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.always_tired said: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.always_tired said: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}##.always_tired said: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.DrDu said: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/always_tired said: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).
Gibbs energy, also known as free energy, is a thermodynamic property used to measure the maximum amount of work that can be extracted from a system. It takes into account both the entropy and enthalpy of a system and is important in determining the spontaneity and equilibrium of chemical and physical processes.
This condition, known as the Gibbs energy criterion, states that a process will be spontaneous if the change in Gibbs energy is negative or equal to 0. This means that at constant temperature and pressure, the system will naturally move towards a state of lower energy and higher stability.
Yes, dG can be greater than 0 at constant temperature and pressure. This would indicate that the process is non-spontaneous and will not occur without an external input of energy.
The equilibrium constant, K, is related to Gibbs energy through the equation ΔG = -RTlnK, where R is the gas constant and T is the temperature. This equation shows that a lower value of ΔG corresponds to a higher equilibrium constant, indicating a greater tendency for the reaction to proceed in the forward direction.
Some common experimental techniques used to measure Gibbs energy include calorimetry, which measures the heat transfer during a process, and electrochemistry, which measures the potential difference between reactants and products. Other methods include measuring equilibrium constants and using thermodynamic tables to calculate ΔG values.