Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 5.45.
Earlier, we worked out the dry adiabatic lapse rate, which is the temperature gradient in dry air at which convection begins, that is, it is the point at which air near the surface achieves a density low enough that it starts to rise adiabatically. However, in the real atmosphere there will always be some water vapour, and as the air rises and cools, some of this water vapour will condense out, forming clouds. When water vapour condenses to liquid, it releases an amount of heat given by the latent heat of vapourization , causing a decrease in the rate at which cooling occurs with height. We can get an estimate of this revised rate of cooling, called the wet adiabatic lapse rate (sometimes the moist adiabatic lapse rate) by combining some of our earlier results.
First, we can use the first law of thermodynamics (conservation of energy) to include the effect of condensation. Suppose we have a mass of air containing moles of air molecules and moles of water vapour (we’ll assume that in what follows). If an amount of water vapour condenses, it releases an amount of heat . As the air rises, it also expands by a volume , doing work , so the net energy change of the air mass is
The first negative sign is because the gas does work (and hence loses energy) as it expands, while the second negative sign is because a decrease in (that is, ) releases heat, thus increasing .
We want to convert this to a formula for the change in temperature in terms of and . From the equipartition theorem, the energy of the air mass is
where is the number of degrees of freedom of an air molecule, which is usually taken to be . We therefore have
We now need to convert the term. My first thought was that, since the gas is supposed to be expanding adiabatically, we can use the relation
where and is a constant. Taking differentials gives
and plugging all this back into 1 gives
Applying the ideal gas law to the first term, we get
This does not agree with Schroeder’s result, which has instead of in the last term.
We can get Schroeder’s answer if instead of using the adiabatic relation 4, we use the ideal gas law and take differentials:
We now get
I’m not entirely sure why the adiabatic relation gives the wrong answer, although it could have something to do with the fact that, strictly speaking, the process isn’t adiabatic, since the condensing vapour injects some heat into the air mass.
In any case, we can proceed with the derivation of the wet adiabatic lapse rate. The next step is to assume that the air remains saturated (100% humidity) at all stages of the expansion. In this case, the ratio satisfies
where is the vapour pressure at which condensation occurs at temperature . Thus, is a function of total air pressure and temperature at a given height . We can now take the total derivative to get
The partial derivatives are
So we get
From the Clausius-Clapeyron equation, we have
where is the latent heat of vapourization per mole, and is the difference in volume between the gas and liquid phases. Since the volume of the liquid is negligible relative to the volume of the gas, we can approximate using the ideal gas law:
[Note that here since in 20 is the latent heat per mole.] This gives
Therefore, 19 becomes
We can now use the barometric equation on the RHS:
where is the mass of one air molecule and is the mass of a mole of air molecules. This gives
If we plug in some numbers we can see the effect of condensation on the lapse rate. We’ll use the earlier expression for and other values from Schroeder’s book:
[The last 3 quantities are in SI units.]
At a temperature of and pressure of 1 bar (, we have
or about 4 K per kilometre. The dry lapse rate is around 9.75 K per kilometre, so condensation has a considerable effect.
At the lower temperature of and pressure of 1 bar we have
Presumably this is because at lower temperatures the air can hold less water vapour, so the air is dryer, which means less vapour is available for condensation.