Polymers, like rubber, 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 $\ell$(see Figure 3.17). 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.

(a) Find an expression for the entropy of this system in terms of N and N_R, the number of links pointing to the right.
(b) Write down a formula for L in terms of N and N_R.
(c) For a one-dimensional system such as this, the length L is analogous to the volume V of a three-dimensional system. Similarly, the pressure P is replaced by the tension force F. Taking F to be positive when the rubber band is pulling inward, write down and explain the appropriate thermodynamic identity for this system.
(d) Using the thermodynamic identity, you can now express the tension force F in terms of a partial derivative of the entropy. From this expression, compute the tension in terms of L, T, N, and $\ell$.
(e) Show that when L << N$\ell$, the tension force is directly proportional to L (Hooke's law).
(f) Discuss the dependence of the tension force on temperature. If you increase the temperature of a rubber band, does it tend to expand or contract? Does this behavior make sense?
(g) Suppose that you hold a relaxed rubber band in both hands and suddenly stretch it. Would you expect its temperature to increase or decrease? Explain. Test your prediction with a real rubber band (preferably a fairly heavy one with lots of stretch), using your lips or forehead as a thermometer. (Hint: The entropy you computed in part (a) is not the total entropy of the rubber band. There is additional entropy associated with the vibrational energy of the molecules; this entropy depends on U but is approximately independent of L.)

The system consists of a chain of N links of length l, each having two possible states, pointing left N_L and right N_R.
Total length of the rubber band =L

Find entropy of system in terms of N and N_R

Multiplicity is given as , $\Omega =\frac{N!}{N_{R,L}!\left(N-N_{R,L}\right)!}$
Entropy, $S=k\ln \Omega$
From Stirling's approximation, $\ln N!=N\ln N-N$
Now calculate
Multiplicity in case of N_R
$\Omega =\frac{N!}{N_R!\left(N-N_R\right)!}$
The entropy in terms of N and N_R is,

$$\begin{gathered} S=k\ln \left[\frac{N!}{N_R!\left(N-N_R\right)!}\right] \\ S=k\left[\ln N!-\ln N_R!-\ln \left(N-N_R\right)!\right] \\ S=k\left[N\ln N-N-N_R\ln N_R+N_R-\left(N-N_R\right)\ln \left(N-N_R\right)+\left(N-N_R\right)\right] \\ S=k\left[N\ln N-N_R\ln N_R-\left(N-N_R\right)\ln \left(N-N_R\right)\right] \end{gathered}$$

The system consists of a chain of N links of length l, each having two possible states, pointing left N_L and right N_R.
Total length of the rubber band =L

Find length in terms of N and N_R.

The total length L of the rubber band is:

$$\begin{gathered} L=\left(N_R-N_L\right)l \\ L=\left(N_R-\left(N-N_R\right)\right)l \\ L=\left(2N_R-N\right)l \end{gathered}$$

The system consists of a chain of N links of length l, each having two possible states, pointing left N_L and right N_R.
Total length of the rubber band =L

Identity appropriate thermodynamic for the system.

When the rubber band is pulled inward the force is F>0
And the rubber band is stretched so the change in length i s
$dL>0$
So the work done is: $F.dL>0$

Thus the required thermodynamic identity is,
$dU=TdS+FdL$

The thermodynamic identity is, $dU=TdS+FdL$
The length of the rubber band, $L=\left(2N_R-N\right)l$
Entropy of the system is, $S=k\left[N\ln N-N_R\ln N_R-\left(N-N_R\right)\ln \left(N-N_R\right)\right]$
Find the tension force in terms of L, T, N and l.

As the energy remains constant, the thermodynamic identity is,

$$\begin{gathered} -TdS=FdL \\ F=-T{\left(\frac{\partial S}{\partial L}\right)}_U \end{gathered}$$
Differentiate the length with respect to N_R we get,
$$\begin{gathered} L=\left(2N_R-N\right)l \\ dL=2ld{N}_R \end{gathered}$$
So, the force is,

$$\begin{gathered} F=-T{\left(\frac{\partial S}{\partial L}\right)}_U \\ F=-\frac{T}{2l}{\left(\frac{\partial S}{\partial N_R}\right)}_U \\ F=-\frac{T}{2l}\left(\frac{\partial \left[k\left[N\ln N-N_R\ln N_R-\left(N-N_R\right)\ln \left(N-N_R\right.\right.\right.}{\partial N_R}\right. \\ F=-\frac{kT}{2l}\left[-1-\ln N_R+1+\ln \left(N-N_R\right)\right] \\ F=-\frac{kT}{2l}\left[-\ln N_R+\ln \left(N-N_R\right)\right] \\ F=-\frac{kT}{2l}\left[\ln \frac{N-N_R}{N_R}\right] \\ F=-\frac{kT}{2l}\left[\ln \frac{N-\frac{1}{2}\left(\frac{L}{l}+N\right)}{\frac{1}{2}\left(\frac{L}{l}+N\right)}\right] \\ F=-\frac{kT}{2l}\ln \frac{\frac{1}{2}\left(N-\frac{L}{l}\right)}{\frac{1}{2}\left(\frac{L}{l}+N\right)} \\ F=-\frac{kT}{2l}\ln \frac{(Nl-L)}{(L+Nl)} \end{gathered}$$

The tensile force $F=-\frac{kT}{2l}\ln \frac{(Nl-L)}{(L+Nl)}$

Shoe the Force is directly proportional to length L.

Calculate force at L<<Nl,

$$\begin{gathered} F=-\frac{kT}{2l}\ln \frac{(Nl-L)}{(L+Nl)} \\ F=-\frac{kT}{2l}\ln \frac{\left(1-\frac{L}{Nl}\right)}{\left(1+\frac{L}{Nl}\right)} \\ F=-\frac{kT}{2l}\ln \left(1-\frac{L}{Nl}\right){\left(1+\frac{L}{Nl}\right)}^{-1} \\ F=-\frac{kT}{2l}\ln {\left(1-\frac{L}{Nl}\right)}^2.................................\left(1\right) \end{gathered}$$

From the equation( 1) we can say that $F\propto L$

Temperature is increased of the system.
The force is given by $F=-\frac{kT}{2l}\ln {\left(1-\frac{L}{Nl}\right)}^2$
Find relationship of force with Temperature

Let's approximate .$\ln \left(1-\frac{L}{Nl}\right)\approx \frac{L}{Nl}$
Use this in the force equation,

$$\begin{gathered} F=-\frac{kT}{l^2}\ln {\left(1-\frac{L}{Nl}\right)}^2 \\ F=-\frac{kT}{N{l}^2}L \\ F\propto T \end{gathered}$$

Rubber band is stretched.

Find the change in temperature

When rubber band is stretched it will get warmer
Because N_R would get closer to its maximum entropy, which will cause the temperature to increase.