859 
1 Or ds 
(ar +s) da + (as+-Bt) dy + > Loe + B Sf dx? + 
a v U 
( ds Ot aa fy os 30 a 
an Zn + ae de dy + = See + | ay dy? +... 
OV OV 
=| v— V, —a— — B— }dP..... eft ae (ed 
Oa Oy ) 
Let us now deduce (21) from (20) after having substituted in (20) 
f= and yi we. find : . 
1 1 
Ta da? + sde dy + Ea tds Ae AAP Ee en ae) 
in which the coefficients of dP.de and dP.dy are nil, whereas 
for the sake of brevity we write the coefficient of dP in (21) 
—(A+C). A and C then have herein the same values as in our 
former equations. Then, however, we assume «=a, y= 8, 2, = 0, 
and N= 0. 
From (22) follows dP of the order dz? and dy’, here from (21) 
at first approximation : 
(ar +- Bs) de + (as + Bt)dy=0 . = .-. » (23) 
In connection with (13) it appears from this that the lquidum 
line passing through point / and the saturation line of /’ under its 
own vapour pressure come into contact with each other. 
If we eliminate dP from (21) and (22) we obtain: 
(ur + Bs) dx + (as + pt)dy + 4 (5 Bs + r+ar ) da? +- 
® 
Ow 
ds ot Os dt oe 
+ {| a HB +s +4 As |dedy El a HB 4+¢+ ut) dy? = 0 (24) 
Ow 0a Oy dy 
af 
im whieh< «== ==. 
A 
For the liquidum line passing through point /” we find: 
(ar + Bs) dw + (as + Bt) dy + 43 (5 + De +--+ ‚) dex? 
a 
v 
dert OE Os Ot is 
ie antie te) dedy + (es +8 +) dyj*=0. 28) 
\ Ow Oa . Ov Oy 
For the sake of brevity we write (24) and (25) as follows: 
aX + bY +4 (e+ aX? A (d HA) XY H ble HM) V7 =0 … (26) 
AKK Ad PSS Or On 
Equation (26) now relates to the saturation line, under its own 
pressure, passing through /’, (27) on the liquidum line of the hete- 
rogeneous region ZG passing through /’. 
