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Lecture 8: Maximum and Minimum Work, Thermodynamic Inequalities

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Lecture 8: Maximum and Minimum Work, Thermodynamic Inequalities Lecture 8: Maximum and Minimum Work, Thermodynamic Inequalities Chapter II. Thermodynamic Quantities A.G. Petukhov, PHYS 743 October 4, 2017 Chapter II. Thermodynamic Quantities Lecture 8: Maximum and Minimum Work, ThermodynamicA.G. Petukhov,October Inequalities PHYS 4, 743 2017 1 / 12 Maximum Work If a thermally isolated system is in non-equilibrium state it may do work on some external bodies while equilibrium is being established. The total work done depends on the way leading to the equilibrium. Therefore the final state will also be different. In any event, since system is thermally isolated the work done by the system: jAj = E0 − E(S); where E0 is the initial energy and E(S) is final (equilibrium) one. Le us consider the case when Vinit = Vfinal but can change during the process. @ jAj @E = − = −Tfinal < 0 @S @S V The entropy cannot decrease. Then it means that the greater is the change of the entropy the smaller is work done by the system The maximum work done by the system corresponds to the reversible process when ∆S = Sfinal − Sinitial = 0 Chapter II. Thermodynamic Quantities Lecture 8: Maximum and Minimum Work, ThermodynamicA.G. Petukhov,October Inequalities PHYS 4, 743 2017 2 / 12 Clausius Theorem dS 0 R > dS < 0 R S S TA > TA T > T B B A δQA > 0 B Q 0 δ B < The system following a closed path A: System receives heat from a hot reservoir. Temperature of the thermostat is slightly larger then the system temperature B: System dumps heat to a cold reservoir. Temperature of the system is slightly larger then that of the thermostat Chapter II. Thermodynamic Quantities Lecture 8: Maximum and Minimum Work, ThermodynamicA.G. Petukhov,October Inequalities PHYS 4, 743 2017 3 / 12 Clausius Theorem Cont'd Entropy balance: dSR > 0 A δQA δQA A dSR < 0 −dSR = ≤ S = dSsys TA TA T > T S S A A TB > TB B δQB δQB B dSR = − ≥ − S = −dSsys T T A B δQBA > 0 B Q 0 δ B < I δQ − dS = −dSA − dSB = ≤ dSA + dSB = ∆S = 0 R R R T sys sys sys I δQ ≤ 0 T Here δQ is the heat absorbed by the system and T is the of the thermostat. Chapter II. Thermodynamic Quantities Lecture 8: Maximum and Minimum Work, ThermodynamicA.G. Petukhov,October Inequalities PHYS 4, 743 2017 4 / 12 Carnot Engine A A A A Carnot Cycle: Isothermal expansion at TH Adiabatic expansion with cooling TH ! TL Isothermal compression at TL Adiabatic compression with heating TL ! TH Chapter II. Thermodynamic Quantities Lecture 8: Maximum and Minimum Work, ThermodynamicA.G. Petukhov,October Inequalities PHYS 4, 743 2017 5 / 12 Carnot's Engine Cont'd Since Carnot cycle is reversible the Clausius equality holds: Z dQ Q Q 0 = = H − L T TH TL because for two adiabatic processes Q1 = Q2 = 0. Efficiency: A Q − Q Q T η = = H L = 1 − L = 1 − L QH QH QH TH The work extracted from the Carnot engine is the maximum work because fro any other engine Z dQ T ≤ 0 ) η ≤ 1 − L T TH Chapter II. Thermodynamic Quantities Lecture 8: Maximum and Minimum Work, ThermodynamicA.G. Petukhov,October Inequalities PHYS 4, 743 2017 6 / 12 Minimum work done by an external medium 2 Work done on the system A 0 12 > Work done by the system A 0 12 < 1 If 1 ! 