# PHYSICS

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=

1. 👍 0
2. 👎 0
3. 👁 114
1. 8:01 final question

1. 👍 0
2. 👎 0
posted by Damon
2. It has finished.

1. 👍 0
2. 👎 0
3. Thank heavens !

1. 👍 0
2. 👎 0
posted by Damon
4. 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

1. 👍 0
2. 👎 0
posted by Damon
5. 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.

1. 👍 0
2. 👎 0
posted by Damon
6. 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.

1. 👍 0
2. 👎 0
posted by Damon
7. dP/dh = .29 n = .29 P /(RT)
( We better start doing this in meters, not km of course )
dP/P = [.29/(RT)] dh

1. 👍 0
2. 👎 0
posted by Damon
8. 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

1. 👍 0
2. 👎 0
posted by Damon
9. 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)

1. 👍 0
2. 👎 0
posted by Damon

## Similar Questions

1. ### Physics Classical Mechanics Help ASAP

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

asked by Anonymous on January 10, 2014
2. ### PHYSICS(HELP)

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

asked by Anonimus on January 12, 2014
3. ### Classical mechanics

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

asked by JuanPro on January 10, 2014
4. ### Physics

A 0.1134kg spherical meteor of diameter 3 cm enters the earth’s atmosphere at an elevation of 121920m with a velocity of 13411.2 m/s. The meteor’s temperature is -73.15˚C. The specific heat at constant volume (Cv) of the

asked by Jack on November 28, 2016
5. ### ENGLISH

Before class, you should look over your notes from the last(9)lecture and take a minute to speculate about what your instructor is going to talk about today. (9) a) No change b) lecture, and c) lecture; and I think it need a comma

asked by nancy on August 4, 2010
6. ### English

In our text discussion, we compared a two-way conversation with a top-down situation, such as that which students encounter when listening to a lecture. What was this comparison intended to illustrate? A. The need to pay attention

asked by clary on April 30, 2015
7. ### fluid mechanics

a gas well containd hydrocarbon gases with an average molecular weight of 24, which can be assumed to be an ideal gas with a specific heat ratio 1.3. The pressure and temperature at the top of the well are 250psig and 708F,

8. ### english

PLEASE CORRECT PUNCTUATION ONLY. THANK YOU! Re: Evans vs. Brown I tried to call you about the above case but learned that you were out of town, and will not return until Wednesday. You wrote to me on June 1 to ask what authority I

asked by April on February 24, 2011
9. ### Thermo

In an isothermal reaction (either reversible or irreversible), DELTA T is constant..hence zero since temperature remains constant, can we also assume that heat is also zero? if so, how then is an isothermal reaction different from

asked by Kim P on March 12, 2010
10. ### english

Please Help correct punctuation & spelling! Thank you! I tried to call you about the above case, but learned that you were out of town and will not return until Wednesday. You wrote to me on June 1 to ask what authority I had or

asked by April on February 27, 2011

More Similar Questions