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This question already has an answer here:

Which of these two equations of state are valid?

$$S_1 = L_0 \gamma (\theta E/L_0)^{1/2} - L_0\gamma\left[\frac{1}{2} \left(\frac{L}{L_0}\right)^2 + \frac{L_0}{L} - \frac{3}{2}\right]$$

$$S_2 = L_0 \gamma e^{\theta n E/L_0} - L_0\gamma\left[\frac{1}{2} \left(\frac{L}{L_0}\right)^2 + \frac{L_0}{L} - \frac{3}{2}\right]$$

where $\gamma$, $\theta$ are constants, $L_0$ is a function of $n$ (which is the mole number). $S$ has to be monotonically increasing and linear as a function of $E$ since it's extensive. Clearly $S_2$ can't fit that description, but $S_1 \propto \sqrt{E}$ which is not linear either, and it also scales that way for the system length too.

This is exercise 1.2 from David Chandler's introduction to statistical mechanics book.

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marked as duplicate by Kyle Kanos, Jim, ACuriousMind, Brandon Enright, Danu Oct 30 '14 at 17:39

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I was aware of the other question before I posted this, and it does not ask the same thing because the asker says he already answered part (a) by himself, which is precisely what this question asks. And so the answers given in the two questions are different. $\endgroup$ – hadsed Oct 31 '14 at 17:35
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To give some context, the question is asking which of the two formula give the correct entropy for a rubber band and why. $S_1$ is the correct formula. Note that entropy is an extensive variable so that if we double the size of the system, the entropy should also double. That is, entropy scales linearly with the size of the system. You didn't mention that $L_0 = n l_0 $ but it doesn't really matter. Let $n \to \alpha n$, $L \to \alpha L$. It follows that since energy is extensive $E \to \alpha E$. Substituting, we get

\begin{align*} &\alpha L_0 \gamma \left(\frac{\theta \alpha E}{\alpha L_0}\right)^{1/2}-\alpha L_0 \gamma \left[ \frac{1}{2} \left(\frac{\alpha L}{\alpha L_0}\right)^2 +\frac{\alpha L_0}{\alpha L} - \frac{3}{2} \right] \\ &= \alpha S_1 \quad \text{after cancelling appropriate $\alpha$'s} \end{align*} So $S_1$ scales linearly with the size of the system. You can see for yourself that this isn't true for $S_2$.

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