dT/dz=−α forz≤11 km
where α=6.5 K/km (Kelvin per km) and z is the height above the sea level. The temperature stays then approximately constant between 11 km and 20 km above sea level.
Assume a temperature of 5 ∘C and a pressure of 1 atm at sea level (1 atm = 1.01325 ×105 N/m^2). Furthermore, take the molecular weight of the air to be (approximately) 29 g/mol. The universal gas constant is R=8.314 JK−1mol−1 and the acceleration due to gravity is g=10 m/s2 (independent of altitude). Assume that air can be treated as an ideal gas.
(a) Under the assumptions above, calculate the atmospheric pressure p (in atm) at z= 10 km above sea level for the case of a linear temperature drop.
p=
(b) The cruising altitude of a commercial aircraft is about 33'000 ft (or 10 km). Assume that the cabin is pressurized to 0.8 atm at cruising altitude. What is the minimal force Fmin (in Newton) per square meter that the walls have to sustain for the cabin not to burst? Use the atmospheric pressure found in (a).
Fmin=
(c) We close a plastic bottle full of air inside the cabin when the aircraft is at cruising altitude of z= 10 km. The volume of the bottle is V1, the pressure and temperature inside the cabin are 0.8 atm and T1=27 ∘C, respectively. Assume that at sea level the atmospheric pressure is 1 atm, and the temperature is decreased by 15 Kelvin with respect to the cabin's temperature.
What is the magnitude of the percentage change in volume of the air inside the bottle when it is brought to sea level? (Enter the magnitude of the percentage change in volume in
∣∣∣ΔV/V1∣∣∣×100=
8:01 exam
To solve these problems, we will use the ideal gas law and the formula for pressure variation with respect to altitude.
(a) To calculate the atmospheric pressure at 10 km above sea level, we need to integrate the pressure variation formula from sea level to 10 km. The given formula for pressure variation is:
dp/dz = -α
Integrating both sides with respect to z and substituting the values:
∫dp = -α ∫dz (from 0 to 10 km)
p - p0 = -α * z (where p0 is the pressure at sea level)
Now we can substitute the given values:
p - 1 atm = -6.5 K/km * 10 km
p = 1 atm - 65 atm
p = -64 atm
As pressure cannot be negative, we need to approximate it to zero (since the problem mentions a temperature drop, which would decrease the pressure). Therefore, the atmospheric pressure at 10 km above sea level is approximately 0 atm.
(b) To calculate the minimal force per square meter (Fmin) that the cabin walls have to sustain, we need to multiply the pressure difference between the cabin and outside air with the area.
Pressure difference:
Δp = 0.8 atm - 0 atm (from part (a))
Δp = 0.8 atm
Area:
A = 1 m^2 (assuming 1 square meter)
Force:
Fmin = Δp * A
Fmin = 0.8 atm * 1 m^2
Fmin = 0.8 atm * 1.01325 × 10^5 N/m^2
Fmin = 8.106 × 10^4 N/m^2
Therefore, the minimal force per square meter that the walls have to sustain is approximately 8.106 × 10^4 N/m^2.
(c) To calculate the percentage change in volume of the air inside the bottle when brought to sea level, we will use the combined gas law.
The combined gas law is given by:
(P1 * V1) / (T1 + 273.15) = (P2 * V2) / (T2 + 273.15)
Where:
P1 = initial pressure inside the bottle = 0.8 atm
V1 = initial volume of the air inside the bottle = V1 (given)
T1 = initial temperature inside the cabin = 27 °C = 27 + 273.15 K
P2 = atmospheric pressure at sea level = 1 atm
V2 = final volume of the air inside the bottle (to be determined)
T2 = final temperature at sea level = T1 - 15 K
Rearranging the equation and substituting the given values:
V2 = ((P1 * V1 * (T2 + 273.15)) / (P2 * (T1 + 273.15)))
Substituting the given values:
V2 = ((0.8 atm * V1 * (27 + 273.15 - 15)) / (1 atm * (27 + 273.15)))
Simplifying the equation:
V2 = (6.88 * V1)
The magnitude of the percentage change in volume is given by:
∣∣∣ΔV/V1∣∣∣×100 = ∣∣∣(V2 - V1) / V1∣∣∣×100
Substituting the values:
∣∣∣ΔV/V1∣∣∣×100 = ∣∣∣((6.88 * V1) - V1) / V1∣∣∣×100
∣∣∣ΔV/V1∣∣∣×100 = ∣∣∣(5.88 * V1) / V1∣∣∣×100
∣∣∣ΔV/V1∣∣∣×100 = ∣∣∣5.88∣∣∣×100
∣∣∣ΔV/V1∣∣∣×100 = 588%
Therefore, the magnitude of the percentage change in volume of the air inside the bottle when brought to sea level is 588%.