Work done for isothermal process in terms of Helmholtz func

[Hello890]

The Helmholtz function differential form for a reversible process is: dF = -SdT - PdV, as for a reversible process δW (by system/here an (ideal) gas) = PdV and dS = δQ/T.

Therefore, for a reversible isothermal process, dT = 0 and hence dF = -PdV. Therefore, the work done by the system is W = -ΔF. As F is a state variable it is path independent. Therefore, this makes the work done path independent as it is only a function of a state variable?

If this is correct, I do not understand why it is said that for irreversible processes W ≤ - ΔF. Surely, as W is now path independent in this case, W = -ΔF both any process, regardless of wether it is reversible or irreversible. Or is the statement W≤-ΔF simply saying the W for an irreversible process equal to F(initial) - F(final) but will be less that (F(initial) - F(final))for a reversible process?

 

[Lord Jestocost]

From “The Principles of Chemical Equilibrium” by Kenneth Denbigh:

w ≤ – (F₂ – F₁) ......... (2-15)

In relation (2-15) the inequality sign refers to an irreversible process and the equality sign to a reversible one. Between the assigned initial and final states the value of F₂ – F₁ is, of course, the same whether the path is reversible one or not, since F is a function of state. Relation (2-15) may thus be interpreted as follows. The work done in a process, in which the initial and final temperatures and the temperature of the heat reservoir are all equal, is either less than or equal to the decrease in F.

 

[Hello890]

From “The Principles of Chemical Equilibrium” by Kenneth Denbigh:

w ≤ – (F₂ – F₁) ......... (2-15)

In relation (2-15) the inequality sign refers to an irreversible process and the equality sign to a reversible one. Between the assigned initial and final states the value of F₂ – F₁ is, of course, the same whether the path is reversible one or not, since F is a function of state. Relation (2-15) may thus be interpreted as follows. The work done in a process, in which the initial and final temperatures and the temperature of the heat reservoir are all equal, is either less than or equal to the decrease in F.

From Finn's Thermal Physics book, it gives a similar situation where the adiabatic work W = U(final) - U(initial), and later states that this relation holds whether the process is reversible or not. It says this is true as W is now simply a function of state variables and therefore is path independent. Therefore, I still don't quite understand why W in any isothermal processes not simply W = F(initial) - F(final)? Why is W not simply equal to the decrease in F, now that W is a function of state variables as well?

 

[Chestermiller]

Why don’t you solve it for two cases (a) reversible and (b) irreversible at constant external pressure (less than the initial pressure) and see how the results compare? In both cases, the system would be in contact with a constant temperature reservoir, equal to the initial system temperature.

Many times, devising and modeling a simple focus problem helps you solidify your understanding.