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Thermal Physics > Air and Internal Energy

  1. Jun 24, 2005 #1
    This Q has me stumped, I'm still flicking through some web pages and my text book, but been unable to find a useful formula to work it out yet:

  2. jcsd
  3. Jun 24, 2005 #2

    For each molecule, the kinetic energy associated with each degree of freedom is [itex]\frac{1}{2}kT[/itex], so the kinetic energy of each molecule in the O2, N2 mixture is [itex]\frac{5}{2}kT[/itex]. Summing over the entire gas gives the internal energy [itex]U = N\frac{5}{2}kT = nN{_\mathrm{A}} \frac{5}{2}kT = \frac{5}{2}nRT[/itex].

    P.S. Not PHYS1901 by any chance?
  4. Jun 24, 2005 #3
    No. PHYS 1001 actually :eek:
    I just found this forum, looks like a useful resource all round.

    Thanks for clearing that up. My main problem I think is remembering all the letters and where they come from.

    BTW, you have U = N*5/2*k*T = n*N_a*5/2*k*T

    The only thing that changes there is N --> n*N_a

    What is "N" if N=n*N_A

    (Haven't worked out how to use the "Latex" code yet)
  5. Jun 24, 2005 #4
    [itex]k[/itex] is defined as [itex]\frac{R}{N_\mathrm{A}}[/itex], N is the number of gas molecules.
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