Molecular Diffusion
A molecule in solution does not sit still. Random thermal motion carries it through the surrounding medium one collision at a time. When many molecules do this together, the net result is a flux from high concentration to low. This chapter establishes Fick's two laws, introduces the diffusion coefficient, and shows how the same resistance framework used in heat conduction applies directly to mass transfer.
The Glove Problem
A chemical plant worker puts on nitrile rubber gloves before handling toluene. The gloves look intact. The container says the gloves are rated for toluene. But after two hours the worker's skin is irritated. What went wrong?
The glove did not tear. Toluene did not leak through a hole. Instead, individual toluene molecules diffused through the solid rubber, one molecular collision at a time, until enough had crossed to reach the skin. This is permeation: mass transfer through a solid membrane driven purely by a concentration gradient. No bulk flow, no pressure difference, just molecular random walks adding up to a net flux.
The same process governs oxygen transport through packaging films, drug release from polymer matrices, solvent uptake in coatings, and barrier performance in multilayer films. Every one of these problems is controlled by the diffusion coefficient $D$ and the concentration gradient across the membrane.
Figure 11.1: Steady-state diffusion through a polymer membrane. The concentration profile is linear when $D$ is constant. Net flux $J$ is directed from high to low concentration.
11.2 Fick's First Law
Adolf Fick observed in 1855 that the rate of mass transfer is proportional to the concentration gradient. For one-dimensional diffusion in the $x$ direction:
The minus sign has the same origin as Fourier's law: molecules move from high to low concentration, which is the direction of decreasing $C$, so the gradient $dC/dx$ is negative when flux is positive. The minus sign ensures $J > 0$ when mass moves in the positive $x$ direction.
Steady-State Flux and Diffusive Resistance
For steady-state diffusion through a flat membrane of thickness $L$, the concentration profile is linear. Substituting the constant gradient gives:
The diffusive resistance $R = L/D$ increases with thickness and decreases with diffusivity. Two membranes placed in series add their resistances directly, just as thermal resistances do:
The Heat-Mass Analogy
| Quantity | Heat Conduction | Molecular Diffusion |
|---|---|---|
| Driving potential | Temperature $T$ [K] | Concentration $C$ [mol/m³] |
| Flux | $q$ [W/m²] | $J$ [mol/(m²·s)] |
| Transport property | Conductivity $k$ [W/(m·K)] | Diffusivity $D$ [m²/s] |
| Resistance (slab) | $R = L/(kA)$ | $R = L/(DA)$ |
| Rate equation | $q = k\Delta T / L$ | $J = D\Delta C / L$ |
Steady-state O₂ permeation through an HDPE film
An HDPE packaging film (thickness $L = 0.50$ mm) separates oxygen at $C_1 = 2.0$ mol/m³ (outside) from a food package at $C_2 = 0$. The diffusion coefficient of O₂ in HDPE is $D = 2.0 \times 10^{-10}$ m²/s. Find the steady-state oxygen flux.
Solution
Convert thickness: $L = 5.0 \times 10^{-4}$ m. Apply Eq. 11.2:
$$J = D\,\frac{\Delta C}{L} = \frac{(2.0\times10^{-10})(2.0)}{5.0\times10^{-4}} = \frac{4.0\times10^{-10}}{5.0\times10^{-4}} = 8.0\times10^{-7}\;\mathrm{mol/(m^2{\cdot}s)}$$This equals 0.80 μmol/(m²·s). For a 1 m² package, about 69 mmol/day of oxygen enters, more than enough to spoil fresh produce within days. A barrier coating that reduces $D$ by 100× extends shelf life proportionally.
11.3 Fick's Second Law
Fick's first law describes the flux at a given instant when the profile is known. To find how the profile itself evolves in time, apply a mass balance on a thin slab of thickness $\Delta x$. In the limit $\Delta x \to 0$:
This is identical in form to the heat conduction equation $\partial T/\partial t = \alpha\,\partial^2 T/\partial x^2$, with the thermal diffusivity $\alpha$ replaced by $D$. Every solution from transient heat conduction applies here with the substitution $\alpha \to D$ and $T \to C$.
