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Why is kinetic energy NOT proportional to degrees F/C

  1. Feb 24, 2008 #1
    Hello all,
    I have a question, that regards a problem I was assigned. The problem is:

    Why is kinetic energy NOT proportional to degrees F/C?

    -I understand it has something to do with absolute zero Kelivn...not C. Thanks

    Any help would be greatly appreciated...I need a response by tomorrow. Thanks again
     
    Last edited: Feb 24, 2008
  2. jcsd
  3. Feb 25, 2008 #2
    [tex]PV = \frac{1}{3}Nmv_{rms}^{2}[/tex]
    [tex]PV = nRT[/tex]

    Therefore:
    [tex]\frac{1}{3}Nmv_{rms}^{2} = nRT[/tex]

    Rearranging:
    [tex]v_{rms} = \sqrt{\frac{3nRT}{Nm}} [/tex]

    Since [tex] N = nN_{a}[/tex] and [tex]R = kN_{a}[/tex] it can be simplified to:

    [tex]v_{rms} = \sqrt{\frac{3nN_{a}kT}{nN_{a}m}} = \sqrt{\frac{3kT}{m}} [/tex]

    Therefore:
    [tex] v_{rms}\propto \sqrt{T}[/tex]

    EDIT: Similar looking results are found if you doing the same with boltzmann curves or whatever else tickles your fancy.
     
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