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Reversible adiabatic expansion -- calculate change in E

  1. Jan 8, 2017 #1
    1. The problem statement, all variables and given/known data

    Calculate the change in internal energy of an ideal gas the expands reverisbly and adiabatically from ##v_i## to ##v_f##

    2. Relevant equations

    ##pV^{\gamma} = constant ##*, where ##\gamma=\frac{c_p}{c_v}##
    for a reversible adiabatic.

    ##E=\alpha p V##, ##\alpha## a constant.

    3. The attempt at a solution

    The solution does

    ##E=\alpha pV##

    ##\implies## ##E_f-E_i=\alpha (p_fV_f-p_iV_i) = W##

    (And then continues to eliminate a variable using * which I am fine with etc)

    MY QUESTION IS

    (by the first law) ##dE=dW=-pdV##

    But I don't understand how we can simply do ##p_fV_f-p_iV_i ## because I thought that, in general, ##p=p(V)## and so we need to integrate over ##p## instead of just looking at the initial and final state.(unless a process is isochoric - ##p(vf-vi)=W##).

    Or in general would this be the case but due to * we do not need to integrate ? (if so how can this be seen)

    Many thanks in advance.
     
  2. jcsd
  3. Jan 8, 2017 #2

    vela

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    You have ##p_iV_i^\gamma = k = p_fV_f^\gamma##, so the work done is
    $$W = -k\int_{V_i}^{V_f} V^{-\gamma}\,dV = -k\frac{V_f^{-\gamma+1}}{-\gamma+1}+k\frac{V_i^{-\gamma+1}}{-\gamma+1}.$$ Replace the first instance of ##k## with ##p_fV_f^\gamma## and the second with ##p_iV_i^\gamma##. What do you end up with?
     
  4. Jan 8, 2017 #3

    TSny

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    You can find W by integrating PdV. Following Vela, you should find that you get the correct result.

    However, the first law tells you that for this process W must equal ΔE. Finding ΔE does not require any integration.
     
  5. Jan 8, 2017 #4
    I have seen this method of doing it, and I understand that, that is fine.

    What I dont understand is what I asked.
     
  6. Jan 8, 2017 #5

    vela

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    If you follow through with my suggestion, you end up with the result you're asking about. How does that not answer your question?
     
  7. Jan 8, 2017 #6
    that tells me that you are doing is correct.the two methods are the same

    it doesn't tell me how or why it is okay to not need a ##p=p(v)## and integrate.
     
  8. Jan 8, 2017 #7

    vela

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    If you're going to calculate the work to find ##\Delta E##, then you do need to do exactly that.

    As TSny noted, the other way is to find ##\Delta E = E_f - E_i## directly. What does the internal energy of an ideal gas depend on?
     
  9. Jan 11, 2017 #8
    temperature
     
  10. Jan 11, 2017 #9

    TSny

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    (1) Integrating by parts gives ##\int_i^fpdV = (pV) |_i^f - \int_i^fVdp##.

    (2) For an adiabatic process, show that ##Vdp = -\gamma pdV##.

    Combine (1) and (2) to show that ##\int_i^fpdV## can be expressed in terms of just ##(pV)|_i^f## and ##\gamma##.
     
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