91影视

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 , $惟=\frac{N!}{N_{R,L}!\left(N-N_{R,L}\right)!}$
Entropy, $S=k\ln 惟$
From Stirling's approximation, $\ln N!=N\ln N-N$
Now calculate
Multiplicity in case of N_R
$惟=\frac{N!}{N_R!\left(N-N_R\right)!}$
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,

$$\begin{gathered} -TdS=FdL \\ F=-T{\left(\frac{鈭�S}{鈭�L}\right)}_U \end{gathered}$$
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,

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

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 $F鈭�L$

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 .$\ln \left(1-\frac{L}{Nl}\right)鈮�\frac{L}{Nl}$
Use this in the force equation,

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

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.