# Homework Help: Entropy of a One Dimensional Polymer

1. Oct 6, 2008

### M@B

1. The problem statement, all variables and given/known data

Polymers are made of very long molecules, usually tangled up in a configuration that has lots of entropy. As a very crude model of a rubber band, consider a chain of N links, each of length l. Imagine that each link has only two possible states, pointing either left or right. The total length L of the rubber band is the net displacement from the beginning of the first link to the end of the last link.

Find an expression of the entropy of this system in terms of N and NR (total number of links pointing right), and an expression for L in terms of N and NR.

2. Relevant equations

S=kln(Ω)

3. The attempt at a solution

As far an entropy goes, I've tried to treat it as a regular two state system (like a coin toss) and used

S=kln(N!/NR!(N-NR)!)

The problem is that, later in the problem one is asked to simplify the derivative of S with respect to L into something that looks like Hooke's Law in terms of L,N,l, and T(temperature). And I'm almost positive my error is somewhere in my formation of the entropy, as I feel it is incomplete and relatively simple.

As for L, I have that L=l(NR - (N - NR)

L=l(2NR-N)
This one seems more correct to me, but I could be wrong.

Any help at all, especially some insight into the entropy of the system is much appreciated.

2. Oct 6, 2008

### Mapes

Hi M@B, welcome to PF. It seems like the differential energy of your system could be written

$$dE=T\,dS+F\,dL$$

where F is a force pulling on the end of a chain that is fixed at one end. At equilibrium, this differential energy would be zero, providing a simple way to calculate $\frac{\partial S}{\partial L}$. Does this help?

3. Nov 22, 2009

### DMcG

Hi! I am currently working on this same question.

I was stumped for a bit at the same point as M@B was. Then I thought I should try:
dS/dL = (dS/dNR)*(dNR/dL) for a constant N.

Does this seem okay?