Entropy of a One Dimensional Polymer

In summary, the homework statement is that polymers are made of very long molecules which have lots of entropy. Polymers are modeled as a rubber band with a chain of links each with two possible states. The total length of the rubber band is the net displacement from the beginning of the first link to the end of the last link. The entropy of the system is found to be in terms of N and NR and an expression for L is found in terms of N and NR.
  • #1
M@B
3
0

Homework Statement



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.

Homework Equations



S=kln(Ω)


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.
Thanks in advance.
 
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  • #2
Hi M@B, welcome to PF. It seems like the differential energy of your system could be written

[tex]dE=T\,dS+F\,dL[/tex]

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 [itex]\frac{\partial S}{\partial L}[/itex]. Does this help?
 
  • #3
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?
 

1. What is entropy in the context of a one dimensional polymer?

Entropy refers to the measure of disorder or randomness in a system. In the case of a one dimensional polymer, it refers to the degree of freedom of the polymer chain to move and adopt different conformations.

2. How does the length of a one dimensional polymer affect its entropy?

The longer the polymer chain, the higher the entropy. This is because longer chains have more possible conformations, leading to a higher degree of disorder and therefore, higher entropy.

3. Is there a relationship between temperature and entropy in a one dimensional polymer?

Yes, there is a direct relationship between temperature and entropy. As the temperature increases, the molecules in the polymer chain have more thermal energy, leading to an increase in their movement and freedom to adopt different conformations, thus increasing the entropy.

4. How does the type of monomer units in a one dimensional polymer affect its entropy?

The type of monomer units in a polymer can affect its entropy in different ways. For example, if the monomer units have bulky side chains, the polymer chain may have less freedom to move and adopt different conformations, resulting in a lower entropy compared to a polymer with smaller side chains.

5. Can the entropy of a one dimensional polymer be measured experimentally?

Yes, the entropy of a one dimensional polymer can be indirectly measured experimentally through techniques such as thermogravimetric analysis or differential scanning calorimetry. These techniques measure changes in energy or mass as a function of temperature, which can be used to calculate the change in entropy.

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