# PHYSICS

dT/dz=−α for z≤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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1. 8:01 exam question

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posted by Damon
2. not asking for solution, just tips and considering this is an open book exam

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3. The pressure is 0.2526 atm

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posted by Phy
4. what about b and c

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5. and if the temperature was 20∘C?

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posted by Greco
6. 0.3343

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posted by shaka
7. no that's my value for 15º

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posted by shaka
8. please guys can you tell formula because I have temp. 10 degree C.

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posted by asha
9. Can anyone please explain the logic of the problem, not a solution a hint.

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posted by Daoine
10. I found two types of formulas for such a problem - the first one is only for the case of isothermal atmosphere, the second one is for this case, in which the temperature is the function of the height - maybe that's why your solutions aren't correct

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posted by MM
11. Can someone tell the formula because I want to calculate by myself?

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12. Can someone teach us how to do b,c?

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posted by A
13. dT/dh= - ƴ/(ƴ-1) Mg/Nk with ƴ/(ƴ-1) as a constant is what I remember of the temperature as a function of time. Still I can't comprehend entirely the problem.

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posted by Daoine
14. I think we have to find out ƴ ant then use dp/p= ƴ/(ƴ-1) dT/T

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posted by Daoine
15. I used this one: p=p0*[T0/(T0+alpha*(h-h0))]^(g*M/(R*alpha))

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posted by MM
16. hey guys can you please tell the method for part a?

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posted by asha
17. Hi MM, what is h and h_0?

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posted by asha
18. sorry, didn't realize - h is z (height)

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posted by MM
19. I used p=p_0*e(-M*g*h/R*T) but its wrong..Any help pls..:(

where I used h=10km=10000m and temp.T= 273+10=283, M=0.029

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posted by asha
20. you cannot use it, because the temperature is not a constant

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posted by MM
21. Hi MM sorry to disturb you, but I just tried according to ur formula and still getting wrong. I used h-h_0= z= 10000m and T_0= 278, alpha=0.0065 and M=0.029 right?

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posted by asha
22. seems to be OK...really do not know where is the problem...

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posted by MM
23. Thank you for your help. I will try it again.

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posted by asha
24. Any hint for b? and c?

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posted by Roy
25. asha which was your pressure value at a?

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posted by Daoine
26. Hi Daoine, My answer is 0.3289 atm, and I have temp. 10 degree c. This time I afraid to check answer because I don't know part b and c too, so I don't have more attempt. Is it right for 10 degree temp?

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posted by asha
27. who has b & c?

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28. WHAT IS THE FORMULA??

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