# 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?

$0.547 \setminus \setminus {\textrm{k \frac{g}{m}}}^{3}$

#### Explanation:

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

$\setminus \rho = \setminus \frac{P}{R T}$

Where $R$ is characteristic gas constant.

Now, the density ${\rho}_{s}$ at sea level at temperature $T = {20}^{\setminus} \circ C = 293 \setminus K$ & a pressure $1 \setminus a t m$ is given as

${\rho}_{s} = \setminus \frac{1}{R \left(293\right)}$

$1.25 = \setminus \frac{1}{293 R} \setminus \ldots \ldots . . \left(1\right)$

Now, the density ${\rho}_{t}$ at top of Mt Everest at temperature $T = - {50}^{\setminus} \circ C = 223 \setminus K$ & a pressure $\frac{1}{3} \setminus a t m$ is given as

${\rho}_{t} = \setminus \frac{\frac{1}{3}}{R \left(223\right)}$

${\rho}_{t} = \setminus \frac{1}{669 R} \setminus \ldots \ldots . . \left(2\right)$

Dividing (2) by (1), we get

$\setminus {\rho}_{t} / 1.25 = \setminus \frac{293 R}{669 R}$

$\setminus {\rho}_{t} = 1.25 \left(\frac{293}{669}\right)$

$= 0.547$

hence the density at the top of Mt.Everest is about $\setminus {\rho}_{t} = 0.547 \setminus \setminus {\textrm{k \frac{g}{m}}}^{3}$