Approximately what fraction of the sea-level air density is found at the top of Mount Everest, a height of 8848 m above sea level?

The given data can be listed below,

The density of air is determined by temperature, pressure, and the amount of water vapour present. The density of air is affected by pressure in the opposite direction.

The density in terms of Boltzmann constant and temperature is given by,

$\rho ={\rho }_{\mathrm{sea}}\exp \left(\frac{-E}{k_{B}T}\right)$

Here,${\rho }_{\mathrm{sea}}$is the density at sea-level, E is the potential energy, $k_B$is Boltzmann constant whose value is $1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}$and T is the temperature.

The potential energy is given by,

E= mgy

Here, m is the mass of the air particle whose value is $\frac{M}{N_A}$, g is the acceleration due to gravity whose value is $9.81\mathrm{m}/{\mathrm{s}}^2$and y is the height.

The fraction in air density is given by,

$\frac{\rho }{{\rho }_{sea}}=\exp \left(\frac{-{\displaystyle \frac{M}{N_A}}gy}{k_{\mathrm{B}}\mathrm{T}}\right)$

Substitute all the values in the above,

$$\begin{gathered} \frac{\rho }{{\rho }_{sea}}=\exp \left(\frac{-\left({\displaystyle \frac{28\mathrm{g}/\mathrm{mol}\left({\displaystyle \frac{1\mathrm{kg}}{1000\mathrm{g}}}\right)}{6.02\times {10}^{23}{\mathrm{mol}}^{-1}}}\right)(9.81\mathrm{m}/{\mathrm{s}}^2}{1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}(293\mathrm{K})}\right) \\ =1.00 \end{gathered}$$

Thus, the fraction of air density at the top of the Everest is 1 .