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March 3, 2015

March 3, 2015

Posted by **christina** on Saturday, September 5, 2009 at 8:47pm.

- college/earth science -
**drwls**, Saturday, September 5, 2009 at 10:00pm23 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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