350 Dr. G. Bakker on the 



The equation, which gives the equality of the thermo- 

 dynamical potentials respectively in the interior of the drop 

 and in the vapour which surrounds the drop, was : 



Oi + ?•/) (jp liq . —pj) = (> vapi - p x ) {v 2 + r inax .). . (20) 



Now we have : i\ = 1'36 cm. 3 and v 2 = 1273*4. The 

 value of y max< can be calculated by 



1 2 3 



- + = — , 



Vi v max , v k 



where v/ c = 3*8. That gives i* max . = 37 cm. 3 



Further, we have for the ordinary vapour-pressure of ether 

 at 0° Celsius : 



lh = ±^ x 1-0133 X 10 G = 2-158 x 10 G dyne per cm. 2 



The equation (20) gives therefore : 



2 x l'30?i iq - 2-458 x 10 5 ) = Q\ ap . -2-458 x 10 5 ) 1310'4 

 or 



1310-4/7 vap .-2-% iq . = 3217-7G8X10 6 . . . (20a) 



The rule of Maxwell-Clausius gives again : 



(Pl—Pmir0( v s-* V l) = (Pvap.—^lX^ — V 3 ), . (21) 



where p m [ n . denotes the ordinate of the point A x in fig. 1. This 

 point lies in our case far below the axis of volume. To find 

 the absolute value of the ordinate of the point A l5 I consider 

 the part of the figure between the axis of volume and the 

 part of the isotherm below this axis, firstly as the sum of two 

 parabolic segments, with a common tangent, and secondly 

 us the value of the integral : 



^pdv. 



Using the equation of state of van der Waals, the segments, 

 cut from the axis of volume by the isotherm, are given by the 

 equation 



RT _ a 



v — b v 2 ' 



For T = 0'585T& I have calculated for these segments : 

 1-49% and 0"326tfo. 



In this way we find for the absolute value of p m i n . : 



Pmin. = 4*45jt?jfc. 



