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Is there an entropy increase?

  1. Dec 5, 2007 #1


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    Suppose we have box with a volume V which is divided down its middle.
    So the box is split in two sides with equal volume (V_1 = V_2) and gas molecules (n_1 = n_2).

    Now, we remove the partition and the gas from both sides mix. Is there an entropy increase in the system?
    The whole process is reversible, since we can re-establish the initial state by sliding down the partition again.
  2. jcsd
  3. Dec 5, 2007 #2
    I think it depends on how large the box is compared to the mean free path of the gas. If the half box width is larger MFP, then entropy is unchanged. If it is smaller then the entropy increases.
  4. Dec 5, 2007 #3


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    Ok, suppose the half box width is larger than MFP.
    So far, we agree.

    Now, what if we have two distinguishable gases in the two sides (for example: helium in V_1 and neon in V_2)? Upon removal of the partition there is a net entropy increase, since both gases expands to the double volume, that is, V = 2V_1 = 2V_2 (from V_1 and V_2, respectively). This process is not reversible, since the gases has been mixed.
    So in this case, we have an entropy increase.
    How do we explain this?
    Last edited: Dec 5, 2007
  5. Dec 5, 2007 #4

    Andrew Mason

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    Are the two sides at the same Temperature? If not, there will be an increase in entropy. If they are at the same temperature, there is no entropy change as there is no heat flow.

    If n1 = n2, then assuming an ideal gas, P1/T1 = n1R/V1 = n2R/V2 = P2/T2

    If [itex]T_2 \ne T_1[/itex] then there is a flow of heat from the hotter gas (say n1): dQ = -n1CvdT. This flows into the gas on the other side (n2): dQ = n2Cv(T2+T1/2)

    The magnitude of the entropy loss of the hotter side (dS = dQ/T) will be less than the entropy gain of the colder side, so the entropy will increase.

  6. Dec 5, 2007 #5


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    Yes, the two side are at the same T. They are perfectly symmetrical.
    The real problem is when the two gases in each side is of two different kinds. In that case, there is an entropy increase.
    Do I need elaborate #3 a bit more?
    Last edited: Dec 5, 2007
  7. Dec 5, 2007 #6
    Sure, in the 2nd case, you mix two different gases into one, it's more chaotic so entropy increases.
    Entropy is somewhat linked with energy : G =H-TS. If you want to separate things out of a mixture, you have to consume energy. In the 1st example, you do not really mix them, and when you separate, the enery is zero because the state is unchanged, or entropy is unchanged.
    It is not easy to explain everything, I hope you get my thinking.
  8. Dec 5, 2007 #7

    Chris Hillman

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    Entropy of mixing

    Right, this is known as the entropy of mixing.

    A Markov chain model similar to the famous Ehrenfest urn models can be useful here. Consider balls on a square subset of the integer lattice, [itex]Z^2 \cap \left[ -10, \, 10 \right]^2[/itex], and let each ball move with equal probability one step up, down, left right at each time (with obvious modifcation of the boundary of the square). Start with an initial state with say five black balls to left of x=0 and five white balls to right, and let the simulation proceed.

    Mathematicians have a lot of fun with questions like this: Consider the positive integers, and place white balls on the primes and black balls on the squares. How long does it take to mix up primes and squares? (The first problem is turn this statement into one which makes sense in ergodic theory.) See for example Mark Kac, Statistical Independence in Probability, Analysis and Number Theory, and then see Walters, Ergodic Theory, Springer, 1981. Note we are talking here about correlation and how quickly that decreases.
    Last edited: Dec 5, 2007
  9. Dec 5, 2007 #8


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    Hi LTP,
    This is the question addressed by "Gibbs Paradox". If the gas on each side is different, there is an increase in entropy. If the same, no increase. Do a search on "Gibbs Paradox".
  10. Dec 6, 2007 #9


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    That is way over my knowledge of the subject, but I may have to look into it. Thanks anyway.

    Thanks for the search term, I was not aware that I had already been adressed by Gibbs.
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