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Ω

_{1}

^{R}-Ω

_{2}

^{R}= E

_{1}- E

_{2}(see eqn. 6.18, p. 99) is transferred out of the composite of a system A and a reservoir R to the weight, where Ω

^{R}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 E

_{1}

^{R}- E

_{2}

^{R}≠ 0 for the case of reversible weight processes?

Because, according to the equation above, E

_{1}

^{R}= E

_{2}

^{R}must be true (see below). Also, (DE

_{12}

^{R})

_{rev}

^{A}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 E

_{1}

^{R}= E

_{2}

^{R}:

Given two states, A

_{1}and A

_{2}, of a system A and a reservoir R. Consider two reversible weight processes, α und β, consisting of two sub-processes, where A

_{0}R

_{0}is the state where A and R are in mutual stable equilibrium. The only difference is the final state of the reservoir (R

_{1}in process β instead of R

_{2}in process α).

α: A

_{1}R

_{1}→ A

_{0}R

_{0}→ A

_{2}R

_{2}

β: A

_{1}R

_{1}→ A

_{0}R

_{0}→ A

_{2}R

_{1}

The first sub-process in each process transfers energy of the amount of available energy to the weight: Ω

_{1}

^{R}. The second sub-process in α and the second sub-process in β both transfer the available energy Ω

_{2}

^{R}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 Ω

_{2}

^{R}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: Ω

_{1}

^{R}-Ω

_{2}

^{R}= E

_{1}- E

_{2}+ E

_{1}

^{R}- E

_{2}

^{R}

β: Ω

_{1}

^{R}-Ω

_{2}

^{R}= E

_{1}- E

_{2}

Comparing both results, one gets that all states of the reservoir have the same energy: E

_{1}

^{R}= E

_{2}

^{R}

So: Ω

_{1}

^{R}-Ω

_{2}

^{R}= E

_{1}- E

_{2}

The last equation corresponds to equation 6.18, p. 99.