533 
dz we sAV,—tAV, : Oy mae sAV,—rA V, ' 
’ 
’ 
Op rt —s? 
From (35) and (46) follows: 
Ox a tQ,—sQ, AN, Oy Sy 7Qy— sQr 
On nies: 4 OT rs: 
Op aCe rt—s* 
T 
Substitution of these values in the left hand side of (5) gives a 
fraction of which the numerator = 0, while the denominator > 0, 
if the equilibrium is stable. The equation (5) is thus established. 
Oa 0: 
We may remark that the separate sums, as a6 Ie = = AREN 
QV AVy 
are not zero except in the special case when — = : 
3. We shall now attempt to establish the general equation (2). We 
assume that we have a liquid phase consisting of one mol of solvent 
and «,y,z,... mols of the dissolved substances. At pressure p and 
temperature 7’ the solution is saturated with respect to these sub- 
stance. We have thus m-++1 components in as many phases and 
have therefore two degrees of freedom at our disposal. 
The equilibrium conditions are (for the notation see above): 
Die oN AN 
nec 
dz 
rme ee Mn IE (6) 
dz 
0 
TR 
The expressions on the left-hand side of these n equations are 
again functions of z,y,z,... p, and 7. The last two we consider 
as independent variables. If we differentiate, first with respect to p 
alone and then with respect to 7’ alone, we obtain the two sets of 
equations : 
WZ 02 OF dy OF dz 
2, + Lj ll EAT == — AV; 
dz? Op dady Op 0x0z Op 
0°Z 0e  07Z dy 0'Z dz 7% 
drdy Op iz dy” Op aE On0z Op "4 y sxe (6a) 
0°Z da 02 Oy 0727 0z 
Ordz Op  dydz Op | Oz? Op 1 
