Given:
\[
\text{SO}_2 \text{ is diffusing through a stagnant air film.}
\]
\[
\text{Thickness} = z_2 - z_1 = 2 \times 10^{-3} \, \text{m},
\]
\[
\text{Total pressure} = P_T = 10^5 \, \text{Pa},
\]
\[
\text{Temperature} = T = 30^\circ \text{C} = 303 \, \text{K}.
\]
At Location (1):
\[
z_1, \quad P_{A1} = 0.15 \, \text{bar}, \quad P_{B1} = P_T - P_{A1} = 0.85 \, \text{bar}.
\]
At Location (2):
\[
z_2, \quad P_{A2} = 0.05 \, \text{bar}, \quad P_{B2} = P_T - P_{A2} = 0.95 \, \text{bar}.
\]
Steady-state flux of \(\text{SO}_2\):
The steady-state flux of \(\text{SO}_2\) can be expressed as:
\[
N_A|_{\text{SO}_2} = \frac{D_{AB} P_T}{RT} \cdot \frac{(P_{A1} - P_{A2})}{P_{B\text{lm}}} \cdot \frac{1}{z_2 - z_1}.
\]
Log mean partial pressure of \( \text{B} \):
The logarithmic mean partial pressure of \( \text{B} \) is:
\[
P_{B\text{lm}} = \frac{P_{B1} - P_{B2}}{\ln\left(\frac{P_{B1}}{P_{B2}}\right)}.
\]
Substituting the given values:
\[
P_{B\text{lm}} = \frac{0.10}{\ln\left(\frac{0.95}{0.85}\right)}.
\]
Substitute into the flux equation:
\[
N_A|_{\text{SO}_2} = \frac{10^{-3} \cdot 10^5}{8.314 \cdot 303 \cdot 2 \times 10^{-3}} \cdot \frac{0.10}{\ln\left(\frac{0.95}{0.85}\right)}.
\]
Simplify step-by-step:
\[
N_A|_{\text{SO}_2} = 0.022 \, \text{mol/m}^2\text{sec}.
\]
Final Answer:
The steady-state flux of \( \text{SO}_2 \) is:
\[
N_A|_{\text{SO}_2} = 0.022 \, \text{mol/m}^2\text{sec}.
\]