A passenger flying in an airplane at 11 km above sea level tightly closes the em

Question

A passenger flying in an airplane at 11 km above sea level tightly closes the empty water bottle. Upon landing in Miami, near the beach, he notices that the bottle shrunk, and its inside volume is only 2/3 of its original volume. What was the pressure of the plane cabin when the bottle was closed? What was the pressure just outside of the plane in flight? If the pressures inside and outside the flying plane are different, how do you think the pressure difference in the plane is obtained and maintained? Assume that T=25C and remains constant. To determine the composition of air at 11 km above sea level and the total pressure outside the plane flying at 11 km above sea level, use the information given in problem 3.

g utminers.utep.edu/lvukovic/teach/Hw1.pdf CHEM 342 or a gas mixture in a gravity field, it can own that of the gases obeyS distribution law independent of the others. For each gas, P poexp-Mgz/RT] where p, is the partial pressure of the ith gas in the mixture at the height z, pio is the partial pressure of the gas at ground level, and M, is the molecular weight of the gas. The approximate composition of the atmosphere at sea level is given in the table below Mole p(atm) at percent 50km Gas Mole Mole Pi(atm) at 100km percent at 50km at 100km 78.09 20.93 0.93 Nitrogern Oxygen Ar Carbon Dioxide0.03 Neon Helium 0.0018 0.0005 0.0001 5x1 8x10 5x 10 on en Xenon ne Total Ignoring the last four components, compute the partial pressure of the others, the total pressure, and the composition of the atmosphere in mole percent, at altitudes of 50 and 100 km (assuming t-25°C). Check out the oxygen! Tot h C 4. Calculate the pressure at 10 km above the sea level, and at 10 km below the sea level (10 km below the water surface). (Hints: the derivation done in class should be helpful 2:06 PM O Type here to search [D 9/14/2017 14

Explanation / Answer

for same temperature conditions
from ideal gas law
P1V1 = P2V2

now, in the airplane
let the pressure be P1
volume V1 = V

then outside
pressure = P2 = 1 atm
V2 = 2V/3

so comparing
P1*V = 1*2V/3
P1 = 2 / 3 atm is the pressure inside the cabin

at 11km altitude, pressure is given by Po = 0.2233 atm
where as pressure inside the cabin = 0.66 atm

so pressure difference in the plane is maintained by employing the structural strength of the plane, which takes care of the extra force created from the hogh pressure inside the cabin

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