Answer
$M_{m}=16.58$ kg/kmol
$V_{w}=1.956$ m$^{3}$
$V_{O2}=0.3118$ m$^{3}$
$P_{He}=63.77$ kPa
Work Step by Step
The total mass of the mixture is $$
m_{\mathrm{it}}=m_{\mathrm{O} 2}+m_{\mathrm{CO} 2}+m_{\mathrm{He}}=0.4+0.7+0.2=1.3 \mathrm{~kg}
$$ The mole numbers of each component are $$
\begin{aligned}
N_{\mathrm{O} 2} & =\frac{m_{\mathrm{O} 2}}{M_{\mathrm{O} 2}}=\frac{0.4 \mathrm{~kg}}{32 \mathrm{~kg} / \mathrm{kmol}}=0.01250 \mathrm{kmol} \\
N_{\mathrm{CO} 2} & =\frac{m_{\mathrm{CO} 2}}{M_{\mathrm{CO} 2}}=\frac{0.7 \mathrm{~kg}}{44 \mathrm{~kg} / \mathrm{kmol}}=0.01591 \mathrm{kmol} \\
N_{\mathrm{He}} & =\frac{m_{\mathrm{He}}}{M_{\mathrm{He}}}=\frac{0.2 \mathrm{~kg}}{4 \mathrm{~kg} / \mathrm{kmol}}=0.05 \mathrm{kmol}
\end{aligned}
$$ The mole number of the mixture is
$$ N_{\mathrm{w}}=N_{\mathrm{O} 2}+N_{\mathrm{CO} 2}+N_{\mathrm{He}}=0.01250+0.01591+0.05=0.07841 \mathrm{kmol}
$$ Then the apparent molecular weight of the mixture becomes $$
M_m=\frac{m_{\text {w }}}{N_m}=\frac{1.3 \mathrm{~kg}}{0.07841 \mathrm{kmol}}=16.58 \mathrm{~kg} / \mathbf{k m o l} $$ The volume of this ideal gas mixture is
$$ V_w=\frac{N_m R_u T}{P}=\frac{(0.07841 \mathrm{kmol})\left(8.314 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kmol} \cdot \mathrm{K}\right)(300 \mathrm{~K})}{100 \mathrm{kPa}}=\mathbf{1 . 9 5 6 \mathrm { m } ^ { 3 }}
$$ The partial volume of oxygen in the mixture is $$
V_{\mathrm{O} 2}=y_{\mathrm{O} 2} V_m=\frac{N_{\mathrm{O} 2}}{N_m} V_m=\frac{0.01250 \mathrm{kmol}}{0.07841 \mathrm{kmol}}\left(1.956 \mathrm{~m}^3\right)=0.3118 \mathrm{~m}^3
$$ The partial pressure of helium in the mixture is $$
P_{\mathrm{He}}=y_{\mathrm{He}} P_w=\frac{N_{\mathrm{He}}}{N_w} P_{\mathrm{wt}}=\frac{0.05 \mathrm{kmol}}{0.07841 \mathrm{kmol}}(100 \mathrm{kPa})=63.77 \mathrm{kPa}
$$
