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Reversible vs. Irreversible change (gases)

  1. Sep 25, 2011 #1
    1. The problem statement, all variables and given/known data

    I know how to do the problem so I don't need help actually solving it, I just don't understand the concepts.

    A sample of 4.5g of methane occupies 12.7L at 310k.
    a) Calculate the work done when the gas expands isothermally against a constant external pressure of 7.7 kPa until its volume has increased by 2.5L
    b) Calculate the work that would be done if the same expansion occurred reversibly.

    2. Relevant equations

    (1) dw=-pdV
    (2) w= nRT ln (V final/V initial)

    3. The attempt at a solution

    For part a you use equation 1. For part b you use equation 2.

    I don't get why part a is not reversible. In my textbook equation 1 is listed with the label "REVERSIBLE EXPANSION WORK". So I use it to calculate the work, but then in part b it's like, now do it it it was reversible, and I just got really confused. If equations 1 and 2 are both reversible equations, when do I use them both?

    Is part a not reversible because the pressure is staying constant? I just found another equation that is w=-p(external) (delta)V. I see that equation 1 and the one I just wrote are different somehow. But my book says:

    "To achieve reversible expansion we set p(ex) equal to p at each stage of the expansion."

    So both processes, reversible and irreversible are isothermal, correct? So the different lies in what is happening with the pressure? How do I tell from the wording of part a that it is not a reversible process?


    EDIT: Is the difference that in the irreversible process they just have an arbitrary external pressure that stays constant, but when it is reversible, the external pressure is constant but always equal to the internal pressure?

    How would you physically set this up? For the second one you could just have a piston that moves to keep the pressure constant. But what about the first? How can you ever have an external pressure that isn't equal to the internal pressure, without it expanding or contracting to match that pressure.... unless that's the crux of it, that the irreversible process expands or contracts to match the constant pressure, which will be the final pressure of the gas.(or maybe I'm just completely off base here)
    Last edited: Sep 25, 2011
  2. jcsd
  3. Sep 25, 2011 #2


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    Yes. For a reversible process, the system must always be in equilibrium with its surroundings. This means that the surroundings have the same pressure as the surroundings at all time.

    Consider a piston containing a compressed gas that is locked in place something to hold the piston down. You remove this piece, allowing the piston to shoot upward until the pressure inside the piston equals the atmospheric pressure.
  4. Sep 25, 2011 #3
    Ok so then my question becomes how would I know that in part (a) it is not reversible? I am just supposed to infer it from the question in part (b)?

    I'm just a little confused because it seems like sometimes they will use dw=-pdV, w=-nRT or w= nRT ln (V final/V initial) for both types of processes.
  5. Sep 26, 2011 #4


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    If you calculate the pressure of the system, you will see that it is much greater than 7.7 kPa. Therefore, the system is not in equilibrium with the surroundings, so the expansion will not be reversible.

    The most general equation for work is dw = - pextdV. Because this expression is a differential, you need to integrate it over the change in volume to obtain the work. For a process occurring at constant pressure, the integration gives you w = -pextΔV. For an isothermal reversible expansion, pext = p = nRT/V. Plugging this into the equation for dw and integrating gives you w = nRT ln(Vfinal/Vinitial).
  6. Sep 26, 2011 #5
    So why then for this question:

    A sample of 1 mol of water is condensed isothermally and reversibly to liquid at 100 degrees C. What is w?

    And they use w= nRT (actually, they use the squiggly equals sign saying it is approx. equal. Does the ln term go away or something? If water is condensing then it would go from a big volume to a small volume so ln(big number/small number)= some random positive number....)

    what happened to ln(Vfinal/Vinitial)? Shouldn't they use the equation that had that term?

    I've been trying to learn this for days now and my quiz is tomorrow.... every time I think I understand it something new pops up.
    Last edited: Sep 26, 2011
  7. Sep 26, 2011 #6


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    Always start from the fundamental equation for work: dw = -pextdV.

    The first question you should ask is whether the external pressure is constant or not. If the external pressure is constant, integrating the expression is easy and:

    w = -pextΔV

    If the pressure is not constant, then you need some way of writing the external pressure as a function of V, so that you can perform the integration.

    For the reversible, isothermal condensation of water at 100oC, the pressure must be constant at 1 atm (since the condensation of water is reversible at 100oC only when the pressure is 1 atm). Therefore, all you need to figure out is the change in volume. Since the volume of the gas is much greater than the volume of the liquid, you can ignore the volume of the liquid and assume that ΔV ~ the volume of the gas.
  8. Sep 26, 2011 #7
    Ok I think I can manage that now.

    One last question. I was just doing a new problem:

    1 mol of Ar is expanded isothermally at 22C from 22.8L to 44.8L A) reversible B) against a constant external pressure equal to the final pressure of the gas C) freely. For the three processes, calculate q, w, ΔU and ΔH.

    What I really don't get is that the solution says that ΔU is 0 for each one because "the internal energy of a perfect gas depends only on it's temperature."

    Uh.... what? I've just spent an entire chapter doing problems with ΔU=q+w where w is expansion work, i.e. if there is a change in volume then the internal energy changes. Now this question is saying a change in volume doesn't change U? I mean, when you calculate w in the problem you get a value... how then can ΔU be 0? Like I said, just when I think I understand everything...

    EDIT: I just read that "the heat supplied by the surroundings is equal to the work done by the system to keep the process isotherma; and these two cancel each other out to give delta U = 0".


    for: A sample of 1 mol of water is condensed isothermally and reversibly to liquid at 100 degrees C.

    When they calculate the change in U, it is not zero. So my next thought was, well if the work to keep it isothermal cancels with the work done by expansion, then maybe ΔU=q required to do the condensation? So you do U=ΔH-w to get just the work that went into changing the internal energy?

    And if it is not isothermal, then a change in volume DOES change internal energy right? I'm just all over the place right now.

    I think my textbook is just intentionally screwing with me.
    Last edited: Sep 26, 2011
  9. Sep 27, 2011 #8


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    (emphasis mine)

    The assumption that internal energy depends only on temperature applies only to perfect gasses. If you have a gas going to a liquid (not a gas), this assumption does not apply, and you cannot make the assumption that internal energy depends only on temperature.

    Why is this? The internal energy of a substance is composed of two parts: the energy from the interactions between the particles composing the substance and the kinetic energy of the particles composing the substance. For a perfect gas, we assume that there are no interactions between particles, so the internal energy depends solely on the kinetic energy of the particles (which is directly proportional to temperature).

    A liquid, however, has fairly strong interactions between the particles, so these interactions contribute greatly to the internal energy of a liquid. Even though the kinetic energy of the particles does not change when you convert a gas to a liquid at a constant temperature, the energy from the interaction between particles does change. Therefore, the internal energy changes.
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