20.5: The Famous Equation of Statistical Thermodynamics is S=k ln W

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    20.5: The Famous Equation of Statistical Thermodynamics is S=k ln W

    A common interpretation of entropy is that it is somehow a measure of chaos or randomness. There is some utility in that concept. Given that entropy is a measure of the dispersal of energy in a system, the more chaotic a system is, the greater the dispersal of energy will be, and thus the greater the entropy will be. Ludwig Boltzmann (1844 – 1906) (O'Connor & Robertson, 1998) understood this concept well, and used it to derive a statistical approach to calculating entropy. Boltzmann proposed a method for calculating the entropy of a system based on the number of energetically equivalent ways a system can be constructed.

    Boltzmann proposed an expression, which in its modern form is:

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

    \(W\) is the number of available microstates in a macrostate (ensemble of systems) and can be taken as the quantitative measure of energy dispersal in a macrostate:

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

    Where \(a_j\) is the number of systems in the ensemble that are in state \(j\) and \(A\) represents the total number of systems in the ensemble:

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

    Equation 20.5.1 is a rather famous equation etched on Boltzmann’s grave marker in commemoration of his profound contributions to the science of thermodynamics (Figure \(\PageIndex{1}\)).

    Black and white portrait of a man with glasses and a full beard, wearing a suit with a high collar.
    A stone monument featuring a bust of a man and an engraved plaque with text, surrounded by greenery.
    Figure \(\PageIndex{1}\): Ludwig Boltzmann (1844 - 1906)
    Example \(\PageIndex{1}\):

    Calculate the entropy of a carbon monoxide crystal, containing 1.00 mol of \(\ce{CO}\), and assuming that the molecules are randomly oriented in one of two equivalent orientations.

    Solution:

    Using the Boltzmann formula (Equation \ref{Boltz}):

    \[S = nK \ln (W) \nonumber \]

    And using \(W = 2\), the calculation is straightforward.

    \[ \begin{align*} S &= \left(1.00 \, mol \cot \dfrac{6.022\times 10^{23}}{1\,mol} \right) (1.38 \times 10^{-23} J/K) \ln 2 \\ &= 5.76\, J/K \end{align*} \]

    Contributors and Attributions

    • Patrick E. Fleming (Department of Chemistry and Biochemistry; California State University, East Bay)

    • Jerry LaRue (Department of Chemistry and Biochemistry, Chapman University)

    20.5: The Famous Equation of Statistical Thermodynamics is S=k ln W is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.