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Homework Help: Microstates of two (similar looking) configuration

  1. Sep 26, 2012 #1
    1. The problem statement, all variables and given/known data
    Question: A box contains 1 mole of a gas. Consider two configuration :
    a) each half of the box contains half the molecules and
    b) each third of the box contains one-third of the molecules.
    Which configuration has more microstate?

    2. Relevant equations
    The only equations to be used here are i think of permutations and combinations (algebra).

    3. The attempt at a solution
    I tried calculating the two as:
    1 mole = 6.2 * 1023
    For a) (6.2 * 1023)!/{(6.2 * 1023 /2)! }2
    = e(0.7* 1023)

    For b) (6.2 * 1023)!/{(6.2 * 1023 /3)! }3
    =e(1.1* 1023)
    (I have simplified the factorials using the stirling's approximation ln N! = N lnN - N)
    Showing that the configuration of b) has more micro states ....and that implies more entropy.
    But when i looked at the question again the two configuration are looked the same, that is they are saying the molecules are evenly spread.... but i guess here lies the error.
    When we say that the each half contains half mole we never did say that they are evenly distributed in the the half volume they have, and hence this configuration does not imply that it is the same as the next one... Am I on the right track?
    (While writing it occurred to me what i wrote in the above paragraph, that solved my doubt to some extent but i wanted to be sure if i am doing it right or not. Also i have to mention please try and keep the answer simple because i have not studied statistical mechanics much and have a rough idea of micro states and entropy)

    But then as far as i know if the initial and final states are equilibrium states then entropy is a state function, hence for the two cases shouldn't the entropy be same?
    Any help is appreciated.
    Last edited by a moderator: Sep 27, 2012
  2. jcsd
  3. Sep 27, 2012 #2
    So what you are calculating here is the number of ways you can pack your particles in 2 or 3 boxes. So if the particles are distinguishable and the position of the particles is not relevant, then you are correct.

    But what I thought when I saw the problem is that putting the particles in boxes constraints their position. So since you know that there are as many particles in each of the boxes, some configurations are not allowed any more, like for example having all particles in one corner of the box.

    So all in all, if you think the relevant degree of freedom is which box does a particle belong to, then increasing the number of boxes increases the degrees of freedom. On the other hand, if the relevant degrees of freedom are particle position and momentum (this is more common for sure), then the degrees of freedom should go down as you increase the number of boxes.
  4. Sep 27, 2012 #3
    I guess that what i did calculate. But since it was about 1 mole of a gas, the particles should be indistinguishable. What effect this have?

    I think putting them in 2 or three boxes constraints the volume N number of particles must occupy or be in.

    I didn't get you on this. Can you be a little more elaborate about what you were trying to say here.
  5. Sep 27, 2012 #4
    Well, if the particles are indistinguishable, then why do you care which particle is in which box? There's no way to tell! Your microstate is just that N/2 particles are in one box and N/2 particles are in the other, so the number of microstates is one!

    Indeed, but your calculation did not care at all about the volume. All you took into account is which box each particle is in (which, as I said above, is not the interesting bit unless you take the particles to be distinguishable)

    Let's simplify. Suppose you have 2 particles and two positions a particle can be in. Call them position a and b. Let's also suppose that the particles are bosons, so they can be in the same position in the same time, and that they are distinguishable. Then there are four microstates:
    x1 = a, x2=a;
    x1 = a, x2=b;
    x1 = b, x2=a;
    x1 = b, x2=b.

    Suppose now we put in the box and require that one particle is in a and one particle is in b. That reduces the number of microstates to
    x1 = a, x2=b;
    x1 = b, x2=a!

    Now the particles are no longer allowed to clump to one side. Same is of course true for a mole of particles and a continuum of possible positions, but then the effect is more subtle.

    So what's the effect of being distinguishable here? Well, suppose you can't tell them apart. Then the microstates x1 = a, x2=b and x1 = b, x2=a collapse into a single states where one particle is at a, and another is in b. Then you initially have three possible states: (a,a), (a,b), (b,b) and adding the boxes reduces this down to just (a,b).

    It might be useful for you to check what happens with 4 particles, and two positions. It should still be fairly straightforward (although the number of permutations grows out of hand fast)
  6. Sep 27, 2012 #5
    So that means the calculations i did was for distinguishable particles...right?

    And how do we take into account the volume or position constraint for such a large system, is it suppose to be the extension of what you have just explained above?
  7. Sep 27, 2012 #6
    Right, because if you couldn't tell the particles apart, then you only had one microstate.

    It does seem pretty difficult to do the calculation for a large number of particles, starting from first principles. If you were given some more information about the gas, you could use thermodynamics to find the change of entropy. However now you do not know anything of the sort. Luckily, you are not asked to calculate the change of entropy. You are simply asked which configuration has more microstates (larger entropy). This you can answer even without calculating it explicitly.
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