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Pressure, temperature and entropy vs. volume graphs

  1. Aug 8, 2012 #1
    1. The problem statement, all variables and given/known data
    "In the following a<b<c are finite positive constants. One mole of an ideal monoatomic gas, initially at volume Vi and temperature 1000K, expands to a final volume cVi in 3 reversible steps: (1) isothermal expansion from Vi to aVi (2) adiabatic expansion from aVi to bVi (3) isothermal expansion from bVi to cVi.

    b)a) Make careful sketches, approximately to scale, of how i) the pressure ii) the temperature and iii) the entropy evolve with volume.



    2. Relevant equations
    pV=nRT,
    W = ∫pdV
    ΔU= Q-W
    W=nRTlnVf/Vi



    3. The attempt at a solution

    I've come up with these after quite a bit of thinking:

    a) i)pV:
    pressure.png

    ii)Temp v pressure.This doesn't look right to me, but my reasoning is, first step temp stays the same, reversible adiabatic expansion has no ΔQ and again isothermal keeps it the same. Right? The only thing I can think is there's a problem in my understanding of a adiabatic reversible expansion- how does the temp change?!

    tempchange.png

    Entropy vs Volume
    entropychange.png

    *Just realized I should have arrows an all isotherms because it's reversible




    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 8, 2012 #2

    Andrew Mason

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    The adiabatic path has a greater rate of decline in pressure with decrease in volume than does the isothermal path. So you need to change the second isothermal path.
    An adiabatic expansion will result in work being done but since there is no heat flow, the work done must be equal to the change in internal energy (use the first law). So internal energy must decrease. For an ideal gas, how is internal energy related to temperature?

    AM
     
  4. Aug 9, 2012 #3
    With the adibatic path do you mean a greater rate of decline in pressure with increase in volume? Or have I misunderstood the concept?

    Also with the adiabatic expansion, because U=3/2 nRT, that a decrease in U, would be a result of a decrease in T since R is a constant and n is kept the same? How steep would I expect the slope to be- is it a rapid decrease (looking like your average adiabatic isotherm) or does it have a shallower slope?

    Thanks for the help, much appreciated!
     
  5. Aug 9, 2012 #4

    Andrew Mason

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    Yes. That is what I meant to say. Sorry for the confusion.
    I am not sure that you mean an adiabatic isotherm. That would be a free expansion which is not a quasi-static process. The adiabatic (quasi-static) curve reflects a greater decrease in pressure as volume increases than the isotherm. Just draw the second isotherm like the first, with a shallower negative slope.

    AM
     
  6. Aug 9, 2012 #5
    Sorry to beat the question to death but I'm a bit confused now. So should my Temperature Volume graph should look like the pV graph with a steeper middle isotherm? Or should it have two flat isotherms (1st at 1000K, the second at the decreased T1) with a steep isotherm between the flat parts where isothermal expansion happens :uhh:
     
  7. Aug 9, 2012 #6

    Andrew Mason

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    There is no "middle isotherm". The second (middle) graph is of an adiabatic process - temperature is not constant during this process, as you have already determined.

    AM
     
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