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Translational Energy

  1. Dec 19, 2009 #1
    1. The problem statement, all variables and given/known data
    1.2 x10^23 molecules of a monatomic ideal gas occupies the volume of 0.63 m3 at
    atmospheric pressure. Find the average translational kinetic energy of one molecule.

    2. Relevant equations

    PV = NRT

    K = 3/2*kB*T

    3. The attempt at a solution

    T = PV/NR

    But.. whats N(moles)?
  2. jcsd
  3. Dec 19, 2009 #2


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    Homework Helper

    One mole of anything = 6.03*10^23 molecules of that thing. BTW, you can also use PV=nkT. That way, you can directly plug in 1.2 x10^23 into "n".
  4. Dec 19, 2009 #3
    In this equation usually moles: here 1.2/6.02= n. The 10^23 on top and bottom cancel.
  5. Dec 19, 2009 #4
    Ok so I get T = 39425 K :S
  6. Dec 19, 2009 #5


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    Homework Helper

    That's correct, although not realistic. 1.2*10^23 of a room-temperature gas would take up around 5 L.
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