Answer:
[tex]5.31\frac{kg}{m^3}[/tex]
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 [tex]\frac{m}{M}[/tex], where m is the mass and M the molar mass of the gas, and the density is [tex]\frac{m}{V}[/tex].
For air [tex]M=28.66*10^{-3}\frac{kg}{mol}[/tex] and [tex]\frac{5}{9}R=K[/tex]
So, [tex]598.59 R*\frac{5}{9}=332.55K[/tex]
[tex]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}[/tex]
