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Change in entropy in an isolated system

  1. Aug 24, 2011 #1
    1. The problem statement, all variables and given/known data

    Derive an equation for the change in entropy that occurs in an isolated (micro-canonical) system containing N particles, if an adiabatic expansion from volume V1 to volume V1 takes place. Show that the number of microstates is given by V^N.


    2. Relevant equations

    Entropy S = K[itex]_{b}[/itex] ln [itex]\Omega[/itex]
    Where [itex]\Omega[/itex] is multiplicity, the number of microstates for distinguishable partciles= N!/[itex]\Pi[/itex][itex]_{i}[/itex]n[itex]_{j}[/itex]!

    3. The attempt at a solution

    Ok i'm not too sure where to start. I know that dQ = 0 as this is an adiabatic expansion.
    Meaning dU = dW = - NK[itex]_{b}[/itex]T ln (V2/V1), but i'm not sure if this helps anything.

    I also know that a microcanonical system is thermally isolated and has a fixed N. So would thermally isolated mean dT = 0? in which case dU = 0 ... confused.

    please help!
     
  2. jcsd
  3. Aug 24, 2011 #2

    lightgrav

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    Homework Helper

    thermally isolated, so Q=0, means all the Work goes into changing the Temperature.
    How much Work is done during an adiabatic expansion? (as a function of Volumes)
     
  4. Aug 25, 2011 #3
    Is my answer in my first post incorrect?

    "I know that dQ = 0 as this is an adiabatic expansion.
    Meaning dU = dW = - NKbT ln (V2/V1)"

    as dU= dW = (integrating from V1 to V2) - pdV
    where P = NKbT / V
     
  5. Aug 25, 2011 #4

    jambaugh

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    Science Advisor
    Gold Member

    Thermally isolated means no heat enters or leaves, thus the expansion will change the temperature.
     
  6. Aug 25, 2011 #5
    Ok thanks, but I still don't understand how what to do for this question. :S
     
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