Camille Bélanger-ChampagneLuis Anchordoqui LehmanMcGill College University City University of New York ThermodynamicsCharged Particle and StatisticalCorrelations Mechanics in Minimum Bias Events at ATLAS • Entropy● Physics changes motivation in cyclic processes Maxwell● Minbias relations event and track selection Thermodynamics III • th Dissipative work as heat supply 11 SeptemberFebruary 2014 26 2012,• ● Azimuthal correlation results Entropy of ideal gas WNPPC 2012 • ● Forward-Backward correlation results • Third Law of Thermodynamics Saturday, September 13, 14 1 @V @V dV = dP + dT (7) @P @T ✓ ◆T ✓ ◆P W = PdV (17) Z (39) CP = CV + nR @P @V C = C + T (42) P V @T @T ✓ ◆V ✓ ◆P nR nRγ C = C = (50) V γ 1 P γ 1 − − δQ = TdS (79) Saturday, September 13, 14 2 ENTROPY Consequence of secondOverview law is existence of entropy: a state function @ thermodynamic equilibrium whose differential is given by δQ = TdS (79) S being a state function ☛ does not change in any reversible cyclic process: δQ =0 T I Since Q is extensive ☛ so is S Units of entropy are [S]=J/K C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 3 CHANGE OF ENTROPY IN IRREVERSIBLE PROCESSES In non equilibrium statesOverview ☛ thermodynamic entropy is undefined If initial and final states of irreversible process are in equilibrium entropy of these states is defined so ☛ entropy change ∆S = S S 12 2 − 1 We can find always reversible process connecting 1 and 2: equivalent reversible process Both processes can be joined into irreversible cyclic process for which Clausius inequality applies and takes the form 2 δQ 1 δQ + reversible 0 (80) T T Z1 Z2 Since reversible integral is related to change of entropy 2 2 δQreversible δQ (81) =∆S12 1 T ≥ 1 T Z δQ Z dS ≥ T (82) C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 4 CHANGE OF ENTROPY ON ISOLATED SYSTEMS If system is isolated ☛OverviewδQ = δW =0 Still ☛ dS 0 due to irreversible processes inside systems ≥ e.g. relaxation to equilibrium Consider isolated system that consists of two subsystems each of them at internal equilibrium but there is no equilibrium between subsystems Changes in reservoir entropies are: δQHOT Q2 (∆S)HOT = = < 0 T − T2 T =Z T2 δQCOLD 1 Q1 (∆S)COLD = = Q = > 0 T T1 − T1 T =Z T1 because hot reservoir loses heat Q 2 to engine and cold reservoir gains heat from engine Q1 C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 5 ENTROPY CHANGES IN CYCLIC PROCESSES Overview Then for the whole system follows Q1 Q2 (∆S) = + 0 (83) TOTAL − T T ≥ ✓ 1 2 ◆ According to second law of thermodynamics heat flows from hot to cold body T1 <T2 Reversible cycle ☛ net change of total entropy of the engine + reservoir is zero Irreversible cycle ☛ increase of total entropy due to spontaneous processes C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 6 MAXWELL RELATION Inserting (17) and (79) into first law of thermodynamics mainOverview thermodynamic identity dU = TdS PdV (84) − (84) is a differential of internal energy as a function of two variables @U @U Correspondingly ☛ T = P = (85) @S − @V !V !S As second mixed derivative does not depend on order of differentiation @2U @2U = @S@V @V@S (86) it follows that @T @P = @V − @S !S !V (87) C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 7 DISSIPATIVE WORK AS HEAT SUPPLY Overview δQ = δQreversible + δQdissipative δQreversible = TdS and (88) δW = δWreversible + δWdissipative δWreversible = PdV Then using the first law (25), (84), and (88) we have δQ δW = TdS PdV = δQ δW (89) − − reversible − reversible It follows that ☛ δQ δW =0 dissipative − dissipative or equivalently ☛ δQ δW = TdS (90) − dissipative This shows that dissipative work is equivalent to heat supply C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 8 VOLUME DEPENDENCE OF ENTROPY AND INTERNAL ENERGY 1 P OverviewdS = dU + dV (91) T T @U @U Use dU = dT + dV @T @V ✓ ◆V ✓ ◆T to obtain differential of entropy as a function of T and V 1 @U 1 @U dS = dT + + P dV (92) T @T T @V !V " !T # Use uniqueness of mixed second derivatives of S to obtain a relation involving derivatives of U @S 1 @U C (92) implies = = V (93) @T T @T T !V !V and @S 1 @U = + P @V T @V (94) !T " !T # C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 9 INTERNAL ENERGY OF IDEAL GAS Uniqueness of secondOverview mixed derivative of S ( T,V ) is expressed as @ 1 @U @ 1 @U = + P @V T @T @T T @V (95) " !V #T ( " !T #)V Performing differentiation 1 @2U 1 @U 1 @2U 1 @P = + P + + (96) T @V@T −T 2 @V T @T@V T @T " ◆T # !V Taking into account uniqueness of mixed second derivative of U ( T,V ) @U @P = T P @V @T − (97) ◆T !V that is (41) For the ideal gas from this formula and equation of state follows (@U/@V )T =0 that is U = U(T ) C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 10 MORE ON MAXWELL RELATIONS Substituting (97) intoOverview (94) @S @P = (98) @V @T !T !V Maxwell relation (98) allows derivation of (97) in a shorter way From (84) it follows that @U @S = T P (99) @V @V − !T !T Substituting in (98) yields (97) C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 11 THERMODYNAMIC COEFFICIENTS FOR ADIABATIC PROCESSES Adiabatic compressibility Overview1 @V = (100) S −V @P !S To express S through experimentally measurable quantities consider @S @S dS = dT + dV (101) @T @V !V !T Setting dS =0 and inserting (93) and (98) C @P 0= V dT + dV (102) T @T !V Combining (102) with (7) @V @V T @P dV = dP dV (103) @P − @T CV @T !T !P !V C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 12 EQUIVALENTLY... @OverviewV T @P @V 1+ dV = dP (104) @T CV @T @P " !P !V # !T @V γdV = dP (105) can be simplified with help of (42) to @P (105) !T T Adiabatic compressibility ☛ S = γ (106) Since C P >C V for all substances S < T is universally valid Adiabatic compression is accompanied by temperature increase that is described by thermodynamic coefficient (@T/@P )S Combining (102) and (105) C @P C @V V dT = V dP − T @T CP @P (107) !V !T C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 13 THIS YIELDS... @T T @P @V (108) Overview= dP −CP @T @P !S !V !T With help of (13) this simplifies to @T T @V VTβ = = (109) dP CP @T CP !S !P For ideal gas T β =1 and this formula gives @T T nR V (110) = = dP CP P CP !S Adiabat equation of ideal gas in the form of (53) can be rewritten as 1 1/γ T = AP − A = const (111) Differentiating this equation @T 1 1/γ 1 T 1 V = A 1 P − = 1 = 1 dP − γ − γ P − γ nR (112) !S ! ! ! C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 14 ENTROPY OF IDEAL GAS Energy of ideal gas depends on temperature only U = U(T ) For perfect gas ☛ CV Overview= const and (93) can be integrated over T (113) S(T,V )=CV ln T + f(V ) Using (98) @S P nR = = @V T V (114) !T Inserting S ( T,V ) into (114) ☛ df/dV = nR/V Integration yields ☛ f = nR ln V + S 0 Then with help of (50) one obtains γ 1 S = CV ln T + nR ln V + S0 = CV ln (TV − )+S0 This formula defines entropy up to an arbitrary constant S0 γ 1 In adiabatic process of a perfect gas TV − = const and entropy does not change C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 15 THIRD LAW OF THERMODYNAMICS Analyzing experimental data Walther Nernst concluded that In the limit T 0 entropyOverview becomes constant independent of other! thermodynamic parameters such as volume and pressure @S @S = =0 (117) @V @P !T 0 !T 0 ! ! Since in thermodynamics entropy is defined up to a constant Planck has suggested to define S(T 0) = 0 (118) ! Explanation of these results is possible only within statistical physics It turns out that statistically defined entropy always satisfies (117) whereas (118) holds for most substances Some materials have a degenerate ground state and in this case entropy tends to a finite constant at T 0 ! C.Luis B.-Champagne Anchordoqui 2 Saturday, September 13, 14 16 CONSEQUENCES OF THE THIRD LAW Overview T C Integrating (93) one obtains ☛ S = V dT + S T 0 (119) Z0 If C V is finite at T 0 entropy logarithmically diverges ! contradicting third law ☛ C (T 0) = 0 V ! (120) Same condition for C P can be proved in a similar way Note that divergence of entropy of ideal gas (115) at T 0 ! only proves that concept of ideal gas breaks down at low temperatures where gases become liquid and solid From (98) ☛ pressure thermal coefficient vanishes at absolute zero @P =0 @T ! ? (121) V ?T 0 Next class we will see that ?thermal! expansion coefficient also vanishes: ? @V ? =0 @T ! ? (122) P ?T 0 ? ! C.Luis B.-Champagne Anchordoqui ? 2 Saturday, September 13, 14 ? 17 2.6. HEAT ENGINES AND THE SECONDTHE LAWSTIRLING OF THERMODYNAMICS CYCLE 39 Consists of two isotherms and two isochores Figure 2.15: A Stirling cycle consists of two isotherms (blue) and two isochores (green). Isothermal ideal gas equation of state ☛ d ( pV )=0 AB: This isothermal expansion is the power stroke.