In the Clausius inequality, T is the temperature of the reservoir, not the system (except at the boundary between the system and the reservoir where the heat flow ##\delta q## occurs). Only for a reversible process is the system temperature necessarily equal to the reservoir temperature during the process.By Clausius' inequality δq - TdS ≤ 0. For a constant T,P process in a closed system and no non-expansion work my text states that dG = dH - TdS = δq - TdS ≤ 0 but this seems incorrect to me. If pressure is constant such that dH = δq, doesn't this mean that δq = δqrev since dH is a state function and subsequently dG = 0, not ≤ 0. This is obviously incorrect as dG can certainly be less than 0 but I'm wondering how you prove it under constant T,P conditions.
This is only the case when chemical reaction is involved.I looked over that book and I'm still confused. It says that a constant T,P and no non-expansion work, dG < 0 for an irreversible process.
The correct result is that ##\Delta G## of the universe is zero if no chemical reaction is involved. So, your math is correct. ##\Delta G## of the universe being less than zero for a closed system is not a criterion for irreversibility. ##\Delta S## of the universe being greater than zero for a closed system, on the other hand, is a criterion for irreversibility. However, if you have a single ideal constant temperature reservoir (at the same temperature as the initial system temperature) in contact with a closed chemically reacting system in contact with a surroundings that are at the same pressure as the initial system pressure, and the system is allowed to chemically react spontaneously, then G for the final thermodynamic equilibrium state of the system will be less than G for the initial state of the system, and G for the final thermodynamic equilibrium state of the universe will be less than G for the initial thermodynamic equilibrium state of the universe.If we look at your example, the entropy of the universe clearly increases and since dG = δq - TdS = -T* (dSsurr + dS), I would expect dG to equal -Tσ. However, if I consider that δq is from a process at constant pressure and temperature, I should conclude that δq = ΔHp = TdS and that dG = 0.
Say you are stirring a paddle in a viscous liquid. Even if you keep temperature and pressure (and thus also H) constant, this will constantly produce heat which will be leaving over the boundary. So ##\delta q>\delta q_\mathrm{rev}##.By Clausius' inequality δq - TdS ≤ 0. For a constant T,P process in a closed system and no non-expansion work my text states that dG = dH - TdS = δq - TdS ≤ 0 but this seems incorrect to me. If pressure is constant such that dH = δq, doesn't this mean that δq = δqrev since dH is a state function and subsequently dG = 0, not ≤ 0. This is obviously incorrect as dG can certainly be less than 0 but I'm wondering how you prove it under constant T,P conditions.
Actually, since ##\delta q## is the heat flow from the surroundings to the system, in this case ##\delta q## is negative. But your point is well-taken. The change in entropy of the system is zero in both cases, but, in the case of the stirring paddle, the change in entropy of the surroundings is positive.Say you are stirring a paddle in a viscous liquid. Even if you keep temperature and pressure (and thus also H) constant, this will constantly produce heat which will be leaving over the boundary. So ##\delta q>\delta q_\mathrm{rev}##.
I had no idea from your posts that you were trying to analyze anything like this. For a comprehensive discussion of this, please see the following PF link: https://www.physicsforums.com/threads/thermochemistry-challenge-problem-chets-paradox.913567/I finally figured it out. Only one of H or S can be defined in terms of ni such that the heat given by ΔH(T,P,n1,n2,...) ≠ TΔS(T,P).