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7.5: Entropy Can Be Expressed in Terms of a Partition Function

  • Page ID
    202919
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    We have seen that the partition function of a system gives us the key to calculate thermodynamic functions like energy or pressure as a moment of the energy distribution. We can extend this formulism to calculate the entropy of a system once its \(Q\) is known. We can start with Boltzmann's (statistical) definition of entropy:

    \[S = k \ln(W) \label{Boltz} \]

    with

    \[W=\frac{A!}{\prod_j{a_j}!} \nonumber \]

    Combining these equations, we obtain:

    \[S_{ensemble} = k \ln\frac{A!}{\prod_j{a_j}!} \nonumber \]

    Rearranging:

    \[S_{ensemble} = k \ln{A!}-k \sum_j{\ln{a_j!}} \nonumber \]

    Using Sterling's approximation:

    \[\begin{split}S_{ensemble} &= k A\ln{A}-k A - k \sum_j{a_j\ln{a_j}} + k \sum_j{a_j} \\ &= k A\ln{A}- k \sum_j{a_j\ln{a_j}}\end{split} \nonumber \]

    Since:

    \[\sum_j{a_j}=A \nonumber \]

    The probability of finding the system in state \(a_j\) is:

    \[p_j=\frac{a_j}{A} \nonumber \]

    Rearranging:

    \[a_j = p_jA \nonumber \]

    Plugging in:

    \[S_{ensemble}=k A\ln{A}- k \sum_j{p_jA\ln{p_jA}} \nonumber \]

    Rearranging:

    \[S_{ensemble}=k A\ln{A}- k \sum_j{p_jA\ln{p_j}}- k \sum_j{p_jA\ln{A}} \nonumber \]

    If \(A\) is constant, then:

    \[k \sum_j{p_jA\ln{A}} = k A\ln{A}\sum_j{p_j} \nonumber \]

    Since:

    \(\sum_j{p_j} = 1\)

    We get:

    \[S_{ensemble}=k A\ln{A}- k \sum_j{p_jA\ln{p_j}}- k A\ln{A} \nonumber \]

    The first and last term cancel out:

    \[S_{ensemble}=- k \sum_j{p_jA\ln{p_j}} \nonumber \]

    We can divide by \(A\) to get the entropy of the system:

    \[S_{system}=- k \sum_j{p_j\ln{p_j}} \label{eq10}\]

    If all the \(p_j\) are zero except for the for one, then the system is perfectly ordered and the entropy of the system is zero. The probability of being in state \(j\) is

    \[p_j=\frac{e^{-\beta E_j}}{Q} \label{eq15}\]

    Plugging Equation \ref{eq15} into Equation \ref{eq10} results in

    \[\begin{align*}S &= - k \sum_j{\frac{e^{-\beta E_j}}{Q}\ln{\frac{e^{-\beta E_j}}{Q}}} \\[4pt] &= - k \sum_j{\frac{e^{-\beta E_j}}{Q}\left(-\beta E_j- \ln{Q}\right)} \\[4pt] &= - \beta k \sum_j{\frac{E_je^{-\beta E_j}}{Q}}+\frac{k\ln{Q}}{Q}\sum_j{e^{-\beta E_j}} \end{align*}\]

    Making use of:

    \[\beta k=\frac{1}{T} \nonumber \]

    And:

    \[\sum{\frac{e^{-\beta E_j}}{Q}}=\sum{p_j}=1 \nonumber \]

    We obtain:

    \[S= \dfrac{U}{T} + k\ln Q \label{20.42} \]


    7.5: Entropy Can Be Expressed in Terms of a Partition Function is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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