Answer:
the rate of diffusion of CO₂ gas in air through the bed of sand is approximately 2.304 × 10^-6 mol/(m²·s).
To calculate the rate of diffusion of CO₂ gas in air through a bed of sand, we can use Fick's law of diffusion:
J = -DAB (dC/dx)
where J is the molar flux of CO₂, DAB is the diffusion coefficient of CO₂ in air, and (dC/dx) is the concentration gradient of CO₂ in the direction of diffusion.
To calculate the concentration gradient, we can use the following equation:
(dC/dx) = (ΔC/Δx)
where ΔC is the difference in partial pressure of CO₂ between the top and bottom of the bed, and Δx is the bed depth.
We are given that the bed depth is 1.25 m and the void fraction is 0.3, which means that the volume of the bed is:
V = (1 - e) A L
where A is the cross-sectional area of the bed and L is the bed depth. Assuming a circular cross-section, we can calculate the area as:
A = π (d/2)²
where d is the diameter of the bed. We are not given the diameter, so we cannot calculate the area.
However, we are given the partial pressure of CO₂ at the top and bottom of the bed, as well as the diffusion coefficient and temperature. We can use these values to calculate the molar flux of CO₂ using Fick's law of diffusion.
First, we need to convert the diffusion coefficient to the appropriate units:
DAB = 0.142 × 10^-9 m²/s
Next, we can calculate the concentration gradient:
ΔC = 2.026 × 10^4 Pa - 0 Pa = 2.026 × 10^4 Pa
Δx = 1.25 m
(dC/dx) = (ΔC/Δx) = (2.026 × 10^4 Pa/1.25 m) = 1.6208 × 10^4 Pa/m
Finally, we can calculate the molar flux of CO₂:
J = -DAB (dC/dx) = -(0.142 × 10^-9 m²/s) (1.6208 × 10^4 Pa/m) = -2.304 × 10^-6 mol/(m²·s)
The negative sign indicates that the molar flux of CO₂ is in the opposite direction of the concentration gradient, which is expected for equimolar counterdiffusion.
Therefore, the rate of diffusion of CO₂ gas in air through the bed of sand is approximately 2.304 × 10^-6 mol/(m²·s).