help on either part will be greatly appreciated and i will ratelifesaver. a.) Us
ID: 685925 • Letter: H
Question
help on either part will be greatly appreciated and i will ratelifesaver.a.) Use the barometric relation ph=po(-Mgh/RT) to calculate the air prssure at aheight of 7,500 ft (1 ft =30.48 cm) assuming that the temperatureremains constant at 288K. Take air to have an average molecularweight of 28.8 g mol-1 , g = 9.807 m s-2 ,and the atmospheric pressure at sea level (h=0) po=1.01325 x 105 Pa.
b.) The warm "Santa Ana" winds in Southern California develop whenair at a high altitude undergoes compressional heating as it movesto a lower level at higher pressure. Assume that the processcorresponds to a reversible adiabatic compression, and that airbehaves as a perfect gas with a molar heat capacity Cp,m = (7/2)R = 29.1 J K-1mol-1 .With the aid of your result in part (a) for the initial pressurePi at 7,500 ft and using an initial temperatureTi of 288K, calculate the final temperatureTf that would occur after the air mass moves into thecoastal plain at sea level with a final pressure Pf of1.01325 x 105 Pa.
Explanation / Answer
a) The barometric relation you have given is wrong (just put in h=0and confirm that po is not equal to 1 ). It should however be ph=poe(-Mgh/RT) R = 8.31432 N·m / (mol·K) (it's the universalgas constant) Always use SI-units when using formulas such as this one, so firstwe convert the values that aren't yet expressed in SI-units. 7500 ft = 7500*0.3048 m = 2286 m 28.8 g mol-1 = 0.0288 kg mol-1 Then you put everything into the formula: ph=101325*e(-0.0288*9.807*2286/(8.31432*288)) This gives you ph = 77377.2133 Pa
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