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Stable Bonds at Room Temperature

  1. Dec 2, 2012 #1
    1. The problem statement, all variables and given/known data
    We know that the average kinetic energy of an ideal gas atom or molecule at Kelvin temperature T is (3/2)kbT. For what temperature T_He does the average kinetic energy of an ideal gas atom or molecule equal the bond energy of the van der Waals bond in He_2 7.9*10^-4 ev?

    2. Relevant equations

    (3/2) kb * T

    3. The attempt at a solution

    7.9*10^-4 ev * 1.602*10^-19 J = 1.265*10^-22 J

    (3/2) * kb * 298 K = 1.265*10^-22 J

    kb = 2.829*10^-25 J

    The answer asks for a temperature and I have an energy. I was really unsure about this one.
  2. jcsd
  3. Dec 2, 2012 #2


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    Hello. kb stands for "Boltzmann's constant" and has a value that you can find in your text. Looks like you assumed room temperature, but the temperature is what you want to determine.
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