(a) Consider a horizontal parcel of air with cross section A and height dh. Le the pressure on the top surface and bottom surface be p and p + dp. If the parcel is in equilibrium, then the net upward force must be balanced by the weight.
i.e. (p + dp) −pA = −ρgAdh [such that, Weight = Density × Volume × g]
or dp = −ρgdh [here, ρ = density of air] (-v) sign indicated pressure decrease with height.
(b) Let the density of air on the earth’s surface be ρ0, then
\(\frac{p}{p_0}\) = \(\frac{ρ}{ρ_0}\)
Or ρ = \(\frac{ρ_0}{p_0}\)p
∴ dp = − \(\frac{ρ_0g}{p_0}\)pdh
[such that, dp = −pgdh]
or \(\frac{dp}{p}\) = \(\frac{ρ_0g}{p_0}\)dh
or \(∫^p_{p_0}\frac{dp}{p}\) = − \(\frac{ρ_0g}{p_0}\)\(∫^h_0dh\)
[at h = 0, r = p0]
[at h = h, p = p]
or In\(\frac{p}{p_0}\) = − \(\frac{ρ_0g}{p_0}\)ℎ
Taking antilog,
Or p = p0 exp(− \(\frac{ρ_0g}{p_0}\)ℎ)
(c) In \(\frac{p}{p_0}\) = − \(\frac{ρ_0gh}{p_0}\)
In \(\frac{1}{10}\) = − \(\frac{ρ_0g}{p_0}\)h0
∴ h0 = − \(\frac{p_0}{ρ_0g}\) In \(\frac{1}{10}\) = \(\frac{-p_0}{ρ_0g}\) In(10)−1 = \(\frac{p_0}{ρ_0g}\) In 10
= \(\frac{p_0}{ρ_0g}\) × 2.303
= \(\frac{1.013×10^5}{1.29×9.8}\) × 2.303 = 0.184 × 105m = 18.4 km
(d) The assumption p ∝ ρ is valid only for the isothermal case which is only valid for small distances.
Temperature remains constant only near the surface of the earth, not at greater heights