Given the Temperature and Pressure at the stagnation point. Density is related to pressure and temperature as $P=\rho RT$. Calculate the density at this point. Given,  $$\begin{array}{l}T = {934^ \circ }R\\p = 7.8\,atm\\Density = \rho  = ?\\P = \rho RT\\\rho  = \frac{P}{{RT}} = \frac{{\left( {7.8 \times 2116} \right)}}{{1716 \times 934}} = 0.0103\,slug/f{t^3}\end{array}$$ Therefore, density is $0.0103\,slug/f{t^3}$.  Calculate ${c_p},{c_v},e\,{\rm{and}}\,h$ for stagnation point conditions, Specific heat at constant pressure $c_p$ and Specific heat at constant volume $c_v$ is related to gamma $\gamma$ and gas constant $R$ as

a)   $${c_p} = \frac{{\gamma R}}{{\gamma  – 1}} = \frac{{1.4 \times 1716}}{{0.4}} = 6006\frac{{ft}}{{slug}}\frac{{lb}}{{{}^ \circ R}}$$ $${c_v} = \frac{R}{{\gamma  – 1}} = \frac{{1716}}{{0.4}} = 4290\frac{{ft}}{{slug}}\frac{{lb}}{{{}^ \circ R}}$$ $$e = {c_v}T = 4290(934) = 4.007 \times {10^6}\frac{{ft\,lb}}{{slug}}$$ $$h = {c_p}T = 6006(934) = 5.610 \times {10^6}\frac{{ft\,lb}}{{slug}}$$ $e$ and $h$ are the internal energy and specific enthalpy of the gas molecules

b) For a  calorically perfect gas, ${c_p}$ and ${c_v}$ are constants, $${c_p} = 6006\frac{{ft}}{{slug}}\frac{{lb}}{{{}^ \circ R}};\,{c_v} = 4290\frac{{ft}}{{slug}}\frac{{lb}}{{{}^ \circ R}}$$ Also at standard sea level $T = {519^ \circ }R$, therefore $$e = 4290 \times 519 = 2.227 \times {10^6}\frac{{ft\,lb}}{{slug}}$$ $$h = 6006\left( {519} \right) = 3.117 \times {10^6}\frac{{ft\,lb}}{{slug}}$$