Diffusion in solids

Steady diffusion through slabs, cylinders and spheres, gas solubility and permeability, porous-solid and Knudsen diffusion, error-function solutions and the Arrhenius temperature dependence.

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Why it matters

Drying of solids, leaching, adsorption, catalyst pellets, gas permeation through polymer membranes, hydrogen leakage through vessel walls and case hardening of steel are all controlled by diffusion inside a solid. Diffusion in solids is slow, so it is frequently the rate-limiting step, and design depends on getting the geometry, the solubility of the solute in the solid and the temperature dependence of D right.

Key ideas

Two broad classes.

  1. Structure-insensitive (Fickian) diffusion. The solute dissolves in the solid and diffuses through it as through a homogeneous medium – gases through polymers or metals, solutes through gels, carbon in iron. Fick's laws apply directly with a solid-phase diffusivity D_AB, and bulk flow is negligible because concentrations are small.
  2. Diffusion in porous solids (structure-sensitive). The solute moves through the fluid filling the pores. The rate depends on porosity, pore size and tortuosity.

Solubility and permeability. For a gas dissolving in a solid, the concentration at the surface is proportional to the partial pressure outside (a Henry's-law-type relation). Solubility S is often quoted as m³ of gas (at STP, 0 °C and 1 atm) per m³ of solid per atm. The permeability is the product D·S; membrane makers quote it in many units, so convert carefully.

Geometry matters in steady state. For a flat slab the flux is uniform. For a cylinder or sphere the area changes with radius, so the total rate (mol/s) is constant but the flux is not; the equations contain ln(r₂/r₁) or (1/r₁ − 1/r₂).

Unsteady diffusion. Fick's second law with suitable boundary conditions gives:

  • a semi-infinite solid with a suddenly fixed surface concentration (carburising, early-time drying) – the error-function solution; penetration depth grows as √(D·t);
  • finite slabs, cylinders and spheres – series solutions, available as charts (Gurney–Lurie type) of unaccomplished change versus D·t/a².

Porous solids.

  • If the pore diameter is large compared with the mean free path of the gas, molecular (bulk) diffusion operates in the pores; the effective diffusivity is D_eff = ε·D_AB/τ, with porosity ε (0.3–0.7 typical) and tortuosity τ (2–6 typical, take from data).
  • If the pore diameter is small compared with the mean free path (fine pores, low pressure), molecules collide mainly with the walls: Knudsen diffusion. D_K depends on pore radius, T and molar mass but not on pressure or on the other gas.
  • In the transition region the resistances add: 1/D = 1/D_AB + 1/D_K (for equimolar counter-diffusion).

Temperature dependence. Diffusion in crystalline solids is activated (vacancy or interstitial jumps) and follows an Arrhenius law. Interstitial diffusion of small atoms (H, C, N in metals) is much faster and has lower activation energy than substitutional (vacancy) diffusion.

Formulas

Steady diffusion through a flat slab of thickness z: N_A = D_AB (C_A1 − C_A2) / z

Through a hollow cylinder (length L, radii r₁ < r₂), total rate: w_A = 2π·L·D_AB (C_A1 − C_A2) / ln(r₂/r₁)

Through a spherical shell, total rate: w_A = 4π·D_AB (C_A1 − C_A2) / (1/r₁ − 1/r₂)

  • N_A – flux, mol/(m²·s); w_A – rate, mol/s; C_A – concentration in the solid, mol/m³; D_AB – m²/s; z, r, L – m.

Gas solubility to concentration: C_A = S·p_A / 22.414 (C_A in kmol/m³, S in m³(STP)/(m³·atm), p_A in atm; 22.414 m³/kmol at STP)

Effective and Knudsen diffusivity: D_eff = ε·D_AB / τ D_K = (2/3)·r·√(8RT/(πM)) = 97·r·√(T/M) (D_K in m²/s, r – pore radius in m, T in K, M in kg/kmol)

Semi-infinite solid, constant surface concentration C_s, initial C₀: (C_A − C₀)/(C_s − C₀) = 1 − erf[x / (2√(D·t))]

Arrhenius dependence: D = D₀·exp(−Q/(R·T)), so D₂/D₁ = exp[−(Q/R)(1/T₂ − 1/T₁)]

  • Q – activation energy, J/mol; R = 8.314 J/(mol·K).

