Question

The summit of mount everest can be at a temperature of -50ºC and the pressure at its summit is roughly 1/3 that at sea level. The density of air at sea level is about 1.25kg/m^3. Calculate the density of air at the top of the mountain?

Answer 1

`0.547\ \text{kg/m}^3`

Explanation:

The density `rho` of air treating an ideal gas, at absolute temperature `T` & pressure `P` is given as

`\rho=\frac{P}{RT}`

Where `R` is characteristic gas constant.

Now, the density `rho_s` at sea level at temperature `T=20^\circ C=293\ K` & a pressure `1\ atm` is given as

`rho_s=\frac{1}{R(293)}`

`1.25=\frac{1}{293 R}\ ........(1)`

Now, the density `rho_t` at top of Mt Everest at temperature `T=-50^\circ C=223\ K` & a pressure `1/3\ atm` is given as

`rho_t=\frac{1/3}{R(223)}`

`rho_t=\frac{1}{ 669R}\ ........(2)`

Dividing (2) by (1), we get

`\rho_t/1.25=\frac{293R}{669R}`

`\rho_t=1.25(293/669)`

`=0.547`

hence the density at the top of Mt.Everest is about `\rho_t=0.547\ \text{kg/m}^3`