Value of "a" of van der Waals' Equation 195 



bon monoxide, which is hopelessly aberrant if carbon and oxy- 

 gen have a higher valence than one, the value of C comes close 

 to the value found in other substances, namely, 2.98 X io~ 37 . 

 It is true that in making the calculation I have, in one or two 

 instances, used other valence numbers than those generally 

 attributed to the elements. Thus chlorine is trivalent. But 

 I shall show in a subsequent paper that chlorine is always 

 trivalent in its organic compounds; nitrogen in the gaseous 

 form is univalent, while it is bivalent in nitrous oxide; 1 but 

 both these valences have already been ascribed to nitrogen. 

 Oxygen cannot have more than two valences in the molecule. 

 The value I have taken for M 2 K is certainly as high as the 

 facts warrant and the critical data seem reliable. I believe 

 that the conclusion is justified that oxygen is monovalent 

 in its gaseous form. The surprising fact is the value for car- 

 bon dioxide. The carbon cannot be more than bivalent 

 here, giving a kind of peroxide formula, if carbon dioxide is 

 to follow the rule. On the whole, the gases agree well with 

 the results already obtained for C. 



In a subsequent paper I shall show that all chlorine com- 

 pounds also follow the rule if the chlorine be taken trivalent; 

 and that there is good reason, from quite independent sources, 

 to attribute a valence of three to chlorine. The valences of 

 sulphur, nitrogen and the argon group will also be considered 

 separately. 



In closing it may be pointed out that the law just estab- 

 lished can be used to determine the number of valences in a 

 molecule if M 2 K of the substance is known, or if its critical 

 data are known, as follows: If the computation is made 

 from the approximate values of "a" given in the Landolt- 

 Bornstein-Meyerhoffer tables, which were computed by the 

 formula: a 27^7(64 X 273' X P c ), proceed by the 

 following formula: (i) Valence number (a) 3/2 X 3.2 X 

 io 5 /(Mol. Wt.). Or, if the calculation is made from the 

 critical data, the following formula will serve: 



1 Possibly nitrogen is univalent in this gas, but the oxygen is quadriva- 

 lent, having two free valences. 



