504 
PROFESSOR J. J. THOMSON ON SOME APPLICATIONS OF 
In this case, for chemical combination to take place, one molecule of A must come into 
collision with two of B simultaneously. The number of such collisions is proportional 
to so that the number of equivalents of A and B destroyed and of C and D 
produced by this combination will be 
where p is a constant. 
In a similar way we can show that the number of equivalents of C-and D destroyed 
and of A and B produced in unit time by the combination of C and D is 
/ y 0. 
q £e~, 
where q is a constant. 
In the state of equilibrium these quantities must be equal, so that 
PW = 
which, so long as the temperature remains constant, agrees with equation (49), but not 
with Guldberg and Waage’s equation. 
In order to illustrate some important points connected with the theory of the 
combination of gases, we will consider the simpler case when two gases, A and B, 
combine to form a third, C, while C again splits up to form A and B. A particular 
case of this is dissociation, which we have already considered. Let £ rj, £, be the 
number of equivalents of A, B, C, respectively, and let m x , m 2 , m 3 , be the masses of the 
molecules of A, B, C. 
Then, with the same notation as the last case, the value of L is 
e (c, + Rj log AT) + m#n (c t + R, log M) 
+ n ‘'H (cs + R 3 AG!)} - w, 
where w is the mean potential energy of the three gases. We have, as before, 
= drj = — d£; 
so that the condition that the value of L should be unaltered when £ is increased 
by cl£ gives 
o\jn x p (cj — Hi) + m 2 q (c a — K 3 ) — m s r (c 3 — R 3 ) + m l p B 1 log 
PoQ 
+ q log ^ - m a r R 3 log 
m^/rj 
Pi q 
dw 
W 
( 52 ) 
For perfect gases 
B 1 «i 1 = Lm, = B 3 ?n 3 = K. 
