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Entropy of a mole of an ideal gas

  1. Jun 7, 2009 #1

    fluidistic

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    1. The problem statement, all variables and given/known data


    Show that the entropy of one mole of an ideal gas is given by [tex]\Delta S = \int \frac{C_v dT+PdV}{T}[/tex].






    2. The attempt at a solution
    [tex]\Delta S= \frac{dQ}{T}[/tex]. From the 1st of Thermodynamics, [tex]\Delta U= \Delta Q - \Delta W[/tex], where [tex]W[/tex] is the work done by the gas.


    Hence [tex]\Delta Q = \Delta U + \Delta W \Rightarrow dQ=dU+dW=C_vdT+PdV[/tex]. So [tex]\Delta S= \int \frac{C_vdT+PdV}{T}[/tex].[tex]\square[/tex].
    But this is cheat. By this I mean that I didn't know that [tex]dU=C_vdT[/tex]. I deduced it because I had to fall over the result. How can I deduce [tex]dU=C_vdT[/tex], for an ideal gas?
    Where did I supposed the 1 mol of the ideal gas?
    Last question, what are the bounds of the integral of the change of entropy?
    Because [tex]\Delta S = \int \frac{C_v dT+PdV}{T}=\int \frac{C_vdT}{T}+ \int \frac{PdV}{T}[/tex] and I'm sure the bounds of the 2 integrals are different. For instance I think that the bounds of [tex]\int \frac{PdV}{T}[/tex] are [tex]\int_{V_1}^{V_2} \frac{PdV}{T}[/tex]. And I guess that the bounds of [tex]\int \frac{C_vdT}{T}[/tex] are [tex]\int_{T_1}^{T_2} \frac{C_vdT}{T}[/tex] but I'm not sure.
     
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  3. Jun 7, 2009 #2

    Mapes

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    By definition,

    [tex]C_X=T\left(\frac{\partial S}{\partial T}\right)_X[/tex]

    where [itex]X[/itex] is some constraint. This is easily combined with [itex]dU=T\,dS-P\,dV[/itex].

    If you define [itex]S[/itex] and [itex]V[/itex] as the molar entropy and molar volume, respectively, then everything works out. You'll just multiply the answer by 1 mole.

    I don't believe it's possible to write a definite integral that contains both [itex]dT[/itex] and [itex]dV[/itex]. Combining them is a useful shorthand that works for the indefinite integral only.
     
  4. Jun 7, 2009 #3

    fluidistic

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    Ok, thank you very much!
     
  5. Jun 7, 2009 #4

    Andrew Mason

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    For a constant volume process, W = 0. From the first law, if W = 0, dQ = dU.

    By definition dQ = nCvdT for a constant volume process. So dU = nCvdT.

    AM
     
  6. Jun 7, 2009 #5

    fluidistic

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    Thanks. I think I understand well now.
     
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