( 524 ) 
normal substances, of which the critical pressures differ as much as 
possible. We take therefore ether and carbon disulphide. The critical 
data are the following: 
CS T, =b48R ip = 76 alm. 
3 
ether | Tr AGT" Te pr 30 atm: 
3 
E v ded 
In order to determine p= —, we remark, that EE 
UV; Y Pi 
FR i . > ay, ” Tal a, 
v, = —.—, as for instance — —=fRT, and —=yp,. We have there- 
Y Ps vy VY, 
fore: 
OE i mencl 8s 
. D . 
that is to say p — Oa, where the proportion ee represented by a. 
st ; 5 
2 
Now for the designed substances 6 = 0,852, a = 2,17, so that 
we find p= 1,85. Since r= py —1, the equation for 2, passes into 
1 Ss 
a Gn er): te Te ORE) 
and hence we find for «a, the value 0,29. Further y/4==0,923, 
io = 1,36, f= 7, and so (8) becomes: 
14 X 0,206 « 1,85 X 0,191 
one Goan eS 
548: (1 + Ad), 
eC 
Tis 
or 
1,019 
ms 
1,94 
We have further: 
0,723 0,0650 
1,555 1,153 
548: (1 + A.) = 288:(1 + Ad. 
= 
A. = 0,412 — 0,412 X 0,409 — 0,169, 
so that we find for 1 + A, the value 1,17. 
Hence 7, becomes 288: 1,17 = 246 = — 27°C. 
The critical point of the chosen substances lies therefore still a 
thirty degrees beneath the common zero of Celsius. And for the 
greater majority of other normal substances we will find for 7; still 
much smaller values — because the critical pressures will differ 
there in most of the cases less than in the case of ether and CS,. 
4. All that precedes now undergoes important modifications, when 
one of the two components is anomalous, specially water, For in 
the first place the critical pressure of the water is very high, not 
