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CaHeK987 [17]
3 years ago
12

A missile flying at high speed has a stagnation pressure and temperature of 5 atm and 598.59 °R respectively. What is the densit

y of air at this point?
Engineering
1 answer:
alexdok [17]3 years ago
7 0

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}

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