Diffusion-film estimate: Pure aniline evaporates through a stagnant air film of thickness 1 mm at 300 K and 100 kPa. Given p_sat(aniline) = 0.1 kPa, total molar concentration = 40.1 mol/m^3, and D_AB = 0.74 × 10^-5 m^2/s, the mass transfer coefficient is 7.4 × 10^-3 in which units?

Introduction / Context:
Mass transfer coefficients can be defined with different driving forces: concentration, mole fraction, or partial pressure. Recognising the correct units for a given numerical value prevents dimensional mistakes in design equations for evaporation, drying, and gas–liquid operations.

Given Data / Assumptions:

• Stagnant air film of thickness δ = 1 mm = 1 × 10^-3 m.
• Binary diffusivity D_AB = 0.74 × 10^-5 m^2/s.
• Reported coefficient magnitude: 7.4 × 10^-3 (value only).
• Film theory with concentration driving force k_c ≈ D_AB / δ applies.

Concept / Approach:
When the driving force is concentration difference (e.g., mol/m^3), the mass transfer coefficient k_c has units of velocity (m/s), because flux N = k_c * ΔC has units mol/m^2·s. Using the film approximation directly yields k_c = D_AB / δ with velocity units.

Step-by-Step Solution:

Verification / Alternative check:
If a pressure driving force were used, k_G would have units mol/m^2·s·Pa. The given numeric matches D/δ, confirming a concentration-based coefficient with velocity units.

Why Other Options Are Wrong:

• cm/s: Same dimension but wrong magnitude if units were converted without changing the number.
• mol/m^2·s·Pa or kmol/m^2·s·Pa: These correspond to pressure-based coefficients, which is not consistent with D/δ.

Common Pitfalls:
Mixing k_c, k_y, and k_G; always match the driving-force form (ΔC, Δy, or Δp) with the corresponding coefficient and its units.

Final Answer:
m/s