Question

Consider the steady, uni-directional diffusion of a binary mixture of \( A \) and \( B \) across a vertical slab of dimensions \( 0.2 \, \text{m} \times 0.1 \, \text{m} \times 0.02 \, \text{m} \) as shown in the figure. The total molar concentration of \( A \) and \( B \) is constant at \( 100 \, \text{mol} \, \text{m}^{-3} \). The mole fraction of \( A \) on the left and right faces of the slab are maintained at \( 0.8 \) and \( 0.2 \), respectively. If the binary diffusion coefficient \( D_{AB} = 1 \times 10^{-5} \, \text{m}^2 \, \text{s}^{-1} \), the molar flow rate of \( A \) in \( \text{mol} \, \text{s}^{-1} \), along the horizontal \( x \)-direction is: \includegraphics[width=0.5\linewidth]{q17 ce ..PNG}

• A. \( 6 \times 10^{-4} \)
• B. \( 6 \times 10^{-6} \)
• C. \( 3 \times 10^{-6} \)
• D. \( 3 \times 10^{-4} \)

Hint:

For diffusion problems, use Fick's law, calculate the concentration gradient carefully, and account for the cross-sectional area to determine the molar flow rate.

Answer 1

The Correct Option is A

Step 1: Use Fick's First Law of Diffusion. The molar flux \( J_A \) is given by Fick's first law: \[ J_A = -D_{AB} \frac{\partial C_A}{\partial x}. \] Step 2: Calculate the concentration gradient. The mole fraction gradient of \( A \) is: \[ \frac{\partial y_A}{\partial x} = \frac{y_{A,\text{right}} - y_{A,\text{left}}}{\Delta x} = \frac{0.2 - 0.8}{0.02} = -30 \, \text{m}^{-1}. \] The total molar concentration is \( C = 100 \, \text{mol} \, \text{m}^{-3} \). The concentration gradient is: \[ \frac{\partial C_A}{\partial x} = C \cdot \frac{\partial y_A}{\partial x} = 100 \cdot (-30) = -3000 \, \text{mol} \, \text{m}^{-4}. \] Step 3: Calculate the molar flux. Substitute \( D_{AB} = 1 \times 10^{-5} \, \text{m}^2 \, \text{s}^{-1} \): \[ J_A = -D_{AB} \frac{\partial C_A}{\partial x} = -\left( 1 \times 10^{-5} \right) \cdot (-3000) = 3 \times 10^{-2} \, \text{mol} \, \text{m}^{-2} \, \text{s}^{-1}. \] Step 4: Calculate the molar flow rate. The molar flow rate is: \[ \dot{N}_A = J_A \cdot A, \] where \( A \) is the cross-sectional area: \[ A = 0.2 \, \text{m} \cdot 0.1 \, \text{m} = 0.02 \, \text{m}^2. \] Substitute \( J_A = 3 \times 10^{-2} \, \text{mol} \, \text{m}^{-2} \, \text{s}^{-1} \): \[ \dot{N}_A = 3 \times 10^{-2} \cdot 0.02 = 6 \times 10^{-4} \, \text{mol} \, \text{s}^{-1}. \] Step 5: Conclusion. The molar flow rate of \( A \) is \( 6 \times 10^{-4} \, \text{mol} \, \text{s}^{-1} \).