Now because μ_s depends on pressure then Γ_s depends on the temperature and pressure. The numerical value for μ_s requires knowing e_s which we can calculate from the Clausius-Clapeyron equation which is coming up next.

However, its far easier to write a simple computer program and generate lines of contant Γ_s in the temperature vs pressure plane and just read off the graph:



For example, at p = 1000 mb and T =288K, Γ_s ~4.7. Note that as T increase Γ_s decreases as there is more available water vapor, e.g. μ_s is bigger.

Finally here is a simple numerical example:

From before we have



dividing through by c_p:

δT = -((g/c_p) * L(δμ_s/c_p))

Remember, δμ_s would be the physical mass of water vapor that condenses.

Constants right units: L ~ 2.5 x 10 ⁶; c_p ~ 1005 ; g/c_p ~ -10K/km

Remember we have assumed L is independent of T but then said that was not actually true, but the dependence is small so we will ignore it.

But for most atmospheric temperatures, the full cubic expression, as determined from a fit to data, is:



So if you had 1 kg of dry air rising from 0 to 3 km that would experience δT = -30K.

We now mix in 6 grams of water vapor per kg of air at z = 0 km and let it rise. The rising air will cool condense; the corresponding heat release would then be:

ΔQ = 2.5 x 10 ⁶ *.006/1005 ~ +15K

so the total change is now only -15K during the rise to 3 km. Hence, water vapor condensation is a MAJOR source of heat input into rising parcels of air.

Note the partial pressure would be:

e =μp/f

p = 1000mb ; f ~0.6; =μ_s =0.006 e ~10mb or about 1% initially; eventually liquid forms and e becomes e_s