1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Entropy change : pure mixing of gases

  1. Jan 25, 2017 #1
    Consider three identical boxes of volume V. the first two boxes will contain particles of two different species 'N' and 'n'.

    The first box contains 'N' identical non interacting particles in a volume V. The second box contains 'n' non interacting particles. The third box is the result of mixing of these two boxes.

    Note that the particles are non interacting and the volumes of all the boxes are identical. The total energy which is the just the kinetic energy is conserved.

    The change in the entropy for such a process is zero from either statistical or thermodynamical calculations.

    The SMI of the first box is:

    If we assume that there are M cells in volume V then the number of configurations are M^N(if the particles are distinguishable) and if the particles are indistinguishable then the no. of configurations are M^N / N!

    hence, SMI ( 1 ) = log M^N / N!

    similarly, SMI ( 2 ) = log M^n / n!

    and , SMI ( 3 ) = log M^( N+n ) (distinguishable particles are present)

    now Δ SMI =SMI(3) - ( SM1( 1 ) + SMI( 2 ) ) = log M^( N+n ) - log M^( N+n ) / N! n! = log N! n!

    And therefore ΔS = K ln 2 (ΔSMI) which turns out to be not zero.

    Why is there a discrepancy?. Is there some conceptual mistake in the SMI calculation.

    In the book, the author explains that the change in SMI is zero since there is no change in volume or energy and hence the locational SMI change and the momenta SMI change are zero...please help.
  2. jcsd
  3. Jan 25, 2017 #2


    User Avatar
    Science Advisor
    Homework Helper
    2017 Award

    If, by 'process', you mean the mixing, then: no.
  4. Jan 25, 2017 #3
    as long as the volume of the box after mixing remains the same and for an ideal gas, the change in entropy is zero. But, if the volume of the third box is larger then ofcourse, entropy increases...

    note: Referred from "Entropy and the second law, interpretations and mis-interpretations " Ariah Ben Naim (2013) pg. no. 144
  5. Jan 25, 2017 #4


    User Avatar
    Science Advisor
    Homework Helper
    2017 Award

    So: what is the 'process' ?
  6. Jan 25, 2017 #5


    User Avatar
    Science Advisor

    "SMI ( 3 ) = log M^( N+n ) (distinguishable particles are present)"
    This is not correct. The particles are not all distinguishable from each other. The Ns are indistinguishable from each other , and so are the ns, but the Ns are distinguishable from the ns. The correct expression is log M^(N+n)/N!n! (as distinct from log M^(N+n)/(N+n)!, which is what it would be if all the particles were indistinguishable).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted