Answer
$V_{m}=289.3$ m$^{3}$
Work Step by Step
The total number of moles is
$$
N_m=N_{\mathrm{O}_2}+N_{\mathrm{CO}_2}=8\ \mathrm{kmol}+10 \mathrm{kmol}=18\ \mathrm{kmol}
$$
Then
$$
\boldsymbol{V}_m=\frac{N_m R_u T_m}{P_m}=\frac{(18 \mathrm{kmol})\left(8.314 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kmol} \cdot \mathrm{K}\right)(290 \mathrm{~K})}{150 \mathrm{kPa}}=\mathbf{2 8 9 . 3 \mathbf { ~ m } ^ { 3 }}
$$
