# (a) Pressure decreases as one ascends the atmosphere. If the density of air is ρ, what is the change in pressure do over a differential height dh?

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(a) Pressure decreases as one ascends the atmosphere. If the density of air is ρ, what is the change in pressure do over a differential height dh?

(b) Considering the pressure p to be proportional to the density, find the pressure p at a height h is the pressure on the surface of the earth is p0.

(c) If p0 = 1.03 × 105 Nm-2, p0 = 129 kgm-3 and g = 9.8 ms-2, at which height will the pressure drop to (1/10) the value at the surface of the earth?

(d) This model of the atmosphere works for relatively small distances. Identify the underlying assumption that limits the model.

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(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$\frac{p}{p_0}$ = − $\frac{ρ_0g}{p_0}$

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

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