# Entropy of a rubber-band (force determined by entropy or energy)?

## Homework Statement

The resultant of a experimental run of a force (N) vs temperature (K) gives me a slope of 0.0039x - 1.0929. The slope has a (∂f/∂T)l = (∂S/∂l)T

## Homework Equations

Equation (11) attached as PNG file

We can obtain all the quantities of the right side of Eq. (11) by studying the force at difference temperatures for given extensions. A plot of f versus T has a slope (∂f/∂T)l and the intercept gives the energy change with length at a given T. As we have seen, the term -(∂f/∂T)l = (∂S/∂l)T gives the entropy change with length at a given T. It is thus possible to investigate whether f is determined principally by (∂U/∂l)T or (∂S/∂l)T

## The Attempt at a Solution

My issue is really just trying to figure out what wins. the intercept i'm assuming is the ∂U of energy they are talking about, and that the slope at a given temperature that overcomes the intercept energy and that would be the tipping point where force is determined by one and not the other. I'm assuming that the negative intercept is our restoring force.

So am I correct in this line of thinking, or am I way off?

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Chestermiller
Mentor
According to Eqn. 11, the slope is $(\frac{\partial f}{\partial T})_l=-(\frac{\partial S}{\partial l})_T$ and the intercept is $(\frac{\partial U}{\partial l})_T$. The question is: which makes a bigger contribution to the force f, $(\frac{\partial U}{\partial l})_T$ or $T(\frac{\partial f}{\partial T})_l=-T(\frac{\partial S}{\partial l})_T$?

I think what you will find is that the entropy term makes a much bigger contribution to the force than the energy term. This is because the configurational entropy change associated with the extension of the polymer molecules when the rubber is deformed relates to the molecules acting like microscopic springs.

ahhh, makes a lot more sense when you think of it like little springs. many thanks, entropy has been a bit tough to wrap my head around it seems. Thanks for the help.