# Violation of the 2nd law of thermodynamics?

According to the second law of thermodynamics, the product of a system's temperature and change in entropy is greater than or equal to the heat supplied to the system, equality if and only if the change is done reversibly. Now, let's say we have an open system (a system exchanging matter with the surroundings), say an ideal gas, that experiences no heat supply (for instance because the temperature of the surroundings is exactly equal to the temperature of the system), but the system loses particles to the surroundings (for instance because of a difference in chemical potential between the system and surroundings). In this case, why doesn't the entropy of the system go down (violating the 2nd law)? The system loses particles, which corresponds to a decrease in entropy. The particles carry some energy as well, such that the internal energy of the system goes down, again corresponding to a decrease in entropy. Now, one idea to resolve this could be that the inequality I mentioned earlier only applies to closed systems (no exchange of matter), and that for open systems there is an extra term corresponding to the entropy supplied to the system as particles (and not just heat). But a few days ago, I asked a question on this forum about whether the Clausius theorem still applies for open systems, and apparently the answer was yes. And since Clausius' theorem forms the basis of the inequality mentioned earlier, this would suggest that the inequality still applies. So what's going on here?

Your rendering of the 2nd law is incorrect (or at least, incomplete). First of all, it applies only to a closed system. Secondly, it does not involve the temperature of the system; it involves the temperature $$T_R$$ of the interface between the system and surroundings. Thirdly, $$Q\leq T_R\Delta S$$ only if the temperature of the interface between the system and surroundings is constant during the process. Otherwise, the 2nd law can only be written as $$\int{\frac{dQ}{T_R}}\leq \Delta S$$
• The correct form that applies in all circumstances is derived by the Continuum Mechanics guys, and, for an open system (control volume), involves flow of entropy in fluid streams across the system boundaries as well asl flux of entropy by heat flow of the form $\int{\frac{-q\centerdot ndA}{T_R}}$ where q is the heat flux at the boundary and n is a unit normal to the boundary. It also involves the time derivative of the specific entropy over the control volume. See Truesdell et al publications. May 1, 2021 at 13:48