289 
Hence @/j, is always negative at 
the critical temperature itself, when 
Awa has such a high value as in 
mercury and similar substances. Then 
the degree of dissociation of the double 
molecules Hg, decreases when the 
volume becomes greater, instead of 
increasing — as it generally does. 
As a(v—b):v’? (a and b assumed 
constant) has its maximum value at 
v = 26, the transition temperature for 
veo oe V->co values of v both about < 26 and 
> 26 will be lower than that (7) at 
v = 26, which latter will be only little higher than that at v = ve 
(about 4 7) *). See fig. 1. 
The decrease of x with increasing volume is of course accounted 
for in this way, that on increase of the degree of dissociation « with 
increasing volume also a becomes greater. But this causes the volume 
to decrease again, in which the decrease in the end exceeds the 
original increase. 
At high temperature the pressure will be comparatively great, 
so that then, in consequence of an increase of a, p + “%/,2 will be 
increased little; v—5, hence also v, will then be lowered compara- 
tively little. For this reason 4*/q will always be positive at high 
1) The righthand branch of the transition curve (dotted in the figure) will get 
more to the left, and 7, possibly slightly lower than would follow from the above 
calculation, because then x can no more be assumed near 0. The intersection with the 
vapour branch of the saturation curve takes place at 7’ = about 0,8 T¢. For 
from pv ='/,(1-+2) RT andp =p eb) where 4,14 is the vapour pressure 
factor f = 1,8 X 2,303 and m = T: Te, follows pe ve © = Ig (1 + 2) MRT. When 
x is put approximately =1/,, then ne” “* becomes = 5/4, ms. (with v = nve and 
RTe:peve = 8). Now in mercury s = 2,62 (see the cited first part), so that finally 
n = 1,965 m ef l4(L/m— 1) (saturation curve). This must now be combined with 
In this Va = 30.102, while Va becomes = 25,10-2 
v v Wa 
with «= 1/,. When at the point of intersection a is put about = 11/4 ac, b =1!/4 be, 
= Lb 20 
Org Semen laabe | with ye Ea RT. == == (see 
: NVe NVe OT be 
; 1,8n—1'/, oe . . 
above) m = 2,5 ——__—— (transition curve). Both equations are satisfied by 
n 
m = 0,778, n = 4,99 (point of intersection), so that this will lie at about 
T=08 7, where v = bve. 