Semi-Infinite Medium: the Error Function Solution
When a polymer slab is much thicker than the diffusion depth, the "semi-infinite" approximation applies. For a constant surface concentration $C_0$ imposed at $x = 0$ for $t > 0$, the solution to Eq. 11.4 is:
At $x = 0$: $C = C_0$ for all $t > 0$ (the boundary condition). Far from the surface, $C \to 0$. As time increases, the profile shifts deeper into the material.
Diffusion Length and Fourier Number
The argument of erfc equals 1 when $x = 2\sqrt{Dt}$. At this depth the concentration has fallen to $\operatorname{erfc}(1) \approx 0.16\, C_0$. The diffusion length is:
The Fourier number for mass transfer compares diffusion depth to a characteristic system length $L$:
Figure 11.2: Concentration profiles $C(x,t) = C_0\,\operatorname{erfc}(x/2\sqrt{Dt})$ at three times. The surface is fixed at $C_0$. The diffusion front deepens as $\sqrt{t}$. The diffusion length $\delta_D$ (shown for $t_2$) marks the depth where $C \approx 0.16\,C_0$.
Diffusion length in an epoxy coating after one hour
A thick epoxy coating is exposed to acetone vapor at $t = 0$. The diffusion coefficient of acetone in the epoxy is $D = 1.0 \times 10^{-11}$ m²/s. Estimate the diffusion length after 1 hour. Is the semi-infinite approximation valid for a 5 mm coating?
Solution
Convert time: $t = 3600$ s. Apply Eq. 11.6:
$$\delta_D = 2\sqrt{Dt} = 2\sqrt{(1.0\times10^{-11})(3600)} = 2\sqrt{3.6\times10^{-8}}$$ $$= 2\times1.90\times10^{-4}\;\mathrm{m} \approx 0.38\;\mathrm{mm}$$After one hour the concentration front has penetrated only 0.38 mm out of a 5 mm coating. Check: $\mathrm{Fo}_m = Dt/L^2 = (10^{-11} \times 3600)/(5\times10^{-3})^2 = 0.0014 \ll 1$. The semi-infinite approximation is fully valid. The thick coating will not be saturated for many hours.
To reach $\delta_D = 5$ mm (the full coating), solve $5\times10^{-3} = 2\sqrt{Dt}$: $t = (2.5\times10^{-3})^2/D = 6.25\times10^{-6}/10^{-11} = 6.25\times10^5$ s $\approx 7$ days. Sixteen times the depth requires 256 times the time, because of the $\sqrt{t}$ dependence.
11.4 The Diffusion Coefficient
The diffusion coefficient $D$ encodes how fast a given molecule moves through a given medium. Its value spans many orders of magnitude, depending on molecular size, medium viscosity, and temperature.
| System | Typical D (m²/s) | Example |
|---|---|---|
| Small molecule in gas | 10⁻⁵ to 10⁻⁴ | O₂ in air |
| Small molecule in liquid | 10⁻⁹ to 10⁻⁸ | Toluene in water |
| Small molecule in polymer | 10⁻¹⁴ to 10⁻¹⁰ | O₂ in HDPE |
| Polymer chain in solution | 10⁻¹⁸ to 10⁻¹⁶ | PEG in water |
| Polymer in glassy polymer | < 10⁻²⁰ | PE chain in PMMA glass |
The Stokes-Einstein Equation
For a spherical particle diffusing in a viscous solvent, Einstein (1905) combined Stokes drag with the equipartition theorem to obtain:
Three key predictions follow directly. First, $D \propto 1/r$: larger molecules diffuse more slowly. A polymer chain with $r = 30$ nm diffuses 100 times more slowly than a solvent molecule with $r = 0.3$ nm in the same medium at the same temperature.
Second, $D \propto T/\mu$: raising temperature increases $D$ directly through the numerator and also decreases liquid viscosity $\mu$, so both effects raise $D$. This is why diffusion through rubbery polymers (above $T_g$) is orders of magnitude faster than through glassy polymers.
Third, $D \propto 1/\mu$: a molecule in glycerol ($\mu \approx 1.5$ Pa·s) diffuses about 1500 times more slowly than in water ($\mu \approx 0.001$ Pa·s).
