erpendicular to the
axis of the cylinder, so that the direction of diffusion is along the
length of the cylinder, and we suppose no external forces, such as
gravity, to act on the system.

The densities of the gases are denoted by [rho]1, [rho]2, their
velocities of diffusion by u1, u2, and if their partial pressures are
p1, p2, we have by Boyle's law p1 = k1[rho]1, p2 = k2[rho]2, where
k1, k2 are constants for the two gases, the temperature being constant.
The axis of the cylinder is taken as the axis of x.

From the considerations of the preceding section, the effects of
inertia of the diffusing gases may be neglected, and at any instant of
the process either of the gases is to be treated as kept in
equilibrium by its partial pressure and the resistance to diffusion
produced by the other gas. Calling this resistance per unit volume R,
and putting R = C[rho]1[rho]2(u1 - u2), where C is the coefficient of
resistance, the equations of equilibrium give

dp1 dp2
--- + C[rho]1[rho]2(u1 - u2)= 0, and --- + C[rho]1[rho]2(u2 - u1)= 0 (1).
dx dx

These involve

dp1 dp2
--- + --- = 0 or p1 + p2 = P (2)
dx dx

where P is the total pressure of the mixture, and is everywhere
constant, consistently with the conditions of mechanical equilibrium.

Now dp1/dx is the pressure-gradient of the first gas, and is, by
Boyle's law, equal to k1 times the corresponding density-gradient.
Again [rho]1u1 is the mass of gas flowing across any section per unit
time, and k1[rho]1u1 or p1u1 can be regarded as representing the flux
of partial pressure produced by the motion of the gas. Since the total
pressure is everywhere constant, and the ends of the cylinder are
supposed fixed, the fluxes of partial pressure due to the two gases
are equal and opposite, so that

p1u1 + p2u2 = 0 or k1[rho]1u1 + k2[rho]2u2 = 0 (3).

From (2) (3) we find by elementary algebra

u1/p2 = - u2/p1 = (u1 - u2)/(p1 + p2) = (u1 - u2)/P,

and therefore

p2u1 = - p2u2 = p1p2(u1 - u2)/P = k1k2[rho]1[rho]2(u1 - u2)/P

Hence equations (1) (2) gives

dp1 CP dp2 CP
--- + ---- (p1u1) = 0, and --- + ---- (p2u2) = 0;
dx k1k2 dx k1k2

whence also substituting p1 = k1[rho]1, p2 = k2[rho]2, and by
transposing