The pilot of an airplane reads the altitude 9,000 m and an absolute pressure of
ID: 2056885 • Letter: T
Question
The pilot of an airplane reads the altitude 9,000 m and an absolute pressure of 25 kPa when Hying over a city. Calculate the local atmospheric pressure in that city in kPa and in mm Hg. Take the density of air and mercury to be 1.15 kg/m3 and 13,600 kg/m3, respectively. The average temperature of the atmosphere in the world is approximated as a function of altitude by the relation: Tatm = 288.15 - 6.5z Where Tatm is the temperature of the atmosphere in K and z is the altitude in km with z = 0 at sea level. Determine the average temperature of the atmosphere outside an airplane that is cruising at an altitude of 12,000 m. Determine the density of air at 12 km if the pressure is known to be 1.8506 x 104 N/m2 Reminders: For air R = 287 J/(kg*K) P*V = R*T v = V/mExplanation / Answer
2) T atm = 288.15 - 6.5 z
For z = 12000 m = 12 km
T atm = 288.15 - 6.5*12 = 210.15 K.
From Gas law,
PV = nRT
PM = RT ; [M = Molecular wt. ; = Density.]
= PM/RT = 1.8506*10^4 * 29*10^-3 / 8.314 * 210.15
= 0.307164 kg/m3....
Molecular wt of air Mole fraction of N2 * Molecular wt of N2 + Mole fraction of O2 * Molecular wt of O2
= 0.79*28 + 0.21*32 = 28.84 29 g = 29*10^-3 kg [Bcoz N2 and O2 are the major constituent]
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