Classical thermodynamics More about irreversibility chap. 6 Isentropic expansion of an ideal gas Sudden expansion of a gas into vacuum cf Kittel and Kroemer end of Cyclic engines cf Kittel and Kroemer chap. 8 Need for 2 heat reservoirs Carnot Efficiency Refrigerator, heat pumps Example: Gas engines 1
Reversibility The thermodynamic identities only apply for reversible processes Otherwise, the pressure, temperature, chemical potentials are undefined! Succession of equilibria! The evolution is so slow that the system has time to relax to equilibrium after each infinitesimal step. This is also called an adiabatic evolution. Of course, an idealization. For an isolated system reversibility <-> constant entropy dσ = 0 σ = constant! dt The system should stay in a configuration of maximum entropy This is an isentropic evolution! At the quantum level, an adiabatic evolution of the boundary conditions does not change the quantum number of an isolated system. Therefore, the entropy stays constant. For an isolated system, adiabatic and isentropic have identical meaning! 2
Heat Capacities Reminder: Two heat capacities! Define C V = dq dt V = τdσ dt V = du C P = dq dt P = τdσ dt P = γ = C P dt V du + PdV dt P = du + d(pv ) dt P = dh dt P C V e.g., non relativistic ideal gas PV = Nk B T U = U K + U int C V = 3/2Nk B + C int C P = 3/2Nk B + Nk B + C int γ = 5/2Nk B + C int 3/2Nk B + C int < 5/3 γ 1 = Nk B 3 / 2Nk B + C int 3
Isentropic Expansion Isentropic expansion: For an ideal gas PdV = Nk BT V dv = C P C V ( )T dv V a good way to change temperature reversibly requires to be isolated from heat bath and slow (i.e. adiabatic) dσ = 0 τdσ = du + PdV = 0 du = ( 3/2Nk B + C int )dt dt T + ( γ 1) dv V = 0 TV γ 1 = Constant PV γ = Constant T γ P 1 γ = Constant Quantum Mechanics: Adiabatic approximation It can be shown that if the boundary conditions change slowly enough, the orbitals keep the same quantum number e.g. in one dimension potential well of width L t ( ) Ψ = sin n xπ x L t ( ) if dl dt is small enough 4
Irreversibility: Examples Two isolated systems at different temperature suddenly put in contact We know from general principles Δσ > 0 W T 1 T 2 We can also compute with thermodynamic identities T from same initial to final states ΔU = f T C 1 dt + f T 1 C 2 dt = 0 T 2 If T f C Δσ = 1 k B T dt T f C + 2 T 1 k B T dt T f C T T > 1 f C T 2 dt + 2 1 < T 2 T 1 k B T f dt = 0 T 2 k B T f <0 <0 => Δσ > 0 This is a special case of: Transport of heat (without work) between two temperatures dσ h dσ l du h = du l dw = 0 < dq l = dq h = dq h dq h dq l < Dissipation: does not need to be brutal! dσ l = dq l > dσ h = dq h dσ > 0 Potential energy => heat Can be mechanical, chemical, electric or magnetic e.g. resistor Motion can be slow ( quasi-static ) and irreversible : e.g. friction 5
Expansion of Ideal Gas into Vacuum Let us go back to our prototype Conceptually Important! V i cf. end of chap. 6 in Kittel & Kroemer Initial Sudden! V f Final Isolated What is the increase of entropy? σ = log g t = NlogV + 3 2 N logu... Δσ = Nlog V f V i ΔS = Nk B log V f V i Use of the thermodynamics identity: Go to final configuration by succession of equilibrium configurations (i.e., replace by reversible process with same initial and final configurations )=> same entropy change. Take e.g., isothermal process (use mechanical energy to heat gas in the same proportion) du = TdS PdV PV = Nk B T = TdS Nk B T dv V = 0 6 ΔS = Nk B log V f V i
Irreversibility (2) Is this a counterexample of dσ=0? No! Potential problem: we claim to have a reversible process and dσ>0! Let be a little more specific: we use a machine to use the mechanical energy to heat gas in the same proportion System dw dq = τdσ Machine = dw > 0 But this was with an external machine! If reversible, machine has to loose entropy to have no total entropy change In general, external machine is not reversible (e.g., generator+resistance) => net entropy change >0 7
