GARVER] THE RANGE OF MOLECULAR ACTION 83 
in liquid density, does not account for the undoubted fact that there zs 
an increase in vapour-density accompanying an increase in pressure. 
It is possible, however, to obtain by analysis a kinetic interpretation of 
the conditions necessary to produce molecular equilibrium expressed 
in terms of the space relations of the molecules, including change in 
liquid density. 
For, consider a spherical globule, or drop, of water in equilibrium 
with its saturated vapour. Kinetic equilibrium must consist merely 
of a balanced interchange between the two phases. Hence, during the 
equilibrium of the vapour and liquid phases of a substance, suppose 
that n, molecules of vapour per unit area of the intervening area of 
liquid surface return to the liquid as many molecules as the n, mol- 
ecules of liquid per unit area emit when the intervening surface is plane. 
Let us suppose that the interchanges taking place extend to an average 
distance, or depth, € on both sides of the interface, or surface, of the 
liquid of area a. The ratio n,/n, will be the same as the ratio of the 
vapour-density, @ to the liquid-density, 9, or 9/p = n,/n,. Now 
let us suppose that when the liquid is in the form of a spherical drop in 
equilibrium with the surrounding vapour the area of the drop is a. 
The vapor at an average distance, ¢ from the surface of the drop will 
lie on a sphere of area, a + Aaand the liquid at an average distance, €, 
on the other side of the surface, inside the drop, will occupy (omitting 
second order differences) a sphere of area a — /\a. Now, the n, mole- 
cules of vapour per unit area multiplied by the area, or n, (a + /\a) 
must maintain equilibrium with the n, («a — /\a) molecules of liquid 
acting through the same area, a. The ratio, when the intervening area 
is plane is n,/n,. When the intervening surface of area a, is spherical, 
the ratio, as just shown, must be n, (a + Aa) / n, (a — Aa) = @'/p’ 
(say). If the liquid does not change density, p = p and the ratio 
a’/o will be (a + /\a) / (a— /\a) which represents the ratio of the 
vapour-density in equilibrium with a spherical drop of area a to the 
vapour-density in equilibrium with an equal plane area a. If we assume 
that the vapour-pressures, ©! and ©, are proportional to the densities, 
oa’ and 6, we shall have 
where r is the radius of the drop and Ar = €, is the average distance, or 
range of molecular action in which the change of phase from liquid to 
vapour and vice versa, takes place. 
