Rob Ellison wrote : (quoting eHow) ‘The energy of the parcel is always the same, no matter what the air is doing. If this energy is stretched out over a wide area, it cannot create heat by being concentrated in a single spot. The farther away from each the molecules moves, the lower the temperature of the parcel will become.’
Read more : http://www.ehow.com/how-does_5220405_warm-air-rise-cool-expands_.html .
This is literally true – energy is conserved in a greater volume assuming there are no radiative losses or gains – which we did to start with. The total is x Joules – and the average is x Joule/V – in a bigger volume the total is x Joule and the average is x Joules/(V + dV).”
First, you are assuming that under adiabatic expansion of the rising parcel of air there is zero external work is done and we therefore hare complete preservation of internal energy. I think that incredible but let me grant you this claim for the sake of argument.
Second, you *hadn’t* initially argued that the average energy density density per volume unit is reduced. You had rather argued that the average *kinetic* energy *of the molecules* is reduced. You have to argue this since you acknowledge that the temperature is dropping. (Remember, you had written: “The expansion is the result of elastic decompression of the air inside the chamber and the reduction in temperature is the result of a reduced average kinetic energy. The same kinetic energy in a bigger volume.”)
But since you are denying that there is any internal energy loss from any external work done, and are again assimilating the case to a process of free expansion, (or Joule-Thompson expansion), then, although there trivially is a reduction in volumetric internal energy density, there isn’t (much of) a drop in average kinetic energy of the molecules. The drop in energy density rather is (mainly) a result of the reduction of the number of molecules per unit volume. It is indeed a well known feature of Joule expansion that it does *not* yield any temperature change. It does yield a small energy change in the case of the Joule-Thomson expansion, but this effect on temperature for the expansion of air is tiny, as some kinetic energy is converted in electrostatic potential energy. But this still contradicts flatly your claim that we have the “same *kinetic* energy in a bigger volume” for the simple reason that I stated (if the average KE per molecules drops, then so does the total KE).
So, in summary, in order to remove the inconsistency in your initial claim, we have to modify it thus: “The expansion is the result of [Joule-Thompson] decompression of the air inside the chamber and the reduction in temperature is the result of a reduced average kinetic energy. [We have however] the same [internal, i.e. kinetic + electrostatic potential] energy in a bigger volume.”
But then the temperature drop for air would be a very small fraction of the drop dictated by the dry adiabatic lapse rate calculated in the normal fashion and, I surmise, would deviate wildly from observed temperature profiles over dry land in mid-afternoon. And the idea that expanding air parcel perform no work on the surrounding is strangely unphysical and unjustified.