Answer
$c_{}=387.2$ m/s
$V=310$ m/s
$ρ=1.867$ kg/m$^{3}$
$T_{0}= 421$ K
$P_{0}= 305k$ Pa
$ρ_{0}= 2.52$ kg/m$^{3}$
Work Step by Step
The speed of sound in air at the specified conditions is $$
c=\sqrt{k R T}=\sqrt{(1.4)(0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(373.2 \mathrm{~K})\left(\frac{1000 \mathrm{~m}^2 / \mathrm{s}^2}{1 \mathrm{~kJ} / \mathrm{kg}}\right)}=387.2 \mathrm{~m} / \mathrm{s}
$$ Thus, $$
V=\operatorname{Ma} \times c=(0.8)(387.2 \mathrm{~m} / \mathrm{s})=\mathbf{3 1 0} \mathrm{m} / \mathbf{s}
$$ Also, $$
\rho=\frac{P}{R T}=\frac{200 \mathrm{kPa}}{\left(0.287 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kg} \cdot \mathrm{K}\right)(373.2 \mathrm{~K})}=1.867 \mathrm{~kg} / \mathrm{m}^3
$$ Then the stagnation properties are determined from $$
\begin{aligned}
& T_0=T\left(1+\frac{(k-1) \mathrm{Ma}^2}{2}\right)=(373.2 \mathrm{~K})\left(1+\frac{(1.4-1)(0.8)^2}{2}\right)=\mathbf{4 2 1} \mathbf{K} \\
& P_0=P\left(\frac{T_0}{T}\right)^{k /(k-1)}=(200 \mathrm{kPa})\left(\frac{421.0 \mathrm{~K}}{373.2 \mathrm{~K}}\right)^{1.4 /(1.4-1)}=\mathbf{3 0 5} \mathbf{k P a} \\
& \rho_0=\rho\left(\frac{T_0}{T}\right)^{1 /(k-1)}=\left(1.867 \mathrm{~kg} / \mathrm{m}^3\right)\left(\frac{421.0 \mathrm{~K}}{373.2 \mathrm{~K}}\right)^{1 /(1.4-1)}=\mathbf{2 . 5 2} \mathbf{k g} / \mathrm{m}^3
\end{aligned}
$$
