In the lecture, we discussed the case of an isothermal atmosphere where the temperature is assumed to be constant. In reality, however, the temperature in the Earth's atmosphere is not uniform and can vary strongly and in a non-linear way, especially at high altitude. To a good approximation, the temperature T drops almost linearly with altitude up to 11 km above sea level, at a constant rate:
dTdz=−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 15 ∘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=
unanswered
(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=
unanswered
(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 20 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
∣∣∣ΔVV1∣∣∣×100=
To solve part (a) of the question, we need to calculate the atmospheric pressure at a height of 10 km above sea level using the linear temperature drop assumption.
We can start by finding the temperature at 10 km using the given temperature drop rate:
dTdz = -α = -6.5 K/km (since the temperature drop is linear and negative)
z = 10 km
The temperature drop can be calculated as:
ΔT = dTdz * Δz
ΔT = -6.5 K/km * 10 km = -65 K
Since the temperature at sea level (T0) is 15°C, we convert it to Kelvin:
T0 = 15°C + 273.15 = 288.15 K
The temperature at 10 km above sea level (T10) is then:
T10 = T0 + ΔT
T10 = 288.15 K - 65 K = 223.15 K
Now, let's use the ideal gas law formula to find the pressure, assuming the air is an ideal gas:
pV = nRT
Where:
p = pressure (unknown)
V = volume (not given, can be assumed constant)
n = number of moles of air
R = ideal gas constant (8.314 J/(mol*K))
T = temperature (223.15 K)
molecular weight of air ≈ 29 g/mol
We need to find the number of moles of air (n) at sea level:
n = mass / molecular weight
mass = 1 atm * seawater level area / (9.8 m/s^2) = 1.01325 × 105 N/m^2 * (10^4 m^2) / (9.8 m/s^2) = 1.033 kg
n = 1.033 kg / (0.029 kg/mol) ≈ 35.6 mol
Now, let's solve for the pressure (p):
pV = nRT
p = nRT / V
Since we're only interested in the atmospheric pressure at 10 km, we can assume that the volume at sea level and 10 km altitude is the same. Therefore, we can cancel out the volume (V) term and solve for p:
p = (nRT) / (V)
Substituting the values:
p = (35.6 mol * 8.314 J/(mol*K) * 223.15 K) / V
Since the volume (V) is canceled out, we don't need to know its value to calculate the pressure (p) at 10 km altitude. Therefore, we can simply provide the expression for the pressure in terms of the given variables.
The atmospheric pressure (p) at 10 km above sea level is given as:
p = (35.6 mol * 8.314 J/(mol*K) * 223.15 K) / V
This is the answer to part (a) of the question.