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Adiabatic Compression

  1. Jan 16, 2012 #1
    Hi, I have a little problem understanding adiabatic compression.

    Let me start with the definition of adiabatic process from wikipedia, "In thermodynamics, an adiabatic process or an isocaloric process is a thermodynamic process in which the net heat transfer to or from the working fluid is zero."

    My problem is, when we compress a certain gas in a closed container, we inject our kinetic energy to decrease the volume of the container, so shouldn't this means there is a net heat change, or a change in the total energy of the system?

    or this kind of injection of energy does not categorize under heat transfer, Im quite confused.

    I give my greatest thanks in advance! :)
     
  2. jcsd
  3. Jan 16, 2012 #2

    Doc Al

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    Staff: Mentor

    When you do work on the gas, you definitely add energy. But that's not 'heat'. Heat is the flow of energy due to temperature difference.

    Read this: What is Heat?
     
  4. Jan 16, 2012 #3
    Thank you for the link, it helped me a lot!

    One more small thing, why is it that the rotation of a diatomic molecule along the atom-atom bond not counted as one of the degree of freedom?
     
  5. Jan 16, 2012 #4

    Ken G

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    Gold Member

    That's actually a quantum mechanical effect. The kinetic energy for something undergoing cyclical motion (vibration or rotation) is characterized by mx2w2, where x is the spatial size scale and w is the frequency. But here's where a big difference between vibrations and rotations appears-- for vibrations, x is a variable, and can be as large as it needs to be to get ~kT of energy into the mode in question. But for rotations, x is fixed by the size scale of the rotating object, the "lever arm" of the appropriate rotation. So when x is extremely small, as in the case you mention, it would require huge w to get kT of energy into that mode, of order w=(kT/m)1/2x-1. However, huge w, coupled with the quantum mechanical minimum action h, means you won't excite that mode, since here hw >> kT, because (kT/m)1/2x-1>> kT/h whenever x << h/(mkT)1/2. So we only exite modes like that when T is very high, and it generally isn't that high in the applications you have in mind.
     
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