3
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Consider the following (alleged) cycle:

  1. Quasi-static adiabatic expansion of a vapor in such a condition that it results in the condensation of part of the mols*;
  2. Out of equilibrium adiabatic compression;
  3. Quasi-static isochoric heating;

Statements:

  1. The latent heat required to change the state of part of the mols during steps 1 and 2 is due to the work done by the vapor;
  2. Each infinitesimal compression dV during step 2 being done farther away from equilibrium than an infinitesimal expansion of same size done in step 1 at the same volume, grants that if the internal energy in both cases were also the same (ie all other factors being equal), a smaller molar fraction would be recovered to vaporous state than it was lost, so that
  3. (lesser mols - lower energy feedback) with lesser mols recovered at each point of step 2, there is lower energy than in step 1, which implies a lower pressure, which implies a lower infinitesimal work done by each infinitesimal compression, which implies lower gain of latent heat, which yields an even lower recovery of vapor mols;
  4. Steps 1 and 2 being done adiabatically, all the difference of work came from the internal energy of the mols. The initial state of step 1 and final state of step 2 have the same volume and different energies, and thus, step 3 can go from the latter to the former simply by means of a quasi-static heating.
  5. We have, thus, performed a closed loop in the space of thermodynamic states, whose only effect is the total transformation of heat into work, in other words, a heat engine with efficiency 1.

The experiment can be more easily conceived by a piston that connects with a separated chamber that has the exact volume of condensed (or sublimed) mol fraction in the liquid (or solid) state. During step 1, the referred mol fraction 'rains' (or 'snows') into that chamber that is then closed so to completely prevent it to be recovered to the vapor fraction during step 2, regardless even of how near the equilibrium it is done. After step 2, the chamber is reopened, thus allowing those mols to be recovered to the vapor state in step 3.

The few experts on thermodynamics whom I have consulted haven't achieved to solve this apparent fallacy, which, I actually came up with before actually having formally studied thermodynamics, in my school days, back in 2008.

*The change of physical state to one of lower energy due to adiabatic cooling does happen, for example, in the formation of some types of clouds and in the sublimation of gaseous CO2 into dry ice in fire extinguishers.

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  • $\begingroup$ I guess that step 2 is forbidden. It must be related to entropy, the out-equilibrium scenario must produce more entropy than the quasi-static one, so it cannot follow the path you are describing. $\endgroup$ – Tony May 22 '17 at 14:27
  • $\begingroup$ What forbids one to adiabatically compress the system resulting from step 1 out of equilibrium, back into the original volume? Moreover, in the first paragraph after the statements enumerations I provide a way to bypass the need for considerations on non-equilibrium thermodynamics. $\endgroup$ – Yuri S VB May 22 '17 at 15:20
  • $\begingroup$ It does not forbid to go back to the original volume. It forbids to go back adiabatically to the original volume and at a lower pressure. It basically says that going back adiabatically, you end up at a greater (or equal) pressure. $\endgroup$ – Tony May 22 '17 at 15:33
  • $\begingroup$ But why, since the mols that were condensed would be prevented, either by non static process or by physical removal, to from joining the vapor fraction again? $\endgroup$ – Yuri S VB May 22 '17 at 15:57
  • $\begingroup$ It is difficult for me to understand your phrasing in points 2 and 3. But one way or another, "a smaller molar fraction would be recovered to vaporous state than it was lost" is not possible. If you are out of equilibrium, you produce more entropy. Or your system may not be isolated from the outside world ? $\endgroup$ – Tony May 22 '17 at 16:02

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