1. C5 Irreversibility, Entropy Changes, and"Lost work" Consider a system in contact with a heat reservoir during a reversible process. If there is heat O absorbed by the reservoir at temperature T, the change in entropy of the reservoir is different temperatures. To analyze these, we can visualize a sequence of heat reservoirs a AS=Q/T In general, reversible processes are accompanied by heat exchanges that occu different temperatures so that during any infinitesimal portion of the cycle there will not be any heat transferred over a finite temperature difference During any infinitesimal portion, heat dOrey will be transferred between the system and one of the reservoirs which is at T. If d@ey is absorbed by the system, the entropy change of the system is The entropy change of the reservoir is The total entropy change of system plus surroundings is dslotal= ds system ds reservoir=0 This is also true if there is a quantity of heat rejected by the system The conclusion is that for a reversible process, no change occurs in the total entropy produced, i.e the entropy of the system plus the entropy of the surroundings: ASo(a=0 We now carry out the same type of analysis for an irreversible process, which takes the system between the same specified states as in the reversible process. This is shown schematically at the right, with I and r denoting the irreversible and reversible processes In the irreversible process, the system receives heat do and does work di The change in internal energy for the irreversible process is du=do-dw(always true- first law) For the reversible pr du= tds-dW Irreversible and reversible state changes Because the state change is the same in the two processes (we specified that it was), the change in internal energy is the same Equating the changes in internal energy in the above two expressions yields dQactual-dWactual=TdS-dw IC-81C-8 1.C.5 Irreversibility, Entropy Changes, and “Lost Work” Consider a system in contact with a heat reservoir during a reversible process. If there is heat Q absorbed by the reservoir at temperature T, the change in entropy of the reservoir is ∆S QT = / . In general, reversible processes are accompanied by heat exchanges that occur at different temperatures. To analyze these, we can visualize a sequence of heat reservoirs at different temperatures so that during any infinitesimal portion of the cycle there will not be any heat transferred over a finite temperature difference. During any infinitesimal portion, heat dQrev will be transferred between the system and one of the reservoirs which is at T. If dQrev is absorbed by the system, the entropy change of the system is dS dQ T system rev = . The entropy change of the reservoir is dS dQ T reservoir rev = − . The total entropy change of system plus surroundings is dS dS dS total system reservoir =+ = 0. This is also true if there is a quantity of heat rejected by the system. The conclusion is that for a reversible process, no change occurs in the total entropy produced, i.e., the entropy of the system plus the entropy of the surroundings: ∆Stotal = 0. We now carry out the same type of analysis for an irreversible process, which takes the system between the same specified states as in the reversible process. This is shown schematically at the right, with I and R denoting the irreversible and reversible processes. In the irreversible process, the system receives heat dQ and does work dW. The change in internal energy for the irreversible process is dU dQ dW = − (Always true - first law). For the reversible process dU TdS dW = − rev . Because the state change is the same in the two processes (we specified that it was), the change in internal energy is the same. Equating the changes in internal energy in the above two expressions yields dQ dW TdS dW actual actual − =− rev . Irreversible and reversible state changes