Winter 2013 Chem 254: Introductory Thermodynamics Chapter 5: Entropy ........................................................................................................................ 46 Calculating ΔS for processes in ideal gas .................................................................................. 48 ΔS as a function of changes in P,T,V in pure substances (not mixtures or reactions) .............. 51 The Second law of thermodynamics......................................................................................... 54 Absolute Entropy (Pure Substances) ........................................................................................ 56 Microscopic (molecular) view of Entropy ................................................................................. 56 Heat Engine ............................................................................................................................... 58 Carnot Cycle .............................................................................................................................. 59 Midterm Review............................................................................................................................ 61 Chapter 5: Entropy Entropy: the second and third law thermodynamics First law: Energy can be converted from one form to another but the total energy for an isolated system is constant (heat-work-chemical energies) U 0 (is constant if all processes are included) Can we obtain a law that describes what can or cannot happen? Mixture of N2 , H 2 (raise T , catalyst) forms NH3 in equilibrium, once it reaches equilibrium, it does not go back to reactants, again we observe an “arrow of time”. For isolated system - Entropy ( S ) S 0 for any possible, irreversible process 46 Winter 2013 Chem 254: Introductory Thermodynamics S 0 for a reversible process - Other quantities (Ch 6) G Gibbs free energy, used when TP, constant A Helmholtz free energy, used when TV, constant Entropy is an abstract quantity for macroscopic systems, thermodynamics History of discovery - discovered by engineer Carnot - improve efficiency of steam engines (1830s) - it is still taught like this (see 5.2 in book) Physical insight into entropy microscopic theory - Quantum Mechanics - Statistical Mechanics (molecular point of view, discussed at Ch. 5 end) Consider a reversible process for an ideal gas rev rev rev rev du q w qw, inexact differentials (depends on path) rev CV dT q P ext dV nRT nRT qrev C dT dV PPrev V V extV gas Differential for qrev qrev : NOT a exact differential nRT CV VTV T V nR 0 V nR 0 therefore qrev is not an exact differential V qrev C nR V dT dV is this an exact differential? TTV CV nR VTTVT V 00 therefore is an exact differential qrev dS entropy, an exact diff, independent of path T Chapter 5: Entropy 47 Winter 2013 Chem 254: Introductory Thermodynamics rev f q SSS entropy is a state function fii T 1 : is called an integrating factor changes inexact diff to exact diff T Eg. From math dF ydx xdy yx yxx y 11 dF is an inexact diff dF y 1 dG dx dy xx22 x y 1 2 yx x x x y 11 dG is an exact diff xx22 1 : integrating factor x2 Calculating ΔS for processes in ideal gas qrev dS (exact differential) T S is a state function STPV ,, of single component systems SSS final initial Like HU, , S independent of process or path How to calculate S rev f q S i T What does this mean? Choose a reversible path: at all stages, the system retains equilibrium (stays on the ideal gas curve) Chapter 5: Entropy 48 Winter 2013 Chem 254: Introductory Thermodynamics Even for irreversible process calculate S for a reversible path with same initial and final states Ideal gas PV nRT a)TVTV1 1 2 2 SSS()()()CAB A. Constant T (reversible isothermal) B. Constant V U q w 0 w 0 Uq f nRT q w () P dV dV q dU C dT ext i V V nRT C q dV dS V dT V T f q nR CV Tf ds dV S dT C ln i V TV TTi f nR Vf S dV nRln i VVi nC nR dSVm, dT dV TV TVff S nCVm, ln nR ln TVii b) PV1 1 PV 2 2 SFSESD()()() Chapter 5: Entropy 49 Winter 2013 Chem 254: Introductory Thermodynamics D. Constant P E. Constant (same as before) f CV Tf q H C dT S dT C ln PP i V TTi C nRT dS P dT Constant Volume , constant T P Tf TPff SCP ln Ti TPii P Tf Ti nRT f since P constant SCV ln VVfi V Pi Vf SCP ln Vi V CC dS nP,, m dV n V m dP VP VPff S nCP,, mln nC V m ln VPii c) PT1 1 PT 2 2 Pf Vi constant T PVif C R dS nPm, dT n dP TP TPff S nCPm, ln nR ln TPii Chapter 5: Entropy 50 Winter 2013 Chem 254: Introductory Thermodynamics d) adiabatic reversible process qrev qrev 0 0 T S 0 TVTV1 1 2 2 adiabatic reversible process TVff nCvm, ln nR ln TVii Cvm, TVff ln ln (see section 2.11 in book) RTVii ΔS as a function of changes in P,T,V in pure substances (not mixtures or reactions) Dependence of S on TP, SS