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Heat capacities and negative temperature

  1. Oct 8, 2013 #1
    Hi everybody,
    I have the following doubt. We know that for a thermodynamic system the following equality holds:
    $$
    C_P-C_V=-T\frac{\left[\left(\frac{\partial P}{\partial T}\right)_V\right]^2}{\left(\frac{\partial P}{\partial V}\right)_T}
    $$

    Now, the mechanical stability of the system requires that the volume decreases with increasing pressure, i.e. [itex](\frac{\partial V}{\partial P})_T<0[/itex]. So this seems to lead to [itex]C_P>C_V[/itex]. Is that always true? What happen if the temperature is negative, [itex]T<0[/itex]?

    Thanks
     
  2. jcsd
  3. Oct 8, 2013 #2
    The absolute temperature can't be negative, and Cp is always greater than Cv.

    chet
     
  4. Oct 8, 2013 #3
    I am not really sure about that. There are quantum systems with negative temperature. For example a system with just two energy levels has negative temperature.
     
  5. Oct 9, 2013 #4
    I never heard of that, but I don't know much about qm. When I studied qm, I did not encountered the concept of negative absolute temperature.
     
  6. Oct 9, 2013 #5
    It depends on the definition. When you work with statistical mechanics the temperature is defined through its relations with entropy, free energy and so on. And it turns out that from this relations it can also be negative.
     
  7. Oct 9, 2013 #6

    UltrafastPED

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  8. Oct 9, 2013 #7

    Vanadium 50

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    Yes, but systems with negative temperature are not ever in thermal equilibrium with gasses, and since you're talking about gasses, you are describing an unphysical system.
     
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