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Work of an isothermal expansion

  1. Nov 9, 2014 #1
    I made a thread on this asking the general question in the other forum, but I don't know how to delete that thread.
    1. The problem statement, all variables and given/known data
    An isothermal expansion from Vi = 10.0 L, Pi = 2.46 atm against a constant external pressure until the Pf = 0.246 atm.
    How much work is done by the gas in joules?

    2. Relevant equations
    [itex]w=-\int P\,dV[/itex]
    PV = nRT
    3. The attempt at a solution
    I tried using the work integral, but that P is the external pressure, and that would mean I'm finding the work of the external gas, right? I'm trying to find the work done by the expanding gas, which is positive, correct? So I just used
    W = nRT ln(Vf/Vi)
    That gives me a positive work of 5743 J.
    Is that correct? Thanks.
     
  2. jcsd
  3. Nov 9, 2014 #2

    stevendaryl

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    Staff Emeritus
    Science Advisor

    Your title says "isothermal expansion", but the problem statement says constant external pressure. Are you sure it's not constant external temperature? Usually, when people talk about reversible isothermal expansions (or contractions), the internal and external pressures are assumed to be the same. If they aren't the same, then the expansion or contraction is irreversible.

    But if the external pressure is really held constant, then the work done by the expanding gas is computed very simply:

    [itex]w=+\int P_{ext} \,dV = P_{ext} (V_f - V_i) [/itex]

    The work done BY the gas is determined by the external pressure, not the internal pressure (and it's a + sign for work done by the gas, and - sign for work done ON the gas).
     
  4. Nov 9, 2014 #3
    To expand on what stevendaryl is saying, this is obviously an irreversible expansion. In such a situation, when they say "isothermal", they mean that the gas temperature is the same in the initial and final equilibrium states of the gas, but not necessarily during the expansion. In fact, during the irreversible change, the temperature and/or the pressure of the gas may not even be spatially uniform within the system, so use of the ideal gas law is not of much use (except at the initial and final equilibrium states). The only place where you can be sure that the gas pressure is equal to the external pressure Pext is at the interface between the system and the surroundings (e.g., the face of the piston). So, irrespective of whether a process is reversible or not, the equation stevendaryl gave always determines the amount of work done by the gas on the surroundings.

    Chet
     
  5. Nov 9, 2014 #4
    Thanks for the responses. Yes, it's an isothermal expansion and the external pressure is constant. So I take it the internal pressure change is irrelevant? I guess that information was only given so that I could use the ideal gas law to find the final volume.

    I forgot to quote part of it. It says at the top that "one mole of an ideal gas undergoes the following 2 changes". The question I asked was about the first change it undergoes. The second asks about the change in entropy, which I did correctly.

    With using the integral, I get 911.8.8 J as my final answer (using 1 atm as the constant external pressure, which is what he told us to use if it's not given). But I think I had that as my answer when I went to the teacher's office and he said it was wrong. He had the key for the homework and he did something with the ideal gas equation. He had an answer, I think, similar to what I had as my answer given in my first post. You guys sure this is correct?
    I hate to get it wrong, but I also hate to use the wrong technique to get his wrong answer, if his answer is indeed wrong.
     
  6. Nov 9, 2014 #5
    To get the entropy change from an initial equilibrium state to a final equilibrium state, you can't use dQ/T for the actual path that the system follows if the process is irreversible, because that will give the wrong answer. You need to dream up a reversible path between the same two equilibrium states, and evaluate dQ/T for that path. This is essentially what you did in your original calculation. But that calculation will not give the correct W for the actual path.

    So, to get the actual work, you have to evaluate the actual path, but this won't give you the correct entropy change. To get the correct entropy change, you have to evaluate a reversible path, but this won't give you the actual work. Only if the process path is reversible can you get both using the actual path.

    Chet
     
    Last edited: Nov 9, 2014
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