Stanton D. answered • 05/11/15

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I'd have to differ on the answer here. Richard P., the adiabatic lapse rate arises from a volume of gas expanding against a gradually decreasing surround pressure. Thus, such a gas packet does work on the surroundings as it rises, and experiences a loss of sensible heat as a result. Here, in spite of perfectly insulating walls, one must expect both a temperature decrease (molecules slow their vertical component as they rise against the gravitational potential) from ~290 K at sealevel to ~220 K at 10 km, and a pressure decrease (from ~760 to ~200 mm Hg). One can immediately appreciate that the temperature decrease in insufficient to account solely for all the pressure decrease; there is a substantial actual density decrease also! (Actual atmosphere data: https://www.avs.org/AVS/files/c7/c7edaedb-95b2-438f-adfb-36de54f87b9e.pdf ). This must be taken into account when integrating for mean potential energy of the molecules.

Actually the above data are for air (mostly N2): the He of the problem will be somewhat less concentrated towards the cylinder bottom than is N2, i.e. the exponential decreases in temperature and pressure will be spread to higher altitudes, but only by a scale factor proportional to (m1/m2)^0.5.

Sorry I can't offer you a better closed-form total solution; the pressure for Earth's atmosphere is considered to follow an exponential decay with 7 km as the 1/e-producing scale factor.(reference: http://www.regentsprep.org/regents/math/algtrig/atp8b/exponentialresource.htm)

Anyway, hope this helps you start to get a grip on the problem!