Posted by christina on Saturday, September 5, 2009 at 8:47pm.
23 C = 300 K
The air density at the initial altitude (assuming sea level) is (29 g)/(24.6 *10^3 cm^3) = 1.18*10^-3 g/cm^3 = 1.18 kg/m^3
The pressure decrease after a 500 m elevation increase is
(density)*g*500 m = 5780 N/m^2
The pressure decrease factor is
(Po - 5780)/Po = 0.990, where
Po is the sea level atmospheric pressure.
Assume an adiabatic isentropic change in air temperature as the pressure decreases, with an air specific heat ratio (gamma) of g = 1.4.
(T2/T1) = (P2/P1)^[(g-1)/g]
= 0.990^(0.286) = 0.997
T2 = 299 K = 22 C
If you need a more complete explanation, see
http://farside.ph.utexas.edu/teaching/sm1/lectures/node56.html
I may have made a computing error somewhere along the way. You should verify the calculations
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