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Introductory Thermodynamics and mathematical consistency?

  1. May 30, 2009 #1
    I have a few extremely simple questions about thermodynamics... I'm trying to self study it, but a few things just don't add up.... I'm currently working on the first law =/

    1. My book defines work as an indefinite integral:
    [tex]w = -\int p_{ext}dV[/tex]
    However, it then states that, if the external pressure remains constant:
    [tex]w = -\int p_{ext}dV = -p_{ext}\int dV = -p_{ext}v^{v_{f}}_{v_{i}} = -p_{ext}\Delta V[/tex]

    It might sound like a stupid question, but I can't quite figure out why the author defined work as an indefinite integral instead of a definite integral. Is this just bad notation or a typo?

    2. The books mentions that work = 0 for free expansion. However, under what circumstances would you actually have free expansion? If a gas expanded into a container with p = 0 atm, the pressure inside the container would rise as the rest of the gas is expanding into it =/

    3. w = work, U = internal energy.
    [tex]\int dw = w,
    but
    \int dU = \Delta U[/tex]
    Why are these two treated differently thermodynamically and mathematically? Am I missing something, or is this just, once again, bad notation?

    That's it for now. Thanks!
     
    Last edited: May 30, 2009
  2. jcsd
  3. May 30, 2009 #2

    atyy

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    Work is a path integral, and the path, not just start and end points must be specified.

    Internal energy is a state variable, and the change in internal energy is path independent.

    In free expansion, the increase in pressure is so fast and variable, that it is not well defined at each point in the container at every time point.
     
  4. May 30, 2009 #3
    3. The delta U has to do with the fact that a number is never put on internal energy itself, which would include forms of energy that are not changing in these problems, such as chemical and nuclear energy. There is only interest in a change in internal energy. This makes it allowable to pay attention only to the kinetic energy of the gas particles. The integral of a differential change in internal energy is the overall change in internal energy.
     
  5. May 31, 2009 #4

    Mapes

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    The limits are implied until the integration is actually performed. This could be viewed as sloppy notation, but I'd disagree: leaving off the limits makes for a simpler, more general expression. There's little chance for misinterpretation.

    Even with a non-zero pressure, no work is done. That's the point of the free expansion thought experiment: work and heat are zero, so an ideal gas doesn't change temperature. It doesn't matter in this interpretation if the pressure in the container isn't exactly zero.

    The notation is correct. Work is a process variable that characterizes energy transfer; energy is a state variable that can change.

    Here's an analogy I've found helpful: energy is like the level of a lake, work the water added by rain, heat the water added by a stream. Without knowing the depth of the lake, all we can measure are changes relative to arbitrary marks on the shore (i.e., [itex]\Delta E[/itex]). The water added by each method isn't a change, though, it's just work and heat (w and q). Finally, once the water is added to the lake, its origin as rain or stream water isn't retained. Similarly, there's no such thing as work or heat within a body; work or heat are classifications of energy transfer.
     
  6. May 31, 2009 #5
    Now the question about work is bugging me too.

    When a gas expands into a vacuum, it doesn't have to do any work on its environment, because it doesn't have any force opposing it, unlike the cases when a gas expands against the opposing pressure of the atmosphere, and unlike the cases when the expanding gas has to displace a piston of an engine which pushes back.

    To rephrase question #2 of the original post: After some of the gas has moved into the new space, doesn't that portion of the gas that is just entering the new space have to do work to overcome the pressure exerted by that portion of the gas that has already moved into the new space? To take a microscopic perspective -- did we choose to neglect the number of molecules that have had sufficient time to reach the opposite wall of the container, bounce back in the reverse direction, and then collide with an entering molecule?

    At first I was tempted to answer it as follows: we must be relying on the ideal gas assumption that molecules have zero volume and therefore have zero probability of colliding with each other, and that all collisions are between molecules and the walls.

    However, now I reject that answer. Regardless of how often they may occur, we can simply forget about collisions between molecules, because it's a zero sum game when all of those collisions are perfectly elastic. The only collisions that aren't zero sum would be those of molecules hitting the wall, which is perceived by the external environment as a single macroscopic property: pressure. The container wall is not being displaced, therefore no work is done by the gas on its environment.

    Summary: Molecules colliding with each other as they fill up the new space will affect how much time it takes for the gas to expand, but it has no effect on the fact that the gas cannot doesn't lose any internal energy by means of doing work on its environment. Conclusion: The expression W=0 is correct.

    I will be grateful to anyone who criticizes what I said.
     
    Last edited: May 31, 2009
  7. Jul 9, 2009 #6

    jmb

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    Sorry for the delay in posting this!...

    scorpion990: The point is that the net work done is zero during free expansion. You are correct that as the container is filled the pressure starts to rise. But it is being filled by the same gas doing the expanding, so the only thing it is doing work against is itself. Parts of it may lose energy (do work) but only at the expense of transferring this energy to other parts of it (the parts being worked upon).

    From this alone it should be clear that the net work done is zero. But technically classical thermodynamics can only be applied to states of thermodynamic equilibrium: so we should consider the case when the system has returned to equilibrium in the new container. At that point its energy will have been redistributed across all its molecules and (since the container is thermally isolated and no energy can enter or leave it) the total internal energy will be the same as in the initial state: from which it is clear (1st law of thermodynamics, or just conservation of energy) that the net work done must be zero.

    It gets more interesting if we consider free expansion into an already present identical gas. Consider a thermally isolated vessel with a movable membrane (initially fixed) dividing it into two parts. Put two identical gases either side of the membrane, let them be of equal internal energy but let one side (call it gas A) be at a higher pressure than the other side (gas B). Now allow the membrane to move: gas A will expand against the membrane doing work against gas B. Since the container is thermally isolated this work must result in an increase of the internal energy of gas B at the expense of a decrease in the energy of gas A until both sides of the membrane reach the same pressure. Now that the pressures are equalised we can just remove the membrane. This will cause the two identical gases to mix and the system will equilbriate by redistributing the total internal energy between all the molecules. However since the total internal energy of the system is unchanged (the loss of A was matched by an equal gain in B) once we reach equilibrium we find that the internal energy of the molecules of gas A and gas B (not that we can actually separate them any more) is the same and back at its original value: no net work has been done. Now the clever part is: the internal energy of the system is a path independent quantity, so we would get the same result if we had removed the membrane before we allowed the pressures on either side to equilibriate: there is still no work done when we allow a gas to freely expand into another identical one.

    We could get more complicated still and say what if one gas expands into a different (distinguishable) gas. What do you think would happen then?


    mikelepore: I broadly agree with your conclusions, but it is more general than that (see my answer to scorpion990 above). In particular it doesn't matter if the collisions between gas molecules are inelastic. Fundamentally if no energy can leave the system (i.e. thermally isolated) then the total internal energy of the system cannot change and so once the gas reaches equilibrium again after the expansion process (no matter how microscopically messy) the internal energy of the molecules will be unchanged --- implying no net work done (1st law).

    The standard formal proof for free expansion basically just consists of setting up a thought experiment that states the above mathematically --- let me know if you want me to write it out here.
     
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