Diffusion in Polymers
In small-molecule fluids, diffusion is fast and nearly temperature-independent. In polymers, it is orders of magnitude slower, exquisitely sensitive to temperature, and sometimes follows rules unlike anything Fick described. This chapter explains why, and shows how that difference is exploited in pharmaceutical patches, food packaging, and membrane separations.
The Nicotine Patch
A nicotine replacement patch is a laminate of polymer films, each a fraction of a millimeter thick. The backing layer is nearly impermeable; the rate-controlling membrane releases active ingredient into the skin at a precisely engineered dose over 24 hours. The difference between these two layers is not chemistry but physics: one polymer sits above its glass transition temperature (rubbery, mobile chains, fast diffusion), the other below it (glassy, frozen free volume, slow diffusion). The ratio of diffusion coefficients across the laminate spans six orders of magnitude, all because of one transition in polymer chain mobility.
Every food package, gas-separation membrane, and protective coating works on the same principle. This chapter gives you the equations to design them.
12.1 Why Polymers Are Different
In Chapter 11, the diffusion coefficient $D$ was treated as a material constant. In low-molecular-weight liquids that is a fair approximation. In polymers, $D$ can vary by ten orders of magnitude depending on:
- Temperature relative to the glass transition $T_g$
- Degree of crystallinity (crystalline lamellae are impenetrable)
- Penetrant size and shape
- Concentration of absorbed penetrant (plasticization)
The central insight is that a small molecule diffuses by jumping into free volume: the unoccupied space between polymer chains. In a rubbery polymer above $T_g$, chain segments are mobile and redistribute free volume continuously, providing frequent jump opportunities. Below $T_g$, the chain segments are frozen, free volume is fixed and scarce, and diffusion is orders of magnitude slower.
Crystallinity Effects
Polymer crystals are close-packed and essentially impenetrable to small molecules. A penetrant must navigate around crystalline lamellae, taking a tortuous path through the amorphous regions only. The effective diffusion coefficient in a semicrystalline polymer is:
Figure 12.1: Tortuous diffusion path through a semicrystalline polymer. Crystalline lamellae (blue) act as impermeable baffles; the penetrant (red dashed path) must navigate around each baffle through the gap on the opposite edge, traveling far more than the direct straight-line distance (gray). Tortuosity $\tau$ quantifies this path lengthening, and oriented lamellae (as in drawn PET) maximize $\tau$ to minimize permeability.
12.2 Temperature Dependence: The Arrhenius Equation
For a rubbery polymer in a moderate temperature range above $T_g$, the diffusion coefficient follows an Arrhenius (activated-rate) form:
Activation Energy and the Arrhenius Plot
Taking the ratio at two temperatures eliminates $D_0$ and yields a practical working form:
Activation energies for diffusion in rubbery polymers are typically 20 to 80 kJ/mol, increasing with penetrant size. A helium atom ($E_a \approx 20$ kJ/mol) is much less sensitive to temperature than an organic molecule like toluene ($E_a \approx 60$ kJ/mol). This has practical consequences for gas-separation membranes, where operating temperature can tune selectivity.
Figure 12.2: Arrhenius plot of $\ln D$ versus $1/T$ for diffusion in a rubbery polymer (blue solid, gentle slope, low $E_a$) and a glassy polymer (red dashed). Above $T_g$ the glassy polymer behaves like a rubbery material (same slope); below $T_g$ the slope steepens sharply because frozen free volume raises the effective activation energy. The slope of each segment equals $-E_a/R$.
Activation energy from two-temperature data
The diffusion coefficient of CO₂ in a rubbery silicone membrane is $D_1 = 1.0 \times 10^{-12}$ m²/s at $T_1 = 300$ K and $D_2 = 1.0 \times 10^{-11}$ m²/s at $T_2 = 350$ K. Find the activation energy $E_a$.
