The Valence of the Argon Group, Etc. 



339 



great number of substances that M 2 K, a factor proportional 

 to "a" of van der Waals' equation, is equal, when expressed 

 in dynes, to 2.98 X io~ 37 (mol. wt. X No. of valences) 2 / 3 . 

 A substance having cohesion cannot, therefore, have zero 

 valence. If it had no valence it could have only gravita- 

 tional attraction between its monatomic molecules, no cohe- 

 sional attraction. These gases have cohesion and they must, 

 therefore, have valence. 



Their valence, n, can be calculated from their critical 

 data by the formula: (i) n = 0.0043 T C 3 /P C 3/2 (mol. wt.); or, 

 n '= a 3/2 X 3.2 X io 5 /(mol. wt.); or (2) n -- 4.21 X io~ 5 

 (V c T c ) 3/2 /(rnol. w t.). The first formula, which is derived 

 from the ordinary formula for computing "a" from the 

 critical temperature and pressure, gives values for "a, " 

 and hence for n, a little lower than the second formula in the 

 case of simple gases. The second formula computes "a" 

 from the surface tension, as shown in my former paper. In 

 Table i, I have given both values and I regard the second 

 as the more correct; but as the critical density of not all the 

 gases is known, I have had to rely on the first formula for a 

 comparison. The results given by the two formulas are not 

 widely different. 



The valence, n, together with the critical data 1 used in 

 the calculation are given in Table i. 



TABLE i COMPUTATION OF THE AVERAGE NUMBER OF VALENCES 

 PER MOLECULE FROM THE MOLECULAR COHESION 



1 In my preliminary paper, Science, N. S., 36, 6 (1912), in Table I a 

 mistake occurred in the computation of helium. The value 2.90 is wrong. 

 The critical density was taken as 0.015 instead of 0.065. 



