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Heat Capacity of Distinguishable Particles

  1. Apr 22, 2016 #1
    1. The problem statement, all variables and given/known data

    A system of N distinguishable particles, each with two energy levels. The lower energy level has energy equal to zero, and the higher energy level has energy ##\epsilon##. The higher energy level is four fold degenerate. Calculate the heat capacity.

    2. Relevant equations


    3. The attempt at a solution

    I have an expression for the heat capacity, but I think I have a problem that it does not go to zero as temperature goes to zero, and was hoping to get some help on this matter.

    The partition function for such a system is given by

    ##Z = \Sigma_s g_s e^{-\beta E(s)} = 1 + 4e^{- \beta \epsilon}##, where ##\beta = \frac{1}{k_B T}##

    The average energy of the system ##\bar{E} = - \frac{1}{Z} \frac{\partial Z}{\partial \beta} = \frac{4 \epsilon e^{- \beta \epsilon}}{1 + 4 e^{- \beta \epsilon}} = \frac{4 \epsilon}{e^{\beta \epsilon} + 4}## and the heat capacity is given by ##C_v = \frac{\partial \bar{E}}{\partial T}##, so substituting back in the expression for ##\beta##

    ##C_v = \frac{\partial}{\partial T} \frac{4 \epsilon}{e^{\frac{\epsilon}{k_B T}} + 4} = \frac{4 \epsilon^2 e^{\frac{\epsilon}{k_B T}}}{k_B T^2 (e^{\frac{\epsilon}{k_B T}} + 4)^2}##

    It looks to me like my expression blows up at T=0.

    Any help is appreciated,

    thanks!
     
  2. jcsd
  3. Apr 22, 2016 #2

    TSny

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    Homework Helper
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    Can you show how you are taking the limit of your expression as T goes to zero?
     
  4. Apr 23, 2016 #3
    Your comment has made me realise I was being stupid, and just plugging zero into my equation.

    I have now used l'hopital's rule and seen that the limit is indeed zero. Thanks for the nudge :)
     
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