1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. May 24, 2013 #1
    1. The problem statement, all variables and given/known data

    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


    2. Relevant 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

    3. 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?
     

    Attached Files:

  2. jcsd
  3. May 24, 2013 #2
    According to Eqn. 11, the slope is [itex](\frac{\partial f}{\partial T})_l=-(\frac{\partial S}{\partial l})_T[/itex] and the intercept is [itex](\frac{\partial U}{\partial l})_T[/itex]. The question is: which makes a bigger contribution to the force f, [itex](\frac{\partial U}{\partial l})_T[/itex] or [itex]T(\frac{\partial f}{\partial T})_l=-T(\frac{\partial S}{\partial l})_T[/itex]?

    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.
     
  4. May 25, 2013 #3
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Entropy of a rubber-band (force determined by entropy or energy)?
Loading...