( 703 ) 



Instead of (1) we get now 



dn, dn, di,^ dij, 



no-r-^n,-—-\-n,-—-\-ti,-~ = 0'). . . . (1«) 

 ax ax ax ax 



dug dfx^ , . 



And as n, —— = x — — remains fmite at .v = 0, viz. RT, from (1") 

 dx dx 



(/(<„ 

 and (2) will now /u>/ follow — — = 0, when ,v = 0. (n, : n, = v. -. i', 



dx 



dT 

 is always satisfied in this case). And consequently — will not be 0. 



dx 



tThat X — ^ continues to have a finite value at ,r^0, follows from this, 

 dx 



m, ■'-• du, dti\ RT RTdN 



that fi, =r ft , 4- Sr hf] — yields -^ = — --i 1^ 'T ' •^®"'^^ 



iV rf.B d.« X N dx 



d^i. I dfi', 22 T (iiV 1 



.t; ^ iïr + A- —r — ~ , in which the expression betweei 



dx [ dx N dx I 



always remains finite. At x = we have therefore x ^ RT 



•' dx 



3. We now proceed to derive an expression for the course of the 

 melting-point curve in the case of increasing excess of one of the 

 products of dissociation in the liquid phase. 



Let us for this purpose suppose, that in this phase there are present 

 (in Gr. mol.) 1 — ,)■ AB and x B, while the 1 — x A 13 is disso- 

 ciated to an amount a. We have then : 



AB A B 



(1 — «) (1 — x) « (1 — ,r) « (1 — x) + X, 



together 1 -)- « (1 — x) molecules. 



We suppose then, that the compound consists of 1 raol. A and 

 1 mol. B, which simplifies the calculations. 



The equilibrium between the solid phase and the non-dissociated 

 molecules in the liquid ]ihase yields: 



ft = (i„ , 

 or (the terms with Tlog T on either side cancel each other) 



(1 - «) (1 - X) 



s — cTz=s, — c„7'-f RTlog 



1 -f « (1 — x) ' 



1) This too is easy to test, when /u',, , i^\, etc. are considered as constant, i 

 that e. g. in 



1 — « oft ( 1 Ö ) 



f'o = Mo + J'^T log — -— , -— becomes = RT \~ _ etc. 



A' ou 1 _ n jV 



