ABSTRACT

Gas diffusion through solids is a relatively simple and useful example of steady-state diffusion that has many commercial applications. Usually, gases diffuse as interstitial molecules in polymers and inorganic glasses and interstitial atoms in metals. Consider a sheet of thickness L, which is small compared to the other two dimensions; that is, the size of the sheet is large compared to the thickness-a window pane or a sheet of poly(methyl methacrylate)—Plexiglas®—for example. Figure 10.1 shows gas pressures of p0(x = 0) and pL(x = L) that determine the concentration of the diffusing species in the solid-hydrogen in palladium, water in polyvinyl chloride, oxygen in SiO2-where C(0) = C0 and C(L) = CL, which are the two necessary boundary conditions for solving the steady-state diffusion equation. Also shown are some transient values of the concentration, C(x, t), that finally lead to the steady-state concentration. The flux or flux density J (Units (J) = mol/cm2 s) as a function of time is sketched in Figure 10.2. The flux through the solid becomes constant when the steady state is reached. For now, the interest is only on the steady state: what happens during the transient is left for Chapter 11. To determine the flux of gas through this plate, it is necessary to begin with the steadystate version of Fick’s second law in one dimension:

∂ ∂

= ∂ ∂

=C t

D C x

2 0 (10.1)

or

d C dx

2 0=

which is now an ordinary differential equation because the change in concentration with time is no longer being considered. This is integrated twice to give C x Ax B( ) = + ; for x = 0, C0 = B and for x = L, A =(CL – C0)/L so the solution to the differential equation with the two fixed boundary conditions is

C x C C L

x CL( ) = −( ) +0 0 (10.2)

which clearly satisfies the two boundary conditions and is the linear function of x-which no longer changes with time-shown in Figure 10.1. So the steady-state flux, J, is given by

J D dC dx

D C C

L D

C C L

L L= − = − −( )

= −( )0 0 (10.3)

which is a positive outward flux because C0 > CL.