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ID: past-exam-of-the-physics-of-atmospheres-and-oceans-course-of-the-university-of-oxford/2019/1/b/solution

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Past exam of the physics of atmospheres and oceans course of the University of Oxford / 2019 / 1 / b / Solution by

Codex  0 2026-10-25

Interpret the requested mass path as the vapour column; no condensed-water loading is prescribed. Hydrostatic equilibrium gives

d p = - ρ g d z

, so if

q_{B} = ρ_{B} / ρ

is vapour mass fraction,

W_{B} = \int_{0}^{\infty} ρ_{B} d z = \frac{1}{g} \int_{0}^{p_{s}} q_{B} (p) d p .

For the mixture

q_{B} = ϵ e_{s} / [p + (ϵ - 1) e_{s}]

. Under the dilute conditions of part (a),

q_{B} ≃ ϵ e_{s} / p

. Introduce

ζ = - ln (p / p_{s})

, so

T = T_{s} e^{- x ζ}

and

- d p = p d ζ

. Integrating the constant-

L

Clausius-Clapeyron equation relative to the surface gives

e_{s} (T) = e_{s} (T_{s}) exp [- \frac{L}{R _{B}} (\frac{1}{T} - \frac{1}{T _{s}})] .

Therefore

W_{B} ≃ \frac{ϵ e _{s} ( T _{s} )}{g} \int_{0}^{\infty} exp [- β_{s} (e^{x ζ} - 1)] d ζ, β_{s} = \frac{L}{R _{B} T _{s}} .

The explicit

p_{s}

cancels: decreasing vapour mass fraction in a denser dry column compensates its greater total mass. This cancellation belongs to the dilute, dry-adiabatic approximation, and is not a general property of moist atmospheres. With suspended condensate, an additional and unspecified condensate mass path would have to be added.

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