( S04 ) 
AEN a ee 
is in inverse ratio to z, and — to a*. So it follows from 
Vn Uv 
this that all the negative isobars starting from the point zr =a,, 
v=0O will have the shape we indicated before. The line p—0 will 
For 
dp ed ee ie 
intersect the line os = 0 at infinite distance. To this isobar, however, 
v 
also the branch belongs starting from the point p =O on the line 
«= .2,, and also the line v=o. For a positive pressure the isobar 
consists of two separate branches. One of them, starting from the 
point e=, v =O remains confined to smaller volumes than p = 0, 
the other starting from a point on the line «=a, arrives some- 
dp 
p 
and 2, has a tangent there parallel to the v-axis, and returns again 
to the line «—wz,, now on the other side of the maximum pressure. 
So we get fig. 10 for the complete course of the isobars. 
where on the vapour branch of =0, with ascending value of v 
13. How will this figure be modified with increase of temperature. 
€ 
Let us consider the temperature which ET of the minimum critical 
_ 
temperature. From equation (1) on page 802 follows that we may expect 
a point of intersection at a volume v = 2/ and a temperature = Ty 
; OF Ks db da 
for the mixture with minimum critical temperature ( where dn 0 5 
> da x 
The line > 0 lying at smaller volumes than the line a= 0 for 
very great values of #, as we saw in 10, there must be another 
point of intersection more to the right. It is clear that these points 
ae orem f ‚dp dp 
of intersection have arisen by a contact of Je 0 and rl 0, and 
that the two points of intersection have moved from this point of 
contact in opposite direction. For as the equations for the points of - 
intersection are of the first degree with respect to, 7’ and wv, it is 
not possible that two points of intersection lying beside each other 
move in the same direction; for then we should find different values 
of ZY for the same value of 2. In the point of intersection lying 
most to the left the pressure is 0, from which it already follows 
that there must be another point of intersection ; for the pressure 
finally verging again to zero towards the right, there must be a 
point between where the pressure has reached its lowest value on 
