When I stretch a rubber band, the fibers straighten out causing lesser randomness and more order. This should imply the entropy of the rubber band has decreased, but how does the entropy of the whole universe increase here? Does this mean a rubber band gets colder on stretching? I have seen an experiment where it gets hotter. Doesn't entropy of the whole universe increase only in spontaneous processes? Here I have to pull it hence its not spontaneous? In a nutshell, I'm really confused.
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3 Answers
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You are treating the rubber band as an isolated system when in fact you are acting upon it. The entropy increase occurs in your body as glucose is oxidize into water and carbon dioxide to produce the energy needed for your muscles to contract to stretch the rubber band.
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The force you are exerting on the rubber to stretch it is in the same direction as the displacement. So you are doing positive work on the rubber, and it is doing negative work on you (its surroundings). So, from the first law of thermodynamics, if the process is adiabatic, the internal energy of the rubber is increasing. The internal energy of rubber is known from observations to be nearly a function only of temperature. So, when you stretch the rubber adiabatically, its temperature should rise. For an adiabatic reversible process, the change in entropy of the system has to be zero. So the decrease in configurational entropy of the rubber must be offset by the increase in entropy resulting from the rise in temperature.
answered Jul 5, 2016 at 12:42
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Let $L$ denote the length of the substance then you can always write
$$\left ( \frac{\partial T}{\partial L} \right )_{S} = - \left ( \frac{\partial T}{\partial S} \right )_{L} \left ( \frac{\partial S}{\partial L} \right )_{T}$$
The term $\left ( \frac{\partial T}{\partial S} \right )_{L} $ is always positive for being the reciprocal of the heat capacity per unit temperature. Thus $\left ( \frac{\partial T}{\partial L} \right )_{S}$ is positive or negative whether $\left ( \frac{\partial S}{\partial L} \right )_{T}$ is negative or positive. Therefore, if entropy decreases upon isothermal stretching then the temperature must increase during adiabatic stretching.
