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

We have, `dp = - rho gdh` and from gas law `rho = (M)/(RT) p`…(1)
Thus `(dp)/(p) = -(Mg)/(RT) dh`
Integrating, we get
or, `int_(p_o)^p (dp)/(p) = - (Mg)/(RT) int_0^h dh` or, `1 n(p)/(p_0) = -(Mg)/(RT) h`,
(where `p_0` is the pressure at the surface of the Earth.)
`p = p_0e^(-M gh//RT)`,
[Under standard condition, `p_0 = 1 atm, T = 273 K`
Pressure at a height of `5 atm = 1 xx e^(-28 xx 9.81 xx 5000//8314 xx 273) = 0.5 atm`.
Pressure in a mine at a depth of `5 km = 1 xx e^(-28 xx 9.81 xx(-5000)//8314 xx 273) = 2 atm`.