Question

Calculate the density of the air at the top of Mt Everest given the pressure is 278 hPa and the temperature is-43.2°C (you may ignore the effect of water vapour on density for this problem).

Answer 1

density of air at the top of Mt Everest is
`\rho = 0.418 kg/m^3`

Step-by-step explanation:

using ideal gas equation

P V = n R T...........(1)

molecular weight of gas

n =
`(m)/(M)`

volume V =
`(m)/(\rho)`

therefore from equation (1)

`P((m)/(\rho))=(m)/(M)RT`

`\rho = (PM)/(RT)`

`\rho = (278* 10^2* 28.8* 10^(-3) kg/mol)/(8.314J/mol.K \ (273-43.2)K)`

`\rho = 0.418 kg/m^3`

hence, density of air at the top of Mt Everest is
`\rho = 0.418 kg/m^3`