The London, Edinburgh and Dublin philosophical magazine and journal of science . he verified, if correct, at the criticalstate, inasmuch as it must give us the value of the criticalpressure pc. In this manner we can test a general expression for thecohesive forces., deduced as follows. V. The cohesive forces causing the Inward Pressure appear tofolloiv a law similar to the inverse square law of massattraction. If c is the Inward Pressure at the volume 1, then atthe volume v, and c = x. W Pl=„2 m* = x . p* (28) m being the mass of a molecule, N the number of moleculesat the volume v0= I, hence
The London, Edinburgh and Dublin philosophical magazine and journal of science . he verified, if correct, at the criticalstate, inasmuch as it must give us the value of the criticalpressure pc. In this manner we can test a general expression for thecohesive forces., deduced as follows. V. The cohesive forces causing the Inward Pressure appear tofolloiv a law similar to the inverse square law of massattraction. If c is the Inward Pressure at the volume 1, then atthe volume v, and c = x. W Pl=„2 m* = x . p* (28) m being the mass of a molecule, N the number of moleculesat the volume v0= I, hence N . m the density p. The factor xis considered a constant peculiar to the nature of a gas.(Compare e. g. J. H. Jeans,, Dynamic Theory of Gases/Cambridge, 1904, pp. 118 & 114.) In the following I intend to show that c is dependent onthe mass or molecular weight M only and x is a constant forall substances. Without discussing how the form ^91 = — has been arrived•at, I will here give my own deduction. Suppose 0 (vide fig. 2) to be a molecule at the surface of Fiff. a fluid attracted by the molecules below the surface in everydirection at angles between -: and 0. Let us consider 0 90 Mr. J. Kam on Molecular Attraction and and every molecule in the stationary motion of Clausiusand let OL be the radius of the sphere of attractionfor 0. Every molecule on the spherical surface RFL will be justthe last to attract 0, and each radius through a molecule ofthat surface is the direction of a force attracting 0. Suppose k to be such a force, and some function of distance,of the radius r, « =/(r). The horizontal components compensate each other, but thevertical components of all forces k form the normal resultantto which 0 is subjected. If <f) is the angle between such a force k and OF, the normalcomponent of k is k = k . cos (29) All forces at the angle <p have this same component h andlie in the surface of the cone with the angle in 0 and gothrough the curve of intersection
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