Solution
Rearrange Eq. 12.3 for $E_a$:
$$E_a = R\,\frac{\ln(D_2/D_1)}{(1/T_1 - 1/T_2)}$$Compute each term:
$$\ln\!\frac{D_2}{D_1} = \ln(10) = 2.303$$ $$\frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{300} - \frac{1}{350} = \frac{50}{300\times350} = 4.76\times10^{-4}\;\mathrm{K^{-1}}$$ $$E_a = (8.314)\,\frac{2.303}{4.76\times10^{-4}} = \frac{19.14}{4.76\times10^{-4}} \approx 40{,}200\;\mathrm{J/mol}$$The activation energy is 40 kJ/mol, typical for a small gas molecule (CO₂) in a rubbery polymer. A tenfold increase in $D$ results from raising temperature by only 50 K, illustrating why warm storage dramatically accelerates gas permeation through elastomers.
12.3 Free-Volume Theory
The Arrhenius equation describes the temperature dependence once the mechanism is established. Free-volume theory explains why the mechanism changes so dramatically at $T_g$ and how diffusion depends on polymer structure.
The fractional free volume $v_f$ is the fraction of total volume not occupied by polymer chains: $v_f = (V - V_0)/V$, where $V$ is the specific volume and $V_0$ is the volume of the chain segments themselves. Doolittle (and later Cohen and Turnbull) showed that diffusion requires a free-volume fluctuation large enough to accommodate the penetrant, leading to:
The Glass Transition Jump
Below $T_g$ the polymer is in a non-equilibrium glassy state: free volume is "frozen in" at its $T_g$ value and increases only slowly on cooling. Above $T_g$ the polymer is in thermodynamic equilibrium (rubbery): free volume increases linearly with temperature at rate $\Delta\alpha = \alpha_{\rm rubber} - \alpha_{\rm glass}$ (the difference in thermal expansion coefficients). This creates a sharp change in the slope of $v_f(T)$ at $T_g$, which appears as a kink in the Arrhenius plot of $\ln D$ vs. $1/T$.
12.4 Permeability and Barrier Design
The steady-state flux of a gas through a polymer membrane depends on how easily it dissolves (solubility $S$) and how fast it moves once dissolved (diffusivity $D$). The combination is the permeability:
The steady-state flux through a film of thickness $L$ under pressure difference $\Delta P$ is:
CO₂ loss through a PET bottle wall
A PET bottle wall has thickness $L = 0.50$ mm. The diffusion coefficient of CO₂ in PET is $D = 2.0\times10^{-12}$ m²/s and the solubility is $S = 2.5\times10^{-3}$ mol/(m³·Pa). The carbonation pressure inside is $\Delta P = 1.0\times10^5$ Pa above ambient. Find the steady-state CO₂ flux.
Solution
Compute permeability using Eq. 12.5:
$$P = D\cdot S = (2.0\times10^{-12})(2.5\times10^{-3}) = 5.0\times10^{-15}\;\mathrm{mol/(m{\cdot}s{\cdot}Pa)}$$Apply Eq. 12.6 with $L = 5.0\times10^{-4}$ m:
$$J = \frac{P\,\Delta P}{L} = \frac{(5.0\times10^{-15})(1.0\times10^5)}{5.0\times10^{-4}} = \frac{5.0\times10^{-10}}{5.0\times10^{-4}} = 1.0\times10^{-6}\;\mathrm{mol/(m^2{\cdot}s)}$$The steady-state CO₂ flux is 1.0 μmol/(m²·s). For a 0.5 L bottle with $\approx$ 0.02 m² of wall area, that is roughly 1.7 mmol/day of lost carbonation. With an initial 10 mmol of dissolved CO₂, the bottle goes flat in about a week at room temperature even if sealed. Barrier coatings (EVOH, diamond-like carbon) that reduce $P$ by 10 to 100 times extend shelf life proportionally.
Selectivity
A membrane that separates two gases A and B has ideal selectivity:
12.5 Non-Fickian (Case II) Diffusion
Chapter 11 assumed Fickian diffusion: diffusion is fast compared to the chain-relaxation dynamics of the polymer. In glassy polymers, this assumption breaks down. Penetrant molecules must wait for chain segments to relax before they can jump, so the relaxation timescale $\tau_r$ competes with the diffusion timescale $\tau_D = L^2/D$.
