Question

s 1.3 x 10 M and CR was zero, the flux of a certain molecule across a microporous membrane was found to be 0.344 pmol cm2s-1. The free diffusion coefficient of the mate measured was 0.6 x 10-6 cm's1. (pmol-pi rial being comole) A. What is the permeability (p) of the membrane to the material? B. If the thickness of the membrane is 15 x 10 m, what is the equivalent relative area available for diffusion of the material (the porous area/total area)?

Answer 1

A.

Considering unit area

Flux(J) = K * ( CL - CR)

J = 0.344 pmol cm^-2 s^-1 = 0.344 x 10^-12 mol cm^-2 s^-1

K = J / (CL - CR) = ( 0.344 x 10^-12 / 1.3 x 10^-6 )

K = 2.65 x 10^-7 cm^-2 s^-1

Now,

P = K * D / (dx)

where, P = permeability of membrane, K = constant, D = diffusion coefficient and dx = thickness

dx = 15 x 10^-6 m = 15 x 10^-4 cm

P = (2.65 x 10^-7 * 0.6 x 10^-5 / 15 x 10^-4 )

P = 1.06 x 10^-9 cm^-1 s^-2

B.

Area available for diffusion ( Ap / Am) = (Flux * thickness) / ( Diffusion coeff * ( CL - CR))

Ap / Am = 0.344 x 10^-12 * 15 x 10^-4 / ( 0.6 x 10^-5 * 1.3 x 10^-6)

Ap/ Am = 6.62 x 10^-5