1) The temperature and pressure at the stagnation point of a high speed missile are \({934^ \circ }\) R and 7.8 atm, respectively.Calculate the density at this point. Solution: \[\begin{array}{l}T = {934^ \circ }R\\p = 7.8\,atm\\Density = \rho = ?\\p = \rho RT\\\rho = \frac{P}{{RT}} = \frac{{\left( {7.8 \times 2116} \right)}}{{1716 \times 934}} = 0.0103\,slug/f{t^3}\end{array}\]2)Calculate \({c_p},{c_v},e\,{\rm{and}}\,h\) for a) The stagnation point conditions given in problem (1). b) Air at standard sea level conditions. Solution: \({c_p},{c_v},e\,{\rm{and}}\,h\) for stagnation point conditions will be 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 ...