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Thermodynamics help

  1. Sep 29, 2013 #1
    Thermodynamics - adiabatic process

    1. The problem statement, all variables and given/known data
    The question is: Consider a hypothetical ideal gas with internal energy U = NkTo(T/T0)α+1, where To and α are positive constants. Show that in an adiabatic process, V*exp[(1+1/α)(T/To)α] = constant.


    2. Relevant equations
    PVγ = constant
    γ = Cp/Cv
    Cp = Cv + Nk


    3. The attempt at a solution
    I'm pretty sure that I'm supposed to show that [(1+1/α)(T/To)α] is equal to γ and since PVγ = constant, V*exp[(1+1/α)(T/To)α] = constant. When I try to solve it though I can't get the solution to come out. I differentiate U to get Cv = Nk(1+α)(T/To)1+α. When I plug that into γ I get γ = 1 + 1/[(1+α)(T/To)α]. Either I'm just not simplifying it enough and the answer is correct, or I solved for λ incorrectly, or my equations are incorrect. I don't know which it is though.
     
  2. jcsd
  3. Sep 29, 2013 #2
    The equation
    PVγ = constant
    is not valid for this problem. It follows from the usual internal energy for ideal gas,
    U(T)=nCvT

    Here you have a different function U(T) and you have to find the relationship between volume and temperature. (for the "usual" ideal gas this will be TVγ-1 = constant )
    You can use the first law in conjunction with the equation of state to do this.
     
  4. Sep 29, 2013 #3

    ehild

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    Is PVγ = constant when Cv depends on T?

    ehild
     
  5. Sep 30, 2013 #4
    I'm not really sure what you mean, can you explain it more?
     
  6. Sep 30, 2013 #5
    I'm guessing it's not, but I don't know what it should be then.
     
  7. Sep 30, 2013 #6

    Chestermiller

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    For an ideal gas,
    dU=CvdT
    PV=RT
    From the first law, for an adiabatic reversible process, how is dU related to PdV?
     
  8. Sep 30, 2013 #7
    dU = -PdV, so Cv = -PdV/dT right?
     
  9. Sep 30, 2013 #8

    ehild

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    U is given as function of T. PV=kNT is valid for the ideal gas, and also the First Law is valid. For an adiabatic process dU=-PdV. Use P=kNT/V, and integrate.

    ehild
     
  10. Sep 30, 2013 #9
    Ok, so I get U = -kNT(ln(V))
     
  11. Sep 30, 2013 #10

    haruspex

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    No, that would not follow, so I don't think you want to show that [(1+1/α)(T/To)α] is equal to γ. Instead, try raising V*exp[(1+1/α)(T/To)α] to the power of γ, using the expression for γ that you derived.
     
  12. Sep 30, 2013 #11
    So that makes (V*exp[(1+1/α)(T/To)α])1+1/(1+α)(T/To)α

    I'm not sure what to do with that
     
  13. Sep 30, 2013 #12
    Well, I don't understand what part you don't understand.:confused:

    But the idea is "forget gamma". And "forget pv^gamma". Does not apply here.

    1. From first law applied to adiabatic process you have:
    dU=pdV
    You have U(T) so find dU.

    2. You have PV=nRT so you can eliminate p on the right hand side:
    pdV= nRTdV/V

    So you will have an equation relating V and T. Integrate (after separating variables) and you'll find that exponential relationship.
     
  14. Sep 30, 2013 #13

    Chestermiller

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    You have the equation of U as a function of T, and you know know that
    [tex]C_v=\frac{\partial U}{\partial T}[/tex]
    Just differentiate the equation for U with respect to T, and write
    [tex]dU=C_vdT=\frac{\partial U}{\partial T}dT=-PdV[/tex]
    Then, just substitute the ideal gas law for P, and integrate.
     
  15. Sep 30, 2013 #14
    This is a "hypothetical" ideal gas.
    Cv=∂U/∂T is valid for a "real" ideal gas. :smile:

    The whole point here is that U(T) is not given by
    dU=CvdT but by that other, more complicated formula.
    If he does what you suggest he'l get just the usual
    [tex]TV^{\gamma -1 }= constant[/tex] and not the formula required by the problem.

    But the method will work. This is what I tried to explain as well.
    Just use
    [tex]dU=Nk(\alpha +1) (T/T_0)^{\alpha} dT[/tex].

    There is no need to introduce Cv or gamma.
     
  16. Sep 30, 2013 #15

    DrClaude

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    I have to disagree. Cv=∂U/∂T follows simply from the definition of heat capacity,
    $$
    C = \frac{Q}{\Delta T}
    $$
    by considering a constant volume (hence ##W=0##), without invoking an ideal gas.
     
  17. Sep 30, 2013 #16
    Did I say anything that seem to contradict your statement? I just meant just that you don't need Cv to solve the problem. It does not appear in this problem.
    Oh, I see. I used partial derivatives.

    I meant that dU=CvdT may not apply to other systems.
    It is valid only for some systems, like ideal gas in the "proper" definition.


    So dU=Nk(α+1)(T/T0)^α dT
    You don't need to define or use a specific heat to solve the problem.
    Sorry for the confusion.
     
    Last edited: Sep 30, 2013
  18. Sep 30, 2013 #17

    Chestermiller

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    This is exactly what I was suggesting. I brought the heat capacity into the picture because I felt the OP would feel more comfortable with it. For this particular ideal gas, Cv is not independent of temperature, but is given by:
    [tex]C_v=Nk(\alpha +1) (T/T_0)^{\alpha}[/tex]
    Are you uncomfortable with an ideal gas heat capacity that varies with temperature. A temperature-dependent heat capacity is part of the definition of an ideal gas that we engineers use.
     
  19. Sep 30, 2013 #18
    Is this a question?
    I don't feel any discomfort about temperature variation of Cv or about Cv in general. Even Cp it's bearable, despite all these pressure variations. :smile:
     
  20. Sep 30, 2013 #19

    Chestermiller

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    Oops. I left out the question mark. Thank you for serving as the grammar police enforcer.

    Getting back to the thread, I think we are (and were) totally in agreement on how this problem should be solved. Of course, for an ideal gas, Cp is also a function only of temperature.
    Chet
     
  21. Sep 30, 2013 #20
    I agree that we are in agreement. :smile:
    It was not intended as police work. Just curious.
     
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