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

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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 $$\PageIndex{1}$$). 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*} \nonumber