724 
(H),=G HLH Hs [Curve ¢,8 figs. 9 and 10] 
Hi), =H LH, [Curve qe figs. 9 and 10). 
Let us imagine the singular equilibrium (J/) = (MD), = H,+ Hz 
in the point g. It appears from fig. 9 that a complex MH, + H‚ can 
not be converted into the invariant equilibrium of the point q viz. 
into G+ L, + H,+ Hz. [We assume that the gas G consists of 
watervapour only, so that point G coincides with W |. The singular 
equilibrium (M)={(M), is, therefore, not transformable into the 
invariant equilibrium q: curve (M/) is consequently bidirectionable 
and does not terminate, therefore, in the point ¢, but it goes through 
that point. 
Let us now imagine the singular equilibrium (M) = (M), in the 
point g,. It appears from fig. 9 that a complex H/, 4-H, may be 
A 
A) 
! 
| 
1, 
Fig. 9. Fig. 10. 
converted into the invariant equilibrium of the point q, viz. into 
GH LH, + Hs. The singular equilibrium (J/)—(M),, is, 
therefore, transformable into the invariant equilibrium q,; conse- 
quently curve (J/) is monodirectionable and terminates in the 
point g,. The (M)-eurve is represented, therefore, in fig. 10 by 
curve 9,90, = g,q0;. 
Further the singular equilibria 
(LE), = H, a H: G and (G); == i i He a L 
start from the point g,; as the (J/)-curve is monodirectionable in 
g;, the three singular curves (M), (ZL), and (G), coincide in the 
same direction. The curves (Z), and (G@), go, therefore, also, starting 
from q, in the direction towards g. 
