Hence substitution in (1) gives: 
dn, On 
4 , © 02! 02’ 
(CC, + 2C,) —| — — |+ &T (— loge, + 2logc,)— 0, 
when 2’ — 2— RT Zn, . log =n,. 
If 8 is the degree of dissociation of the double molecules, we have: 
02 0Ddn, | OBdn, _ AZ a’ 
ican de on Oa) Sane 
because n, = 1— B, n, = 28, so that we may write: 
xy Y 0$2' r ase A 
(— C, + 2C,) — —— + RT log — = 0,7 
dg rane 
or as: 
1—p 28 
err Es 
148 148 
also : 
; 0&2! 
= 230) oa 
nee MG = d a 0p 
Dap RT - 2 «2° 2 sn 
de 
So we must determine the value of Fn From: 
t 
ae + 8) RT a 
a Pv 
Den) v 
follows: 
je (1 + p) RT log (v — b) == ri ) 
rv 
so that we find for 2’ (Xn, =1 +9): 
1) In this integration the quantity @ must namely be kept constant, because an 
eventual equilibrium between the components exerts no influence on the determination 
oe 
of the values of 2, ay and 5 for the two components. In the calculation 
n, n 
2 
of the thermodynamical potentials of the different components of an arbitrary 
mixture we have namely not to take account of a later eventual occurrence of an equili- 
brium. Thus we calculate here gj and uz quite independently, and simply introduce 
for the equilibrium the additional condition — gj +2u,=0. So we must imagine the 
integration [oa for a perfectly arbitrary ratio of mixing (?, which quantity 6 
does not become the degree of dissociation of equilibrium until after the introduc- 
tion of the condition 4; = 2g. (see also Arch. Teyler p. 4). 
