A water layer carries ethanol out of a solvent, a rubber sheet slowly lets carbon dioxide through, and a rubber hose leaks hydrogen even with no hole in it. All three are steady diffusion, and all three use Fick’s law. Each adds one new detail: what the total concentration is in a liquid, what sets the concentration at the surface of a solid, and what area to use when the area changes along the path.

In this lesson you will work the liquid film, the solid membrane and the curved wall in turn (Figure 1). The equations come from Geankoplis, Transport Processes and Unit Operations, 3rd edition, Sections 6.3 and 6.5.

1. Liquid film$c_{av},\ x_{BM}$average c, log mean of B
2. thensolids
3. Solid$c_A = \dfrac{S\,p_A}{22.414}$dissolve, then diffuse
4. thencurved walls
5. Curved wall$W_A = N_A A$molar flow is constant

Learning Objectives

After this lesson, you should be able to:

1. Calculate the steady flux of a solute through a liquid film, with the average total concentration and the log-mean mole fraction of the stagnant liquid.
2. Find the dissolved concentration at the face of a solid from the solubility, and calculate the flux through a solid sheet.
3. Explain why the molar flow, not the flux, is constant across a curved wall.
4. Calculate the leak rate through a cylindrical or spherical wall with the right mean area.

Key Ideas

1. Diffusion in Liquids

A liquid obeys the same two flux forms as a gas, but its total concentration is not $\frac{P}{RT}$. Density and mean molar mass change with composition, so Geankoplis uses the average of the two ends (Eq. 6.3-2):

For equimolar counterdiffusion (Eq. 6.3-1):

The common liquid case is A diffusing through B that does not move, for example ethanol through a water film whose far side is an organic solvent that water does not enter. Then $N_B = 0$ and A rides a small bulk flow (Eqs. 6.3-3 and 6.3-4):

Here $x_{BM}$ is the log mean of the mole fractions of B at the two faces. It is less than 1, so dividing by it raises the flux. For a dilute solution $x_{BM} \to 1$ and the flux reduces to $N_A = \frac{D_{AB}(c_{A1}-c_{A2})}{z_2-z_1}$ (Eq. 6.3-5). Liquid diffusivities can change markedly with concentration, so use a value measured for the range you are working in.

Key idea: in a liquid, use $c_{av}$ from the densities and molar masses at both faces, and divide by the log mean $x_{BM}$ when B is stagnant.

Worked Example 1: Ethanol Through a Stagnant Water Film

Problem: Ethanol (A) diffuses at steady state through a water (B) film 1.5 mm thick at 293 K. On one side the solution is 16.8 wt% ethanol with density 972.8 kg m⁻³; on the other it is 6.8 wt% ethanol with density 988.1 kg m⁻³. Water does not diffuse ($N_B = 0$). Use $D_{AB} = 0.740 \times 10^{-9}\,\text{m}^{2}\,\text{s}^{-1}$, $M_A = 46.05$ and $M_B = 18.02\,\text{kg}\,\text{kmol}^{-1}$. Find $N_A$. The data are from Geankoplis Example 6.3-1 (p. 399, diffusivity from Treybal, 1963), which uses a 2.0 mm film.

Step 1. Ethanol mole fractions. Take 100 kg of solution.

Step 2. Mean molar mass at each face.

Step 3. Average total concentration (Eq. 6.3-2, p. 398).

Step 4. Log mean of the water mole fractions.

Step 5. Substitute. Write the thickness in m.

Final Answer: $N_A \approx 1.19 \times 10^{-6}\,\text{kmol}\,\text{m}^{-2}\,\text{s}^{-1}$

Check: leaving out $x_{BM}$ gives a flux 5.1% low, so bulk flow matters even at 7 mol% ethanol. Using pure water, $\frac{998.23}{18.02} = 55.40\,\text{kmol}\,\text{m}^{-3}$, in place of $c_{av}$ overstates the flux by 9.7%.

Note: this lesson uses the log mean $x_{BM}$ of Eq. 6.3-4. Geankoplis’s Example 6.3-1 uses the arithmetic average, here $0.9495$ against $0.9493$, a difference of about 0.02%. With the book’s own 2.0 mm film and the arithmetic average, carried without rounding to the end, the same method gives $N_A = 8.9505 \times 10^{-7}\,\text{kmol}\,\text{m}^{-2}\,\text{s}^{-1}$. Geankoplis reports $8.99 \times 10^{-7}$ because of small slips in its printed intermediate values (for example $M_2$ printed as 18.75 against 18.80 here).

Your turn: ethanol through a water film

Same steps as Worked Example 1: mole fractions, mean molar masses, $c_{av}$, the log mean $x_{BM}$, then substitute with the thickness in m. The densities given are from Perry’s Handbook at 20 °C. Enter the flux as, for example, 1.23e-6.

