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Total translational kinetic energy

  1. May 11, 2008 #1
    [SOLVED] Total translational kinetic energy

    1.A 0.03m3 vessel contains helium (monatomic) gas at 0.0C and 1.00 atm. The total translational kinetic energy of the gas molecules is (in KJ).

    2. 3/2KbT


    3. pV=nRT

    where p is the pressure, V is the volume, n is the number of molecules present, R is the gas constant (8.31J/(mol*K)), and T is the temperature in Kelvins (273K = 0ºC)
    The other equation is that the average translational kinetic energy K of a single molecule is

    K = (3RT)/(2N)

    where R and T are from the first equation and N is Avogadro's number (6.022E23).
    Just sub in numbers:
    K = (3 * 8.31 * 273) / ( 2 * 6.022E23) = 5.651E-21 Joules. This is the kinetic energy of one atom of helium at 0º C. Change the first equation around to get n = (pV) / (RT) and then multiply 5.65E-21 by n


    The answer in the book shows its suppose to be 4.5kJ, but I'm yet to get that, I'm getting a way off number. Thanks for any help given :+)
     
  2. jcsd
  3. May 11, 2008 #2

    hage567

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    Homework Helper

    This doesn't look right. It should be [tex] K = \frac{3Nk_BT}{2}[/tex] where N is the number of molecules in the vessel.

    Again, you are missing the N in the numerator. This equation is just the one above with a substitution for k_B.

    If you have the N in the numerator, you will have a ratio of [tex]\frac{N}{N_A}[/tex] which is the number of moles of gas, n.

    So just start over with the right equation and you should come out to the right answer.
     
  4. May 11, 2008 #3
    Awesome, Thanks for showing me where I made my mistake :+)
     
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