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Why the change in internal energy is considered same?

  1. Feb 9, 2013 #1

    Elz

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    1.While studying the thermodynamic properties of perfect gases I saw that the change in internal energy for constant volume process is mCvdT. but it is also the same for const pressure, const temp, const entropy and other reversible non-flow processes. Why is it so?? why it is not mCpdT for const pressure process? The value of Cv and Cp are not same. And if we consider the same system working once for const volume and again for constant pressure then will the temp difference be same? I mean will both the process have same initial and final temperature? If the are same then how can be the value of dU be same as the values of Cv and Cp are different?
    does it work like if i heat CO2 at constant pressure then the value of change in internal energy and if i heat the same amount of CO2 at const volume or const entropy then the value of change in internal energy will be same? I mean has it been experimentally established?




    2. dU=mCvdT

    Why not Cp for const pressure process?
     
  2. jcsd
  3. Feb 9, 2013 #2

    Simon Bridge

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    Welcome to PF;
    What determines the change in internal energy: where does it come from?
     
  4. Feb 9, 2013 #3

    rude man

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    First, C = dQ/dt, not dU/dt. So you have no basis in assuming C = dU/dt, whether C is constant pressure, constant volume, or anything else.

    The proof that dU = CVdT for an ideal gas rests on the following:

    1. Definition of an ideal gas. That definition comprises pV = nRT and also ∂U/∂p|T = 0.

    From those two equations you can show that U must be a function of temperature only (and not of p or V).

    Then, invoking the 1st law,
    dU = dQ - pdV so that
    dU/dT = dQ/dT - pdV/dT.

    However, having shown that U = U(t) only, we know that pdV/dt = 0 and conclude that

    dU/dT = CV for an ideal gas, independent of p or V.
     
  5. Feb 10, 2013 #4

    Elz

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    thanks :)


    Thanks a lot for replying. can u please explain how ∂U/∂p|T = 0 and pdV/dt = 0 is derived?
     
  6. Feb 10, 2013 #5

    Elz

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    Thanks a lot for replying. can u please explain how ∂U/∂p|T = 0 and pdV/dt = 0 is derived?
     
  7. Feb 10, 2013 #6

    rude man

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    ∂U/∂p|T = 0 is as I said part of the DEFINITION of an ideal gas.

    Then we show that U ≠ U(V) as follows:
    dU/dV|T = dU/dp|T * dp/dV|T
    and dp/dV = -nRT/V2 = -p/V ≠ 0 but dU/dp|T = 0,
    dU/dV|T = 0 and U ≠ U(V).

    So both derivatives of U wrt to p and V are zero, so U ≠ U(p) ≠ U(V).
    (Replace all "d" above with "∂").

    dQ = dU + pdV = dU at constant volume. So
    CV = ∂Q/∂T|V = ∂U/∂T|V

    But for the special case of an ideal gas we know U = U(T) only,
    so ∂U/∂T|V = dU/dT. QED
     
  8. Feb 10, 2013 #7

    Elz

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    Thanks a lot....... It explained everything. All my doubts are clear... :)
     
  9. Feb 13, 2013 #8
    My friend Rude Man urged me to submit the following repsonse, which, for completeness, provides an alternate perspective on the answer to the OP's question.

    The heat capacity at constant volume Cv and the heat capacity at constant pressure Cp are physical properties of a material, and are not wedded to the heat transferred in any specific process. These properties are precisely defined mathematically as:
    [tex]C_v=(\frac{\partial U}{\partial T})_v[/tex]
    [tex]C_p=(\frac{\partial H}{\partial T})_p[/tex]
    where H is the enthalpy. For an ideal gas, U and H are functions only of temperature.

    Here, for the case of a gas that is not necessarily ideal, is how these heat capacities can be applied in practice:
    [tex]dH=TdS+VdP[/tex]
    If we regard the molar enthalpy as a function of T and P, then

    [tex]dH=(\frac{\partial H}{\partial T})_PdT+(\frac{\partial H}{\partial P})_TdP[/tex]

    From these two equations, we have:

    [tex](\frac{\partial H}{\partial T})_P=C_p=T(\frac{\partial S}{\partial T})_P[/tex]

    and [tex](\frac{\partial H}{\partial P})_T=T(\frac{\partial S}{\partial P})_T+V[/tex]

    Now, we need to evaluate [itex](\frac{\partial S}{\partial P})_T[/itex]. We obtain this from the equation for the differential change in Gibbs free energy:

    [tex]dG=-SdT+VdP=(\frac{\partial G}{\partial T})_PdT+(\frac{\partial G}{\partial P})_TdP[/tex]
    From this, it follows that:
    [tex]S=-(\frac{\partial G}{\partial T})_P[/tex]
    [tex]V=(\frac{\partial G}{\partial P})_T[/tex]
    If we take the partial of the first of these equations with respect to P and the partial of the second with respect to T, and sum the equations, we obtain:
    [tex](\frac{\partial S}{\partial P})_T=-(\frac{\partial V}{\partial T})_P[/tex]
    Substituting back into the equation for dH, we have
    [tex]dH=C_P+(V-T(\frac{\partial V}{\partial T})_P)dP[/tex]

    Note that the coefficient of dP in this equation is equal to zero for an ideal gas, but it is not equal to zero beyond the ideal gas region. This equation provides us with what we need to include the effect of pressure on enthalpy. But, to evaluate the second term, we need to know the equation of state (EOS) of the material, beyond the ideal gas region.
     
  10. Feb 14, 2013 #9

    Elz

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    Thanks a lot Chestermiller,and give my thanks to Rude Man too...
     
  11. Feb 15, 2013 #10
    Oops. There is a glitch in the final equation in my recent reply. The final equation should read:

    [tex]dH=C_PdT+(V-T(\frac{\partial V}{\partial T})_P)dP[/tex]

    Sorry for the error.

    Chet
     
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