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Homework Help: A simple question in thermodynamics

  1. Sep 24, 2012 #1
    I have a cylinder with gas in it. I can make it expand in two ways: spontaneously isobaric process or reversible isothermal process.

    I understand W and Q for each process are different, but is ΔE the same? If not - why?
  2. jcsd
  3. Sep 24, 2012 #2

    Simon Bridge

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    There are more than two ways (mechanical processes) for expanding the gas.
    I think you need to more carefully describe the system you are looking at.
    As it stands, it is not clear what you are asking.
  4. Sep 25, 2012 #3
    I'm sure there are more than two ways, but I'm asking about these two.

    First is to relase the piston at once (getting a new constant external pressure).
    Second is to release it in very small steps (thus temprature remains constant).

    As far as I see we have the same Ef and Ei, because it should not matter in which way we reached those states, so delta E should be the same. I know I'm wrong, question is why.
  5. Sep 25, 2012 #4

    Simon Bridge

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    See .. you said nothing about a piston before ;)

    The first is a free expansion - the gas is not in thermal equilibrium with itself through the expansion. The second is just where care is taken to stay in a well-defined state throughout the expansion.

    So what was the question?
  6. Sep 25, 2012 #5
    The spontaneous isobaric process occurs in almost two seperate phases, in which the gas first rapidly expands, and, at the end of this phase its temperature has dropped. After this first phase, if the cylinder is being held in contact with a temperature bath at the initial temperature, the gas will heat up at essentially constant pressure to the initial temperature. The final state will be the same as the isothermal reversible expansion, so the change in internal energy will be the same for both processes. But the heat added and the work done will not be the same for the two processes. Only their difference will be the same.
  7. Sep 26, 2012 #6
    But on both processes I start with same Pi and Vi, and end up with same Pf and Vf.
    I thought E is a state function, so it should not matter how I reached from initial to final state.
    If it's an ideal gas, then for same P and V I must get same T, regardless of any bath...
    What am I missing?
  8. Sep 26, 2012 #7
    By a spontaneous isobaric expansion process, I've been assuming you mean that the pressure of the surroundings is held fixed at Pf.

    In the isobaric process, you need to add heat to get to the same final volume and final temperature (equal to the initial temperature). The amount of heat you have to add is Pf Δ V. This has to come from outside the system, and can be provided by a heat bath at temperature T. This heat added is just equal to the work done on the surroundings, so ΔU = 0.

    In the reversible isothermal process, the amount of work you do is the integral of P dV. This is RT ln (Vf/Vi). This is also the amount of heat you have to add to maintain isothermal operation, and has to come from outside the system, and can be provided by a heat bath at temperature T. In this situation, ΔU is likewise equal to zero.

    So, in both cases, the change in internal energy is zero.
    Last edited: Sep 26, 2012
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