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Adiabatic process and heat exchange

  1. Dec 11, 2005 #1
    All the places I've searched I find that the change in the internal energy for the adiabtic process is

    [tex] dU = C_V\,dT [/tex]

    But I don't understand why they use C_V, neither the volume nor the preassure is constant during the adiabtic process.. I know that

    [tex] dU = C_V\,dT = C_p\,dT - p\,dV [/tex]

    But why can that expression for dU be used for the internal energy. In the book "Physics for Scientists and Engineers", they say that because U is only a function of temperatur, then

    [tex] U(T)=\int_0^{T} C_V\,dT [/tex]

    Without explaining why.. I hope someone can explain this. Thx in advance.
  2. jcsd
  3. Dec 11, 2005 #2

    Physics Monkey

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    Adiabatic processes are those in which no heat is exchanged (entropy is constant), thus the change in internal energy is given by [tex] dU = - p \,dV [/tex] from the first law. Now, as you correctly indicate, neither the pressure, temperature, or volume is constant in an adiabatic process, but the combination [tex] p V^\gamma [/tex] is. This condition, together with the ideal gas law, enables you to integrate the above expression.

    From an entirely different point of view, the internal energy is a state function and for an ideal gas it depends only on temperature. This means that irrespective of any other considerations, the change in internal energy for any process starting at [tex] T_i [/tex] and ending at [tex] T_f [/tex] is simply given by [tex] \Delta U = \int^{T_f}_{T_i} C_V \,dT [/tex].

    Hope this helps.
  4. Dec 11, 2005 #3
    I'm familiar with all the things you said. So I don't think that answers my question, which is why can the change in internal energy be expressed like

    [tex] \Delta U = \int^{T_f}_{T_i} C_V \,dT [/tex]

    by using the heat kapacitet for constant volume [itex]C_V[/itex].
    In another forum, a guy says, that because

    [tex]C=\frac{\partial Q}{\partial T}=\frac{\partial U}{\partial T}+p\,\frac{\partial V}{\partial T}[/tex]

    Then for constant volume

    [tex]C_V=\frac{\partial U}{\partial T}[/tex]

    And he says, that this generel, and doesn't depend on the state of the system. But with an adiabatic process you can't say that

    [tex] p\,\frac{\partial V}{\partial T} = 0 [/tex]

    Or am I completly wrong?
  5. Dec 11, 2005 #4
    Oh, and by the way, I want to know, because I wanna prove [itex] p V^\gamma [/itex], so I can't use this as you say.
  6. Dec 11, 2005 #5

    Physics Monkey

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    From the first law, the heat capacity at constant volume is just [tex] \frac{\partial U}{\partial T} [/tex], that's all. The other guy was wrong, or at least misleading, "heat" isn't a state function so you can't take partial derivatives of it. Now, the fact that the heat capacity at constant volume doesn't relate to the heat exchanged (none) in an adiabatic process is irrelevant, it is still always the partial derivative of internal energy with respect to temperature. Therefore, when the internal energy depends on temperature alone, you can always use the heat capacity at constant volume to calculate the change in internal energy.
  7. Dec 11, 2005 #6
    Ahh ok now I see. Then for constant preassure,

    [tex]\frac{\partial U}{\partial T} = -p\,\frac{\partial V}{\partial T}+C_p[/tex]

    And because [itex]C_p = C_V + nR[/itex] and that [itex]p\,\frac{\partial V}{\partial T}=nR[/itex] (is the last expression correct?), then

    [tex]\frac{\partial U}{\partial T} = -p\,\frac{\partial V}{\partial T}+C_V+nR = C_V[/tex]

    But again this is proved by using an isobaric and isochoric process, what is the argument that this will hold, if neither the volumen nor the preassure are kept constant (adiabatic)?
    Last edited: Dec 11, 2005
  8. Dec 11, 2005 #7

    Physics Monkey

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    Yes, at constant pressure, the ideal gas law clearly states that [tex] p \frac{\partial V}{\partial T} = n R [/tex].

    You can also get the result even more directly, consider the first law for a constant volume process: [tex] dU = \delta Q [/tex]. By definition, [tex] \delta Q = C_V\, dT [/tex] at constant volume so clearly [tex] C_V = \frac{\partial U}{\partial T} [/tex]. This is the definition of [tex] C_V [/tex] so it must always be true, and there really isn't much more to it than this. The internal energy is a state function and so is its partial derivative with respect to temperature. Now, you can regard your work above as a proof of the statement that [tex] C_P = C_V + nR [/tex] for constant pressure processes, but since [tex] C_P [/tex] and [tex] C_V [/tex] are both state functions, this equation must actually hold for all processes.
    Last edited: Dec 11, 2005
  9. Dec 11, 2005 #8
    So by showing that

    [tex] C_V = \frac{\partial U}{\partial T} [/tex]

    Holds for constant preassure and constant volume, it is argument enough to say that it most hold, even if both of those thermodynamic variables p and V changes? That is for an adiabatic process.
  10. Dec 11, 2005 #9

    Physics Monkey

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    Yes, [tex] C_V [/tex] is a state function, and once you show that two state functions are equal, no matter what kind of process you used to show it, the equality is valid in general.
  11. Dec 11, 2005 #10

    Andrew Mason

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    Be careful here. Entropy change is not 0 for all adiabatic processes. Just reversible ones.

  12. Dec 11, 2005 #11

    Physics Monkey

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    We were clearly talking in the context of the quasi-static limit here, in this case the two are one and the same.
  13. May 11, 2010 #12
    Hey dudes,
    quick question, and i'm basically to lazy to figure out right now,
    In an Otto cycle PV diagram what is held constant in the adiabatic processes, tempreture?
    odviously volume for isochors
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