 Solution

ID: past-exam-of-the-physics-of-atmospheres-and-oceans-course-of-the-university-of-oxford/2021/3/b/solution

Let

e_{v}

e_{v} / p

is constant. Its dew point

T_{D}

is defined by

E (T_{D}) = e_{v}

, where

E

\frac{L _{v}}{R _{v} T _{D}^{2}} \frac{d T _{D}}{d z} = \frac{d l n e _{v}}{d z} = \frac{d l n p}{d z} ≃ - \frac{g}{R _{d} T} .

The last approximation uses hydrostatic balance, the ideal gas law, and an environmental temperature close to the parcel's. Rearrangement yields

\frac{d l n T _{D}}{d z} = - \frac{g R _{v}}{L _{v} R _{d}} \frac{T _{D}}{T} .

For a modest initial dew-point depression,

T_{D} / T ≃ 1

in absolute temperature, so

T_{D} (z) ≃ T_{D, s} e^{- z / H}, H = \frac{L _{v} R _{d}}{g R _{v}} .

With

L_{v} ≃ 2.5 \times 1 0^{6} J k g^{- 1}

R_{d} ≃ 287 J k g^{- 1} K^{- 1}

, and

R_{v} ≃ 461.5 J k g^{- 1} K^{- 1}

H ≃ 1.58 \times 1 0^{5} m

. This dew point scale is much longer than the atmospheric pressure scale height.

The exponential function is an approximation, not the exact consequence of a dry adiabatic process. Retaining

T = T_{s} - Γ_{d} z

gives

p / p_{s} = (T / T_{s})^{c_{p} / R_{d}}

and the useful comparison

\frac{1}{T _{D} ( z )} = \frac{1}{T _{D, s}} - \frac{R _{v}}{L _{v}} \frac{c _{p}}{R _{d}} ln \frac{T ( z )}{T _{s}} .

It also shows explicitly that expansion lowers the dew point even before any water vapor condenses.

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