Answer
$P_{N2}=195$ kPa
$P_{O_2}=60$ kPa
$P_{CO_2}=45$ kPa
Work Step by Step
For convenience, consider $100 \mathrm{kmol}$ of mixture. Then the mass of each component and the total mass are
$$
\begin{aligned}
& N_{\mathrm{N}_2}=65 \mathrm{kmol} \longrightarrow m_{\mathrm{N}_2}=N_{\mathrm{N}_2} M_{\mathrm{N}_2}=(65 \mathrm{kmol})(28 \mathrm{~kg} / \mathrm{kmol})=1820 \mathrm{~kg} \\
& N_{\mathrm{O}_2}=20 \mathrm{kmol} \longrightarrow m_{\mathrm{O}_2}=N_{\mathrm{O}_2} M_{\mathrm{O}_2}=(20 \mathrm{kmol})(32 \mathrm{~kg} / \mathrm{kmol})=640 \mathrm{~kg} \\
& N_{\mathrm{CO}_2}=15 \mathrm{kmol} \longrightarrow m_{\mathrm{CO}_2}=N_{\mathrm{CO}_2} M_{\mathrm{CO}_2}=(15 \mathrm{kmol})(44 \mathrm{~kg} / \mathrm{kmol})=660 \mathrm{~kg} \\
& m_m=m_{\mathrm{N}_2}+m_{\mathrm{O}_2}+m_{\mathrm{CO}_2}=1820 \mathrm{~kg}+640 \mathrm{~kg}+660 \mathrm{~kg}=3120 \mathrm{~kg}
\end{aligned}
$$ Then the mass fraction of each component (gravimetric analysis) becomes
$$
\begin{aligned}
\mathrm{mf}_{\mathrm{N}_2} & =\frac{m_{\mathrm{N}_2}}{m_m}=\frac{1820 \mathrm{~kg}}{3120 \mathrm{~kg}}=0.583 \text { or } \quad \mathbf{5 8 . 3} \% \\
\mathrm{mf}_{\mathrm{O}_2} & =\frac{m_{\mathrm{O}_2}}{m_m}=\frac{640 \mathrm{~kg}}{3120 \mathrm{~kg}}=0.205 \text { or } \quad \mathbf{2 0 . 5} \% \\
\mathrm{mf}_{\mathrm{CO}_2} & =\frac{m_{\mathrm{CO}_2}}{m_m}=\frac{660 \mathrm{~kg}}{3120 \mathrm{~kg}}=0.212 \text { or } \quad \mathbf{2 1 . 2} \%
\end{aligned}
$$ For ideal gases, the partial pressure is proportional to the mole fraction, and is determined from $$
\begin{gathered}
P_{\mathrm{N}_2}=y_{\mathrm{N}_2} P_m=(0.65)(300 \mathrm{kPa})=195 \mathrm{kPa} \\
P_{\mathrm{O}_2}=y_{\mathrm{O}_2} P_m=(0.20)(300 \mathrm{kPa})=60 \mathrm{kPa} \\
P_{\mathrm{CO}_2}=y_{\mathrm{CO}_2} P_m=(0.15)(300 \mathrm{kPa})=\mathbf{4 5} \mathrm{kPa}
\end{gathered}
$$
