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Thermodynamics: find internal energy given the equation of state

  1. Jun 25, 2013 #1
    Given the following equation of state and [itex]C_p = 3R[/itex], find the equation of the internal energy in terms of T and V

    p(V-b) = RT

    Any clues? I can use also the specific heat at constant pressure or volume. Thanks!
    Last edited: Jun 26, 2013
  2. jcsd
  3. Jun 25, 2013 #2

    Simon Bridge

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    Have you tried applying what you know about "internal energy" from your coursework so far?
    Same with specific heats.
  4. Jun 26, 2013 #3
    Observation: I changed a little the statement of the problem, as I noticed I had additional constraints.

    Now to the problem: I have found the equation (which I didn't remember...)

    [itex] dU =C_{V}dT +\left[T\left(\frac{\partial p}{\partial T}\right)_{V} - p\right]dV [/itex]

    So with the equation of state it gives:

    [itex] \dfrac{\partial p}{\partial T} = \dfrac{\partial}{\partial T}\left(\dfrac{R}{V-b}T\right) = \dfrac{R}{V-b}[/itex]

    And including it in the previous equation:

    [itex] dU = C_{V}dT +\left[T\dfrac{R}{V-b} - p\right]dV = C_{V}dT +\left[T\dfrac{R}{V-b} - \dfrac{R}{V-b}T \right]dV = C_{V}dT[/itex]

    Now, to put [itex]C_{V}[/itex] in terms of [itex]C_{P}[/itex] we have the relation:

    [itex]C_p - C_V = T \left(\frac{\partial p}{\partial T}\right)_{V} \left(\frac{\partial V}{\partial T}\right)_{p}[/itex]

    The terms of which we can find with the equation of state:

    [itex]\frac{\partial p}{\partial T} = \dfrac{R}{V-b} \qquad \frac{\partial V}{\partial T} = \dfrac{R}{p}[/itex]


    [itex]C_p - C_V = \cdots = R \Rightarrow C_V = C_p - R = 2R[/itex]

    And finally [itex]dU = C_{V}dT= 2R dT[/itex] so [itex]\Delta U = 2R \Delta T[/itex]

    Is it alright? Any problem with not determining [itex]U[/itex] but [itex]\Delta U[/itex]?
  5. Jun 26, 2013 #4

    Simon Bridge

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    I dunno - its your course.

    Note: you had $$\frac{dU}{dT}=2R$$ ... so why not treat it like an initial value problem? What is U when T=0K?
  6. Jun 26, 2013 #5
    Well, I'm not sure of the initial condition, that's why I asked ;)
    I guess it must be zero internal energy when temperature is zero, so then we can drop de deltas.
  7. Jun 26, 2013 #6

    Simon Bridge

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    There you go.

    All you need for an "initial" condition is any known value.
    The problem statement will not always include all the information you are expected to use.

    Note: U=0 @ T=0 would be the classical interpretation - IRL you still have zero-point energy corresponding to the QM ground state.
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