Answer
$72.4$ kPa
Work Step by Step
The stagnation temperature of air is determined from $$
T_0=T+\frac{V^2}{2 c_p}=238 \mathrm{~K}+\frac{(325 \mathrm{~m} / \mathrm{s})^2}{2 \times 1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}}\left(\frac{1 \mathrm{~kJ} / \mathrm{kg}}{1000 \mathrm{~m}^2 / \mathrm{s}^2}\right)=290.5 \cong \mathbf{2 9 1 K}
$$ Other stagnation properties at the specified state are determined by considering an isentropic process between the specified state and the stagnation state,
$$ P_0=P\left(\frac{T_0}{T}\right)^{k /(k-1)}=(36 \mathrm{kPa})\left(\frac{290.5 \mathrm{~K}}{238 \mathrm{~K}}\right)^{1.4 /(1.4-1)}=72.37 \mathrm{kPa} \cong 72.4 \mathrm{kPa}
$$
