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Determining pressure in Interstellar space

  1. Sep 18, 2011 #1
    1. The problem statement, all variables and given/known data
    Interstellar space, far from any stars, is filled with a very low density of hydrogen atoms (H, not H2). The number density is about 1atom / cm3 and the temperature is about 3 K. Estimate the pressure in interstellar space. Give your answer in Pa and in atm.


    2. Relevant equations
    Average translational kinetic energy per molecule: Eavg = 1.5KbT = .5mv2
    p = F/A = (1/3)(N/V)mVrms2
    Eth = 1.5nRT
    pV = nRT

    3. The attempt at a solution
    V = 10-6 m3
    T = 3K
    m = 1u = 1.66x10-27
    n = 1.66x10-24 (not sure if this is right?)

    pV = nRT --> p = 4.14x10-17
    Where am I going wrong, or is this correct?

    Thank you for any help/advice!
     
  2. jcsd
  3. Sep 18, 2011 #2

    D H

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    Different approach, I get the same number:

    [tex]
    \frac{10^6 \, \text{atom}} {\text{m}^3} \cdot
    \frac{1 \, \text{mole}} {6.0221415 \times 10^{23} \, \text{atom}} \cdot
    8.3144621 \, \text{J}/\text{K}/\text{mole} \cdot
    3 \, \mbox{K} \approx 4.142 \times 10^{-17} \, \text{pascal}
    [/tex]
     
  4. Sep 18, 2011 #3
    The only formula you need is PV = nRT
     
  5. Sep 18, 2011 #4

    D H

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    That is exactly one of the formulae that pdonovan, Spinnor.
     
  6. Sep 18, 2011 #5
    Thank you very much, that was the correct answer.

    Now, how would I got about finding Vrms?

    I know p = (1/3)(N/V)mVrms2
    So, p = 4.14x10-17
    N = 1
    V = 10-6
    m = 1u = 1.66x10-27
    And found Vrms = .86m/s, but this is incorrect. I think my m or v might be incorrect.
     
  7. Sep 18, 2011 #6

    D H

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    Your value for m is correct (assuming units of kilograms; always carry units around). Your math is wrong somewhere. Show your work.
     
  8. Sep 18, 2011 #7
    Now I have...

    4.14x10-17pa = (1/3)(1/10-6)mVrms2

    1.242x10-22 = mVrms2

    74819.28 = Vrms2
    Vrms = 275.64m/s

    So something is definitely wrong, because the atom should be moving very slowly.
     
  9. Sep 18, 2011 #8

    D H

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    Yeah, your answer is too large. About 2 m/s too large. sqrt(74819.28) is about 273.53.

    As a sanity check, you can always compute sqrt((3 * boltzmann's constant * 3 kelvin) / (1 amu)). You will get the same answer.
     
  10. Sep 18, 2011 #9
    Then which values are wrong in the Vrms = sqrt(3KbT / m) formula? I have T = 3 and if m = 1, then Vrms = 1.11 x 10 ^ -11 which is wrong. And if m =1660x10^-27g it is too big (around 8.6).
     
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