( 711 ) 



AB A BC 



(1— «)(1— .'•) «(1— .1-) «(1— .s) .r, 



together again l-{-ct{l — ,r) molecules. 

 Hence the dissociation isotherm becomes : 



ail-,v) a{l-.v) (l_«)(l-,r) 



N ^ N N 



or 



«' 1— .c 



1 — « 1+«(1— .r) 



K (10) 



Now « does not decrease witii ,v, but increase. Tlie added indilTerent 

 substance C may viz. be considered as "diluent", whereas in the 

 preceding question the addition of one of the products of dissociation 

 depresses the degree of dissociation «. 



If we solve from (10) again a, we llnd in this case : 



K K 



o? (1— ,1') H «.)• = 0. 



By putting ,i' z= 0, it appears again that j=:«/, so that we 



l + A 



must solve « from 



«'(1 — x\ 4- «/«.v — «„' = 0, 



in which «„ is again provisionally assumed to be independent of T. 

 (Cf. k 3). 



Now we find : 



« = «0 [- '1^^ + »/",«„'-f' + (!-.«]: (1- .^•). 

 so 



(i-«) (!-■'■) = (1—') - «0 L- ',V'o''^ ^v\ \ 



1 + a (1-.C) = 1 + «„ [- \;«„.i' 4- I/'] i 



Tlie quantity occurring in (5) under the sign log becomes then : 



(l-.t-)+ 'Uii^x-ay 



Now l/V,«„'.c' + (1^ = l_» '^.t--7^(l— «/).c= so that the 



above fraction passes into 



1 - «„ - ',; (1 - «„) (2 + a,y + v,«„(i - «/)..^ . ■ ■ 

 1 + «0 - V,«o(i + «o)-'^ - Vs«o(i - «o')-^' • • • • 



i. e. into 



(1 - jQV 1 - ■/,(2 + «J.. + '/,»„( I + a^-^ ■ . .] ^ 

 (1 + «„)[1 - 7,«,,. - V3«„(l - «„).«^ . . . ] 

 or into 



