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 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 20 ∘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=

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  1. This is an easy MITX 8.01 final exam problem, it is a shame that you have no honor and found it necessary to cheat

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  2. First Solve for the function T(z)

    T = (-alpha)*z + T(initial)
    T = -6.5z+293

    Now plug that in the hydrostatic pressure equation

    dP/P = (-M*g/RT)dz

    integrate that

    and you will get

    P(final) = P(initial)*(1-(alpha*z/293))^(Ma*g/(R*alpha))

    Part B is easy, get the answer from part a then subtract, don't forget to convert to Newtons per m^2 or pascal cause the given is in atm.

    part C. use the following relationship:

    P1V1/T1 = P2V2/T2 and you are good

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  3. have you solve the problem?

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  4. In the equation of P(final), what does ''a'' mean???
    Because integrating the first relation, I obtained:

    P(final) = P(initial)*e^(-Mgz/RT)

    and what is the value of M ???

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    posted by Hawk
  5. Please some help with this exercise....

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    posted by dew
  6. the value of m isn't the molecular weight of the gas? in this case 29g/mol?

    Somebody with the steps, I can't understand this one!

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    posted by koala
  7. Anonymous, I can't solve the problem.....

    Please some hint.

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    posted by Hawk
  8. some help....anybody????

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    posted by Hawk
  9. Please help with this exercise, is the only one that I need......please.

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    posted by fred
  10. 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!!!

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    posted by fred
  11. c) 24 percent

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    posted by abc
  12. fred
    for temp of 5 degrees celsius
    b) 55465.76
    c) 24 percent

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    posted by abc
  13. Thanks a lot abc, and what about a) and b)

    I did it again and now I obtained

    a)0.83 atm
    b)3061.5 Pa

    But it's my last chance and I am not sure......

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    posted by fred
  14. P(final) =P(initial)*(1-(alpha*z/293))^(M*g/(R*alpha))

    green tick

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    posted by Greco
  15. but what is the answer for b and in this problem
    a)po=0.263

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    posted by Greco
  16. b and c

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    posted by Greco
  17. some help....anybody????

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    posted by Greco
  18. c) 25.33 percent

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    posted by vivipop
  19. If you have a)
    you only need to rest from b)

    ..... but I am continue suffering with a)
    did you replace all the values without converting to other unid system???

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    posted by fred
  20. i need b and c for the problem above (20oC etc) as it is i have only one left
    i found a ) p=0.263

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    posted by Greco
  21. give your values and i ll tell you the answer

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    posted by Greco
  22. fred to the same system but be careful with the units, give me temperatutr and i ll tell you a)

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    posted by Greco
  23. it is midnight here in Greece and i'm feeling realy tired ....last call for help for b and c as i have only one try left..

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    posted by Greco
  24. for a temperature of 20c, a)0.260269. b)54688.24358 and c) 25.33%

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    posted by vivipop
  25. and vivipop rescues the day (or better night!)

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    posted by Greco
  26. thx a lot vivipop!!!

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    posted by Greco
  27. fred
    h=10km=10000m and temp.T= 273+20=293, M=0.029 alpha=0.0065 and M=0.029
    and the formula for a) is
    P(final) = P(initial)*(1-(alpha*z/293))^(Ma*g/(R*alpha))

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    posted by Greco
  28. glad i could help :)

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    posted by vivipop
  29. P(final) = P(initial)*(1-(alpha*z/293))^(M*g/(R*alpha))

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    posted by Greco
  30. Greco, thanks a lot!!

    I obtained this results for the problem above.

    a) 0.26
    b) 54715.5
    c) I am not sure

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    posted by fred
  31. Goodnight and thank you all for your help !!!

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    posted by Greco
  32. Thanks a lot!!!

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    posted by fred
  33. what if temp = 15 C, plz help.

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    posted by gordan
  34. a) 0.256

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    posted by Greco
  35. ok, thx. i understand a. what about for b and c? i am stuck on those.

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    posted by gordan
  36. nvmd got it! thx for ur help. :D

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    posted by gordan

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