( 136 ) 



indeed still two realisable plaitpoints on this spinodal line, but that 



the course of the nodal lines in all this longitudinal plait is as was 



the ease in the upper halt' of the above discussed longitudinal plait 



— so that in both plaitpoints the tangent p-line and the tangent 



g-line descend to the right. All over this longitudinal plait v t ]> v u 



(Pip 



if v a represents the right-hand point of coexistence. But if = 



dx* 



dp 



remains restricted to volumes larger than those of — = 0, the course 



dx 



of' the nodal lines is such that v 1 <^v 1 , and the plaitpoint has such 



dv dv 



a situation that — and — is negative for the p-Yme and the o-line 

 dx p </.!•,, 



which pass through the plaitpoint. I speak of the plaitpoint, because 



I think 1 can demonstrate that then there cannot be question of two 



realisable plaitpoints, nor of a detached closed longitudinal plait. 



dv dv 

 ror when a spinodal line splits up, not only — = — in this split- 



d'v d*v 

 ting point, but also —- = -—=0. Properly speaking I have already 



discussed this point (These Proc. April 26, 1907 p. 848), but on 

 account of the great importance of the question further elucidation 

 is perhaps not uncalled for. The following remarks may serve for 

 this purpose. 



Let us in the first place consider a mixture represented by a 

 region of the general p-figure lying on the right side, and so much 



to the right that the point where — = has minimum volume, no 



dx 



longer occurs, or lies at very small value of a'. Then the point where 



d*ip 



-j— = disappears at T=T g , because it must lie on the line 



d^p dp 



-— = 0, lies at smaller volume than those of the line — = 0; and 



dx' dx 



if this curve is still found at temperatures below T g , the points in 



dp 

 which this curve intersects the line — = 0, lie in the region where 



do 

 dp 

 — is negative. If we now suppose that the temperature rises, and 



the spinodal line might split up, this splitting point must lie between 



d*ip dp 



the larger volumes of = and the volumes of — = 0, so also 



dx* dv 



■ , d P 



in the region in which — is negative. Now the question is it in this 

 dx 



