Evaporation through a stagnant film (units check):\nFor pure aniline evaporating through a 1 mm stagnant air film at 300 K and 100 kPa, the computed rate of evaporation is 2.97 × 10^-4 in consistent SI units. What are the correct units of this flux quantity?

Difficulty: Medium

Correct Answer: mole/m2·s

Explanation:


Introduction / Context:
In mass transfer problems involving diffusion through a stagnant gas film, the calculated “rate” is often a flux, i.e., moles transferred per unit area per unit time. Distinguishing between total rate and areal flux prevents dimensional errors when scaling from laboratory calculations to equipment design, such as evaporators or drying operations.


Given Data / Assumptions:

  • Stagnant air film thickness: 1 mm.
  • Temperature: 300 K; total pressure: 100 kPa.
  • Vapor pressure of aniline: 0.1 kPa; total molar concentration: 40.1 mol/m^3.
  • Diffusivity of aniline in air: 0.74 × 10^-5 m^2/s.
  • Computed mass transfer coefficient: 7.4 × 10^-3 m/s (from D_AB/δ).
  • Reported “rate” magnitude: 2.97 × 10^-4 in SI terms.


Concept / Approach:
The standard film theory gives a molar flux N_A in mol/m^2·s using N_A = k_c * (C_A,s − C_A,∞) or its equivalent for dilute species, where k_c has units of m/s and concentration difference has units of mol/m^3. Multiplying m/s by mol/m^3 yields mol/m^2·s. Thus, a numerical value near 10^-4 with the provided parameters corresponds to a flux, not a total rate integrated over area.


Step-by-Step Solution:

Use k_c = D_AB / δ with δ = 1 × 10^-3 m → k_c ≈ 0.74 × 10^-2 m/s.Compute driving force: C_A,s − C_A,∞ ≈ (p_A,s/RT − ~0) ≈ 0.1 kPa / (R T) → consistent with given overall C of 40.1 mol/m^3 and low vapor pressure.Form units: (m/s) * (mol/m^3) = mol/m^2·s.Hence the reported 2.97 × 10^-4 must carry units of mol/m^2·s.


Verification / Alternative check:
Dimensional analysis alone confirms the unit: any mass transfer coefficient based on concentration difference delivers areal flux units. Converting to kmol/m^2·s would move the decimal by 3 orders of magnitude; cm^2 would change the numerical value by 10^4.


Why Other Options Are Wrong:

  • mol/s: a total rate, not normalized by area.
  • mol/cm^2·s: would require value scaled by 10^4.
  • kmol/m^2·s: different magnitude by 10^3.
  • kg/m^2·s: mass flux; would need molar mass conversion.


Common Pitfalls:
Confusing total evaporation rate with molar flux; always check whether an area normalization is implied by the governing equations.


Final Answer:
mole/m2·s.

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