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Extremely challenging thermodynamics equation

  1. Sep 22, 2010 #1
    hi,
    i actually posted it in physics section but no one knows so i think engineering experts would be stronger to this kind of equations.
    1.
    [tex]\color{blue}dq=dU+PdV[/tex]
    so
    [tex]C_V = \left(\frac {\partial q}{\partial T}\right)_V= \left(\frac {\partial U}{\partial T}\right)_V[/tex]
    leads to
    [tex]dU=C_{V}dT[/tex]
    so
    [tex]\color{red}\left(\frac {\partial U}{\partial V}\right)_T = 0[/tex]
    meaning that internal energy of NONideal gas is a sole function of T?
    2.
    As
    [tex]\color{blue}dq=dU+PdV[/tex]
    so
    [tex]\left(\frac {\partial U}{\partial V}\right)_T= \left(\frac {\partial V}{\partial T}\right)_P \left(C_P-C_V\right) - P[/tex]
    why? even for ideal gas the change in volume results in change in internal energy:
    [tex]\color{red}\left(\frac {\partial U}{\partial V}\right)_T \not= 0??[/tex]

    thanks for answering.
     
  2. jcsd
  3. Sep 22, 2010 #2
     
    Last edited by a moderator: May 4, 2017
  4. Sep 22, 2010 #3
  5. Sep 22, 2010 #4

    boneh3ad

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    [tex]
    dq=dU+PdV
    [/tex]

    [tex]
    \frac{1}{\partial T} \left(dq = dU+PdV \right)_V
    [/tex]

    [tex]
    \left(\frac{\partial q}{\partial T}\right)_V = \left(\frac{\partial U}{\partial T}\right)_V + \left(\frac{\partial (P dV)}{\partial T}\right)_V
    [/tex]

    [tex]
    \left(\frac{\partial q}{\partial T}\right)_V = \left(\frac{\partial U}{\partial T}\right)_V + P \left(\frac{\partial V}{\partial T}\right)_V
    [/tex]

    [tex]
    \left(\frac{\partial V}{\partial T}\right)_V \equiv 0
    [/tex]

    [tex]
    C_V \equiv \left(\frac {\partial q}{\partial T}\right)_V
    [/tex]

    [tex]
    C_V =\left(\frac {\partial U}{\partial T}\right)_V
    [/tex]

    [tex]
    \partial U = C_V \partial T
    [/tex]

    Maybe I missed something in my haste, but I don't know how he got from here to:

    [tex]
    \left(\frac {\partial U}{\partial V}\right)_T = 0
    [/tex]

    The only thing I can think of is if he then held T constant and differentiated with respect to V, but that doesn't really make sense. You would then be holding both T and V constant, so nothing is changing.

    Also I edited this to make it correct but it didn't update the post for some reason. Weird.
     
    Last edited: Sep 22, 2010
  6. Sep 25, 2010 #5
    Sorry for late response.
    Because it is deduced from "nonideal-gas-equation"[itex]dU=C_{V}dT[/itex] and obtaining that internal energy is only sole function of temperature and independent of volume.
     
  7. Sep 25, 2010 #6
    you also say that [itex]
    \partial U = C_V \partial T
    [/itex]so why not [itex]\left(\frac {\partial U}{\partial V}\right)_T = 0[/itex]
     
  8. Sep 25, 2010 #7
    Yes i know that 1st law+kinetic theory leads to [itex]U=\frac{3}{2}RT[/itex], but the last equation i derived in the 1st post omitted kinetic theory and a strange result comes.
     
  9. Sep 25, 2010 #8

    boneh3ad

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    No it isn't. That isn't the "non-ideal gas equation." It is merely a derived relationship between U and T if V is held constant.

    Because that step doesn't make sense. Start with:

    [tex]\partial U = C_V \partial T[/tex]

    This equation is all you need to determine that internal energy is a function of only temperature at constant volume. Differentiating with respect to V is trivial and give you no new information. You can just look at this result and see that there are no pressure terms because you held them constant so they fell out when differentiating earlier.

    Differentiating by V while holding T constant tells you no new information.
     
  10. Sep 25, 2010 #9
    So why whenever we see dU, we always substiture it by CvdT with no regard to the nature of process? How can we be so sure that change of pressure or change in temperuture does not change internal energy?
     
  11. Sep 25, 2010 #10

    boneh3ad

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    I am not sure what you mean here. The equation [itex]dU = C_V dT[/itex] is derived straight from the physical laws, as you know. It gets substituted in because you can't directly measure [itex]dU[/itex], but you can directly measure [itex]dT[/itex], so it makes any equations more useful.

    Well clearly, changing temperature does change [itex]dU[/itex] since [itex]dT[/itex] shows up in the equation. For example, if you integrate that, you get:

    [tex]\Delta U = C_V \left(T-T_0\right)[/tex]

    We know, then, that pressure can change the internal energy in this case because a change in pressure would change the temperature (remember, you held V constant).
     
  12. Sep 26, 2010 #11
    why we have to hold V constant? Can't we change P,V,T at the same time and determine what happens in internal energy?
    Can i just allowing small change in volume so that change in P and T is still in the same direction?
     
  13. Sep 26, 2010 #12

    boneh3ad

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    You can't allow small changes in volume and still use [itex]C_V[/itex]. The definition is that volume is constant.
     
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