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Example 16.2: Determination of \(ΔS\)

Consider the system shown here. What is the change in entropy for a process that converts the system from distribution (a) to (c)?

Solution

We are interested in the following change:
The initial number of microstates is one, the final six:

\(W_{\mathrm{a}}\) \(= 1\)


\(W_{\mathrm{c}}\) \(= 6\)



\(R_{\mathrm{gas}}\) \(= 8.314\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}}\)


\(N_{\mathrm{A}}\) \(= 6.022\times 10^{23}\frac{1}{\mathrm{mol}}\)


\(k_{\mathrm{B}}\) \(= \dfrac{R_{\mathrm{gas}}}{N_{\mathrm{A}}}\)

\(\ \ \ =\dfrac{8.314\ \frac{\mathrm{J}}{\mathrm{mol}\ \mathrm{K}}}{6.022\times 10^{23}\frac{1}{\mathrm{mol}}}\)

\(\ \ \ =1.3806\times 10^{-23}\ \frac{\mathrm{J}}{\mathrm{K}}\)


\(ΔS\) \(= k_{\mathrm{B}} \cdot \mathrm{ln}(\dfrac{W_{\mathrm{c}}}{W_{\mathrm{a}}})\)

\(\ \ \ =1.3806\times 10^{-23}\ \frac{\mathrm{J}}{\mathrm{K}} \cdot \mathrm{ln}(\dfrac{6}{1})\)

\(\ \ \ =1.3806\times 10^{-23}\ \frac{\mathrm{J}}{\mathrm{K}} \cdot \mathrm{ln}(6)\)

\(\ \ \ =1.3806\times 10^{-23}\ \frac{\mathrm{J}}{\mathrm{K}} \cdot 1.7917594692280550\)

\(\ \ \ =2.4737\times 10^{-23}\ \frac{\mathrm{J}}{\mathrm{K}}\)


The sign of this result is consistent with expectation; since there are more microstates possible for the final state than for the initial state, the change in entropy should be positive.