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Chapter 3. Second and Third Law of Thermodynamics

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Chapter 3. Second and Third Law of Thermodynamics Chapter 3. Second and third law of thermodynamics Important Concepts Review Entropy; Gibbs Free Energy • Entropy (S) – definitions Law of Corresponding States (ch 1 notes) • Entropy changes in reversible and Reduced pressure, temperatures, volumes irreversible processes • Entropy of mixing of ideal gases • 2nd law of thermodynamics • 3rd law of thermodynamics Math • Free energy Numerical integration by computer • Maxwell relations (Trapezoidal integration • Dependence of free energy on P, V, T https://en.wikipedia.org/wiki/Trapezoidal_rule) • Thermodynamic functions of mixtures Properties of partial differential equations • Partial molar quantities and chemical Rules for inequalities potential Major Concept Review • Adiabats vs. isotherms p1V1 p2V2 • Sign convention for work and heat w done on c=C /R vm system, q supplied to system : + p1V1 p2V2 =Cp/CV w done by system, q removed from system : c c V1T1 V2T2 - • Joule-Thomson expansion (DH=0); • State variables depend on final & initial state; not Joule-Thomson coefficient, inversion path. temperature • Reversible change occurs in series of equilibrium V states T TT V P p • Adiabatic q = 0; Isothermal DT = 0 H CP • Equations of state for enthalpy, H and internal • Formation reaction; enthalpies of energy, U reaction, Hess’s Law; other changes D rxn H iD f Hi i T D rxn H Drxn Href DrxnCpdT Tref • Calorimetry Spontaneous and Nonspontaneous Changes First Law: when one form of energy is converted to another, the total energy in universe is conserved. • Does not give any other restriction on a process • But many processes have a natural direction Examples • gas expands into a vacuum; not the reverse • can burn paper; can't unburn paper • heat never flows spontaneously from cold to hot These changes are called nonspontaneous changes. It is useful to predict whether a physical/chemical change will go spontaneously in the forward or backward direction. The second law and entropy, S, a state function of the system, is what tells us this. Carnot, investigating heat engines, was the first to answer the question of spontaneity. Nicolas Léonard Sadi Carnot 1796-1832 "father of thermodynamics" The Carnot Engine and Carnot Cycle A heat engine is an engine that uses heat to generate mechanical work by carrying a "working substance" through a cyclic process. A quantity of heat qh is withdrawn from the hot reservoir (the boiler) A quantity of heat qc (negative by convention) is discharged to the cold reservoir (the condenser) the work done by the engine is w (negative by convention). The Carnot Engine and Carnot Cycle http://galileoandeinstein.physics.virginia.edu/mor e_stuff/flashlets/carnot.htm Carnot Heat Engines Step 1: Reversible, isothermal expansion from A to B at Th ∆U1 = 푞ℎ + 푤1 = 0 (isothermal, ideal gas) Heat= 푞ℎ 푉퐵 Work 푤1 = −푅푇ℎ푙푛 = −푞ℎ 푉퐴 Step 2: Reversible, adiabatic expansion from B to C; Th→Tc Heat= 푞 = 0(adiabatic) 푇푐 Work 푤2 = ∆푈2 = ׬ 퐶푣푑푇 푇ℎ 푇푐 퐶 푚 푉 න 푣 푑푇 = −Rln 퐶 푇 푉퐵 푇ℎ Step 4: Reversible, adiabatic compression from D to Step 3: Reversible, isothermal compression from C to D at Tc A; Tc→Th Heat= 푞 = 0(adiabatic) ℎ ∆U1 = 푞푐 + 푤3 = 0 (isothermal, ideal gas) Work 푤4 = ∆푈4 = ׬ 퐶푣푑푇 Heat= 푞 (negative because heat is leaving the system) 푇푐 푐 푇ℎ 푉퐷 퐶 푚 푉 Work 푤3 = −푅푇푐푙푛 = −푞푐 푣 퐴 푉퐶 න 푑푇 = −Rln 푇 푉퐷 푇푐 Efficiency of Heat Engine ∆U퐶푌퐶퐿 = 0 = 푞ℎ+푞퐶 + 푤1 + 푤2 + 푤3 + 푤4 = 푞푐푦푐푙+푤퐶푦푐푙 −푤퐶푦푐푙= 푞푐푦푐푙 = 푞ℎ + 푞퐶 푤표푟푘 푝푒푟푓표푟푚푒푑 −푤 푞 푞 푞 Efficiency: = 퐶푦푐푙= ℎ+ 퐶 = 1 − 퐶 ≡ 휂. ℎ푒푎푡 푠푢푝푝푙푒푑 푞ℎ 푞ℎ 푞ℎ 푇 퐶 푚 푉 푉 Comparing adiabatic steps: ׬ ℎ 푣 푑푇 = −Rln 퐵 = −Rln 퐴 푇푐 푇 푉퐶 푉퐷 푉 푉 푉 푉 푉 푉 Ratios of volumes are equal: 퐵 = 퐴, so 퐶 = 퐵, so −ln 퐷 = ln 퐵 푉퐶 푉퐷 푉퐷 푉퐴 푉퐶 푉퐴 푉퐵 푉퐷 푉퐵 Can not rewrite heats: 푅푇ℎ푙푛 = 푞ℎ ; 푅푇푐푙푛 = 푞푐=−푅푇푐푙푛 푉퐴 푉퐶 푉퐴 푞 푞 푞 푞 Therefore, 푐 = − ℎ 푎푛푑 푐 + ℎ = 0 푇푐 푇ℎ 푇푐 푇ℎ Means two things: 1) Even though q is not a state function, q/T is! 푑푞 We define this as entropy 푑푆 ≡ 푟푒푣 푇 Although a change in S must be calculated along a reversible path, entropy can be defined for any process 푇 푇 −푇 2) Can use T in efficiency calculation. 