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

`5.31(kg)/(m^3)`

Step-by-step 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
`(m)/(M)`, where m is the mass and M the molar mass of the gas, and the density is
`(m)/(V)`.

For air
`M=28.66*10^(-3)(kg)/(mol)` and
`(5)/(9)R=K`

So,
`598.59 R*(5)/(9)=332.55K`

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