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I Entropy of ideal gas: Volume it *has* or is *allowed* to have?

  1. Jan 26, 2017 #1
    Hi.

    If an ideal gas of ##N## particles is allowed to expand isothermically to double its initial volume, the entropy increase is
    $$\Delta S=N\cdot k_B \cdot \log\left(\frac{V_f}{V_i}\right)=N\cdot k_B \cdot \log\left(\frac{2V}{V}\right)=N\cdot k_B \cdot \log\left(2\right)\enspace .$$
    This can also be derived with Boltzmann's entropy formula ##S=k_B \cdot \log(W^N)##, assuming the particles are independent and all have the same number of microstates ##W## which doubles as ##V## does. So after the expansion, the number of microstates of the whole gas is ##(2W)^N## and
    $$S=k_B \cdot \log\left(\left(2W\right)^N\right)=N\cdot k_B \cdot \log\left(2W\right)\enspace ,$$
    also leading to the entropy difference ##\Delta S=N\cdot k_B \cdot \log\left(2\right)##.

    Now there will be fluctuations where more than ##N/2## particles are in the initial half of the containment (Poincaré's recurrence theorem even states there will be moments where all particles are back there). I read that this implies a (temporary) reduction in entropy. This is somehow what
    $$\Delta S=N\cdot k_B \cdot \log\left(\frac{V_f}{V_i}\right)$$
    says if ##V_i## and ##V_f## are taken as the volumes the gas actually has at a moment in time – and not the volumes the gas is allowed to have.

    This gets even more confusing if we use Boltzmann's formula: The number ##(2W)^N## contains all microstates in ##2V##, including the ones where there's an imbalance of particle numbers in the halves. Hence those microstates are part of the macrostate with entropy
    $$S=N\cdot k_B \cdot \log\left(2W\right)\enspace ,$$
    so it will not change even when all particles are in one half of the containment.

    As mentioned before, I think this discrepancy results in using the volume the gas actually has at a point in time versus the volume it is allowed to have. But is the first even a well-defined property, especially in thermodynamic equilibrium that is needed for entropy to make sense?
     
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  3. Jan 26, 2017 #2

    Stephen Tashi

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    Entropy is only defined for equilibrium states, so, for such a fluctuation to occur, the majority of particles would have to be in the initial half on the containment and also be in equilibrium for the gas to have a defined entropy in such a state. Unless you slammed a door shut behind them after the majority of particles entered the initial half of the containment, they wouldn't be in equilibrium. (A Maxwell's Demon could close the door.)

    If your definition of "at a point in time" is that you know the exact position and velocity of each gas molecule at that time, it isn't clear how to define the volume of a gas in that situation To take an extreme example, suppose the gas consists of 3 molecules in a 1 cubic meter chamber. If they are at known locations then how is the volume of the gas defined? Is it 1 cubic meter regardless of whether the molecules are near the walls of the chamber? Or is it the volume of a flat triangular shaped box, with one molecule near each corner of the box?

    One could speak of a gas "at a point in time" and keep the model that exact position and velocity of each gas molecule has some probability distribution instead of being precisely determined. Classical thermodynamics tries to juggle two contradictory concepts. On the one hand, a gas is conceptually a set of particles with definite positions and velocities, and on the other hand, the major theorems of thermodynamics do not apply to any one particular set of such particles, but rather to "ensembles" of sets of particles. So when you speak of a "gas", you must decide whether you mean an ensemble or a particular set of particles.
     
  4. Jan 29, 2017 #3
    Ok thanks, your answer agrees with my understanding of a reasonable notion of entropy. But wasn't it exactly non-equilibrium situations like the ones I mentioned that made Zermelo and many others question the second law of thermodynamics? I don't think Zermelo simply forgot that entropy is only defined in thermodynamic equilibrium.
     
  5. Jan 29, 2017 #4

    Stephen Tashi

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    I'm not familiar with Zermelo's questions. Are they technical questions or philosophical questions?
     
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