I understand the second law of thermodynamics in terms of the improbability of the high--and-low-velocity particles of a gas to separate themselves so that at one moment they are at different sides of a container. But I can't yet understand how this relates to the efficiency of heat-to-work conversion.

I have in mind the following scenario: a single container with two volumes, separated by a movable wall.

You heat one side and the pressure increases because the molecules gain kinetic energy and move faster. They bump into the movable wall and transfer their own kinetic energy to it. The movable wall moves towards the cooler volume, until the pressures are equal. You did work (you moved the wall). Assume no friction.

Where does the second law come in? What does it mean to say that the entire extra kinetic energy of the particles in the heated volume can't be converted into work, in terms of molecules and transfer of kinetic energy?


In your setup with a moving wall the state of the system after heating is not simply the same as the state of the system at the start other than the fact that the wall has moved. In particular the particles in the gas are moving faster, i.e. the gas has a higher internal energy (I am assuming that the other than the heat you added, no other heat transfer took place). This means that not all of the energy you added went into doing work. Now if the gas was in some very organised configuration, say where the particles nearer the wall have on average a higher energy than those further away, then the wall would move further and you would do more work. This is, however, obviously incredibly improbable.

Essentially Kelvin's statement of the Second Law comes from the fact that if you put energy into a system in a random manner (i.e. you heat it), the odds of all that energy going where you want it to go are tiny.

  • $\begingroup$ Thanks. One further question to clarify things: in a heat engine, for the same temperature difference between the heat source and the heatsink (tH - tC), why is the efficiency greater the closer the heatsink is to absolute zero? The Carnot cycle is 100% efficient if tC = 0K. And what would be the analogous situation in my simplified scenario? $\endgroup$ – user413 Apr 18 '15 at 14:20
  • $\begingroup$ To achieve the Carnot limit, every process must occur reversibly. This means that when heat is transferred from the working substance to the heat sink they must have the same temperature. So if the heat sink is at absolute zero, so is the working substance. 0K is the state of minimal entropy, so everything is nicely ordered and no energy is going anywhere unexpected. Additionally at $T=0$ the working substance is in its ground state, so there is no energy to extract. $\endgroup$ – By Symmetry Apr 18 '15 at 14:41

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