Answer
$Ma^{}= 0.620$
$c= 320.7$ m/s
$V=199$ m/s
Work Step by Step
The stagnation pressure of air at the specified conditions is
$$
P_0=P+\Delta P=54+16=70 \mathrm{kPa}
$$ Then, $$
\frac{P_0}{P}=\left(1+\frac{(k-1) \mathrm{Ma}^2}{2}\right)^{k / k-1} \longrightarrow \frac{70}{54}=\left(1+\frac{(1.4-1) \mathrm{Ma}^2}{2}\right)^{1.4 / 0.4}
$$ It yields $$
\mathrm{Ma}=\mathbf{0 . 6 2 0}
$$ The speed of sound in air at the specified conditions is
$$ c=\sqrt{k R T}=\sqrt{(1.4)(0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(256 \mathrm{~K})\left(\frac{1000 \mathrm{~m}^2 / \mathrm{s}^2}{1 \mathrm{~kJ} / \mathrm{kg}}\right)}=320.7 \mathrm{~m} / \mathrm{s}
$$ Thus, $$
V=\operatorname{Ma} \times c=(0.620)(320.7 \mathrm{~m} / \mathrm{s})=199 \mathrm{~m} / \mathrm{s}
$$