The Deborah number for mass transfer quantifies this competition:
Case II Transport
In Case II, penetrant is absorbed in a narrow swollen zone just behind the advancing front. The front position $x_{\rm front}$ grows linearly with time:
Figure 12.3: Concentration profiles for Fickian diffusion (left, blue) and Case II transport (right, red). Fickian diffusion produces a smooth, spreading erfc profile. Case II produces a sharp step: behind the front the polymer is fully swollen at concentration $C_0$; ahead it is dry. The front advances at constant velocity $v$ rather than the $\sqrt{t}$ rate of Fickian diffusion. This behavior is common in glassy polymers exposed to organic solvents.
Case II front penetration in an epoxy adhesive
Methanol diffuses into a crosslinked epoxy via Case II transport with front velocity $v = 2.0\times10^{-7}$ m/s. A bonded joint requires the front to stay below 1.0 mm from the surface for at least 5.0 hours. Does it meet this requirement?
Solution
Convert time: $t = 5.0 \times 3600 = 18{,}000$ s. Apply Eq. 12.9:
$$x_{\rm front} = v\,t = (2.0\times10^{-7})(18{,}000) = 3.6\times10^{-3}\;\mathrm{m} = 3.6\;\mathrm{mm}$$The front reaches 3.6 mm in 5 hours, far exceeding the 1.0 mm safety limit. The joint does not meet the requirement. To keep $x_{\rm front} \le 1.0$ mm, the maximum allowable time is $t_{\rm max} = 10^{-3}/(2.0\times10^{-7}) = 5000$ s $\approx$ 83 minutes. A coating or surface treatment that reduces $v$ by at least threefold would extend this to the required 5 hours.
Interactive: Arrhenius Explorer
Adjust the activation energy $E_a$ and the reference diffusion coefficient $D_0$ to see how $D$ changes with temperature. The plot shows $\ln D$ vs. $1/T$ (the Arrhenius plot) with the slope highlighting $-E_a/R$.
Concept Check
Five question pools, three variants each. The pool advances after a correct answer so every student sees a different question.
A polymer film has 40% crystallinity. Compared to a fully amorphous film of the same polymer, the effective diffusion coefficient $D_{\rm eff}$ is:
Why does stretch-blowing a PET preform into a bottle improve its barrier properties beyond what crystallinity alone predicts?
The tortuosity factor $\tau$ in the effective diffusivity expression $D_{\rm eff} = D_{\rm am}/(\tau\,\phi_{\rm am}^{-1})$ accounts for:
Crystalline regions in a polymer are impermeable to small-molecule penetrants. They reduce $D_{\rm eff}$ through two effects: reducing the amorphous fraction available for diffusion ($\phi_{\rm am}$) and forcing a tortuous path ($\tau > 1$). Biaxial orientation of lamellae, as in stretch-blown PET, maximizes $\tau$ in the transport direction, providing barrier enhancement beyond what crystallinity fraction alone predicts.
A polymer has activation energy $E_a = 40$ kJ/mol and $D = 1.0\times10^{-12}$ m²/s at 300 K. At 350 K, $D$ is approximately:
On an Arrhenius plot of $\ln D$ vs. $1/T$, which polymer has the steeper slope?
A refrigerator stores polymer packaging at 4°C instead of 25°C. For a gas with $E_a = 50$ kJ/mol, refrigeration reduces the permeation flux by approximately what factor?
The Arrhenius equation $D = D_0\exp(-E_a/RT)$ predicts that $\ln D$ is linear in $1/T$ with slope $-E_a/R$. A larger $E_a$ gives a steeper slope, meaning $D$ is more sensitive to temperature. For a given temperature change, the ratio $D_2/D_1 = \exp[(E_a/R)(1/T_1 - 1/T_2)]$. With $E_a = 40$ kJ/mol over a 50 K increase from 300 K to 350 K, $D$ increases by a factor of $e^{2.30} = 10$.
Polymer X and Polymer Y have identical permeabilities $P$ to oxygen. Polymer X has $D_X = 10\,D_Y$. Which statement about their solubilities is correct?
