Answer
$V_{N_2}=0.322$ m$^{3}$
$V_{O_2}=2.065$ m$^{3}$
$P_{w}=204$ kPa
Work Step by Step
The volumes of the tanks are
$$
\begin{aligned}
& V_{\mathrm{N}_2}=\left(\frac{m R T}{P}\right)_{\mathrm{N}_2}=\frac{(2 \mathrm{~kg})\left(0.2968 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kg} \cdot \mathrm{K}\right)(298 \mathrm{~K})}{550 \mathrm{kPa}}=0.322 \mathrm{~m}^3 \\
& V_{\mathrm{O}_2}=\left(\frac{m R T}{P}\right)_{\mathrm{O}_2}=\frac{(4 \mathrm{~kg})\left(0.2598 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kg} \cdot \mathrm{K}\right)(298 \mathrm{~K})}{150 \mathrm{kPa}}=2.065 \mathrm{~m}^3 \\
& V_{\text {lioal }}=V_{\mathrm{N}_2}+V_{\mathrm{O}_2}=0.322 \mathrm{~m}^3+2.065 \mathrm{~m}^3=2.386 \mathrm{~m}^3
\end{aligned}
$$ Also, $$
\begin{aligned}
& N_{\mathrm{N}_2}=\frac{m_{\mathrm{N}_2}}{M_{\mathrm{N}_2}}=\frac{2 \mathrm{~kg}}{28 \mathrm{~kg} / \mathrm{kmol}}=0.07143\ \mathrm{kmol} \\
& N_{\mathrm{O}_2}=\frac{m_{\mathrm{O}_2}}{M_{\mathrm{O}_2}}=\frac{4 \mathrm{~kg}}{32 \mathrm{~kg} / \mathrm{kmol}}=0.125 \mathrm{kmol} \\
& N_a=N_{\mathrm{N}_2}+N_{\mathrm{O}_2}=0.07143 \mathrm{kmol}+0.125 \mathrm{kmol}=0.1964\ \mathrm{kmol}
\end{aligned}
$$ Thus, $$
P_w=\left(\frac{N R_v T}{V}\right)_w=\frac{(0.1964 \mathrm{kmol})\left(8.314 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kmol}-\mathrm{K}\right)(298 \mathrm{~K})}{2.386 \mathrm{~m}^3}=204\ \mathrm{kPa}
$$
