1001 
complex A; in order to realise the equilibrium of a temperature 
TdT, we have to take then a complex of another composition. 
When the latter is the case and when 7’, is the temperature and 
P, the pressure of the invariant point, then we can not obtain with 
a same complex A an equilibrium of the temperature 7, and 
T,+dT or of the pressures P, and PP, + dP; the rules of the 
isovolumetrical and of the isentropical reaction, therefore, are then 
not applicable. 
Jn the first case we have: 
the two indifferent phases have the same sign; the equilibrium 
(M) is, therefore, transformable. 
(4 
4 (6) ae 
it 
Me 
/ pd 
es 1/M) 
1 (/) 
Fig. 1. Fig. 2. 
The stable parts of the curves (/,) and (#41) go in the same 
direction, starting from the point 7; then we obtain P,7-diagrams 
as in figs. 1 and 2. 
(In those and the following figures only the stable part of the 
curves (F,) and (#4) is drawn; the metastable part of the (J/)- 
curve is dotted.| In fig.1 the one part of the (J/)-curve is stable, 
the other part metastable; in fig. 2 the (J/)-curve is only stable in 
the point 4. 
In the second case we have: 
the two indifferent phases have opposite sign, the equilibrium (J/) 
is therefore, not transformable. 
The stable parts of the curves (/,) and (4,41) proceed in opposite 
direction, starting from the point 7; then we obtain P,7-diagrams 
as in the figs. 3, 4 and 5. In fig. 3 the (M)-curve is bidirection- 
able, in fig. 4 monodirectionable, in fig. 5 it is metastable, except in 
the point 2. 
[In a following communication we shall show that the (M)-curve 
can also have a turning-point. When this is casually situated in the 
point 7, then the diagrams under consideration will be changed by 
this in some respect. | 
