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Change in entropy, quasistatic, isothermal expansion

  1. Oct 23, 2014 #1
    1. The problem statement, all variables and given/known data
    I am to show that ΔS=Q/T for the isothermal expansion of a monoatomic ideal gas, when the expansion is so slow that the gas is always in equilibrium.

    2. Relevant equations
    1. law: ΔU=Q+W (We mustn't use dQ and dW - our teacher hates that :( ).
    Ideal gas law: PV=NkT
    We need the equation: ΔS=Nk*ln(V_final/V_initial)
    And that quasistatic expansion work is W=-PΔV

    3. The attempt at a solution
    -I think I am to start with: ΔU=Q+W⇔Q=ΔU-W, where ΔU=0 since its isothermal.
    -I know that it is quasistatic expansion work, so W = -PΔV, so Q = -(-PΔV) = PΔV
    I think I want to get something from the ideal gas law in here: P=(NkT)/V, so

    Q=(NkTΔV)/V

    But then I kind of get stuck there...

    Hope someone can help. I thinks it is really easy, but I kind wrap my head around it.
     
    Last edited: Oct 23, 2014
  2. jcsd
  3. Oct 23, 2014 #2

    RUber

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    Looking at what you are supposed to show, take Q/T and compare that to what you have for ##\Delta S##.
    So it seems like the question is: does ##\frac {\Delta V}{V} = \ln \frac{V_{final}}{V_{initial}}##?
     
  4. Oct 23, 2014 #3
    Yeah - something like that?
     
  5. Oct 23, 2014 #4

    RUber

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    What do you know about the function for V and/or ##\Delta##V if you are given an initial and final state?
    Also, in the equations you provided for the ideal gas law, you change between R and k, are these different parameters?
     
  6. Oct 23, 2014 #5
    I don't know anything but what I have written unfortunately. No, sorry - that's my mistake. It should have been k all along.
     
  7. Oct 23, 2014 #6
    You really had the right idea in your initial post. Nice job. Now, just express the heat in differential form:
    [tex]dQ=PdV=\frac{NkT}{V}dV[/tex]
    Then integrate between the initial and final volumes.

    Chet
     
  8. Oct 24, 2014 #7
    Great - thanks a lot :)
     
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