Figure 11.3: The Stokes-Einstein equation predicts $D \propto 1/r$. A polymer chain with hydrodynamic radius 10 to 100 times larger than a small molecule diffuses 10 to 100 times more slowly in the same solvent at the same temperature.
Stokes-Einstein comparison: solvent vs. dissolved polymer
A solvent molecule has hydrodynamic radius $r_s = 0.30$ nm. A dissolved polymer chain in the same solvent at the same temperature has hydrodynamic radius $r_p = 3.0$ nm. Find the ratio $D_s/D_p$.
Solution
From Eq. 11.8, $D = k_BT/(6\pi\mu r)$. Since $T$, $\mu$, $k_B$, and $6\pi$ are identical for both species:
$$\frac{D_s}{D_p} = \frac{r_p}{r_s} = \frac{3.0\;\text{nm}}{0.30\;\text{nm}} = \mathbf{10}$$The solvent molecule diffuses 10 times faster than the polymer chain. In practical terms, solvent swelling fronts advance far ahead of any polymer redistribution, which is why solvent uptake kinetics are always governed by solvent diffusion, not polymer chain motion.
Interactive: Diffusion Profile Explorer
This widget plots $C(x,t) = C_0\,\operatorname{erfc}\!\left(x/(2\sqrt{Dt})\right)$ for adjustable $D$ and $t$. The red dashed line marks the diffusion length $\delta_D = 2\sqrt{Dt}$.
Concept Check
Each pool has three variants. Answer incorrectly and a fresh version appears. After all variants are exhausted the worked solution is shown.
In Fick's first law $J = -D\,dC/dx$, the minus sign appears because:
Across a membrane, concentration decreases from left ($C_1 = 8$ mol/m³) to right ($C_2 = 2$ mol/m³). The diffusive flux direction is:
You increase the concentration difference $\Delta C$ across a membrane by a factor of 4, keeping $D$ and $L$ fixed. What happens to the steady-state flux $J$?
Fick's first law $J = -D\,dC/dx$ has the same structure as Fourier's law. The minus sign enforces the physical direction: diffusion is from high to low concentration (down the gradient). For a flat membrane at steady state, $J = D\Delta C/L$. Flux is linear in $\Delta C$, $D$, and $1/L$: doubling any one doubles the flux.
Two membranes in series: $R_1 = L_1/D_1 = 500$ s/m, $R_2 = L_2/D_2 = 1500$ s/m. What fraction of the total resistance does Membrane 2 contribute?
You double a membrane's thickness from $L$ to $2L$, keeping $D$ and $\Delta C$ unchanged. The steady-state flux:
A composite film has two layers. Layer A: $L_A = 0.1$ mm, $D_A = 1\times10^{-10}$ m²/s. Layer B: $L_B = 0.05$ mm, $D_B = 5\times10^{-12}$ m²/s. Which layer has more diffusive resistance?
Diffusive resistance $R = L/D$ combines thickness and diffusivity. A thin layer with very low $D$ can dominate the total resistance of a composite film even though it appears thin. This is the principle behind high-barrier packaging: a thin EVOH or aluminium layer controls permeation far more than the thick PE structural layers. Resistances in series add directly: $R_\text{total} = \sum L_i/D_i$.
Fick's second law $\partial C/\partial t = D\,\partial^2 C/\partial x^2$ describes:
The Fourier number for mass transfer, $\mathrm{Fo}_m = Dt/L^2$, physically represents:
With $D = 1.0\times10^{-11}$ m²/s and $t = 1000$ s, the diffusion length $\delta_D = 2\sqrt{Dt}$ equals:
Fick's second law $\partial C/\partial t = D\,\partial^2 C/\partial x^2$ governs transient diffusion. The Fourier number $\mathrm{Fo}_m = Dt/L^2$ measures how far the diffusion front has advanced relative to system size. The diffusion length $\delta_D = 2\sqrt{Dt}$ is the characteristic penetration depth. Since $\delta_D \propto \sqrt{t}$, doubling the penetration depth requires four times the time. Since $\delta_D \propto \sqrt{D}$, doubling $D$ increases depth by only $\sqrt{2}$.
