on the branch RMR’ with an increase and on branch Rm’ with 
a decrease in volume. 
In the points of contact themselves where both branches amalga- 
mate, the case sub A 3 now occurs. Let us take the two-phase com- 
plex F+ liquid A. We now see that, on increase as well as on 
reduction in pressure, the conjugation line /-liquid B gets out- 
side the new three-phase triangle so that no vapour can be formed. 
Let us now see what happens in a similar point of contact FR if 
the pressure changes but infinitesimally. At this infinitesimal change 
of pressure, the liquid then moves at an infinitesimal rate along 
the tangent /’R either towards or away from /. The only thing 
what happens is that in the liquid a little / is dissolved, or else 
crystallised from the same, without any vapour being formed. 
If now a substance /’ melts with increase in volume and, there- 
fore, in this case also dissolves with increase in volume, it will 
erystallise out on increase in pressure and get dissolved on reduction 
of the same. This also is in harmony with the change in pressure 
along the saturation line under its own vapour pressure in the point 
R of fig. 12 (1): on elevation of the pressure the liquid moves, starting 
from F, from the point /’; this signifies that solid matter is being 
deposited. On reduction of pressure tbe liquid moves from 7? towards 
the point /’; this means that solid matter is being dissolved. 
The fact that in a point of contact A no vapour takes part in 
the reaction may be also demonstrated in the following manner. 
We again take at the pressure P a system SS consisting of: 
n quantities F+ m quantities L + q quantities G; 
at the pressure P+ dP is formed thereof the system S’ consisting 
of: 
(n+dn) quantities /'+-(m-+-dm) quantities L’-+-(¢+dq) quantities G’ 
From the three relations already employed for this and which 
indicate that the quantity of each of the three components remains 
the same in this conversion we can deduce: 
Edn = — m {(y,—y)da — (a,—2a)dy} — q \(y,—y)da, — («, —2)dy,} 
Edq = = m{\(8—y)da — (a -a)dy} + q((8—y)da, — (a—«)dy;} 
Edm= — m (y, —-B) dw — (a,— a)dy} + gq \(y, —B)dz, — («,—a)dy,} 
in which all the letters.have again the same meaning as before. 
If now we proceed at the pressure P from the system /’-+ L we 
must call ¢g=0; we then obtain: 
Edn = — m{(y,—y)dz — (w, —x)dy} 
Edq = m\(3—y)dx — (a—«)dy} 
Fdm= m{(y,—B)de — (w,—a)dy}. 
