280 
1 99 O1, kO A WZ 
Des 57 ie 143,5. 105 pd 28 = 248,6 = 249 afm: 
With the now assumed values of a, and b. these would be the 
values of the critical temperature and pressure of perfectly undis- 
sociated Hg, at the critical point. They would give for 
4,766 
—_—_______________ — 4,00, 
248,6X479,2.10-5 al 
which value properly corresponds to r= 2. 
The equation of state is, of course, also identically satisfied. For 
dn ARE 
De is calculate 574,1.10-8 
: (Pe + Gc/V-?) becomes: 
8 SHI. pete With, = 24, the values == 
= 1740 atm., so that v'—é'..= RT: 
4,766 
(2300-1198) JOP 
248,6 + 1740,3 
i.e. 239,6 . 10-5 = 239,6. 10-5. 
We may state here that a,/v,” is also properly =(f.—1) pe = Ip. 
because /, = 8 corresponds to r= 2. 
But all these values are totally changed, when only, in the second 
place, the slightest dissociation of the double molecules exists at 7, 
(which we shall further develop theoretically in the second part of 
this paper). 
Let us suppose for convenience that then m remains = 2 (a will 
possibly be 0,01, so that strictly speaking n= 2:(1+ 2) would 
become 1,98, but in the calculation of 6 and a above we have also 
left n=), and further that in this case in consequence of the 
slight dissociation at 7, the value of » would have become 1,8 
instead of 2 (this too will be further elucidated in the second part); 
then v, becomes == 1,8  119,8* . 10—5 — 215,7.10—5. Further with 
6 = 1,260 and a = 4,268 (see above) RT = 6,228, 7 — 1700 abs., 
pe = 1100 atm., al! of them being the values from which we have 
started for the calculation of the factors 6 and a, and which will 
6,228 
1100 x 481,4.10-5 
give back the value 1,312 for s’ = 
28 
Accordingly the value of 7. has become 27 xXx 1,26 == 1,31-times 
28 
greater, that of p‚ 57 X 4,268 = 4,43-times greater, and that of s’ 
3,05-times smaller. 
The equation of state becomes in this case: 
