Answer:
density of air at the top of Mt Everest is [tex]\rho = 0.418 kg/m^3[/tex]
Explanation:
using ideal gas equation
P V = n R T...........(1)
molecular weight of gas
n = [tex]\dfrac{m}{M}[/tex]
volume V = [tex]\dfrac{m}{\rho}[/tex]
therefore from equation (1)
[tex]P(\dfrac{m}{\rho})=\dfrac{m}{M}RT[/tex]
[tex]\rho = \dfrac{PM}{RT}[/tex]
[tex]\rho = \dfrac{278\times 10^2\times 28.8\times 10^{-3} kg/mol}{8.314J/mol.K \ (273-43.2)K}[/tex]
[tex]\rho = 0.418 kg/m^3[/tex]
hence, density of air at the top of Mt Everest is [tex]\rho = 0.418 kg/m^3[/tex]
