It seems that on stretching a rubberband, entropy or the degree of disorder is likely to increase. Is this the correct answer? My text says that on stretching, the arrangement of particles become more 'ordered' and hence entropy 'decreases'. How?


1 Answer 1


Think of it this way, when an elastic polymer is in its relaxed state, these molecules are all tangled up with each other and have no particular direction to them, but when you apply a force, i.e stretch the rubber band, you end up disentangling some of the polymer molecules.

We are not stretching the actual polymer molecules, just sort of changing the way they are organised. This disentanglement corresponds to a reduction in micro-states that the system occupies and thus, a decrease in entropy.

In simpler words, there's more ways for a polymer chains to jumbled up with each other (more microstates) than for them to be aligned.

More simply-- If the chain were totally stretched out, it would only have one possible conformation. By coiling itself, it increases the number of conformations it can have, and therefore, its entropy.

A more rigorous treatment can be found if you read up about the ideal chain, which serves as a model for such systems.

Thus, assuming that no other energy exchanges takes place, the tendency of the elastic band to shrink back to its original conformation is due to something called an entropic force, i.e a tendency of the system to move to a configuration that maximises entropy.

Some Details

Note: I am not quite confident with this material and might've made some mistakes--if that be the case, I am sure someone here will correct me

Consider the following model system: The chain is made of $n$ rigid segments of size $b$; Thus, the length of the chain is $L = nb$. Also, assume that the chain has one fixed end, and one free end and the distance between the two is denoted as $r$.

Then, one can derive the probability of finding the chain ends a distance $r$ apart is given by the following Gaussian distribution:

$$ P(r,n)\mathrm dr = 4\pi r^2 \left(\frac{2nb^2\pi}{3}\right)^{\frac{-3}{2}}\exp\left(\frac{-3r^2}{2nb^2}\right)\mathrm dr$$

The classical entropy is approximately $ S = k_\mathrm b\ln(P(r,n))$

From hereon, one can go on to show that for an ideal chain, maximising its entropy means reducing the distance between its two free ends (i.e reducing $r$)

  • 2
    $\begingroup$ Nice job of articulating the concept of configurational entropy as it applies to polymer molecules. $\endgroup$ Commented Mar 20, 2016 at 20:18

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