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Adiabatic expansion

  1. Jun 2, 2009 #1
    1. The problem statement, all variables and given/known data

    http://img30.imageshack.us/img30/9954/72990922.jpg [Broken]


    3. The attempt at a solution

    I would say the answer to (a) was no but that would be a guess. Does it have to do with the fact that it's done instantaneously, whereas to achieve anything like an irreversible process in reality you have to perform it very slowly?

    For (b), Q=0 so the change in internal energy is equal to the work done. This is

    [tex]W=\frac{C_{V}}{R}(p_{1}V_{1}-p_{2}V_{2})[/tex]

    But I only know the volumes, not the pressures. Is there some way to use the fact that it expands into a vacuum?

    I think the entropy change will be zero because no heat is going in or out.

    Not sure about the temperature change bit. I would have expected the temperature to change in both. For an ideal gas, if there is some work done then the temperature definitely must change, but how does this differ for a real gas?

    Thanks.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 2, 2009 #2
    Let's look at this problem in a diferent way. Let's forget about the entropy and (ir)reversibility for the moment. Suppose you have a box that is perfectly insulated and its volume is fixed. Then inside the box 2/3 of the volume is vacuum and 1/3 is filled with a gas. The gas is separated from the vacuum by a boundary that can be removed.

    Then this system has some total internal energy. Now, if we remove the boundary, then what does conservation of energy tell you about the total internal energy of the contents of the box?
     
  4. Jun 2, 2009 #3
    That's exactly what it is. If you had a different process where someone in the external environment could control the macroscopic variables of the gas, manipulate the temperature, pressure and volume however you wished, like many of the problems involving a "movable piston" apparatus, then you would have the ability to make the process occur very slowly. But the process in this problem will occur fast no matter what you do. There are a couple common textbook examples of such a spontaneous and fast, and therefore irreversible, processes. Another example that all of the textbooks cite is the case of a hot solid object and a cold solid object suddenly brought into thermal contact, so that a transfer of heat begins once. There's another case where the spontaneous process cannot be controlled by someone in the environment poking in with a manipulation of macroscopic variables.
     
    Last edited: Jun 2, 2009
  5. Jun 2, 2009 #4
    Then you have to consider whether the insulation surrounding the system tells the whole story. It's true that the addition or removal of heat will change the entropy of a system, but - is the reverse also true, that as long as a system is insulated then its entropy in constant? Or are there some other events that will change the entropy?
     
  6. Jun 2, 2009 #5

    Andrew Mason

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    Entropy change is determined by the integral of dQ/T over the reversible path between two states. This is not a reversible path, as you have pointed out. Find a reversible path between these two states and calculate the integral over that path.

    AM
     
  7. Jun 2, 2009 #6

    Hao

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    An abstract (not necessarily better) way to think about this starts from
    dU = T dS - p dV.

    For an ideal gas, dU = 0 (Things are a little more complicated for real gases).

    Hence, T dS = p dV , and
    [tex]\frac{dS}{dV} = \frac{p}{T}[/tex]
    As pressure is positive, and temperature is positive, we can see that Entropy increases as volume increases.

    Inserting the ideal gas equation for p will allow us to integrate for S.
     
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