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B Entropy change when the number of particles decreases

  1. May 27, 2017 #1

    I found following exercise in a high school textbook:
    "Compute the entropy change in following process:"
    The solution is
    "The number of particles decreases from ##N_1## to ##N_2=N_1/2##. Hence the entropy decreases by
    $$\Delta S=-k\cdot N_1\cdot \ln{2}\enspace ."$$

    I can't quite follow the argument here. Assuming the particles in each picture are non-interacting and have the same number of microstates ##\Omega##, I get
    $$S_1=N_1\cdot k\cdot \ln{\Omega}$$
    $$S_2=N_2\cdot k\cdot \ln{\Omega}=\frac{N_1}{2}\cdot k\cdot \ln{\Omega}=\frac{S_1}{2}$$
    from which I can see that the entropy decreases, but cannot compute the difference since I don't know ##\Omega##.
  2. jcsd
  3. May 30, 2017 #2


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    Science Advisor

    I have doubts that the exercise makes much sense. E.g. if this is meant to be a chemical reaction in an isolated container, the system will heat up and entropy will increase rather than decrease.
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