( 126) 



negative. This may be considered to be sufficiently self-evident and 

 not to require an elaborate proof. But when the q-Vme has divided 

 into two separate parts, and when the p-line runs as is represented 

 in fig. 21, we meet with another difficulty, which indeed, calls for 

 a somewhat closer discussion. The joined branches c and g form a 

 curve which cuts the branches (/, e and ƒ, which have joined to a 

 loop-like curve, in two points, but such a point of intersection must 

 really be considered as two altogether different points. Such a point 

 of intersection represents two perfectly different phases according as 

 it is considered as point of c, g or of d, e, f. Hence when drawing 

 the straight line we must bear in mind that the point of intersection 

 of c and d and of e and g does not represent the same phase, and 

 if the line is drawn as in fig. 21, where the two hatched areas are 

 ecpial, the points at the extremities of this line are not points of the 

 binodal line. To see how the straight line must be drawn in such 

 cases we revert to the general equation : 



dM^j = vdp — xdq. 



Now to get from one point to the point with which it coexists, 

 we can no longer follow one q-\me, but we shall partly have to 

 follow a way which joins the two separate branches of the g-line, 

 and for this we choose the isobar of the point of intersection that 

 the branches c, g and d, ƒ have in common. We obtain then the 

 equation : 



(Mjl^c — (Mil* x )e = I vdp — I xdq, 

 e e 



where in I vdp the value of v must be taken which corresponds to the 



chosen value of (/, and in I xdq the value of a; which corresponds to 



that p-line that passes through the point of intersection. Let us call the 

 value of the volume of the point of intersection v s and the values 

 of x for the points where the isobar of the point of intersection cuts 

 the two branches of the q-linè, x t and x t . The above equation assumes 

 then the following form : 



(il/jfijc — (M.iije = p (v c —v e ) — jpdv — q («,— «g — \qdx . 



e 1 



Now if (A^ftjc must be = (üf 1 f*i) e , then p (v c — ?y) — I pdv is 