2. Diffusion Through Solids

A gas such as $\text{H}_2$ or $\text{CO}_2$ first dissolves in the face of a polymer or rubber, then diffuses through it. The dissolved gas is dilute, so the bulk-flow term is small and is dropped, and Fick’s law with constant $D_{AB}$ gives the slab result (Eqs. 6.5-1 and 6.5-2):

The concentration just inside each face is set by the solubility $S$, in m³ of gas at STP (0 °C and 1 atm) per m³ of solid per atm of partial pressure (Eq. 6.5-5). Dividing by 22.414 m³(STP) kmol⁻¹ turns it into kmol m⁻³:

The concentration jumps at each face: the gas and the solid hold different amounts of A, and only their ratio is fixed (Figure 2). Combining the two equations gives the flux in terms of the permeability $P_M$ (Eqs. 6.5-8 and 6.5-9):

For layers of different materials in series, the resistances $\frac{L_i}{P_{Mi}}$ add (Eq. 6.5-12):

Table 6.5-1 (p. 411) lists $D_{AB}$, $S$ and $P_M$ for several gases in rubber. These are separately measured data, so the tabulated $P_M$ differs from the product $D_{AB}S$ by 0.6% to 3.4% for the hydrogen, oxygen, nitrogen and carbon dioxide rows. This lesson always uses $D_{AB}S$. For gases such as $\text{H}_2$ and $\text{O}_2$ in metals such as Ni, Cd and Pt, the flux is approximately proportional to $\sqrt{p_{A1}}-\sqrt{p_{A2}}$, so Eq. 6.5-8 does not hold.

Key idea: in a solid, find the face concentration from the solubility, $c_A = \frac{S\,p_A}{22.414}$, then use Fick’s law for a slab. Permeability is $P_M = D_{AB}S$.

Worked Example 2: Carbon Dioxide Through a Rubber Sheet

Problem: A vulcanized rubber sheet 1.0 mm thick has $\text{CO}_2$ at a partial pressure of 1.5 atm on one side and none on the other, at 298 K. From Geankoplis Table 6.5-1, $D_{AB} = 0.11 \times 10^{-9}\,\text{m}^{2}\,\text{s}^{-1}$ and $S = 0.90\,\text{m}^{3}(\text{STP})\,\text{m}^{-3}\,\text{atm}^{-1}$. Find the steady flux $N_A$. Use $P_M = D_{AB}S$ from these two values, not the tabulated $P_M$.

Step 1. Dissolved concentration at each face.

Step 2. Fick’s law across the slab.

Step 3. Check with the permeability. Here $P_M = D_{AB}S$, not the tabulated value.

Final Answer: $N_A \approx 6.63 \times 10^{-9}\,\text{kmol}\,\text{m}^{-2}\,\text{s}^{-1}$

Check: that is about $5.35 \times 10^{-4}$ m³(STP) of $\text{CO}_2$ per m² per hour. The tabulated $P_M = 1.01 \times 10^{-10}$ gives $6.759 \times 10^{-9}\,\text{kmol}\,\text{m}^{-2}\,\text{s}^{-1}$, 2.0% higher, only because the table’s $D_{AB}$, $S$ and $P_M$ are separately measured; this lesson uses $D_{AB}S$ throughout.

Your turn: a gas through a rubber sheet

Same steps as Worked Example 2: find $c_A$ at both faces from $S$ and the partial pressures, then divide the difference by the thickness in m.

3. Curved Shells: Cylinders and Spheres

In a tube wall or a spherical shell the area grows with radius: $A = 2\pi r L$ for a cylinder and $4\pi r^2$ for a sphere. At steady state no A builds up anywhere, so the molar flow $W_A = N_A A$ is the same at every radius, while the flux $N_A$ falls as the area grows (Figure 3). Integrating Fick’s law with $W_A$ constant gives, for a cylinder of length $L$ (Geankoplis Eqs. 6.5-3 and 6.5-4):

For a sphere, $A = 4\pi r^2$ and the same integration gives

The second form comes from putting the two reciprocals over a common denominator:

So the mean area of a spherical shell is $4\pi r_1 r_2$. A line of algebra shows that this is the square root of the inner area times the outer area:

The same $4\pi r_1 r_2$ appears in heat conduction through a hollow sphere (Geankoplis Eq. 4.2-14, p. 222; Perry’s Eq. 5-8, p. 5-4). For example, the tube wall of Worked Example 3 (radii 5 and 8 mm) made into a sphere, with the same hydrogen and rubber, has

The choice of mean matters only for thick shells (Figure 4). At $\frac{r_2}{r_1} = 1.1$ the log-mean area is within 0.08% of the arithmetic mean and the sphere area within 0.5%; at $\frac{r_2}{r_1} = 2$ the gaps are 3.8% and 20%.

If a gas A diffuses outward from a sphere through a stagnant gas B, the log term of the stagnant film carries over unchanged (Geankoplis Eq. 6.2-31, p. 391, written with $c = \frac{P}{RT}$ and $y_A = \frac{p_A}{P}$):

Key idea: in a curved wall the molar flow $W_A$ is constant, not the flux. Use the slab equation with the log-mean area $A_{lm}$ for a cylinder and $4\pi r_1 r_2$ for a sphere.

