Water vapour diffuses through a stagnant air film of thickness $\delta = 4.0\text{ mm} = 0.004\text{ m}$ above the surface of an open evaporation tank. The molecular diffusivity of water vapour in air is $D_{AB} = 2.4 \times 10^{-5}\text{ m}^2\text{/s}$. The molar concentration of water vapour at the water surface is $C_{A1} = 1.50\text{ mol/m}^3$ and at the outer edge of the film is $C_{A2} = 0.30\text{ mol/m}^3$. Under steady state dilute diffusion, the molar mass flux of water vapour in $10^{-3}\text{ mol/(m}^2\cdot\text{s)}$ is ________ (round off to 1 decimal place).
Answer Key:
7.2
Step-by-Step Solution & Explanation
From Fick's first law of steady-state molecular diffusion across a stagnant film:
$J_A = D_{AB} \frac{C_{A1} - C_{A2}}{\delta}$
$J_A = 2.4 \times 10^{-5}\text{ m}^2\text{/s} \times \frac{1.50 - 0.30\text{ mol/m}^3}{0.004\text{ m}} = 2.4 \times 10^{-5} \times 300 = 7.2 \times 10^{-3}\text{ mol/(m}^2\cdot\text{s)}$
Therefore, in $10^{-3}\text{ mol/(m}^2\cdot\text{s)}$, the value is $7.2$.