( 803 ) 
is zero; also for v = + mand zr =— o. The equation pj; = ee being 
of the fourth degree in 2, no more than 4 values of 2 can ever be 
found for a certain value of py. So no maximum or minimum of 
the critical. pressure can occur on the right of 5. 
Here we interrupt the train of our reasoning for a moment, to 
show that in the case considered a minimum critical temperature 
must occur. As the equation: 
| de — iB 
gives an equation of the third degree, we might expect that 3 values 
of « which make 7%, stationary, could be found for every system in 
the complete diagram of isobars. But as for very great values of « 
da db 
always 0 oa a,b,z°, one root appears always to lie at 
v Hij 
infinity, and there are at most two roots for finite 7. One of them 
lies between a, and a,, where 7), becomes = 0, the other lies on 
the right of b,. For in our case we may write the equation for 
i 
a and 5: 
a= 0, (a — 2, — a, 6= be? — b, 
sO 
da aa. ( db 21 
— Za, (2 — 4 —2b. wv. 
‘dx ; ne da > 
k ; da db 
—— has the sign of 6 —— a —, and so of: 
dx da dx 
2a, 6, «? (« — x,) — 2a, b, 2 (« — wv) = 2a, b, av @, — 2a, b, xx’, 
aly . aie : 
so —— is positive for high values of x, which proves the presence 
lid 
of a minimum critical temperature in connection with the value 
+o for b—= 0. 
12. Let us now return to our diagram of isobars. We can now 
represent it fully for low temperatures, now that we have seen 
; ’ ; dp dp 
that there will be no intersection of ——0O and — = 0 in this case. 
v av 
We have only to add tbe observation that the value of the pressure 
. dp . 
on the line rs =— 0 approaches indefinitely to zero for very great 
av 
value of rz, however small the value of 7’ is, if only not quite zero. 
