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I thought of the following situation:

Suppose there are two separate perfectly rigid vessels enclosing fixed and euqal volumes of the same gas at some state A. An amount E of heat is supplied in an internally reversible manner to the volume of gas in the first vessel, causing the gas to reach a state B. The second vessel, however, contains a wire filament with a certain resistance that crosses its boundary twice and is hooked to an external voltage source. The mass of the filament is negligible. Electrical work in the same amount E is supplied to the gas in the second vessel in a likewise internally reversible manner -- that is, the wire continuously raises the temperature of the gas in a reversible process.

Since the same net amount E of energy was transferred to both vessels -- the control volumes -- the internal energy of each volume of gas must increase by the same amount. The volumes remain the same throughout each process and the initial state is the same. Therefore, the end states of each gas must coincide.

The heat transfer to the first vessel is accompanied by an entropy transfer in the amount $\int \frac{\delta Q}{T}$, by definition. Since the energy transfer to the second vessel was carried out entirely through work and in an internally reversible manner, the entropy transfer has to be zero. Also, given that the process was internally reversible, no entropy was generated within the system boundaries.

This scenario gives rise to an apparent paradox. The first law ensures that the final states are the same: equal increases in internal energy and same constant volume. Through this reasoning, the entropy of both systems must be the same. However, the process relative to the first vessel involved no internal entropy generation but net entropy transfer to the enclosed gas. No transfer occured to the second vessel and no entropy was generated within its boundaries.

So the first law assures that both volumes of gas are at the same state while the second law calls for a difference in entropies, which would prescribe different end states.

Where is the fault in my thought process?

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  • $\begingroup$ The heating of a wire through electrical resistance is not reversible. Resistance can be thought of as analogous to friction for electrons. $\endgroup$ – Señor O Dec 20 '17 at 23:11
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This all depends on whether the wire inside the vessel is considered part of the "system" in the second case or not. If the wire is not considered part of the system, then the situation in the second case is identical to the first case, and the entropy change is the same (transfer from the surroundings to the system). However, if the wire is included as part of the system, then entropy is generated within the wire by the resistance heating, and this entropy is then transferred to the rest of the system. So, if the wire is not considered part of the system, then entropy is transferred from the surroundings (the wire) to the system (the gas). But, if the wire is included within the system, then no entropy is transferred from the surroundings to the system, but entropy is generated within the system. So, either way, the final entropy change of the system is the same.

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"Since the energy transfer to the second vessel was carried out entirely through work and in an internally reversible manner […]"

I think not. The electrical work done on the filament is no more reversible than the work I would do if I kept rubbing the outside of the cylinder containing the gas. Consider this criterion for reversibility: can the process be made to go in reverse by means of an infinitesimal change in conditions (such as gas temperature)? Can we get the electrical work back again by such an infinitesimal change. I think not.

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