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PAPB 
F(p)—=0 the value p =p, = ———— 
PA APB — Pi 
and solve 7}. 
This furnishes: 
i} 
EET oe [ =e VA P'AP°B —p’a] 
pea + p°B + pape 
PB 
Now it is clear that only the positive sign is possible, else we 
could get a pj >pB, and as pp =pa-+2'(pp—pa) we get for the 
composition of the mixture that separates case II from case III: 
Yp Bp A— pa 
yw = 1—- 
P°B— PA 
Therefore the inflection-point will appear for every mixture, for 
which Ys<e' <12’. Hence we see that the region which remains 
for the entirely concave curve, depends only on pa/ps. 
In order to examine this dependence more closely, I have inserted 
in my thesis a needlessly elaborate calculation, which I shall not 
repeat here, the more, because it contains an error of calculation 
which I could not correct anymore. In the formula for g”(y) on 
pag. 153 9y° ought to be substituted for 1046, In consequence of 
this fig. 16 is not quite accurate. Therefore I should like to sub- 
stitute what follows for the passage of my thesis which applies 
to this. 
If we introduce in the formula 2’: 
_ PA 
PB 
y 
we get: 
velen en En As l 
SSS Oe OS ———————————EEEETETEE 
; ZO 1+vy 
dje! a Vy 
d= 2 (ly) 
From this formula follows that for y= 0, so p4= 0, je! = 1. So 
isothermals which are concave throughout their course do not exist. As 