A gas permeates a 1.0 mm polymer film under a pressure difference of $2.0\times10^5$ Pa. The permeability is $P = 5.0\times10^{-15}$ mol/(m·s·Pa). The steady-state flux is:
To halve the permeation flux through a film of fixed area, which change is most straightforward from an engineering standpoint?
Permeability $P = D \cdot S$ combines diffusivity (how fast) and solubility (how much). Two polymers with the same $P$ can have very different $D$ and $S$ values, with implications for selectivity between gases. The steady-state flux $J = P\Delta P/L$ is directly proportional to $P$ and $\Delta P$, and inversely proportional to thickness $L$. Doubling the film thickness halves the flux, regardless of material choice.
A researcher plots solvent uptake versus time for a glassy polymer and observes that absorbed mass grows linearly with time, not as $\sqrt{t}$. This is most consistent with:
Case II transport occurs when the Deborah number $\mathrm{De}_m = \tau_r/\tau_D$ is:
A solvent shows Case II transport in Polymer A at 25°C but Fickian diffusion at 80°C (well above $T_g$). The most likely reason for the change is:
Case II transport occurs in glassy polymers where chain segment relaxation ($\tau_r$) is much slower than diffusion ($\tau_D = L^2/D$), giving $\mathrm{De}_m \gg 1$. The experimental signature is mass uptake proportional to time $t$ (not $\sqrt{t}$) and a sharp, step-like concentration profile advancing at constant velocity. Above $T_g$, relaxation is fast and Fickian behavior is recovered. The transition between regimes passes through an intermediate anomalous (non-Fickian) regime near $\mathrm{De}_m \approx 1$.
A packaging engineer wants to minimize oxygen permeation. Rank these strategies in order of effectiveness, from most to least:
A multilayer film has an LDPE layer ($P_1 = 1.0\times10^{-14}$, thickness $L_1$) and an EVOH layer ($P_2 = 1.0\times10^{-17}$, same thickness $L_2 = L_1$). The effective permeability of the laminate is dominated by:
Free-volume theory predicts that adding a plasticizer to a glassy polymer will:
Effective barrier design exploits multiple mechanisms. Material selection (polymer class) is the most powerful lever, since $P$ varies by over ten orders of magnitude across polymer families. Within a class, maximizing crystallinity (tortuosity) provides moderate improvement. Multilayer structures allow a thin, high-barrier layer (EVOH, PVDC) to control transport while structural layers (PE, PP) provide mechanical performance. Free-volume theory explains why plasticizers and temperature increases dramatically raise $D$: both increase $v_f$, making the exponential in $D = A\exp(-B/v_f)$ less negative.
Problem Set : Chapter 12
Part A : Short Answer
A polymer film is used below its $T_g$. Explain why adding a low-molecular-weight diluent (plasticizer) could dramatically increase the oxygen permeation rate, even at the same temperature.
Two polymers have identical permeabilities to water vapor but different diffusion coefficients ($D_X = 5\,D_Y$). Which polymer has the higher water solubility? By what factor? Explain briefly.
Part B : Calculations
The diffusion coefficient of N₂ in a silicone rubber is $D_1 = 1.0\times10^{-12}$ m²/s at $T_1 = 300$ K and $D_2 = 1.0\times10^{-11}$ m²/s at $T_2 = 350$ K. Find the activation energy $E_a$ in kJ/mol.
Use the two-temperature form: $\ln(D_2/D_1) = (E_a/R)(1/T_1 - 1/T_2)$. Rearrange to isolate $E_a$.
$\ln(D_2/D_1) = \ln(10) = 2.303$. Compute $1/T_1 - 1/T_2 = 1/300 - 1/350 = 50/(300\times350)$.
$E_a = R \cdot \ln(D_2/D_1)/(1/T_1 - 1/T_2)$. Divide by 1000 to convert J/mol to kJ/mol.
A PET film (thickness $L = 0.50$ mm) has $D = 2.0\times10^{-12}$ m²/s and $S = 2.5\times10^{-3}$ mol/(m³·Pa) for CO₂. The carbonation pressure difference is $\Delta P = 1.0\times10^5$ Pa. Find the steady-state CO₂ flux in μmol/(m²·s).
