( 655 ) 
This may be found by combination of the two conditions 
Ou, —=0 ; ois — 0 
Ow i : Ow? DT je 
of of Ov =o ; 
eestor (ear aan an ie | (7) 
when / represents the second member of (6). Indeed, this second 
leading to 
member has in all points of the spinodal curve on the y-surface the 
same value, so that we have, by passing along an element of that curve : 
of 
Ou 
But in the plaitpoint we may regard an element of the spinodal 
curve also as an element of the connodal curve, that is to say as the 
oF 
de +—~—dv=0, 
Ov 
line which joins the two tangent-points of a double tangent-plane, 
when the tangent-points have approached each other to an infinitely 
small distance. And as in these two tangent-points the pressure has 
the same value, the latter does not vary, when at the plaitpoint we 
pass along the considered element of the spinodal curve. Consequently 
] Oe: 
n= a, 
1 ae he z 
which yields immediately equation (7). 
we have: 
For shortness, we will write in the following >,/a, — b‚Wa, = 2, 
by which the second member of (6) passes into 
f== 
2 KE (1—2) |= +a | +a oy | ; 
Ov - 
The value of 5 will be found from (4), viz. 
JpT k 
L 
‘ 
2 
7 
2aVa (v—b)? 
ES Es IRL py? 
Ow pT are 1 2 &/..(v—b)? 
RT 0 
And since the denominator of this expression cannot become o, 
(7) passes into 
2 @/,(v—b)?\ Of Za Va (v—b)*) Of 4 
1 — ——— 2 Sie 4 (7 
( RT 0 Je T (8 RT 4 Ip pes rou 
Now we have: 
