Answer
$0.0156\ \mathrm{kmol} / \mathrm{m}^3$
$0.00780\ \mathrm{kmol} / \mathrm{m}^3$
Work Step by Step
Noting that $500\ \mathrm{kPa}=5\ \mathrm{bar}$, the molar densities of oxygen and nitrogen in the rubber wall are determined to be $$
\begin{aligned}
C_{\mathrm{O}_2, \text { solidside }}(0) & =\mathrm{S} \times P_{\mathrm{O}_2, \text { gas side }} \\
& =\left(0.00312 \mathrm{kmol} / \mathrm{m}^3 \text {.bar }\right)(5 \mathrm{bar}) \\
& =\mathbf{0 . 0 1 5 6}\ \mathbf{k m o l} / \mathrm{m}^3 \\
C_{\mathrm{N}_2 \text {, solidside }}(0) & =\mathrm{S} \times P_{\mathrm{N}_2, \text { gasside }} \\
& =\left(0.00156 \mathrm{kmol} / \mathrm{m}^3 \text {. bar }\right)(5 \mathrm{bar}) \\
& =\mathbf{0 . 0 0 7 8 0}\ \mathbf{k m o l} / \mathrm{m}^3
\end{aligned}
$$ That is, there will be $0.0156\ \mathrm{kmol}$ of $\mathrm{O}_2$ and $0.00780\ \mathrm{kmol}$ of $\mathrm{N}_2$ gas in each $ \mathrm{m}^3$ volume of the rubber wall.
