532 
This equation is the same as (1). It is therefore obvious that 
Braun’s law is a particular case of (2). 
2. We shall first of all establish (2) for the simple case that the 
solution is saturated with respect to two solid substances only; and 
we assume further that these two substances together with the solvent 
are independent components in the sense of the phase rule, so that 
two degrees of freedom are at our disposal, namely, pressure and 
temperature. The ¢-function of the liquid phase is represented by Z, 
those of the solid phases (per mol) by §, and &,. The equilibrium 
conditions are: 
me to Oi elo venus ©) 
0Z rue 4 
Sls one Bo es (4) 
The left hand side of these equations are functions of «,y,p, and 
T, of which, however, only two may be varied independently. 
By differentiating these equations with respect to p with 7’ constant 
there results: 
OZ 02 OZ dy av 
0x? Op dxdy Op a & ve) ; (3a) 
OZ Ox 072 dy OV 
ddy Op | ay? Op (5 ~) ee 
The notation here requires no explanation. It may be remarked 
that the right-hand side of (83a)=—AV, and that of (4a) — — AV, 
If we differentiate the two equations with respect to 7, keeping 
p constant, we obtain: 
0°Z Ox 07Z òy of 
Se On” aay OT (5 ns): G2) 
eZ de Zy (oH 
Òzdy OT | dy? OF (5. ww) (28) 
The right-hand sides of these equations (36) and (40) are equal to 
“ and a respectively. 
From these four equations the following relation is derived: 
Oz Qx Ow Oy’ Qy oy 
sd ve ae (NAT =O 
(=) fil (; = 5 (= Ti (; 7) ae 2) 
P02 077 
Proof: From (8a) and (4a) follows, on substituting == 
wv 
F5, 
dy ras 
VT == Sy 
Ody 
