( 602 ) 
and we have to prove that this value is larger than W5,b,. If we 
now put 6, = 7°b,, the condition which is to be satisfied, is: 
n'+3n*+3n?+ 1 : 
8 
or 
n° + 3n* — Bn? + 3n7 + 1>0 
or 
n= Vi (nt A Bn? Gn? An JD) 0: 
It is clear that for positive values of m this condition is always 
fulfilled, so that 6,,2 > 6,6,, and the equation: 
b, (1 — a)? + 2b,,2(1 — 2) + 5,27 = 0 : 
has always real roots. / 
/ 
4. It has now been assumed in the general diagram of isobars 
(loe eit.) that these roots always lie on the leftside of that value of 
da 
x for which 7 —0. To what change will this diagram be subjected 
v 
in the opposite case? We begin with determining the course of 
dp : ‘ da , EN, ; 
ea in this case. 5O as 7 is positive, according to our suppo- + 
DH HH 
sition, for that value of w for which 6=0O, we can always think 
the temperature so low that for this value of z, which we shall 
call 2°: 
dp 
Then we get for the course of 7 = 0 in the neighbourhood of «, 
Ak 
v—b 
Ss Senor tl = ny =DE 
v 
Now the value of 5 is positive for somewhat higher value of z 
than z,, whereas ) becomes negative for somewhat smaller value of 
d 
x. So we see that two branches of == pass through the point 
Lv 
a= 2,, v=. The two branches lie on either side of the line v = b, 
and have both positive v fore >>, negative v for «<a. Neither 
of the branches touch the line v=, but as follows from (2), both 
form an angle with it, which is the greater as m approaches closer 
to 1. This last result may be verified by a direct determination of 
the direction. For : 
