Question

Let us assume air is under standard conditions close to the earth’s surface. Presuming that the acceleration due to gravity, the temperature and the molar mass of air are independent of height, find air pressure at a height $5km$ over the surface and in a mine at a depth of $5km$ below the surface.

Answer 1

Hint: The pressure in a fluid depends on the density of the fluid, acceleration due to gravity and height in the fluid. For an infinitesimally small change in pressure, the height change is also infinitesimally small. Integrating the equation we can determine a relation known as a barometric formula and use it to calculate pressure at a certain height.

Formulas used:
$P=\rho gh$
$PV=RT$
$P={{P}_{0}}{{e}^{-\dfrac{Mg}{RT}h}}$

Complete answer:
Pressure is the force applied per unit area. Its SI unit is pascal ($P$).
We know that,
$P=\rho gh$
Here, $P$ is the pressure in a fluid
$g$ is acceleration due to gravity
$h$ is the height
From the above equation,
$dP=\rho gdh$ - (1)
Here, $dP$ is the infinitesimally small change in pressure and $dh$ is the infinitesimally small change in height.
From the ideal gas equation,
$PV=RT$ - (2)
Here, $V$ is the volume
$n$ is number of moles of the gas
$R$ is the gas constant
$T$ is the temperature
Therefore, from the above equation,
$\begin{align}
  & \rho =\dfrac{M}{V} \\
 & \Rightarrow V=\dfrac{M}{\rho } \\
\end{align}$
Substituting the above equation in eq (2),
$P\dfrac{M}{\rho }=RT$
$\Rightarrow \rho =\dfrac{PM}{RT}$ - (3)
Substituting eq (3) in eq (1), we get,
$\begin{align}
  & dP=-\dfrac{PM}{RT}gdh \\
 & \Rightarrow \dfrac{dP}{P}=-\dfrac{Mg}{RT}dh \\
\end{align}$
We integrate on both sides of the equation, we get,
$\begin{align}
  & \int{\dfrac{dP}{P}=-\int{\dfrac{Mg}{RT}dh}} \\
 & \Rightarrow \int\limits_{{{P}_{0}}}^{P}{\dfrac{dP}{P}=-\dfrac{Mg}{RT}\int\limits_{0}^{h}{dh}} \\
 & \Rightarrow \left[ \log P \right]_{{{P}_{0}}}^{P}=-\dfrac{Mg}{RT}\left[ h \right]_{0}^{h} \\
 & \Rightarrow \left[ \log P-\log {{P}_{0}} \right]=-\dfrac{Mg}{RT}\left[ h-0 \right] \\
 & \Rightarrow \log \dfrac{P}{{{P}_{0}}}=-\dfrac{Mg}{RT}h \\
 & \Rightarrow \dfrac{P}{{{P}_{0}}}={{e}^{-\dfrac{Mg}{RT}h}} \\
\end{align}$
$\therefore P={{P}_{0}}{{e}^{-\dfrac{Mg}{RT}h}}$ - (4)
In standard conditions, ${{P}_{0}}=1atm$, $T=273K$
For first condition, $h=5km$
Substituting given values in the above equation we get,
$\begin{align}
  & P=1{{e}^{-\dfrac{28\times 10}{8.314\times 273}\times 5\times {{10}^{-3}}}} \\
 & \Rightarrow P=0.55atm \\
\end{align}$
Therefore, the atmospheric pressure at a height $5km$ over the surface is $0.55atm$.
For the second condition, $h=-5km$
Substituting values for second condition in eq (4), we get,
$\begin{align}
  & P={{P}_{0}}{{e}^{-\dfrac{28\times 10}{8.314\times 273}\times -5\times {{10}^{-3}}}} \\
 & \Rightarrow P=1.83atm \\
\end{align}$
Therefore, the atmospheric pressure at a height $5km$ below the surface is $1.83atm$.
Therefore, atmospheric pressure above the surface is $0.55atm$ and the atmospheric pressure below the Earth’s surface is $1.83atm$.

Note:
The negative sign in the relation between density, acceleration due to gravity and height with pressure indicates that as height increases the pressure decreases and vice versa. The number of moles is taken as one for standard conditions. The average molar mass of mixture of gases, i.e. air is taken as$28gm\,mo{{l}^{-1}}$.