Non isolated systems Consider at the same time reservoir(s). Reversibility <=>Σ dσ Reservoirs + dσ system =0 all quantities as input to system (reservoir) considered Signs can be confusing! If irreversible add Advice: Use positive quantities and make two drawings: one for energy, one for entropy dσ Irrev 0 Reservoir h dw in Reservoir h dq h System du s dσ h System dσ s dq l Reservoir l dw out dσ l Reservoir l Look for: Uncontrolled motions Discontinuity in temperature Dissipation 8
Engines: Heat=> mechanical work Cyclic system: Cyclic Engines Comes back regularly to same configuration. Use Δ for change over one cycle. Not energy storage = 0 No entropy storage ΔU cycle Δσ cycle = 0 This does not imply reversibility An thermally isolated cyclic system cannot work! Δσ Irrev ΔW ΔU Δσ Reversible: depletion of energy => Not cyclic ΔU = ΔW < 0 cycle cycle Irreversible: in addition storage of entropy Δσ > 0 cycle 9
Cyclic Engines(2) One thermal reservoir is not sufficient! Reservoir ΔU = ΔQ Reservoir Δσ Δσ Irrev ΔW Reversible: ΔU = ΔQ = τδσ = ΔW Δσ > 0 No sink for entropy: not cyclic Irreversible: Even worse Δσ = ΔQ τ + Δσ Irrev. > 0 >0 10
We need two reservoirs! to dump entropy! Cyclic engine (3) Reservoir h ΔQ h ΔUh Δσ Irrev. Reservoir h Δσ h ΔQ l ΔU l = ΔQ l + ΔW Δσ l Reservoir l Reversible ΔW Reservoir l ΔU h = ΔQ h = ΔU l = ΔQ l + ΔW Δσ h = ΔQ h = Δσ = ΔQ l h ΔW = ΔQ h ΔQ l = ΔQ h 1 τ l Carnot efficiency η c = ΔW ΔQ h = 1 < 1 11
Cyclic Engines (4) Irreversible Δσ l = ΔQ l = Δσ h + Δσ Irrev. = ΔQ h + Δσ Irrev. >0 Lower Efficiency! Entropy dumped in low temperature reservoir has to be larger => ΔQ l larger, ΔW smaller! ΔW = ΔQ h ΔQ l = ΔQ h 1 τ l τ Δσ l Irrev. = ΔQ h 1 τ Δσ l Irrev. ΔQ h η= ΔW ΔQ h < η c =1 >0 >0 12
Carnot Cycle One remaining question: Can we invent a cycle that gives the Carnot efficiency? Carnot cycle σ σ h σ l 3 4 2 1 τ Idealization: an internal fluid in which we can store energy and from which we can extract work : we alternatively put it in contact with hot and cold sources. All parts of cycle have to be reversible. 1->2 Start with fluid at high temperature Increase its entropy from σ L to σ H 2->3Decrease isentropically its temperature from to. 3->4Dump entropy into cold source: decrease fluid entropy from σ H to σ l at constant temperature. 4->1 Increase isentropically its temperature from to. Total work: >0 Cyclic: entropy by construction energy: no storage beyond 1 cycle ΔW = δq h δq l = σ h σ l η = ΔW ( = ) ( σ h σ l ) = δq h ( σ h σ l ) 13 ( ) δq h = σ h σ l δq l = σ h σ l ΔW = δw 12 +δw 23 +δw 34 +δw 41 <0 ( ) ( σ h σ l ) = ( )( σ h σ l ) ( ) = η c ( ) ΔU = ΔQ h ΔQ l ΔW = 0 Carnot efficiency!
Ideal Gas: Carnot Engine PV γ P 1 = Constant 4 PV = N PV = N 2 PV γ 3 = Constant 1->2 Isothermal expansion in contact with hot source δu = 0 2 δw 12 = δq h = PdV = N log V 2 1 V 2->3 Go on with isentropic expansion in order 1 to cool gas ( ) = 3/2N( ) δw 23 = U 3 U 2 3->4 Connect with cold source and compress isothermally 4 δw 34 = δq l = PdV = N log V 4 3 4->1 Go on with isentropic expansion in order Vto 3 heat gas V δw 41 = ( U 1 U 4 ) = 3/2N( ) Clapeyron But V 4 V 1 = 1 γ 1 = V 3 V 2 V 3 V 4 = V 2 V 1 Total work ΔW = δw 12 +δw 23 + δw 34 + δw 41 = N( )log V 2 V 1 η = ΔW ΔQ h = = η c 14
Cyclic! Refrigerator/Heat Pump ΔW Δσ Irrev. ΔQ l ΔQ Δσ l h Δσ h Cold Hot Cold Hot ΔU in = ΔW + ΔQ l = ΔU out = ΔQ h Reversible Δσ l = ΔQ l = Δσ h = ΔQ h ΔW = ΔQ h ΔQ l = ΔQ l Efficiency as a refrigerator: γ = ΔQ l ΔW = ΔQ h = ΔQ l = ΔQ h = 1 for = 2 can be smaller or greater than 1! Large when temperature difference is small. Efficiency as an heat pump: ε = ΔQ h ΔW = >1 15
Irreversible Refrigerator/Heat Pump Δσ h = ΔQ h >0 Lower Efficiency! = Δσ l + Δσ Irrev. = ΔQ l + Δσ Irrev. ΔQ h = ΔQ l + Δσ Irrev. Efficiency as a refrigerator: γ irrev = ΔQ l ΔW < γ Efficiency as an heat pump: Increase of the work necessary to transport entropy Δσ Irrev. + ΔQ l ΔW = ΔQ h ΔQ l = ΔQ l Δσ Irrev. + ΔQ h = ΔQ h >0 >0 ε irrev = ΔQ h ΔW < ε 16