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 10 ∘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 10 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=
yet another 8:01 final question
I tried in the next way:
first I integrated the relation dT/dz =-α
and I obtained:
T = -6.5z+288 (T in kelvin)
then, integrating this relation:
dP/P = -Mg/RT dz
I obtained:
P(final)=P(initial)*e^((-Mgz)/RT)
But here, I don't know which values I must replace for Mg (Molecular weigh) and also I am not sure if the expression is correct.
???
Damon some help please
some help....anybody???
I obtained:
a) 0.86 atm
b) 5503.4 Pa
c) 7%
But all are wrong and I don't know why!!!
And I have only one submission more.
Please help!!!
To calculate the atmospheric pressure at a specific altitude, we need to consider the linear temperature drop and the assumptions given. Let's solve each part of the question step by step.
(a) To calculate the atmospheric pressure (p) at 10 km above sea level, we need to use the ideal gas law:
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.
At a constant temperature, the ideal gas law can be rearranged as:
p = (nRT) / V.
Since we have the molecular weight of air, we can calculate the number of moles (n) using the given volume and pressure. Then substitute it into the equation to find p.
First, let's convert 10 km into meters:
z = 10 km = 10,000 m.
Next, we need to find the temperature (T) at 10 km above sea level. Given that the temperature drops linearly with altitude up to 11 km, we can use the formula:
dTdz = -α.
where dTdz is the rate of temperature change with altitude (gradient) and α = 6.5 K/km.
Since the temperature is constant between 11 km and 20 km, we can assume it is the same as the temperature at 11 km. Thus,
T = T0 + α * (z - z0) = 10 °C + 6.5 K/km * (10 km - 0 km).
Now, we have the temperature and can calculate the pressure using the ideal gas law:
p = (nRT) / V.
Input the values:
p = [(mass of air) / (molecular weight of air)] * R * T / V,
where:
mass of air = density of air * volume,
density of air = pressure / (RT), and
volume = A * z (A is the cross-sectional area).
Since the pressure at sea level is 1 atm = 1.01325 × 10^5 N/m^2, we can use this value and the given temperature and pressure at 10 km above sea level to find the atmospheric pressure at that altitude.
(b) To calculate the minimal force (Fmin) per square meter that the cabin walls have to sustain, we need to consider the pressure difference between the inside and outside of the cabin.
Given that the cabin is pressurized to 0.8 atm at cruising altitude, the force per square meter can be calculated using the formula:
Fmin = (p - p0) * A,
where p is the pressure inside the cabin, p0 is the atmospheric pressure at that altitude (calculated in part (a)), and A is the area of the cabin walls.
(c) To find the magnitude of the percentage change in volume of the air inside the bottle when it is brought to sea level, we can use the ideal gas law:
p1V1 / T1 = p2V2 / T2,
where p1 and T1 are the pressure and temperature inside the cabin at cruising altitude, and p2 and T2 are the pressure and temperature at sea level. The volume of air inside the bottle is V1.
We need to find V2, the volume at sea level, and then calculate the percentage change using the formula:
∣∣∣ΔV / V1∣∣∣ * 100, where ΔV = V2 - V1.
Substitute the given values into the equation to find the magnitude of the percentage change in volume.