**(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 ρ_{0}, 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 = p_{0}]

[at h = h, p = p]

or In\(\frac{p}{p_0}\) = − \(\frac{ρ_0g}{p_0}\)ℎ

**Taking antilog,**

Or p = p_{0} exp(− \(\frac{ρ_0g}{p_0}\)ℎ)

**(c)** In \(\frac{p}{p_0}\) = − \(\frac{ρ_0gh}{p_0}\)

In \(\frac{1}{10}\) = − \(\frac{ρ_0g}{p_0}\)h_{0}

∴ h_{0} = − \(\frac{p_0}{ρ_0g}\) 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 × 10^{5}m = 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