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=
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shaka - I took 8:01 as a course 8 major in 1955 but have no idea what 8:01x is. I could easily tell by the nature of the questions exactly where they were all coming from but what does the "x" mean? Is it an online version?
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I had a suspicion something like that was happening. Thank you.
To solve these problems, we will use the Ideal Gas Law, which states:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the universal gas constant, and T is the temperature in Kelvin.
For part (a):
Given:
dT/dz = -α
α = 6.5 K/km
z = 10 km
To find the atmospheric pressure, we can use the barometric formula:
P = P₀ * e^(-Mgz/RT)
where P₀ is the pressure at sea level, M is the molar mass of air, g is the acceleration due to gravity, R is the gas constant, and T is the temperature.
First, let's convert the given atmospheric pressure at sea level from 1 atm to Pascals (Pa). Since 1 atm = 1.01325 ×10^5 N/m^2, we have:
P₀ = 1.01325 ×10^5 N/m^2 = 1.01325 ×10^5 Pa
Next, we need to find the molar mass of air. The molecular weight of air is approximately 29 g/mol, so:
M = 29 g/mol
Now, we can calculate the pressure at z = 10 km:
P = P₀ * e^(-Mgz/RT)
P = (1.01325 ×10^5 Pa) * e^(- (0.029 kg/mol) * (10 m/s^2) * (10000 m) / (8.314 J/(K mol)) * (273.15 K + 5 °C))
P = (1.01325 ×10^5 Pa) * e^(- 2.86457)
P ≈ 0.717 atm
So, the atmospheric pressure at z = 10 km is approximately 0.717 atm.
For part (b):
Given:
Pressure inside the cabin (P_cabin) = 0.8 atm
Atmospheric pressure at z = 10 km (P_10km) = 0.717 atm
To find the minimal force per square meter that the walls have to sustain, we need to consider the pressure difference between the cabin and exterior.
ΔP = P_cabin - P_10km
ΔP = 0.8 atm - 0.717 atm
ΔP = 0.083 atm
The force per square meter is given by:
F = ΔP * A
where A is the area of the walls. Since the area is not given, we cannot calculate the exact value for Fmin without this information. However, we can provide you with the formula and steps on how to calculate it once you have the area.
For part (c):
Given:
Pressure inside the cabin (P_cabin) = 0.8 atm
Temperature inside the cabin (T_cabin) = 27 °C
Temperature decrease at sea level (ΔT) = -15 K
To find the percentage change in volume, we can use the ideal gas law assuming constant pressure:
V₁/T₁ = V₂/T₂
where V₁ and T₁ are the initial volume and temperature, and V₂ and T₂ are the final volume and temperature.
First, let's convert the temperature to Kelvin:
T₁ = 27 °C + 273.15 K = 300.15 K
Next, let's calculate T₂ at sea level:
T₂ = T_cabin + ΔT
T₂ = 300.15 K + (-15 K) = 285.15 K
Now, let's calculate the percentage change in volume:
ΔV/V₁ = (V₂ - V₁)/V₁
Since pressure is constant, we can rewrite the equation as:
ΔV/V₁ = ΔT/T₁
ΔV/V₁ = (285.15 K - 300.15 K)/300.15 K
ΔV/V₁ = -0.04995
Lastly, let's calculate the magnitude of the percentage change in volume:
|ΔV/V₁| × 100 = |-0.04995| × 100
|ΔV/V₁| × 100 ≈ 4.995%
Therefore, the magnitude of the percentage change in volume when the bottle is brought to sea level is approximately 4.995%.