# Equation of state not linear in energy or system size [duplicate]

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.

## marked as duplicate by Kyle Kanos, Jim, ACuriousMind♦, Brandon Enright, DanuOct 30 '14 at 17:39

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 $S_2$.