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Translational kinetic energy

  1. Apr 12, 2006 #1
    Compute the total translational kinetic energy of one liter of oxygen gas at atmospheric pressure.

    <E>=3/2 kT

    assuming that the T is about the temperature of Earth's surface, 300K, then:

    [tex] \frac{1}{2} mv^2 = \frac{3k(300K)}{2} [/tex]

    what I am wondering is, is this the total translational kinetic energy? I think the book defines it as the average energy but does not give a formula for total translational kinetic energy.
     
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  3. Apr 12, 2006 #2

    Hootenanny

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    That indeed is the formula for average translational kinetic energy, however, I have not come across any formula for the total kinetic energy of a gas.
     
  4. Apr 12, 2006 #3

    dav2008

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    Would it not be just 3/2NkT?
     
  5. Apr 12, 2006 #4

    Hootenanny

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    Ahh indeed I forgot, I've just dug out my old physics textbook now:biggrin:
     
  6. Apr 12, 2006 #5
    what is N?
     
  7. Apr 12, 2006 #6

    dav2008

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    N is the number of molecules.

    I mean it makes sense that if the average translational kinetic energy of each molecule is 3/2kT then to get the total energy you would just multiply by the total number of molecules.
     
  8. Apr 12, 2006 #7
    how would I determine how many molecules are in 1 liter of oxygen? and how would one take into account the temperature when it is given inpressure?
     
    Last edited: Apr 12, 2006
  9. Apr 12, 2006 #8

    dav2008

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    Are you familiar with the ideal gas law?

    PV=NkT

    P=pressure in pascals, V=volume in m3.

    Since you know the pressure and the volume, and you are looking for the term "3/2NkT" I think you can see how you can get it.
     
  10. Apr 12, 2006 #9
    PV=NkT

    1 liter= 0.001m3

    [tex] E= \frac{3}{2} (101.3kPa)(0.001m^3) [/tex]

    would ths be correct?
     
  11. Apr 13, 2006 #10

    Hootenanny

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    This is what i got;

    [tex]pV=NkT[/tex]

    [tex]\frac{2}{3}E = kT[/tex]

    [tex]pV = N\frac{2}{3}E[/tex]

    [tex]E = \frac{3pV}{2N}[/tex]

    I think you've just missed out the number of molecules

    Regards
    -Hoot
     
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