Gibbs Energy Confusion

  • #1
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.
 

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  • #2
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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.
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.
 
  • #3
I'm not sure I understand how that can show dG < 0. Can you write out the equations? Thanks.
 
  • #4
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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?
 
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  • #5
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Please look over section 2.4 of Denbigh, The Principles of Chemical Equilibrium, in which he covers the change in G for a closed (possibly chemical reacting) system initially at temperature and pressure T and P that is in contact with a single ideal reservoir at constant temperature T and in contact with surroundings at constant pressure P, and tell me what you think. In this same section, he also considers the application of G to a steady-flow process, (open system) in which fluid enters the system at temperature T and P and is in contact with a single ideal reservoir at constant temperature T and with an outlet pressure of p. Please tell me what you think about this also.
 
  • #6
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. I just don't understand how ΔHp can differ for a reversible and irreversible process since its a state function.

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.
 
  • #7
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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.
This is only the case when chemical reaction is involved.
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.
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.
 
  • #8
DrDu
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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.
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}##.
 
  • #9
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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}##.
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.
 
  • #10
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).
 
  • #11
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