Sarthaks Test
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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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