( '10 ) 



dn da d.e m 



Now -—=---. And trom 1 — .r = ii, r ^= m follow? .v ^ , 



«H ax du m-\-}i 



dx in da da 



hence — = — =: — x. So -— = — *' t~ • 



dji {in-\-n) dx dx 



Dividing by dn, we find therefore for the total quantity of heat, 



absorbed per Gr. mol.: 



da 

 Q = q + a).-x{\-x)—X. 

 dx 



da 

 Substitution of — - from {a) yields then after a slight transformation (9). 

 dx 



Let us now put x = 0, then we find from (8) on account of the 



factor x: 



fdT\ 



If a is very small, this horizontal course does not continue long. 

 For with small x we may write: 



dT _ RT"" X 



lb;~~ Q~ 'x^ia ' 



As soon therefore as x becomes so large thet 2rt is small with 



X X 



respect to x, the fraction approaches — = 1, and the normal 



^ x-\-2a X V 



course is restored. The greater therefore «, the longer the almost 

 horizontal course will maintain itself in the neighbourhood of T„. 



If a absolute ^0, then — , ' ,^ may be replaced by — =J 



X'\-2a(\ — X) X 



from the beginning, and we have immediately the normal course, 



given by 



dT _ 1 RT-" 



dx 1 — X q 



yielding : 



/dT\ ^ _ rt;- 



\dx), q 



Also To and 7),/, tlien coincide. 



5. In fig. 1 also the line T^B has been drawn. This would be the 

 melting-point line, when instead of an excess of one of the products 

 of dissociation, an excess of an indifferent substance 6' was added. 



The equation (5) remains then the same. But now (6) becomes 

 different. We have now namely : 