휂 = 1 − 푐 = ℎ 푐 푇ℎ 푇ℎ Heat Engines and Refrigerators Operate Heat Engine in Reverse Supply work to system Remove heat from cold reservoir, and add it to hot reservoir Work required to furnish heat qh with a heat pump 푇ℎ − 푇푐 푤 = −푞ℎ 푇ℎ Ratio of work required to remove heat qc with a/c or refrigerator 푤 푇ℎ − 푇푐 = 푞푐 푇푐 푞푐 푇푐 Coefficient of performance 푐 = = 푤 푇ℎ − 푇푐 Definition of entropy (S) Thermodynamic definition – entropy is related to the heat flow f dS= dq /T , DS 1 T dq rev rev i e.g. DS for isothermal reversible expansion of ideal gas: DU = q+w = 0, so, qrev = -wrev = nRT ln(Vf/Vi) DS = qrev/T = nR ln(Vf/Vi) Statistical definition of entropy (S) Statistical definition – entropy is related to the probability Entropy is a measure of the number of possible microscopic states of a system. S=k ln W (k: Boltzmann constant, W: number of microstates W) W is a measure of disorder of the system in a sense. e.g. There is only one way to have 100 coins with all heads up: W=1 There are 100891344545564193334812497256 ways to have half of the coins are heads up: W=100891344545564193334812497256 – a lot more probable. http://www.bbc.com/earth/story/20150309-why-does-time-only-run-forwards Inequality of Clausius The second law discusses the properties of entropy. • Start with First Law dU dq PexdV • For a reversible process, P = Pex & dq = TdS • Gives dU TdS PdV (true for any process, reversible or not since S and U are state variables) • Now use dU = dq + dw • equate 2 equations for dU TdS PdV dq PexdV TdS dq P Pex dV If P > Pex , sys. will expand, dV > 0, 2nd term is + If P < Pex , sys. will contract, dV < 0, 2nd term is + If P = Pex , 2nd term = 0 (equilibrium) Therefore, TdS dq Second Law • If the process is reversible, the equality applies Rudolf Julius • If any part of the process is irreversible, the inequality applies Emanuel Clausius 2 It is impossible for dS dq T January 1822 – 24 August 1888) In an isolated system, dq = 0 DS > 0 spontaneous, irreversible process DS = 0 reversible process DS < 0 impossible process! Second law of thermodynamics - direction of spontaneous change Spontaneous changes occurs in the direction where the energy of isolated system is dispersed in a random, chaotic way. 2nd Law: Total entropy of an isolated system increases in the course of a spontaneous change. DStot >0 You can't break even. dU w q Can't convert all energy to work. q TdS Even in a perfectly efficient system, lose some to entropy. For a reversible system, dU TdS w http://www.bbc.com/earth/story/20150309-why-does-time-only-run-forwards Entropy of System and Surroundings The second law does not say that the entropy of a system can never spontaneously decrease. It does say that the entropy of the universe can never decrease. So, if in a spontaneous process, the entropy of the system goes down, then the entropy of the surroundings must go up. dSsur= dqsur, rev/T , dSsur= dqsur/T To predict spontaneity in a system that is not isolated, must calculate both entropy of system and surroundings. • Surroundings typically very large, and V or P is constant. • Heat transferred to the surrounding is dU or dH, state function – same regardless of reversible path or not. • Entropy change in surroundings can be simply calculated from the heat transferred to the surroundings and its temperature. dSsys = dqsys, rev/T • For reversible process, heat can be used to calculate the entropy change. • For irreversible process, find a reversible path having the same initial and final state. Example of entropy change system & surroundings (reversible case) Reversible isothermal compression of ideal gas 22.4 L of gas compressed to 11.2 L at 273K • For system, DU = 0, qrev = -w = nRTln(Vf/Vi) = 8.314 x 273 ln(1/2) = -1573J DS = qrev /T, -1573/273 = -5.76 J/K (entropy of system decreased) • For surroundings, Heat is added to the surrounding from the system, qsur = - qrev DSsur = qsurv /T, 1573/273 = 5.76 J/K (entropy of surroundings increased) • Total entropy change DStotal = DSsys + DSsur = -5.76 J/K + 5.76 J/K = 0 J/K Total entropy change is zero for a reversible process ! Example of entropy change system & surroundings (irreversible case) Isothermal compression of ideal gas (P1V1=P2V2) initially at 1 atm 22.4 L of gas compressed to 11.2 L at 273K with external pressure same as final pressure • For system, Calculate heat for the reversible path with same initial and final states. Entropy change in the system is the same as in the previous page. -5.76 J/K • For surroundings, must know how much heat is transferred DSsur = qsur /T = - qsys/T, qsys can be calculated by looking at system. DU = 0, qsys = -w = Pext DV = 2 atm x101.325 J/L-atm· (-11.2 L)= -2270J DSsur = 2270/273 = 8.31 J/K • Total entropy change DStotal = DSsys + DSsur = -5.76 J/K + 8.31 J/K = 2.55 J/K Total entropy change is positive for an irreversible process ! Calculating Entropy Changes dU PdV 푑푈 = 푑푞 + 푑푤 to get dS T I.
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