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Average number of air molecules.

  1. Oct 25, 2011 #1
    1. The problem statement, all variables and given/known data

    Hey guys. Hope you can help me out on this one. so heres the question:

    At an altitude of 50km the average atmospheric temperature is 0^oC. What is the average number of air molecules per cubic centimeter at this altitude.


    2. Relevant equations


    M = 28.97 g mol^-1
    g = 9.81 ms^-2
    R = 8.314 J K^-1 mol^-1
    Ph = pressure at any given height
    P0 = pressure at sea level

    Ph = P0e^-Mgh/RT


    3. The attempt at a solution

    Ph = P0e^-Mgh/RT

    p @ 50km = 75.944Pa

    75.944 x 1 x 10^-6
    Ph = -------------------
    8.314 x 273

    = 3.346 * 10^-8. And then I need to divide this by 6.022 * 10^23

    = 5.556 * 10^-32. Is this correct?

    Cheers :)
     
  2. jcsd
  3. Oct 26, 2011 #2

    Borek

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    Staff: Mentor

    What are you calculating here? Ph? That is - pressure?
     
  4. Oct 26, 2011 #3
    Ph = pressure at any given height.
    M = 28.97 g mol^-1
    g = 9.81 ms^-2
    R = 8.314 J K^-1 mol^-1
    P0 = pressure at sea level

    for

    ....... 75.944 x 1 x 10^-6
    Ph = -------------------
    ...........8.314 x 273
     
  5. Oct 26, 2011 #4

    Borek

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    So what is 75.944?
     
  6. Oct 26, 2011 #5
    I did write down pressure at 50km = 75.944 Pa. But this is incorrect. I may have used the equation n = PV but n = number of moles and would be useless, wouldn't it?
    ---
    RT

    Ok. Ive just found another equation in my handout I have been given. It states that "if air pressure at sea level is taken to be 1atm then the equation simplifies to:

    log(Ph) = -[Mgh*10^5/2.303RT]

    so using this equation

    log(Ph) = - [(29.97*9.81*50000*10^5)/(2.303*8.314*273)]

    log(Ph) = -[1.47*10^12)/(5227.169)]

    log(Ph) = -[281222971]

    Ph = 10^-281222971

    Ph = 0

    This is obviously not correct :/ did I miss out a conversion factor?
     
  7. Oct 26, 2011 #6

    Borek

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    And then you used this pressure again to calculate pressure Ph? That's what you wrote.

    Note - using your equation and numbers I didn't get the same 75.944 for the pressure at 50 km.

    Would it? Knowing number of moles in cm3 you can easily calculate number of molecules. You almost did it using Avogadro's number, you just got it the wrong way.

    As fast as I can tell you missed one and added one that is not necessary. 28.98 is in grams, and I don't understand what is 105. Besides, now your pressure is in atm, previously it was in Pa. It is hardly a simplification, it is just a unit conversion (or a lousy wording in the handout).
     
  8. Oct 26, 2011 #7
    Surely I must have got that the wrong way round of P0 and Ph. Which equation do I use? Do I use "log(Ph) = -[Mgh*10^5/2.303RT]" or "Ph = P0e^-Mgh/RT" I have attached a copy of the handout for you to have a little look at for that last equation. I noticed the teacher took log instead what I think should be ln.

    Using n=pV I don't have a value for V. So that wouldn't work.

    Thanks for all of your help Borek. Your a real help!
     

    Attached Files:

  9. Oct 26, 2011 #8

    Borek

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    Doesn't matter which equation you use - in both cases the final result will be the same. Use whichever equation you want to calculate the pressure at 50km, that's your step number one. This is simple plug and chug, nothing more complicated than keying several numbers into your calculator.

    No such equation as n=PV, more like n=PV/RT. Yes, you know the volume. You are asked about a number of molecules in 1 cm3.
     
  10. Oct 26, 2011 #9
    right.

    using n=PV/RT

    V=1cm^3 = 1*10^-6m^3
    R = 8.314 JK-1mol-1
    T = 273K
    P @ 50km = ?

    Have to find Ph

    Plug in values into

    Ph = P0e^-Mgh/RT

    Ph = 101325e^-(28.97*9.81*50000/8.314*273)

    I keep getting a value of 0 for this.
     
  11. Oct 26, 2011 #10

    Borek

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    I already told you were the problem is - please reread my earlier posts.
     
  12. Oct 26, 2011 #11
    I know you said that 28.97 g mol-1 is not in its SI units? I dont know what to convert it to. Also the equation is Ph = P0e(-Mgh/RT) and not Ph = P0e^-Mgh/RT isn't it?
     
  13. Oct 26, 2011 #12

    Borek

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    Everything else is in kg, so molar mass must be in kg/mol as well.

    No idea what you mean - please elaborate on what e(-x) is.

    Looks to me like you should brush up your math and units skills, otherwise you will be completely lost.
     
  14. Oct 26, 2011 #13
    right I understand the kg part. And the equation i mean is it e to the power or e-. ie e^(-x) or e(-x).

    Edit: I will be back on later tonight.
     
    Last edited: Oct 26, 2011
  15. Oct 26, 2011 #14

    Borek

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    It still doesn't make sense. e to power -x is e-x. You are suggesting you see two different things, but I see only one.
     
  16. Oct 29, 2011 #15
    What I wrote seems fairly understandable to me. But I emailed the teacher and it is e^(-x) so ignore that one.

    Ok so the answer I have is 193.544.

    Then

    n=PV/RT

    n= (193.544*1)/(8.314*273)

    n= 0.086 air molecules per cubic centimeter of air @ 50km?
     
  17. Oct 29, 2011 #16

    Borek

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    No.

    Watch your units - you put volume as 1, but 1 of what?

    Then, what you are trying to calculate is number of moles, this is not yet number of molecules.
     
  18. Oct 29, 2011 #17
    Ah I want volume in m^3 yet it is in cm^3. so thats a factor of 10^-6.

    so 1cm^3 = 1*10^-6 m^3

    n= (193.544*1*10^-6)/(8.314*273)

    n= 8.53 * 10^-8

    So to work out number of molecules I have to times the number of moles by 6.022x10^23

    number of molecules = 8.53 * 10^-8 * 6.022x10^23 = 5.135*10^8 number air molecules per cubic centimeter of air @ 50km
     
  19. Oct 30, 2011 #18

    Borek

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    This is the right approach, but I am not convinced if your units are correct - and if they aren't, your answer is still orders of magnitude off. You ignore units all the time, so they are hard to follow and I am not going to browse the thread back to try to decipher them.

    As a general rule: the final step of solving a question should be always check if the units of the answer are correct.
     
  20. Oct 30, 2011 #19
    Ok. Thanks for all of your help
     
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