( 823 ) 
1, 2 and 3 would have to lie in a straight line in the 2,v-diagram. 
If we follow the course of the isobar separately for the condition 
v‚,= 0, the possibility for this condition to be satisfied at temperatures 
above 7), follows at once. In the same way separately for the con- 
dition v,,—0. But for these two conditions to be fulfilled simultane- 
ously the isobars would again have to possess points of inflection, 
which we should certainly have to consider as quite abnormal cases. 
: 9 dp 
But there is more. Then —— would have to be = oo, for the deno- 
123 
minator is then equal to zero; and if we wish to avoid this, we 
should have to make again an entirely arbitrary supposition, viz. 
that then also 2: would be —~*—“". Therefore I do not hesitate 
t,— &, Egt, 
to reject the thesis that v,,—O coincides with v,, — 0 as entirely erro- 
neous. Then there remains no other possibility than to assume that 
the sheet of equilibrium 2,1 when contracting with rising tempera- 
ture, passes through the critical point of contact of the sheet of 
equilibrium 3,1, before having got hidden under this sheet. Then the 
third phase passes from the lower part to the upper part of the 
sheet 3,1. But this implies that at still somewhat higher temperature 
for a value of x which is somewhat higher than that of the critical 
point of contact of 3,1 a vertical line can intersect the whole sheet 
of saturation 4 times. And if the circumstances are chosen in such 
a way that the chosen value of w is also smaller than that of the 
plaitpoint of the equilibrium 2,1, retrograde condensation must occur 
twice. 
In fig. 41e I have drawn the p‚z-line in the neighbourhood of 7%, 
schematically. The discontinuity in the two vapour branches inter- 
secting each other in the point 1 is such that the branch which 
belongs to the equilibrium 3,1 descends more rapidly than that 
which belongs to the equilibrium 2,1. 
2 ) G ) 
d, de,” d, de” 
Now (jr ‚ and also & I= EN 
vy 2 
1 Voy 
oe vt, 
Vay dp 
If has become 0, 758 has got equal to —o, and 
a“ 31 
Es, 1 
dp : : 
— has then a large negative value. Or in other words if 
21 
31 — 9, v,, has still negative value, though it be a small one. But 
then it is exactly this that follows from the known course of the 
