139 



values of T and p. (ƒ between two observations supposed to be 

 constant). And ior f -\- log pi- tlie values 4,789, 4,760, 5,022,4,905; 

 evidently tiie value 4,880 for f -\- log pk corresponds with the mean 

 value 3080 found just now for /7\.. 



The correction quantity A is represented (compare also loc. cit. 



27 fry 



p. 4) by the expression / = — — - I — :^ . And as y, the coefili- 



cient of direction of the straight connecting line between /)jt and Z)^, 

 is given by the approximative formula 2y = 1 -[- 0,04 P^T^., we find, 

 with about t/J^. = 34,5, for 2y the value 2,38 (which at the same 

 time indicates the ratio b/c'.ba), i.e. J, 19 for y. Therefore A beco- 

 mes = 0,936, i. e. somewhat smaller than we assumed formerly 

 (0,964, see above). 



If now .^• is the dissociation degree of the Hg,-molecules at Tk, 

 there are on an average n = 2 : (1 -[- .t) atoms available per mole- 

 cule (the association degree of Hgj), and we have evidently : 



8 (11(1— .r) 4-86.?;)' . 10— » 2 



n = 273 X --1. X 0,936 X ~^—v~-v~-. X 



27 ^ 150. 10-5 l+.iv 



1 (ll(l-.c) + 36.?;)'. 10-4 



Pk = — X 0,936 X — 



^ 27 (150)'. 10-10 



because the quantity y^ajc is = 11 . JO— ^ per Gr. molecule for Hg 

 molecules bound to Hg, ; much higher on the other hand for the 

 free mercury atoms, viz. 36.10-'- (cf. IV in These Pvoc. Vol. XIX, 

 p. 317, where we found for bismuth, which stands in the same 

 horizontal row as meicury, 35,6), so that l^rt^ becomes on an 

 average =: 11(1 — .v) -{- 36,f per Gr. atom. 

 Hence we find : 



(ll(l-.t-)4 36.if 1 



7', = 10,10 X 



1-f ^ '; 



logpk = 0,1877 4- 2 log (11(1— .t') + 36.7;) ) 

 with /Tk = 3080, ƒ + log pic = 4,880 therefore (see above) : 



f— 4,692 - 2 log (ll(l-.?;) + 36^0. 

 so that X can be found from 



10,10( y _ 3080 



" 1-f"!^ "~ 4,692—2 log ( )' 

 or from 



[4,692 2%(ll(l-.iOf 36.6)] X (1 1(1— .r) + 36.tO' : (1 f.'O = 305. 

 If now x = for Tk. i- e. everything bimolecular, this becomes 

 with 10^ .\^ak = ll: 



316 = 305, 



10* 



