Step 1: Case 1 - B is Non-Diffusing ($N_B = 0$)
When B is non-diffusing, we have unimolecular diffusion of A. The flux equation is:
$$N_{A1} = -\frac{cD_{AB}}{1-x_A} \frac{dx_A}{dz}$$
For steady-state diffusion between two points:
$$N_{A1} = \frac{cD_{AB}}{L} \ln\left(\frac{1-x_{A2}}{1-x_{A1}}\right)$$
where $L$ is the distance between the two locations.
Substituting values:
$$N_{A1} = \frac{cD_{AB}}{L} \ln\left(\frac{1-0.1}{1-0.8}\right) = \frac{cD_{AB}}{L} \ln\left(\frac{0.9}{0.2}\right)$$
$$N_{A1} = \frac{cD_{AB}}{L} \ln(4.5) = \frac{cD_{AB}}{L} \times 1.504$$
Step 2: Case 2 - Equimolar Counter-Diffusion ($N_A = -N_B$)
For equimolar counter-diffusion, Fick's first law applies directly:
$$N_{A2} = -cD_{AB} \frac{dx_A}{dz}$$
For steady-state between two points:
$$N_{A2} = \frac{cD_{AB}}{L}(x_{A1} - x_{A2})$$
Substituting values:
$$N_{A2} = \frac{cD_{AB}}{L}(0.8 - 0.1) = \frac{cD_{AB}}{L} \times 0.7$$
Step 3: Calculate the Ratio
$$\frac{N_{A1} - N_{A2}}{N_{A1}} = \frac{\frac{cD_{AB}}{L} \times 1.504 - \frac{cD_{AB}}{L} \times 0.7}{\frac{cD_{AB}}{L} \times 1.504}$$
$$= \frac{1.504 - 0.7}{1.504} = \frac{0.804}{1.504}$$
$$= 0.5346$$
Step 4: Calculate the Final Answer
$$\frac{N_{A1} - N_{A2}}{N_{A1}} \times 100 = 0.5346 \times 100 = 53.46$$
Rounded to one decimal place: 53.5
Answer
The quantity $\frac{(N_{A1} - N_{A2})}{N_{A1}} \times 100$ is 53.5 (rounded to one decimal place).
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