Worked examples

Example 1 – hydrogen through a polymer membrane (standard). Given: H₂ at 1 atm on one side of a neoprene membrane 0.5 mm thick, effectively zero on the other; 17 °C; D = 1.03 × 10⁻¹⁰ m²/s; solubility S = 0.051 m³(STP)/(m³·atm). Find the flux.

  1. Surface concentration: C_A1 = S·p_A/22.414 = 0.051 × 1/22.414 = 2.275 × 10⁻³ kmol/m³; C_A2 = 0.
  2. N_A = D (C_A1 − C_A2)/z = 1.03 × 10⁻¹⁰ × 2.275 × 10⁻³ / 0.5 × 10⁻³.
  3. N_A = 4.69 × 10⁻¹⁰ kmol/(m²·s).

N_A = 4.69 × 10⁻¹⁰ kmol/(m²·s) = 4.69 × 10⁻⁷ mol/(m²·s)

Example 2 – carburising time (GATE level). Given: steel with initial carbon 0.20 wt% is exposed to an atmosphere holding the surface at 1.20 wt%. D of C in austenite = 1.6 × 10⁻¹¹ m²/s. How long until the carbon content 0.5 mm below the surface reaches 0.60 wt%?

  1. (C − C₀)/(C_s − C₀) = 1 − erf[x/(2√(D t))] → (0.60 − 0.20)/(1.20 − 0.20) = 0.40.
  2. erf[x/(2√(D t))] = 0.60 → argument η = 0.595 (from erf tables).
  3. √(D t) = x/(2η) = 0.5 × 10⁻³/(2 × 0.595) = 4.20 × 10⁻⁴ m.
  4. t = (4.20 × 10⁻⁴)²/1.6 × 10⁻¹¹ = 1.10 × 10⁴ s.

t ≈ 1.10 × 10⁴ s ≈ 3.1 h. Doubling the depth would need four times the time.

Example 3 – temperature effect (quick). D = 2 × 10⁻¹² m²/s at 500 K and Q = 100 kJ/mol. At 600 K: D₂/D₁ = exp[−(100 000/8.314)(1/600 − 1/500)] = exp(4.009) = 55.1, so D₂ = 1.10 × 10⁻¹⁰ m²/s.

Common mistakes

  • Using the slab equation for a thick-walled tube or sphere – the area changes with radius.
  • Forgetting to convert solubility from volume at STP to moles (divide by 22.414 m³/kmol) or using atm and Pa inconsistently.
  • Using Celsius in the Arrhenius equation.
  • Applying a pressure correction to a Knudsen diffusivity (D_K is independent of pressure) or forgetting it for molecular diffusion in large pores (D_AB ∝ 1/P).
  • Reading erf tables backwards: the solution gives 1 − erf, not erf.
  • Using the true pore diffusivity where the problem expects D_eff (porosity and tortuosity omitted).

For GATE CH

Typical questions: steady flux or leakage rate through a slab, tube wall or spherical container; effective and Knudsen diffusivity of a gas in a catalyst pore; penetration depth or time scaling with √(D t); Arrhenius ratio of diffusivities. Practise ratio problems (how does time change if depth doubles, or temperature rises) – they need no tables.

Quick check

  1. In a hollow cylinder at steady state, which is constant along the radius – the flux or the rate?
  2. Does Knudsen diffusivity depend on total pressure?
  3. If case depth is to be tripled at the same temperature, by what factor does carburising time change?
  4. Write the expression for effective diffusivity in a porous solid.

