or, making use of the equation of state and the fact that du =0 for an isothermal process AS= NR In This is the entropy change that occurs for the free expansion as well as for the isothermal reversible expansion processes -entropy changes are state changes and the two system final and end states are the same for both processes For the free expansion: NR In There is no change in the entropy of the surroundings because there is no interaction between the system and the surroundings. The total entropy change is therefore +As =NR In > There are several points to note from this result 1)As>0 so the process is not reversible de-o the equ ity between AS and is only for a reversible process iThere is a direct connection between the work needed to restore the system to the original state and the entropy change W=NRrn(≌)=TAs The quantity TAs has a physical meaning as"lost work"in the sense of work which we lost the opportunity to utilize. We will make this connection stronger in Section 1. C For the reversible isothermal expansion: The entropy is a state variable so the entropy change of the system is the same as before. In this case, however, heat is transferred to the system from the surroundings(Q odmg<0)so that The heat transferred from the surroundings, however, is equal to the heat received by the system surrounding Nm(≌ The total change in entropy(system plus surroundings) is therefore QQ 0 The reversible process has zero total change in entropy 1B-109T S T S  V   V  − or, making use of the equation of state and the fact that dU = 0 for an isothermal process, ∆S = NR ln  2  .  V1  This is the entropy change that occurs for the free expansion as well as for the isothermal reversible expansion processes—entropy changes are state changes and the two system final and end states are the same for both processes. For the free expansion: ∆Ssystem = NR ln   V2   ; ∆Ssurroundings = 0  V1  There is no change in the entropy of the surroundings because there is no interaction between the system and the surroundings. The total entropy change is therefore, ∆Stotal = ∆Ssystem + ∆Ssurroundings = NR ln   V2   > 0.  V1  There are several points to note from this result. i) ∆Stotal > 0 so the process is not reversible 2 dQ ii) ∆Ssystem > ∫1 T = 0; the equality between ∆S and dQ is only for a reversible T process iii) There is a direct connection between the work needed to restore the system to the original state and the entropy change: W = NRT ln  2  = ∆ 2 1  V1  The quantity ∆ has a physical meaning as “lost work” in the sense of work which we lost the opportunity to utilize. We will make this connection stronger in Section 1.C. For the reversible isothermal expansion: The entropy is a state variable so the entropy change of the system is the same as before. In this case, however, heat is transferred to the system from the surroundings ( Qsurroundings < 0) so that ∆Ssurroundings = Qsurroundings < 0. T The heat transferred from the surroundings, however, is equal to the heat received by the system: Qsurroundings = Qsystem = W . ∆Ssurroundings = Qsurroundings = −W = - NR ln   V2   . T T  V1  The total change in entropy (system plus surroundings) is therefore ∆Stotal = ∆Ssystem + ∆Ssurroundings = Q − Q = 0. T T The reversible process has zero total change in entropy. 1B-109