Question

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?

Short answer

The fraction of air density at the top of the Everest is 1.

Step-by-step solution

Identification of given data

The given data can be listed below,

• The height of mount Everest above sea level is $h=8848\mathrm{m}$.

Concept/Significance of air density

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.

Determination of the fraction of the sea-level air density is found at the top of Mount Everest

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 .