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Environmental Chem Homework

  1. Jan 31, 2015 #1
    1. The problem statement, all variables and given/known data
    At an altitude of 50 km, the average atmospheric temperature is 0C. what is the average number of air molecules per cm^3 of air at this atmosphere?

    Given formula:
    P(sub h) =
    P(sub 0) x e^(-Mgh/RT) =
    -(Mgh x 10^5)/2.303RT
    Assuming P = 1 atm at sea level and h in km

    M=28.97g/mol=28.97cm^3/mol
    g = 9.81 m/s^2 = .0981 km/s^2
    h = 50 km
    R = 8.314 J/K•mol =
    82.06cm^3•atm/K•mol
    T = 0 C = 273.15 K

    I'm having trouble because using the scale height form of the equation (second form) I get lost in the units, and I'm receiving an error on my TI using the first.

    3. The attempt at a solution

    Using second equation:


    Log P(sub h) =
    (-Mgh x 10^5)/2.303RT =
    -(28.97)(.0981)(50)(10^5)/
    (2.303)(82.057)(273.15) =
    -275.28
    10^(-275.28) = 5.23... But the units after the conversion don't add up ... Suggestions?
     
  2. jcsd
  3. Jan 31, 2015 #2
    *5.23x10^-276
     
  4. Feb 1, 2015 #3

    SteamKing

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    Whew! What a dog's breakfast of units! Let's see if we can clean this mess up a bit.

    R can be expressed in a variety of different units:

    http://en.wikipedia.org/wiki/Gas_constant

    -Mgh/RT should be dimensionless. Unfortunately, you picked km as your distance unit, which really doesn't fit in with your units for R. Meters would be the better choice.

    The molar mass M for air should be expressed in proper units, which is kg/mol. It's not clear how you obtained M = 28.97 cm3/mol.

    You use the formula:

    P(sub h) =
    P(sub 0) x e^(-Mgh/RT) =
    -(Mgh x 10^5)/2.303RT

    where did the last part -(Mgh x 10^5)/2.303RT) come from? That's not equivalent to raising the number e to the power of (-Mgh/RT).

    See what you can do to fix this and we can talk again.
     
  5. Feb 1, 2015 #4
    Okay so, this second equation was derived from the first using an 8 km scale height and is true under the conditions of atm pressure units, according to my textbook, that's whereni acquired all of my information, I've attached a photo of the page, maybe you can see what I'm not understanding... Or rather misunderstanding. image.jpg
     
  6. Feb 1, 2015 #5
    The site autorotated the photo, here's an adjustment.
     

    Attached Files:

  7. Feb 1, 2015 #6

    SteamKing

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    The textbook image answers a lot of questions.

    log (Ph) = (-Mgh x 105) / 2.303 RT, where log here is the base-10 log, not the natural log. ln (10) ≈ 2.303

    The factor 105 converts altitude in kilometers into altitude in centimeters.

    The molecular weight of air M = 28.97 grams / mol, as specified in the text.
    g = 981 cm / s2, as specified in the text
    R = 8.314 * 107 erg / K-mol, as specified in the text.

    If we use h = 8 km as the scale height, then the pressure at this altitude is

    log (Ph) = (-28.97 * 981 * 8 x 105) / (2.303 * 8.314 * 107 * 288)

    log (Ph) = -0.4124

    Ph = 10-0.4124

    which means Ph = 0.3869, or about 39% of the pressure at sea level, which is as described in the text.

    Now, you should be able to work the original HW problem for an altitude of 50 km.
     
  8. Feb 1, 2015 #7
    Thank you!!!
     
  9. Feb 1, 2015 #8
    Log P(sub h) =
    (-28.97*981*50*10^5)/
    (2.303*[8.314x10^7]*273.15)
    LogP(sub h) = -2.72
    p(sub h) = 0.00192 atm

    n=PV/RT=(.00192atm*1cm^3)/ (273.15K*82.057cm^3atm•K*-1•mol^-1)

    n = 8.566 x 10^-8 mol
    = 5.16 x 10^16 molecules,

    Yes?
     
  10. Feb 1, 2015 #9

    SteamKing

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    Your calculation seems to be OK.
     
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