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## Main Question or Discussion Point

Why does free expansion represent the limit of irreversibility at which all of the "potential" work is degraded to heat.

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Why does free expansion represent the limit of irreversibility at which all of the "potential" work is degraded to heat.

Andrew Mason

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Entropy measures irreversibility, or loss of ability to do work. The greater the net increase in entropy, the greater the irreversibility.asdf1 said:Why does free expansion represent the limit of irreversibility at which all of the "potential" work is degraded to heat.

The free expansion of a gas itself does not necessarily result in a total loss of the gas' ability to do work. It causes the gas to lose thermodynamic equilibrium. The temperature of an expanding gas ball is not defined, due to the loss of equilibrium. Contrary to general belief, the free expansion of gas actually does work - on itself. If you think of a sphere of gas as concentric shells of gas, the outer shell does no work, but the inner shells push out and accelerate the outer shells. The freely expanding gas has kinetic energy and, therefore, an ability to do work.

To maximize the entropy change, you would have to take all the internal energy of the gas (U = PV = nRT) and release it as heat to a reservoir that is arbitrarily close to absolute 0 degrees K (ie 0+dT). The change in entropy of the gas would be [itex]\Delta S_{gas} = -Q/T[/itex] and the change in entropy of the reservoir would be [itex]\Delta S_{res} = Q/(0+dT)[/itex]. The entropy change of the universe is sum of these changes:

[tex]\Delta S_{univ} = \Delta S_{gas} + \Delta S_{res} = Q/(0+dT) - Q/T[/tex].

The gas and reservoir at close to absolute 0 has no ability to do work. So the change in entropy is maximum.

AM

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i see~

thanks! :)

thanks! :)

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