( 394 ) 
do hy? 
ag Her a fig si GEE (= ) (4 Lay (iy) dh, 
0 0 
We put fi w (hj) dh; = ej and fi? w (hj) dh; = e3, therefore the ex- 
0 
pression for the molecular pressure is transformed into : 
dg C9 (2 ): 
2 El el 
BN dh? 2 \dh 
If we add to this molecular pressure the external pressure pj, 
we may equate this sum to pag’, if p represents the pressure, 
which belongs to a homogeneous phasis with the density g. 
a rg /d 
pi + av? + ¢ 9 —> v —2(4 
2 
EE a 0e 
dh? =) Pee 
d°g Cy & y 
1 — 9 = — 00 0 —— — — | — 
4 Î 2 dh 2 \dh 
This relation is the same as has been deduced by Prof. VAN 
DER WAALS. 
Lb. We shall now calculate the molecular pressure in the direction 
of the surface of the liquid. For this purpose we suppose a plane 
laid through A, normal to the capillary layer, and in A a cylinder 
with a thickness do normal to 
that plane. The matter in the 
cylinder has everywhere the 
same density. The unity of 
mass in ¢ acts on the unity of 
mass in S with a force p (r), 
if r represents the distance Sc. 
The component of this force 
in the direction normal to PQ 
dr . 
is p(r) cos Sca or p (Dice. if we 
ter a 
Fig. 2. 
call the direction ab the direction z. 
The material cylinder ab L PQ, with a thickness do’ and itl a 
