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Consider a (very long) one-dimensional chain of $N$ moleculs, which can be in either of the energy states $\alpha$ or $\beta$. The configurations have length $a$ or $b$ respectively.

Show that the entropy of the chain with respect to the length $L$ of the chain is $$S(L) = N\ln N - n_\alpha\ln n_\alpha - n_\beta\ln n_\beta$$ where $n_\alpha = (L-bN)/(a-b)$ is the number of molecules in state $\alpha$ and $n_\beta = (L-aN)/(b-a)$ is the number of molecules in state $\beta$. (Use Sterlings formula.)

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I have been thinking about this exercise for a while now and I cannot combine the length of the chain and the number of microstates.

If we have $n_\alpha$ molecules in the state $\alpha$ and $n_\beta$ in state $\beta$ we have $$\Omega(n_\alpha, n_\beta,N ) = \frac{N!}{n_\alpha ! n_\beta!}$$ microstates. But I don't know how to get from this to an entropy that's depending on the length of the whole chain.

My guess is that I only have to substitute $n_\alpha$ and $n_\beta$ by the expressions given above, but I don't really understand why they count the number of molecules in state $\alpha$ and $\beta$ respectively. Perhaps someone can help me with this? Thanks!

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    $\begingroup$ The total energy of the molecule should be fixed. This means that n_alpha and n_beta are going to be kept constant. You can easily see that any length has a unique n_alpha and n_beta, so the entropy can indeed be taken as a function of L. In general, if there were additional states gamma, delta etc. and n_gamma, n_delta, etc. this logic would not necessarily work, you could have that the same length could be realized by states with different total energy. So, in general entropy for a closed system is a function of energy, in this case you can take it to be a function of the length L. $\endgroup$ Oct 20, 2014 at 16:04
  • $\begingroup$ Okay, thanks. Your answer really helped me understand why it is okay to assume the chain has a fixed length. I kept wondering why we can even assume that n_alpha and n_beta are fixed numbers! What I still don't understand, though, is why the number of molecules in each state can be calculated by the mentioned formulas for n_alpha and n_beta. $\endgroup$
    – dinosaur
    Oct 20, 2014 at 16:14

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Since you have the expression of $\Omega$ your work is almost done. First remember that the entropy for a micro-canonical (fixed energy) system at thermal equilibrium is given the very famous Boltzmann's formula :

$$S=k_B\,ln(\Omega)$$

Then, simply use the Stirling's approximation to evaluate $ln(N!)\approx Nln(N)-N$ (because $N>>1$, i.e. very long chain). Simplify using $N=n_{\alpha}+n_{\beta}$ and it's done.

To express $n_{\alpha}$ and $n_{\beta}$ as a fuction of $L$, recall that $L=n_{\alpha}a+n_{\beta}b$ and $N=n_{\alpha}+n_{\beta}$ (solve the system).

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  • $\begingroup$ Wow, this is much easier than I thought. I didn't even think about putting the conditions for $n_\alpha$ and $n_\beta$ into a linear system. Thank you very much! $\endgroup$
    – dinosaur
    Oct 20, 2014 at 16:24

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