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 diffusing through the tube 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 length of the tube. The mass flow rate is $ \dot{m} = J \cdot A $, where $ A $ is the cross-sectional area of the tube. Substituting the given values and converting the time to seconds, the mass of water vapor diffused in one hour is approximately 4.6 × 10^-5 kg.

A Lower Tank Contains Water While an Upper Tank Contains