n 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=
(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 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=
8:01 final question
It has finished.
Thank heavens !
273 + 15 = 288 K at sea level
T = 288 - 6.5 z for z <11 km
when z = 11 km, T = 288-6.5(11) = 216.5 K
T remains at 216.5 K until z = 20 km
At z = 10 km
T = 288 - 65 = 223 K
The column of air above ground 1 meter in cross section weighs 1.01235 *10^5 N at sea level (given)
To get the weight of the column above z = 10 km we must either find the mass between z = 0 and z = 10 or from z = 10 to z = infinity.
Let's try from 0 to 10 km
To do this we need the density as a function of altitude.
that means we first need n/V in PV = nRT
n/V = P/RT
the weight of a cubic meter is then
W/V = n*.029 kg/mol*10 m/s^2 = .29n Newtons
Now at sea level how much does a cubic meter weigh ?
P = 1.01315*10^5
V = 1 m^3
T = 288 K
R = 8.314
so
n = PV/RT = 1.01315*10^5 /(8.314*288)
= 4.23 * 101 = 42.3 moles/m^3
W = .29 n = 12.27 Newtons
so my pressure decreases by 12.27 N/m^2 when I go up one meter
Of course I could repeat that process for ever meter up from ground to 10,000 meters but that would be tiresome.
what in general is my change in pressure per meter up? It is the weight of that cubic meter of gas.
That should get you started.
dP/dh = .29 n = .29 P /(RT)
( We better start doing this in meters, not km of course )
dP/P = [.29/(RT)] dh
Oh, that is negative of course
dP/dh = -.29 n = -.29 P /(RT)
( We better start doing this in meters, not km of course )
dP/P = -[.29/(RT)] dh
Unfortunately T is a function of h
if h is meters h = 10^3 z with z in km
T = 288 - 6.5 * 10^-3 h
so
dP/P = -[.29/(R(288-.0065h)] dh
integrate that
ln p + c = -(.29/.0065R) ln(.0065h-288)
or
ln p + c = ln -(.0065 h -288)^(.29/.0065R)
e^p = const e^ -(.0065 h -288)^(.29/.0065R)
p = constant(288-.0065h)^.29/.0065R
put in initial conditions of p and T
p = 1.01325 *10^5 (1 -.0065h/288)^(.29/.0065R)
(a) To calculate the atmospheric pressure, we can use the ideal gas law, which states:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
First, we need to calculate the number of moles of air at sea level. We can use the molecular weight of air to convert grams to moles:
molecular weight = 29 g/mol
mass of air = weight of 1 mole = 29 g
Next, we calculate the number of moles of air by dividing the mass by the molecular weight:
moles = mass / molecular weight
Now, let's express the volume in terms of the height above sea level:
V = A * h
where A is the cross-sectional area of the atmosphere and h is the height above sea level.
Using the given data, we can substitute the values into the equation:
P * A * h = nRT
Rearranging the equation to solve for pressure:
P = (nRT) / (A * h)
For the linear case, the temperature varies linearly with altitude below 11 km, so for z max of 10 km:
∆T = -α * ∆z
where ∆T is the change in temperature, α is the constant rate of temperature drop, and ∆z is the change in altitude.
Given that α = 6.5 K/km, ∆T = -6.5 K and ∆z = 10 km, we can calculate the new temperature at 10 km:
T = T0 + ∆T
Substituting the known values:
T = 15 °C + (-6.5 K)
Now we need to convert the temperature to Kelvin:
T = 15 + (-6.5 + 273)
Now we have the temperature at 10 km, which we can use to calculate the pressure:
P = (nRT) / (A * h)
(b) To calculate the minimal force, we need to consider the pressure difference inside the cabin and outside:
ΔP = P_outside - P_inside
In this case, P_outside is the atmospheric pressure at 10 km above sea level found in part (a) and P_inside is the pressure inside the pressurized cabin (0.8 atm).
The force per unit area (stress) F/A exerted on the walls of the cabin is given by:
F/A = ΔP = (P_outside - P_inside)
The minimum force F_min that the walls have to sustain is when the difference in pressures is maximum, which occurs at the highest altitude.
Therefore, the minimum force F_min per square meter is given by:
F_min/A = ΔP_max = (P_outside - P_inside)
We can now substitute the values of P_outside and P_inside:
F_min/A = ΔP_max = (P - P_inside)
(c) To find the magnitude of the percentage change in volume V1, we can use the combined gas law:
(P1 * V1) / T1 = (P2 * V2) / T2
where P1 and T1 are the pressure and temperature at cruising altitude (0.8 atm and 27 °C), and P2 and T2 are the pressure and temperature at sea level (1 atm and 7 °C). We need to find the new volume V2 at sea level.
Rearranging the formula:
V2 = (P1 * V1 * T2) / (P2 * T1)
Substituting the given values and solving for V2:
V2 = (0.8 atm * V1 * (7 °C + 273)) / (1 atm * (27 °C + 273))
The magnitude of the percentage change in volume is given by:
|ΔV/V1| * 100 = |(V2 - V1) / V1| * 100