Theory of Surface Forces. 147 



We had : 



(Consequently : 



_ PvzP, v 1 + v ld v 1± v ? p^p, Vi + V, 



2 2 v 2 — vi 2 v 2 — vi * 



We have 



Pi-P=^h(Pi-P2) and ^2-^=-3-(/>i-p 2 ), 



and we get therefore : 



ffi— jp 2 fvjdvi + vidv2 Vi + v 2 7 1 -rV| 

 + 2 1 tfc-*! + 2 %-tfJ' 



Further we have : 



v 2 dvi -|- i\dv 2 t?i + i>2 7 « 1 + ^2 2 v 2 2 ^i — 2 v^dv 2 _ ■, 2v x v 2 



V 2 — V X I Vz — Vx (^2 — ^l) ^2—^1 



So the equation of energy of the capillary layer becomes: 

 TrfW« + tf»-HiS + &=2<*{ **!L- V *±* t*. (30) 



7TT 



In the equation (28) the differential quotient -^ is 



partial, the second parameter being constant. In the equa- 

 tion (30) we must conceive R is the constant. The state of 

 the plane capillary layer is completely determined by one 

 parameter. This parameter is in the present consideration 

 the temperature, and we may now state for the plane capillary 

 layer : 



Td v = d€ + pdv-KdS, .... (31) 



in this case : p x z=p 2 = ordinary vapour pressure. 



The equation (31) is also easily to be deduced directly. 

 For if f is the thickness of the plane capillary layer and 



* The general equation of energy of the capillary layer becomes : 

 \aK/ T 2 lv 2 —t\ v. 2 -v 



L2 