dS dT dP exact differential TP PT CP V dT dP TT P Why is this true? qrev C 1) At constant P dS dTP dT TT SV 2) Maxwell relation, see later PT TP Liquids and solids: V V T P TP ffCP S dT V dP TPiiT T f CP dT V P Ti T Gases: V a little effort T P nRT V nR V (ideal gas) P TPP Chapter 5: Entropy 51 Winter 2013 Chem 254: Introductory Thermodynamics Dependence of on TV, SS dS dT dV TV VT CV P dT dV TT V qrev C 1) At constant V , dSV dT TT SP 2) Maxwell Relation S VT TV PPV TVT VTP Liquids and solid: TV ffCV S dT dV TViiT T f CV dT V Ti T Gases: P Vf P Calculate dV V T V i T V nRT P nR Ideal gas P VTV V VVffP dV dV nR VV iiTVV C R dS nPm, dT n dP TVffTPCvm, nR S n dT dV TViiTV TPff S nCPm, ln nR ln TVff TPii S nCvm, ln nR ln (as we saw before) TVii S for phase change (single substance) sl sg lg Chapter 5: Entropy 52 Winter 2013 Chem 254: Introductory Thermodynamics Reversible process: example o Boiling point of water, 100 C at 1 atm rev o q vap H373 TTvap o Vaporization vap H373 0 o vapH T Sn moles Tvap o o subH T fusH T Sn moles Sn moles Tsub Tfus o o Example: Raise T of water at 1 atm from 25 C to 125 C 298 K 373 K heating liquid At 373 K liquid gas 373 K 398 K heating gas liquid gas 373CCP,, mHvap 398 P m S n dT n n dT 298 373 TTTvap Other example Sliquid freezing + cooling qbath qsystem Sbath TTbath bath qHsystem system (constant P ) OR qUsystem system (constant V ) qqsurrounding system Ssurrounding (constant Tsurrounding ) TTsurrounding surrounding SSsystem surrounding 0 for any possible process. (see later) sl sg lg Chapter 5: Entropy 53 Winter 2013 Chem 254: Introductory Thermodynamics The Second law of thermodynamics Postulate by Rudolph Clausius (~1865) - A Postulate is something you cannot derive but take to be true, because large number of experiments support it. dq 0 for any possible physical process T dq (If 0 for any reproducible process we would have to rewrite T thermodynamics and all of its rules.) Anything derived from postulate should be true also Consequences of Postulate: 2 dq a) S for any process 12 1 T rev 2dq 2 dq 1TT 1 rev 2dq 1 dq dq 0 1TTT 2 rev 2dq 2 dq 0 1TT 1 b) For isolated system S 0 for a possible physical process 2 dq S For isolated system q 0 1 T So c) We can always create a system and environment such that total system is isolated SSStotal system surrounding 0 If environment is kept at constant T qenv qsys Senv TTenv env We can calculate S for ANY process (allowed and unallowed) If we calculate Stotal for isolated system then: S 0 irreversible, possible spontaneous (ice forming at -5oC) Chapter 5: Entropy 54 Winter 2013 Chem 254: Introductory Thermodynamics o S 0 reversible, possible (ice equilibrium at 0 C) o S 0 impossible (ice forming at 5 C) Example: q 0 TT show it? qq dStotal 0 TT qq TT TT Heat flows from hot to cold (another formulation of the second law) Vf Stotal nRln 0 (at constant T ) Vi U w q 0 VV final inital dq 0 T irrev rev 2dq 1 dq 0 1TT 2 Chapter 5: Entropy 55 Winter 2013 Chem 254: Introductory Thermodynamics Absolute Entropy (Pure Substances) Molar entropy, all 3 phases, all H and CPm, can be measured S l g TTTfusCCCP,,, mHHfus vap P m vap f P m S( T ) S (0 K) dT dT dT mm 0 TT fus vap TTTTTtus vap SSTSm m( ) m (0) Nernst and Richards (~1910) Sm (0 K) 0 for perfect crystalline solids. Today, we can calculate absolute entropy from first principles, quantum and statistical mechanics for ideal gas. Confirms Sm (0 K) 0 Non crystalline substances Perfect arrangement, but in practice we can get random arrangements In this case Sm (0 K) 5.76 J/mol ( Rln 2) Microscopic (molecular) view of Entropy Diffusion of gas: Laws of microscope physics are time reversible? Why do processes occur in particular direction? Chapter 5: Entropy 56 Winter 2013 Chem 254: Introductory Thermodynamics Difficult (impossible..) to get the initial conditions to reverse such reactions Calculate for diffusion of gas Prfull 1 1 1 1 ...... Pr 111 NA half 1 222 2 Na Prfull 1 KKKNRBBBAln ln ln 2 ln 2 Prhalf 2 Prfull Thus KRB ln ln 2 (for 1 mol) Prhalf Vf S for VV12nRln R ln 2 Vi Similar idea for the crystal, only 2 orientations Pr 1 N ordered SKKN ln 2 A ln 2 Na BBA Prdisordered 1 2 NA Fully ordered state has zero entropy, SK B ln 1 0 Quantum and statistical mechanic picture of entropy: Quantum: E j quantum energy levels Statistical: Probability to find molecule at particular state (Boltzmann distribution) eEKTjB/ PT() STKPP( ) B j ln j j EKTjB/ e j j This is our best understanding of entropy Chapter 5: Entropy 57 Winter 2013 Chem 254: Introductory Thermodynamics Heat Engine History: Carnot investigated efficiency of heat (steam) engines ~1830 Carnot: run each cycle in reversible fashion to get max efficiency, recall that grain of sand compression is more efficient than irreversible process.