Answers: 1. the rate (mol/s); 2. no; 3. nine times; 4. D_eff = ε·D_AB/τ.

Interview questions

All Mass Transfer interview questions

Try answering each one aloud before you open it.

  1. 1.What is diffusion in solids?Concept

    Diffusion in solids refers to the process by which atoms, ions, or molecules move through a solid material. This movement is driven by concentration gradients and occurs at a much slower rate compared to diffusion in liquids or gases due to the tightly packed structure of solids.

  2. 2.What factors affect the diffusion rate in solids?Concept

    The diffusion rate in solids is affected by several factors, including temperature, the nature of the diffusing species and the host material, the concentration gradient, and the presence of defects or impurities in the solid. Higher temperatures generally increase diffusion rates, while defects can either enhance or hinder diffusion depending on their nature.

  3. 3.Why is diffusion in solids generally slower than in liquids or gases?Application

    Diffusion in solids is slower than in liquids or gases because the atoms or molecules in a solid are more tightly packed and have less freedom to move. The energy barriers for movement are higher, and the pathways for diffusion are more restricted compared to the more fluid structures of liquids and gases.

  4. 4.Explain the role of temperature in the diffusion process in solids.Application

    Diffusion in crystalline solids proceeds by thermally activated jumps of atoms into vacancies or between interstitial sites, so it follows an Arrhenius law, D = D₀ exp(−Q/RT). Raising temperature increases the fraction of atoms with enough energy to jump and also the equilibrium vacancy concentration, so D rises very steeply – often by orders of magnitude over a few hundred kelvin. That is why case hardening and sintering are done at high temperature.

  5. 5.What happens to the diffusion rate if a solid has a high concentration of defects?Application

    If a solid has a high concentration of defects, the diffusion rate can be significantly affected. Defects such as vacancies or dislocations can provide pathways that facilitate the movement of atoms or molecules, potentially increasing the diffusion rate. However, certain types of defects might also trap diffusing species, which could reduce the diffusion rate.

  6. 6.Why is diffusion important in the process of alloying metals?Application

    Diffusion is important in alloying metals because it allows the atoms of different metals to mix and form a homogeneous alloy. The diffusion process enables the distribution of alloying elements throughout the metal matrix, which is essential for achieving the desired mechanical and chemical properties in the final alloy.

  7. 7.Calculate the diffusion flux if the diffusion coefficient is 1.5 x 10^-10 m²/s and the concentration gradient is 2 x 10^6 mol/m³·m.Numerical

    Using Fick's first law of diffusion, J = -D (dC/dx), where D = 1.5 x 10^-10 m²/s and dC/dx = 2 x 10^6 mol/m³·m, the diffusion flux J = -(1.5 x 10^-10 m²/s) * (2 x 10^6 mol/m³·m) = -3 x 10^-4 mol/m²·s.

  8. 8.A solid has a diffusion coefficient of 2 × 10⁻¹² m²/s at 500 K. What is the diffusion coefficient at 600 K if the activation energy for diffusion is 100 kJ/mol?Numerical

    From the Arrhenius law, D₂/D₁ = exp[(Q/R)(1/T₁ − 1/T₂)] = exp[(100 000/8.314)(1/500 − 1/600)] = exp(4.009) = 55.1. So D₂ = 55.1 × 2 × 10⁻¹² ≈ 1.1 × 10⁻¹⁰ m²/s. A 100 K rise increases D about fifty-fold, showing how strongly solid-state diffusion depends on temperature.

  9. 9.What is the significance of the diffusion coefficient in the context of solid-state diffusion?Concept

    The diffusion coefficient is a measure of how easily atoms or molecules can move through a solid. It is a key parameter in predicting the rate of diffusion and is influenced by factors such as temperature, the nature of the diffusing species, and the host material. A higher diffusion coefficient indicates a faster diffusion process, which is crucial for processes like sintering, alloying, and phase transformations.

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