Let us assume that air is under standard conditions close to the Earth's surface. Presuming that the temperature and the molar mass of air are independent of height, find the air pressure at the height `5.0 km` over the surface and in a mine at the depth `5.0 km` below the surface.

Explanation

`:.dP=-rhog dh`………..1
We know that `P=(rhRt)/M`
`:.rho=M/(RT)P` ……2
Substituting the value fo `rho` from equation 2 in equation 1 we get
`dP=-(Mg)/(RT) . Pdh`
or `(dP)/P=-(Mg)/(RT)dh` ……..3
Integrating equation 3 we have `int_(P_(e))^(P)(dP)/P=-(Mg)/(RT)int_(e)^(n)dh`
AT the surfaceof earth,h is zero while the pressure is `P_(o)`. At a height h the pressure is P.
In `(P//P_(o))=-(Mg)/(RT)h`
`P=P_(o)e^(-(Mgh//RT))` .........4
Under standard conditions
`P_(o)=1` atm and `T=273K`
At a height of 5 km
`P=1xxe^(-28xx9.81xx5000//8314xx273)=0.55` atm
At a depth of 5 km
`P=1xxe^(-28xx9.81xx(-5000)//8314xx273) = -1.83` atm
Remember that `P=P_(0)e^(-(Mgh//RT))` is known as Barometric formula.