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Entropy of the universe

  1. Aug 9, 2008 #1

    A model universe comprises 100 atoms in system 1 and 1500 atoms in system 2. Compute the entropy for the universe when there are 3 atoms on in system 2 and 97 atoms on in system 1 (using sterlings approximation).

    3. The attempt at a solution
    I am able to find the entropy of systems 1 and 2 by initially finding the number of microstates and then the equation:

    entropy = boltzmanns * ln(microstates)

    Just wondering how I would get the entropy of the universe though.
  2. jcsd
  3. Aug 10, 2008 #2


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    Well, if you know how many microstates there are for systems 1 and 2, then how many microstates are possible for the combined system (1+2)?
  4. Aug 10, 2008 #3
    I was unsure of the terminology. So taking universe as systems 1 and 2:

    I can find number of microstates using Ω = C(N,n)
    (i.e. the number of ways of moving n atoms from N sites)

    Thus for system 1:
    Ω = C(1500,3)?
    This is a massive number!

    Am I on the right track?
  5. Aug 10, 2008 #4


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    Yup that's the correct way of doing the calculation. The calculation may be made easier by finding a formula for C(N,n) in terms of factorials, then applying Stirling's approximation.
  6. Aug 11, 2008 #5
    sterlings approximation only helps once I get a value of Ω though correct?
    I cannot even compute C(1500,3)=1500!/3!1497! due to overflow!
  7. Aug 11, 2008 #6


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    Try to calculate ln(Ω) instead.
  8. Aug 12, 2008 #7
    Ok, so I managed to compute C(1500,3) and C (100,97).
    I got 561375500 and 646800 respectively.

    (I wish to keep in terms of boltzmanns)
    Therefore for system 1:
    entropy = K*ln(561375500) = 20.15K

    and system 2:
    entropy = K*ln(646800) = 13.38K

    total = 33.53K?

    Where does Sterling's Approximation come into this?
  9. Aug 12, 2008 #8


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    ln(Ω) = ln(1500!/3!1497!) = ln(1500!) - ln(3!) - ln(1497!) = 1500ln(1500) - 1500 - ln(6) - 1497ln(1497) + 1497

    Here's where sterling's approximation saves you from evaluating horrible factorials.
  10. Aug 13, 2008 #9
    yeah i completed it, thanks for help

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