Worked Example 3: Hydrogen Leaking Through a Rubber Tube

Problem: Hydrogen at 2.0 atm flows through a vulcanized rubber tube 2.0 m long with inner radius 5 mm and outer radius 8 mm, at 298 K. The partial pressure of hydrogen outside is zero. From Geankoplis Table 6.5-1, $D_{AB} = 0.85 \times 10^{-9}\,\text{m}^{2}\,\text{s}^{-1}$ and $S = 0.040\,\text{m}^{3}(\text{STP})\,\text{m}^{-3}\,\text{atm}^{-1}$. Find the leak rate $W_A$.

Step 1. Dissolved concentration at the inner face.

Step 2. Inner, outer and log-mean areas.

Step 3. Leak rate.

Step 4. Compare with a flat wall. The leak scales with the area used, so divide each other choice by $A_{lm}$:

So the inner area alone gives a leak 21.7% low, the outer area 25.3% high, and the arithmetic mean of the two areas 1.83% high.

Final Answer: $W_A \approx 8.11 \times 10^{-11}\,\text{kmol}\,\text{s}^{-1}$, about $2.92 \times 10^{-4}$ mol h⁻¹

Check: the first form of the cylinder equation in Key Idea 3, with $2\pi L$ and the logarithm of the radius ratio, gives the same 8.111. With $\frac{r_2}{r_1} = 1.6$ the wall is thick, so the area you use matters far more than which mean you take.

Your turn: a gas leaking from a hollow sphere

Same idea as Worked Example 3, for a sphere: find $c_{A1}$ from $S$, use the mean area $4\pi r_1 r_2$ with the radii in m, then multiply by $\frac{D_{AB}(c_{A1}-c_{A2})}{r_2-r_1}$.

Common Mistakes

Common mistake: Using the dilute form for a concentrated film. With $x_{BM}$ left out, Worked Example 1 comes out 5.1% low. Divide by the log mean of the B mole fractions whenever B is stagnant.

Common mistake: Using pure water for the total concentration. $\frac{998.23}{18.02} = 55.40\,\text{kmol}\,\text{m}^{-3}$ is pure water. The solution has a different density and molar mass at each face: use $c_{av}$.

Common mistake: Treating the partial pressure as a concentration. A gas must dissolve before it diffuses: $c_A = \frac{S\,p_A}{22.414}$, not $p_A$.

Common mistake: Forgetting the 22.414. $S\,p_A$ is in m³(STP) per m³ of solid. Divide by 22.414 m³(STP) kmol⁻¹ to get kmol m⁻³.

Common mistake: Leaving lengths in mm. Write every thickness and radius in m: 1 mm is $10^{-3}$ m. Otherwise the flux is 1000 times too small.

Common mistake: Using the inner or outer area as if the wall were flat. In Worked Example 3 that is 21.7% low or 25.3% high. Use $A_{lm}$ for a cylinder and $4\pi r_1 r_2$ for a sphere.

Common mistake: Mixing up the cylinder and sphere means. The log-mean area belongs to the cylinder, $4\pi r_1 r_2$ to the sphere. For a thick sphere with $\frac{r_2}{r_1} = 2$ the arithmetic mean area is 25% too large.

Common mistake: Rounding too early. Keep 4 significant figures in the mole fractions, $c_{av}$, $x_{BM}$ and the areas; round only the final answer to 3.

Key Takeaway

Liquids, solids and curved walls all use Fick’s law. In a liquid, use $c_{av}$ and divide by $x_{BM}$ when B is stagnant. In a solid, the face concentration is $\frac{S\,p_A}{22.414}$ and the permeability is $D_{AB}S$. In a curved wall the molar flow is constant, so use the slab equation with the right mean area: $A_{lm}$ for a cylinder, $4\pi r_1 r_2$ for a sphere.

Practice: randomized problems

Each problem below uses fresh numbers every time and mixes the three kinds of problem in this lesson: work it out, check your answer, then open the worked solution to compare your steps.

Your results

This summary collects every practice problem you try on this page and suggests what to review next.

Figures: original work by Dr. Ray’s Learning Center, licensed CC BY 4.0. Methods and data: C. J. Geankoplis, Transport Processes and Unit Operations, 3rd edition (1993): Eqs. 6.3-1 to 6.3-5 (p. 398) for liquids; Example 6.3-1 (p. 399, diffusivity from Treybal, 1963) for the ethanol and water film; Eqs. 6.5-1 to 6.5-12 (pp. 408 to 410) and Table 6.5-1 (p. 411) for solids; Eq. 6.2-31 (p. 391) for a sphere in a stagnant gas; Eq. 4.2-14 (p. 222) for the sphere mean area. Perry’s Chemical Engineers’ Handbook, 9th edition: Eq. 5-8 (p. 5-4) for the sphere mean area and Table 2-58 (p. 2-108) for ethanol and water densities. The practice problems use illustrative conditions and are audited; see the lesson audit log.