Question

A missile flying at high speed has a stagnation pressure and temperature of 5 atm and 598.59 °R respectively. What is the density of air at this point?

Answer 1

Answer:

$5.31\frac{kg}{m^3}$

Explanation:

Approximately, we can use the ideal gas law, below we see how we can deduce the density from general gas equation. To do this, remember that the number of moles n is equal to $\frac{m}{M}$, where m is the mass and M the molar mass of the gas, and the density is $\frac{m}{V}$.

For air $M=28.66*10^{-3}\frac{kg}{mol}$ and $\frac{5}{9}R=K$

So, $598.59 R*\frac{5}{9}=332.55K$

$pV=nRT\\pV=\frac{m}{M}RT\\\frac{m}{V}=\frac{pM}{RT}\\\rho=\frac{pM}{RT}\\\rho=\frac{(5atm)28.66*10^{-3}\frac{kg}{mol}}{(8.20*10^{-5}\frac{m^3*atm}{K*mol})332.55K}=5.31\frac{kg}{m^3}$