A lower tank contains water while an upper tank contains air with a negligible concentration of water vapor.A tube with a circular cross section and radius 0.050 m is inserted between the two tanks.The length of the tube is 0.25 m.The diffusion coefficient of water vapor through air is 2.4 × 10^-⁵ m²/s.The concentration of water vapor just above the surface of the water is 1.7 × 10^-² kg/m³.If the water vapor is removed from the upper tank so that the concentration there remains nearly zero,what mass of water vapor diffuses through the tube each hour?

Question

Quizplus · Accepted Answer

The mass of water vapor diffused per unit time can be calculated using Fick's first law of diffusion: $J = -D \frac{\Delta C}{\Delta x}$, where $J$ is the diffusion flux, $D$ is the diffusion coefficient, $\Delta C$ is the concentration difference, and $\Delta x$ is the distance over which diffusion occurs. Given $D = 2.4 \times 10^{-5} \, \text{m}^2/\text{s}$, $\Delta C = 1.7 \times 10^{-2} \, \text{kg/m}^3$, and $\Delta x = 0.25 \, \text{m}$, the area $A$ through which diffusion occurs is the cross-sectional area of the tube, $A = \pi r^2 = \pi (0.05)^2 \, \text{m}^2$. The mass diffused per hour can be calculated by integrating the flux over the area and converting seconds to hours: $m = J \cdot A \cdot t = \left(-D \frac{\Delta C}{\Delta x}\right) \cdot A \cdot 3600 \, \text{s}$, leading to the correct answer after calculations.

A Lower Tank Contains Water While an Upper Tank Contains