The normal airflow over the Rocky Mountains is west to east. The air loses much

Question

The normal airflow over the Rocky Mountains is west to east. The air loses much of its moisture content and is chilled as it climbs the western side of the mountains. When it descends on the eastern side, the increase in pressure toward lower altitudes causes the temperature to increase. The flow, then called a chinook wind, can rapidly raise the air temperature at the base of the mountains. Assume that the air pressure p depends on altitude y according to p = p0 exp(-ay), where p0 = 1.00 atm and a = 1.15 10-4 m^-1. Also assume that the ratio of the molar specific heats is 4/3. A parcel of air with an initial temperature of 5.00

Explanation / Answer

p1 = po*e^(-a*y1) = 1*e^(-4087*1.15*10^(-4))

= 0.625 atm

p2 = po*e^(-ay2) = 1*e^(-1537*1.15*10^(-4))

= 0.8379 atm

p1V1^(gamma) = p2V2^(gamma)

ln(v2) = ln(p1/p2)/gamma

e^(v2) = e^ln(p1/p2)/gamma =

which gives V2 = 0.79

Now using the expression we get

T2 = 278/ (0.79)^(gamma - 1)

= 278/(0.79)^(1/3) = 278/0.9244 = 300.73 K

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