Answer
$0.0039\ \mathrm{kmol}$ or $ 0.1092 \mathrm{~kg/m}^3$ of $\mathrm{N}_2$ gas in each $\mathrm{m}^3$ volume of iron adjacent to the interface
Work Step by Step
Noting that $250\ \mathrm{kPa}=2.5\ \mathrm{bar}$, the molar density of nitrogen in the rubber at the interface is determined to be
$$ \begin{aligned}
C_{\mathrm{N}_2, \text { soludside }}(0) & =\mathrm{S} \times P_{\mathrm{N}_2, \text { gasside }} \\
& =\left(0.00156 \mathrm{kmol} / \mathrm{m}^3 . \mathrm{bar}\right)(2.5 \mathrm{bar}) \\
& =\mathbf{0 . 0 0 3 9} \mathbf{k m o l} / \mathbf{m}^3
\end{aligned}
$$ It corresponds to a mass density of $$
\begin{aligned}
\rho_{\mathrm{N}_2, \text { solidside }}(0) & =C_{\mathrm{N}_2, \text { solidside }}(0) M_{\mathrm{N}_2} \\
& =\left(0.0039 \mathrm{kmol} / \mathrm{m}^3\right)(28 \mathrm{~kg} / \mathrm{kmol}) \\
& =\mathbf{0 . 1 0 9 2}\ \mathrm{kg} / \mathrm{m}^3
\end{aligned}
$$ That is, there will be $0.0039\ \mathrm{kmol}$ (or $0.1092 \mathrm{~kg}$ of $\mathrm{N}_2$ gas in each $\mathrm{m}^3$ volume of iron adjacent to the interface.