Compute permeability: $P = D \cdot S = (2.0\times10^{-12})(2.5\times10^{-3})$.
Apply $J = P\,\Delta P/L$ with $L = 5.0\times10^{-4}$ m.
$J$ comes out in mol/(m²·s). Multiply by $10^6$ for μmol/(m²·s).
A 1.0 mm thick polymer membrane has $D = 2.0\times10^{-12}$ m²/s. The characteristic diffusion time $\tau_D = L^2/D$ governs how long it takes for a concentration front to traverse the membrane. Compute $\tau_D$ in seconds (give your answer in units of $10^5$ s, i.e., enter 5.0 if $\tau_D = 5.0\times10^5$ s).
Convert: $L = 1.0$ mm $= 1.0\times10^{-3}$ m. Then $L^2 = 1.0\times10^{-6}$ m².
$\tau_D = L^2/D = 10^{-6} / (2.0\times10^{-12})$.
Acetone undergoes Case II transport in a glassy epoxy with front velocity $v = 2.0\times10^{-7}$ m/s. Find the front position after 5.0 hours in mm.
Convert: $t = 5.0 \times 3600$ s. Apply $x_{\rm front} = v \cdot t$.
$x_{\rm front}$ comes out in meters. Multiply by 1000 to get mm.
A membrane has CO₂ permeability $P_{\rm CO_2} = 3.0\times10^{-14}$ mol/(m·s·Pa) and N₂ permeability $P_{\rm N_2} = 6.0\times10^{-16}$ mol/(m·s·Pa). Compute the ideal CO₂/N₂ selectivity $\alpha$.
$\alpha_{\rm CO_2/N_2} = P_{\rm CO_2}/P_{\rm N_2}$. Divide the two values.
Chapter Summary
Crystalline lamellae are impermeable. They reduce $D_{\rm eff}$ by decreasing the amorphous volume fraction and increasing the tortuosity of the diffusion path. Oriented lamellae (drawn PET) maximize tortuosity in the transport direction.
$D = D_0\exp(-E_a/RT)$. The ratio form $\ln(D_2/D_1) = (E_a/R)(1/T_1 - 1/T_2)$ is used for calculation. Larger $E_a$ means greater temperature sensitivity. Typical values: 20 to 80 kJ/mol for gases in polymers.
$D = A\exp(-B/v_f)$. Diffusion requires free-volume fluctuations large enough for the penetrant. At $T_g$, the temperature dependence of $v_f$ changes slope, causing a kink in the Arrhenius plot and a dramatic change in $D$.
$P = D \cdot S$ combines diffusivity (kinetic) and solubility (thermodynamic). Flux: $J = P\,\Delta P/L$. In a multilayer film, layers act as resistances in series; the lowest-$P$ layer dominates. The Robeson upper bound limits $P$ vs. selectivity tradeoffs.
$\alpha_{A/B} = P_A/P_B = D_A S_A/(D_B S_B)$. High selectivity is required for membrane separations. Rubbery polymers tend to be solubility-selective; glassy polymers tend to be diffusivity-selective (small molecules pass preferentially).
In glassy polymers with $\mathrm{De}_m = \tau_r/\tau_D \gg 1$, a sharp front advances at $x_{\rm front} = v\,t$ (linear in time, not $\sqrt{t}$). Signature: mass uptake linear in $t$; step-like concentration profile. Occurs in organic solvents swelling glassy polymers.
- Arrhenius: $D = D_0\exp(-E_a/RT)$
- Two-T form: $\ln(D_2/D_1) = (E_a/R)(1/T_1 - 1/T_2)$
- Free volume: $D = A\exp(-B/v_f)$
- Permeability: $P = D\cdot S$
- Flux: $J = P\,\Delta P/L$
- Selectivity: $\alpha_{A/B} = P_A/P_B$
- Diff. timescale: $\tau_D = L^2/D$
- Case II front: $x_{\rm front} = v\,t$