ChemicalMass TransfereasyFE ExamCoursework

Steady Diffusion Rate Through a Film

A species diffuses at steady state through a stagnant film of thickness $L = 0.010\ \text{m}$ and area $A = 0.10\ \text{m}^2$ .

The diffusivity is $D = 1.0\times10^{-5}\ \text{m}^2/\text{s}$ and the concentration difference across the film is $\Delta C = 2.0\ \text{mol/m}^3$ .

Find the molar flow rate through the film.

Given

D	0.00001 m^2/s	Diffusivity
A	0.1 m^2	Cross-sectional area
L	0.01 m	Film thickness
ΔC	2 mol/m^3	Concentration difference

Hint 1

Fick's law is directly analogous to Fourier conduction.

Hint 2

Flux is DΔC/L; the question asks for a total flow rate.

Hint 3

N = DAΔC/L — check your units come out as mol/s.

Worked solution — try the problem first

Fick's first law, integrated across a film of uniform thickness at steady state:

$N = \frac{DA\Delta C}{L}$

$= \frac{(1.0\times10^{-5})(0.10)(2.0)}{0.010} = \frac{2.0\times10^{-6}}{0.010} = 2.0\times10^{-4}\ \text{mol/s}$

$\boxed{N = 2.0\times10^{-4}\ \text{mol/s}}$

Flux versus flow. The molar flux is $J = D\Delta C/L = 2.0\times10^{-3}\ \text{mol/m}^2\text{s}$ — per unit area. Multiplying by area gives the total molar flow, which is what was asked for. Confusing the two is the most common error here, and the units distinguish them.

Analogy. This has the same form as Fourier conduction ( $Q = kA\Delta T/L$ ) and Ohm's law ( $I = \Delta V/R$ ): a driving force divided by a resistance $L/(DA)$ .

Concepts:Fick's lawMolecular diffusionMass transfer

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