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_{rev}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.

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In the Clausius inequality, T is the temperature of the reservoir,_{rev}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.

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I'm not sure I understand how that can show dG < 0. Can you write out the equations? Thanks.

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If the initial and final states of the system are T,P, then the change in entropy of the system has to be zero: ##\Delta S_{syst}=0##. From the Clausius Inequality, if the process was irreversible, $$\Delta S_{syst}=\int{\frac{dq}{T_{surr}}}+\sigma$$where ##\sigma## is the entropy generated within the system during the irreversible process. Therefore, $$\int{\frac{dq}{T_{surr}}}=-\sigma$$ But, assuming that the surroundings consists of a set of ideal reservoirs, the left hand side of this equation is just minus the change in entropy of the surroundings: $$\Delta S_{surr}=-\int{\frac{dq}{T_{surr}}}=+\sigma$$This means that all the entropy that was generated within the system by the irreversible process is transferred to the surroundings, and: $$\Delta S_{syst}+\Delta S_{surr}=\sigma$$The change in free energy of the system is, of course, $$\Delta G_{syst}=(H_f-H_i)-(T_fS_f-T_iS_i)=0$$

If you are quoting your book correctly, it appears that your doubt about the book's derivation is well-founded and that the change in free energy of the system is zero (since all the generated entropy is transferred to the surroundings). Are you absolutely sure your book wasn't considering a system in which chemical reaction can occur?

If you are quoting your book correctly, it appears that your doubt about the book's derivation is well-founded and that the change in free energy of the system is zero (since all the generated entropy is transferred to the surroundings). Are you absolutely sure your book wasn't considering a system in which chemical reaction can occur?

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If we look at your example, the entropy of the universe clearly increases and since dG = δq - TdS = -T* (dS

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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 closedIf we look at your example, the entropy of the universe clearly increases and since dG = δq - TdS = -T* (dS_{surr}+ 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 = ΔH_{p}= TdS and that dG = 0.

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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}##._{rev}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.

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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}##.

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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}such that the heat given by ΔH(T,P,n_{1},n_{2},...) ≠ TΔS(T,P).

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