u= Mol. AC — Mol. BD. 
The composition of the phase represented by point p may, therefore, 
be deduced in two ways. 
1. From the situation of p in regard to one of the four triangles 
whose angles represent the solid substances. The length of the sides 
of the square is then called 1. We then obtain the composition 
expressed in those three substances which form the angles of the 
observed triangle. 
2. From the situation of p in regard to the two diagonals of the 
square. The length of the half diagonals is then taken as 1. 
If we now add a fourth component, this may be placed on an 
axis in the point O perpendicularly to the plane of the square; if 
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