IN STUDYING QUESTIONS OF CHEMICAL MECHANICS: 385 
ing to the abscissa 8, we have 
B 
[eJ=A+t p 
B+C 
The operation becomes very easy if we employ, as given, the 
value of [a] furnished by the observation of the pure tartaric 
solution. In fact, 6 being then null, this value of [¢] should be 
equal to the constant A which is determined by this condition. 
A being thus known, I put the equation in the form 
{[4]—A}6+{[4]—-A} C= BB. 
Let us designate by [2],, B., [#]5, 63, the simultaneous values 
of [«] and of £, relative to two other observations of the same 
series, in which they have been determined experimentally. 
Each of these pairs should satisfy the preceding relation, which 
will furnish the two conditions 
{[e]o—A} 6. +{[e],—-A}C=B£,; 
{[z]; A} 6, + {[#]; -A} C = B&,. 
All is known in these equations excepting B and C ; and these 
quantities both enter in the linear form. We may therefore 
easily deduce them by elimination ; and on associating them with 
the value already obtained of A, we obtain the three constants 
of the series. We then calculate the values of [a] by those of 8 
for the other observations made ; and on comparing them with 
the [«] observed, we shall prove the accuracy of the hyperbolic 
law which connects them. 
69. But, when we do not wish to employ the observation of 
the pure tartaric solution to determine directly the constant A, 
in order to deduce from the law itself the corresponding value 
of [«], the equation (H.) must be subjected to a preliminary 
transformation, so as to avoid radicals. For this purpose, I 
change the product B £ into its equivalent B (6 + C — C); and 
on effecting the divison by 6 + C on the two first terms, I 
obtain 
(H.) 
apo a ye Bee 
Then substituting new letters to represent the coefficients thus 
grouped, I make 
A+Bsa; BC=48; Cc ='6: 
