Sebi123
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In what follows I refer to the ideas of "Thermodynamic: Foundations and Applications" by Gyftoploulos and Beretta. The abbreviated form of my question is: In a reversible weight process,
Ω1R-Ω2R = E1 - E2 (see eqn. 6.18, p. 99) is transferred out of the composite of a system A and a reservoir R to the weight, where ΩR is the available energy of a system A with respect to reservoir R and E is the energy of system A.
When the above equation is true, why does the book occasionally (problem 6.4, example 6.4 on p. 94) calculates the energy change of the reservoir as E1R - E2R ≠ 0 for the case of reversible weight processes?
Because, according to the equation above, E1R = E2R must be true (see below). Also, (DE12R)revA should always be 0, p. 108/9. However, this quantity is used extensively in the discussion of entropy as if it was non-zero.
My proof of E1R = E2R:
Given two states, A1 and A2, of a system A and a reservoir R. Consider two reversible weight processes, α und β, consisting of two sub-processes, where A0R0 is the state where A and R are in mutual stable equilibrium. The only difference is the final state of the reservoir (R1 in process β instead of R2 in process α).
α: A1R1 → A0R0 → A2R2
β: A1R1 → A0R0 → A2R1
The first sub-process in each process transfers energy of the amount of available energy to the weight: Ω1R. The second sub-process in α and the second sub-process in β both transfer the available energy Ω2R to the weight, because the stable equilibrium state is unique (see proofs on adiabatic availability on p. 76 and regard the adiabatic availability of A+R as the available energy von A) and Ω2R is independent of the initial (or end state in this case) of the reservoir (statement 3 on p. 89). The energy balances for both processes are:
α: Ω1R-Ω2R = E1 - E2 + E1R - E2R
β: Ω1R-Ω2R = E1 - E2
Comparing both results, one gets that all states of the reservoir have the same energy: E1R = E2R
So: Ω1R-Ω2R = E1 - E2
The last equation corresponds to equation 6.18, p. 99.
Ω1R-Ω2R = E1 - E2 (see eqn. 6.18, p. 99) is transferred out of the composite of a system A and a reservoir R to the weight, where ΩR is the available energy of a system A with respect to reservoir R and E is the energy of system A.
When the above equation is true, why does the book occasionally (problem 6.4, example 6.4 on p. 94) calculates the energy change of the reservoir as E1R - E2R ≠ 0 for the case of reversible weight processes?
Because, according to the equation above, E1R = E2R must be true (see below). Also, (DE12R)revA should always be 0, p. 108/9. However, this quantity is used extensively in the discussion of entropy as if it was non-zero.
My proof of E1R = E2R:
Given two states, A1 and A2, of a system A and a reservoir R. Consider two reversible weight processes, α und β, consisting of two sub-processes, where A0R0 is the state where A and R are in mutual stable equilibrium. The only difference is the final state of the reservoir (R1 in process β instead of R2 in process α).
α: A1R1 → A0R0 → A2R2
β: A1R1 → A0R0 → A2R1
The first sub-process in each process transfers energy of the amount of available energy to the weight: Ω1R. The second sub-process in α and the second sub-process in β both transfer the available energy Ω2R to the weight, because the stable equilibrium state is unique (see proofs on adiabatic availability on p. 76 and regard the adiabatic availability of A+R as the available energy von A) and Ω2R is independent of the initial (or end state in this case) of the reservoir (statement 3 on p. 89). The energy balances for both processes are:
α: Ω1R-Ω2R = E1 - E2 + E1R - E2R
β: Ω1R-Ω2R = E1 - E2
Comparing both results, one gets that all states of the reservoir have the same energy: E1R = E2R
So: Ω1R-Ω2R = E1 - E2
The last equation corresponds to equation 6.18, p. 99.