2 ! 1 is a reversible process then A12 + A21 = 0. Therefore (A12)max = − (A21)min Let us consider a system in a thermostat, which is so large that any changes in the system will not affect the macroscopic state of the thermostat (T0; p0). Question: What is the minimum work that should be done by an external source to transform the body into the state (T; p) that is not in equilibrium with the medium? Chapter II. Thermodynamic Quantities Lecture 8: Maximum and Minimum Work, ThermodynamicA.G. Petukhov,October Inequalities PHYS 4, 743 2017 7 / 12 Minimum Work cont'd The work done on the body: A = ∆E + ∆E0 ) ∆E = A − ∆E0 = A − T0∆S0 + p0∆V0 Here ∆E and ∆E0 are changes in the energy of the body and the thermostat respectively. As ∆(S + S0) ≥ 0 and ∆(V + V0) = 0 then A = ∆E + T0∆S0 − p0∆V0 ≥ ∆E − T0∆S + p0∆V Therefore, for a reversible process when "=" : Amin = ∆(E − T0S + p0V ) For any irreversible process A > Amin. If a body is in equilibrium at every moment of time then dE = T dS − pdV and dAmin = (T − T0)dS − (p − p0)dV Chapter II. Thermodynamic Quantities Lecture 8: Maximum and Minimum Work, ThermodynamicA.G. Petukhov,October Inequalities PHYS 4, 743 2017 8 / 12 Minimum Work cont'd Particular cases: I V = const and T = T0 Then Amin = ∆(E − TS) = ∆F Minimal work equals to the change of the Helmholtz Free energy I T = T0, p = p0 Then Amin = ∆(E − TS + pV ) = ∆Φ Minimal work equals to the change of the Gibbs Free energy Chapter II. Thermodynamic Quantities Lecture 8: Maximum and Minimum Work, ThermodynamicA.G. Petukhov,October Inequalities PHYS 4, 743 2017 9 / 12 Thermodynamic inequalities Again, consider a system in a thermostat with T = T0 and p = p0. Under these conditions the Gibbs free energy of the body has a minimum. By virtue of the previous consideration the minimum work required to bring the body from an equilibrium to a non equilibrium state equals to ∆Φ > 0. Thus δE − T0δS + p0δV > 0 But @E @E δE(S; V ) = δS + δV + @S 0 @V 0 2 2 2 1 @ E 2 @ E @ E 2 2 (δS) + 2 δSδV + 2 (δV ) 2 @S 0 @S@V 0 @V 0 > T0δS − p0δV Chapter II. Thermodynamic Quantities Lecture 8: Maximum and Minimum Work, ThermodynamicA.G. Petukhov,October Inequalities PHYS 4, 743 2017 10 / 12 Thermodynamic Inequalities cont'd @E @E Using @S 0 = T0 and @V 0 = −p0 we obtain 2 2 2 @ E 2 @ E @ E 2 2 (δS) + 2 δSδV + 2 (δV ) > 0 (1) @S 0 @S@V 0 @V 0 Inequality (1) must be valid for arbitrary δS and δV . Setting δV = 0 we get @2E 2 > 0 @S 0 In addition we can calculate the discriminant of the quadratic form (1) which yields @2E @2E @2E 2 2 · 2 − > 0 @S 0 @V 0 @S@V 0 Chapter II. Thermodynamic Quantities Lecture 8: Maximum and Minimum Work, ThermodynamicA.G. Petukhov,October Inequalities PHYS 4, 743 2017 11 / 12 Thermodynamic Inequalities cont'd The meaning of these thermodynamic inequalities can be established as follows: @2E @ @E @T 1 @T T 2 = = = T = > 0 @S @S @S V V @S V T @S V cV Thus cV > 0 By using Jacobians (see L&L) it is possible to show that the second condition is equivalent to @p < 0 @V T Finally one can show that cp > cV > 0 Chapter II. Thermodynamic Quantities Lecture 8: Maximum and Minimum Work, ThermodynamicA.G. Petukhov,October Inequalities PHYS 4, 743 2017 12 / 12.
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