Answer
$c^{}= 1368.72$ ft/s
$V^{}=958$ ft/s
$ρ=0.086568$ lbm/ft$^{3}$
$T_{0}=856$ R
$P_{0}=34.678$ psia
$ρ_{0}=0.109$ lbm/ft$^{3}$
Work Step by Step
First, $$T=320+459.67=779.67 \mathrm{~K}.$$ The speed of sound in air at the specified conditions is
$$
c=\sqrt{k R T}=\sqrt{(1.4)(0.06855 \mathrm{Btu} / 1 \mathrm{bm} \cdot \mathrm{R})(779.67 \mathrm{R})\left(\frac{25,037 \mathrm{ft}^2 / \mathrm{s}^2}{1 \mathrm{Btu} / 1 \mathrm{bm}}\right)}=1368.72 \mathrm{ft} / \mathrm{s}
$$ Therefore, $$
V=\mathrm{Ma} \times c=(0.7)(1368.72 \mathrm{ft} / \mathrm{s})=958.10 \cong 958 \mathrm{ft} / \mathrm{s}
$$ Also, $$
\rho=\frac{P}{R T}=\frac{25 \mathrm{psia}}{\left(0.3704 \mathrm{psia} \cdot \mathrm{ft}^3 / \mathrm{lbm} \cdot \mathrm{R}\right)(779.67 \mathrm{R})}=0.086568 \mathrm{lbm} / \mathrm{ft}^3
$$ Then the stagnation properties are determined from $$
\begin{aligned}
& T_0=T\left(1+\frac{(k-1) \mathrm{Ma}^2}{2}\right)=(779.67 \mathrm{R})\left(1+\frac{(1.4-1)(0.7)^2}{2}\right)=856.08 \mathrm{R} \cong 856 \mathbf{R} \\
& P_0=P\left(\frac{T_0}{T}\right)^{k /(k-1)}=(25 \mathrm{psia})\left(\frac{856.08 \mathrm{R}}{779.67 \mathrm{R}}\right)^{1.4(1.4-1)}=34.678 \mathrm{psia} \cong 34.7 \text { psia } \\
& \rho_0=\rho\left(\frac{T_0}{T}\right)^{1 / k-1)}=\left(0.086561 \mathrm{bm} / \mathrm{ft}^3\right)\left(\frac{856.08 \mathrm{R}}{779.67 \mathrm{R}}\right)^{1 /(1.4-1)}=0.10936 \mathrm{lbm} / \mathrm{ft}^3 \cong 0.109 \mathrm{lbm} / \mathrm{ft}^3
\end{aligned}
$$