In the semi-infinite solution $C(x,t) = C_0\,\mathrm{erfc}(x/(2\sqrt{Dt}))$, the concentration at the surface $x = 0$ equals:
If the diffusion coefficient $D$ doubles (all else equal), the diffusion length $\delta_D = 2\sqrt{Dt}$ becomes:
A thick polymer film is exposed to solvent vapor at $t = 0$. After a very long time, the concentration throughout the film approaches:
In the semi-infinite solution $C = C_0\,\mathrm{erfc}(x/2\sqrt{Dt})$: at the surface $x = 0$, erfc(0) = 1, so $C = C_0$ always. Far from the surface, erfc → 0, so $C \to 0$. As $t \to \infty$ in a finite slab with one exposed side, the concentration approaches $C_0$ uniformly. The diffusion length $\delta_D = 2\sqrt{Dt}$ grows as $\sqrt{t}$ and scales as $\sqrt{D}$, so doubling $D$ increases the penetration depth by only $\sqrt{2}$.
According to $D = k_BT/(6\pi\mu r)$, if you triple the molecule's hydrodynamic radius (holding $T$ and $\mu$ fixed), the diffusion coefficient becomes:
At higher temperature, the diffusion coefficient of a small molecule in a liquid solvent generally increases. The primary reasons are:
In the same solvent at the same temperature: solvent molecules have $r_s = 0.3$ nm, dissolved polymer chains have $r_p = 30$ nm. The ratio $D_s/D_p$ is approximately:
$D = k_BT/(6\pi\mu r)$. Key scalings: $D \propto 1/r$ (larger molecules diffuse more slowly), $D \propto T$ (direct), $D \propto 1/\mu$ (lower viscosity increases D). For polymer solutions, the ratio $D_\text{solvent}/D_\text{polymer} = r_\text{polymer}/r_\text{solvent}$. Polymer chains (r ~ 10–100 nm) diffuse 10–1000× more slowly than solvent molecules (r ~ 0.2–0.5 nm) in the same medium.
Problem Set : Chapter 11
Part A : Short Answer
Explain in one paragraph why the diffusion of a small molecule through a glassy polymer is so much slower than through a rubbery polymer of the same composition, using the concept of free volume and the glass transition temperature $T_g$.
A coworker says: "To halve the oxygen permeation rate through our packaging film, we just need to double the thickness." Evaluate this claim using Fick's first law. Is the coworker correct? Are there any practical trade-offs?
A drug-loaded polymer microsphere releases its payload by diffusion. The release rate decreases over time even though the interior drug concentration is still significant. Explain this using Fick's second law and the concept of diffusion length. How would you modify the particle design to maintain a more constant release rate?
A multilayer packaging film consists of an LDPE outer layer (2 mm, $D = 5\times10^{-12}$ m²/s for O₂), an EVOH barrier layer (0.05 mm, $D = 1\times10^{-15}$ m²/s), and an LDPE inner layer (2 mm). Calculate the diffusive resistance of each layer and identify which controls the barrier performance. Does the EVOH layer need to be thicker to further improve the barrier?
A nitrile glove (thickness 0.3 mm) has a toluene breakthrough time of 30 minutes. Using the diffusion length concept, estimate the apparent diffusion coefficient. Compare to a literature value of $D \approx 3\times10^{-12}$ m²/s. What simplification does the breakthrough-time approach make?
Part B : Calculations
Oxygen diffuses through a 0.50 mm thick HDPE film. The concentration on the outside is $C_1 = 2.0$ mol/m³ and the inside is maintained at $C_2 = 0$. The diffusion coefficient is $D = 2.0\times10^{-10}$ m²/s. Find the steady-state oxygen flux in μmol/(m²·s).
Use $J = D\Delta C/L$. Convert $L = 0.50\text{ mm} = 5.0\times10^{-4}$ m. Then $J = (2.0\times10^{-10} \times 2.0)\,/\,(5.0\times10^{-4})$.
$J$ comes out in mol/(m²·s). Multiply by $10^6$ to get μmol/(m²·s).
