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 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=
eh eh eh honor code :D
but by being here are you not violating the honor code, my fellow civ5 player
I'm on edx on touring mode not going for the cert :). im using as a learning to to a final writen exam !
well what makes you think op isn't
got a point there !
(a) To calculate the atmospheric pressure at z = 10 km above sea level, we can use the ideal gas law and the relationship between pressure and temperature with altitude.
Since the temperature drop is linear, we can express the temperature at any height above sea level (z) as:
T = T0 - α*z
where T0 is the temperature at sea level (given as 5 degrees Celsius).
To convert the temperature to Kelvin, we add 273.15:
T(K) = 5 + 273.15 = 278.15 K
Using the ideal gas law:
p1V1 / T1 = p2V2 / T2
where p1 and T1 are the initial pressure and temperature (sea level conditions, given as 1 atm and 278.15 K) and p2 and T2 are the final pressure and temperature (at z = 10 km above sea level). We can assume the volume remains constant.
Substituting the values and rearranging the equation:
p2 = (p1 * T2) / T1
p1 = 1 atm
T2 = 278.15 - 6.5 * 10 = 212.65 K
T1 = 278.15 K
p2 = (1 * 212.65) / 278.15 = 0.765 atm
Therefore, the atmospheric pressure at z = 10 km above sea level is approximately 0.765 atm.
(b) To calculate the minimal force per square meter that the walls of the cabin have to sustain, we can use the difference in pressure between the inside and outside of the cabin.
The pressure inside the cabin is given as 0.8 atm, and the atmospheric pressure at z = 10 km above sea level (found in part a) is 0.765 atm.
The pressure difference is:
Δp = p_inside - p_atm = 0.8 atm - 0.765 atm = 0.035 atm
To convert the pressure difference to Pascals (N/m^2), we multiply by the conversion factor:
Δp = 0.035 atm * 1.01325 × 10^5 N/m^2/atm = 3551.875 N/m^2
Therefore, the minimal force (Fmin) per square meter that the walls have to sustain is approximately 3551.875 N/m^2.
(c) To calculate the percentage change in volume of the air inside the bottle when brought to sea level, we can use the combined gas law.
The combined gas law states:
(p1 * V1) / T1 = (p2 * V2) / T2
where p1 and T1 are the initial pressure and temperature (inside the cabin, given as 0.8 atm and 27 degrees Celsius) and p2 and T2 are the final pressure and temperature (sea level conditions).
The volume inside the bottle (V1) remains constant.
Substituting the values and rearranging the equation:
(p2 * V2) / T2 = (p1 * V1) / T1
(p2 * V2) / T2 = (0.8 atm * V1) / 27°C
Since the temperature at sea level is decreased by 15 Kelvin with respect to the cabin's temperature:
T2 = 27°C - 15°C = 12°C = 12 + 273.15 K = 285.15 K
Substituting the values:
(p2 * V2) / 285.15 K = (0.8 atm * V1) / 27°C
To find the percentage change in volume, we can rewrite the equation as:
ΔV/V1 = (V2 - V1) / V1 = ((p2 * V2) / (285.15 K * V1)) - 1
Rearranging the equation:
ΔV/V1 = (0.8 atm * V1 - 285.15 K * V1) / (285.15 K * V1)
Simplifying further:
ΔV/V1 = (0.8 - 285.15 / 285.15) = -0.9972
Taking the absolute value and multiplying by 100 to obtain the magnitude of the percentage change:
∣∣∣ΔV/V1∣∣∣ × 100 = ∣∣∣-0.9972∣∣∣ × 100 = 99.72%
Therefore, the magnitude of the percentage change in volume of the air inside the bottle when it is brought to sea level is approximately 99.72%.