Question

The density of atmosphere on earth's surface is about $1.3{\text{kg m}}^{-3}$ and the pressure is $1\text{atm}$. If the atmosphere had uniform density, same as that observed from the earth's surface, what would be the height of the atmosphere to exert the same pressure as on the surface?

Would this height be taller or shorter than the Mt. Everest?
(Take necessary assumptions and approximations) $\left[5\text{marks}\right]$

Answer 1

Given: $\left[1\text{mark}\right]$

The atmospheric pressure on the earth's surface, ${\text{P}}_0=1\text{atm}\approx {10}^5\text{Pa}$

The density of the atmosphere, ${\rho }_0=1.3{\text{kg m}}^{-3}$

Let the acceleration due to earth's gravity be $10{\text{m s}}^{-2}$.

The pressure of the atmosphere at a height $\text{h}$ is given by ${\text{P}}_0={\rho }_{0}g\text{h}$ $\left[1\text{mark}\right]$
$⟹\text{h}=\frac{{\text{P}}_0}{{\rho }_{0}g}=\frac{{10}^5}{1.3\times 10}\approx 7692\text{m}$ $\left[2\text{mark}\right]$

Therefore, up to $7692\text{m}$, the atmospheric pressure would be felt the same as that on the earth's surface.

The height of the Mt. Everest is about $8848\text{m}$.
So, the height to which the atmospheric pressure would be felt the same as that on the earth's surface, is actually shorter than the Mt. Everest. $\left[1\text{mark}\right]$