A polymer slab of thickness $L = 0.10$ mm is exposed to a solvent at $t = 0$. The diffusion coefficient is $D = 1.0\times10^{-10}$ m²/s. How many seconds does it take to reach a Fourier number of $\mathrm{Fo}_m = 0.10$?
Rearrange $\mathrm{Fo}_m = Dt/L^2$ to get $t = \mathrm{Fo}_m \cdot L^2/D$. Convert $L = 0.10\text{ mm} = 1.0\times10^{-4}$ m.
$t = 0.10 \times (1.0\times10^{-4})^2 \,/\, (1.0\times10^{-10}) = 0.10 \times 10^{-8} \,/\, 10^{-10} = 0.10 \times 100$.
A chemical worker wears a nitrile glove (thickness $L = 2.0$ mm). An industrial solvent creates a concentration difference $\Delta C = 9.0$ mol/m³ across the glove. The diffusion coefficient is $D = 5.0\times10^{-11}$ m²/s. Find the steady-state flux in nmol/(m²·s).
$J = D\Delta C/L$ with $L = 2.0\times10^{-3}$ m, $D = 5.0\times10^{-11}$ m²/s, $\Delta C = 9.0$ mol/m³. Result in mol/(m²·s); multiply by $10^9$ for nmol.
A thick polymer film is exposed to moisture vapor. The diffusion coefficient of water vapor in the film is $D = 1.0\times10^{-10}$ m²/s. How many seconds does it take for the diffusion length $\delta_D = 2\sqrt{Dt}$ to reach 1.2 mm?
Set $\delta_D = x$: $2\sqrt{Dt} = x$. Solve for $t = x^2/(4D)$. Use $x = 1.2\times10^{-3}$ m.
$t = (1.2\times10^{-3})^2 \,/\, (4 \times 1.0\times10^{-10}) = 1.44\times10^{-6} \,/\, 4\times10^{-10}$.
A solvent molecule has hydrodynamic radius $r_s = 0.30$ nm. A dissolved polymer chain in the same solvent at the same temperature has hydrodynamic radius $r_p = 3.0$ nm. Using the Stokes-Einstein equation, find the ratio $D_s/D_p$.
$D = k_BT/(6\pi\mu r)$. With same $T$, $\mu$, and constants: $D \propto 1/r$. Therefore $D_s/D_p = r_p/r_s$.
Chapter 11 Summary
Key Ideas
- Fick's first law: $J = -D\,dC/dx$. Flux is proportional to concentration gradient, directed from high to low $C$. Structurally identical to Fourier's law.
- Diffusive resistance: $R = L/D$. Layers in series add directly. The layer with the lowest $D$ controls the total resistance, even if thin.
- Fick's second law: $\partial C/\partial t = D\,\partial^2 C/\partial x^2$. Governs transient diffusion; identical form to the heat equation with $\alpha \to D$.
- Diffusion length: $\delta_D = 2\sqrt{Dt}$. Penetration advances as $\sqrt{t}$: doubling depth requires 4× the time.
- Stokes-Einstein: $D = k_BT/(6\pi\mu r)$. Larger molecules diffuse more slowly ($D \propto 1/r$). Temperature and viscosity both matter.
Key Equations
- Fick's first: $J = -D\,dC/dx$
- Steady-state: $J = D(C_1-C_2)/L$
- Resistance: $R = L/D$
- Fick's second: $\partial C/\partial t = D\,\partial^2 C/\partial x^2$
- Semi-infinite: $C = C_0\,\mathrm{erfc}(x/2\sqrt{Dt})$
- Diffusion length: $\delta_D = 2\sqrt{Dt}$
- Fourier number: $\mathrm{Fo}_m = Dt/L^2$
- Stokes-Einstein: $D = k_BT/(6\pi\mu r)$
Next : Chapter 12
Chapter 12 examines diffusion specifically in polymer matrices. Free-volume theory explains why $D$ for small molecules drops by orders of magnitude in glassy polymers. The Arrhenius temperature dependence and Case II (non-Fickian) diffusion are introduced, with applications to barrier design and drug release from polymer matrices.