Answer
$T^{*}= 458$ K
$P^{*}= 184.9$ kPa
$ρ^{*}= 1.46$ kg/m$^{3}$
$k= 1.40$
Work Step by Step
Analysis The gas constant of the mixture is
$$
\begin{aligned}
& M_m=y_{O_2} M_{O_2}+y_{N_2} M_{N_2}=0.5 \times 32+0.5 \times 28=30 \mathrm{~kg} / \mathrm{kmol} \\
& R_m=\frac{R_u}{M_m}=\frac{8.314 \mathrm{~kJ} / \mathrm{kmol} \cdot \mathrm{K}}{30 \mathrm{~kg} / \mathrm{kmol}}=0.2771 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}
\end{aligned}
$$ The specific heat ratio is 1.4 for nitrogen, and nearly 1.4 for oxygen. Therefore, the specific heat ratio of the mixture is also 1.4. Then the critical temperature, pressure, and density of the mixture become
$$
\begin{aligned}
& T^*=T_0\left(\frac{2}{k+1}\right)=(550 \mathrm{~K})\left(\frac{2}{1.4+1}\right)=458.3 \mathrm{~K} \cong 458 \mathrm{~K} \\
& P^*=P_0\left(\frac{2}{k+1}\right)^{k /(k-1)}=(350 \mathrm{kPa})\left(\frac{2}{1.4+1}\right)^{1.4(1.4-1)}=184.9 \mathrm{kPa} \cong 185 \mathrm{kPa} \\
& \rho^*=\frac{P^*}{R T^*}=\frac{184.9 \mathrm{kPa}}{\left(0.2771 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kg} \cdot \mathrm{K}\right)(458.3 \mathrm{~K})}=1.46 \mathrm{~kg} / \mathrm{m}^3
\end{aligned}
$$ Discussion If the specific heat ratios $k$ of the two gases were different, then we would need to determine the $k$ of the mixture from $k=c_{p, m} / c_{v, \mathrm{~m}}$ where the specific heats of the mixture are determined from
$$
\begin{gathered}
C_{p, m}=\operatorname{mf}_{O_2} c_{p, O_2}+\operatorname{mf}_{N_2} c_{p, N_2}=\left(y_{O_2} M_{O_2} / M_m\right) c_{p, O_2}+\left(y_{N_2} M_{N_2} / M_m\right) c_{p, N_2} \\
C_{v, m}=\operatorname{mf}_{O_2} c_{v, O_2}+\operatorname{mf}_{N_2} c_{v, N_2}=\left(y_{O_2} M_{O_2} / M_m\right) c_{v, O_2}+\left(y_{N_2} M_{N_2} / M_m\right) c_{v, N_2}
\end{gathered}
$$
where $\mathrm{mf}$ is the mass fraction and $y$ is the mole fraction. In this case it would give
$$
\begin{aligned}
& c_{p, \mathrm{~m}}=(0.5 \times 32 / 30) \times 0.918+(0.5 \times 28 / 30) \times 1.039=0.974 \mathrm{~kJ} / \mathrm{kg} . \mathrm{K} \\
& c_{p, \mathrm{~m}}=(0.5 \times 32 / 30) \times 0.658+(0.5 \times 28 / 30) \times 0.743=0.698 \mathrm{~kJ} / \mathrm{kg} . \mathrm{K}
\end{aligned}
$$ and $$
k=0.974 / 0.698=1.40
$$
