0 for reversible > 0 for irreversible process

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1 Last class: 2 nd law summary The entropy of an isolated system increases in an irreversible process and remains unchanged in a reversible process. The total entropy can never decrease. 0 for reversible S Ssys Ssurr { } > 0 for irreversible process A system develops towards the state with the greatest probability. Heat cannot flow from a colder body to a warmer one without external aid.

2 What is include in the quizzes /exams / final??? Final Exam 1 Exam n Quiz 1 Quiz n n+1 Time (weeks)

3 Plan today Example (as a motivation) 3 rd law of thermodynamics More examples

4 3 rd law: What about the temperature dependence of the entropy?

5 Examples What about the temperature dependence of the entropy? Motivation: Calculate the value of S in heating n moles of a monoatomic ideal gas from T i to T f at constant pressure? Similar to example 4.9 in your book Molar heat capacities C p -C v = R C p = C v + R C v = f/2r, (f = degree of freedom) C p = f/2r + R = 3/2R + 2/2R = 5/2R (for an atom) That may raise the question: how large is the entropy at zero Kelvin?

6 Laws in Thermodynamics? 0 th law Definition of the temperature considering the equilibrium of systems. 1 st law U q w 0 for reversible > 0 for irreversible process 2 nd law It is impossible to build a cyclic machine which converts heat into work with an efficiency of 100 %. S Ssys The total Sentropy surr is { increasing in any irreversible process S } 0. 3rd law????????????????? Considering the example it has to do with the temperature dependence of the entropy.

7 What is the reference point for the entropy? Page 95 in your textbook.

8 Bad handwriting ok Reference point for entropy is this: Perfect crystal: -- no impurities, -- only one possible crystal structure, -- i.e., most simple case Only one possible state at zero Kelvin. What is the entropy for this kind of system? Only one state w = 1 S = k B ln(w) = k B ln(1) = 0 Plank s version: S of a pure, perfect crystal is zero at zero Kelvin.

9 Laws in Thermodynamics? 0 th law Definition of the temperature considering the equilibrium of systems. 1 st law U q w 0 for reversible > 0 for irreversible process 2 nd law It is impossible to build a cyclic machine which converts heat into work with an efficiency of 100 %. S Ssys The total Sentropy surr is { increasing in any irreversible process S } 0. 3rd law lim ST ( ) 0 T 0K

10 lim ( S) 0 T 0K Unattainability of absolute zero temperature. Plank s version: S of a pure, perfect crystal is zero at zero Kelvin. Nernst's heat postulate: It is impossible for any process, no matter how idealized, to reduce the entropy of a system to its absolute zero value in a finite number of operations. proof : can absolute zero be reached?

11 Walther Hermann Nernst German physicist Worked on chemical affinity third law of thermodynamics Physical chemistry Electrochemistry Thermodynamics Solid state physics Nernst equation 1920 Nobel Prize in chemistry Watch this one at home

12 Is there a 4 th law (5 th law)? The temperature in Fargo, North Dakota, will always be too low or too high. Nernst commented that it took 3 people to formulate the 1 st law, 2 for the 2 nd law, but that he had to do the 3 rd one all by himself. Therefore, by extrapolation, there could never be a 4 th law. Lars Onsager's 4 th law. /famous-scientists/physicists/lars-onsagerinfo.htm

13 Example What happens when a phase change occurs? Motivation: Determine entropy, enthalpy, and internal energy changes for a process that includes a phase change. For example liquid water is heated to its boiling temperature. Parameters: T 1 = 373 K (25C) and T 2 = 100C H vap = 41 kj/mol and c = 75 J/(Kmol) both per mole Similar to example 4.8 in your text book.

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15 Absolute entropies entropy S vaporization S fusion 0 0 freezing point boiling point temperature 3 rd law See Fig. 4.7 in your textbook

16 C p S T > T experimental Measure heat capacity (experimentally) qrev cdt p solid liquid gas T T < T experimental Extrapolate to 0 Kelvin H V S V T T f Tb T C H C H C f V S S0 dt dt dt T T 0 f T T T b T T f solid liquid gas b B solid liquid Liquid gas H f : fusion (solid liquid) T f : freezing

17 Bad handwriting so I typed also this. Some properties of the entropy (S) S is a state function S is extensive vap S >0 and fus S > 0 (S does not depend on the path chosen: initial final state) (S tot = S 1 + S 2 + ) S gas > S liquid > S solid (per mole) S ~ size of a molecule (Why?: # of degrees of freedom increase with number of atoms) P 0 then S (Why? S = nrln(v 2 /V 1 ) V 1 : reference volume V 2 for P 0 ) -3.5 S 0 (Why: T = 0K w = 1 (perfect crystal) S = k B ln(w) = 0, note that w >= and S increases with temperature ) T f Tb T H f S increases with temperature H V 0 0 f T f b TB S diamond < S graphite (Why?: High degree of order reduced S) C C C S S dt dt dt T T T T T ln(x) x

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19 Some properties of the entropy (S) S is a state function S is extensive vap S >0 and fus S > 0 (S does not depend on the path chosen: initial final state) (S tot = S 1 + S 2 + ) S gas > S liquid > S solid (per mole) S ~ size of a molecule (Why?: # of degrees of freedom increase with number of atoms) P 0 then S (Why? S = nrln(v 2 /V 1 ) V 1 : reference volume V 2 for P 0 ) S 0 (Why: T = 0K w = 1 (perfect crystal) S = k B ln(w) = 0, w > 1) S increases with temperature T f Tb T H f H V C C C S S0 dt dt dt T T T T T S diamond < S graphite (Why?: High degree of order reduced S) 0 f T f b TB

20 Some properties of the entropy (S) S diamond < S graphite (Why?: High degree of order reduced S) More complex systems have larger S (S ozone > S O2 ) S increases with the volume of a gas (Why?: S ~ ln(v 2 /V 1 ) Large disorder large entropy S(N 0.1 atm) > S(N 1 atm) Entropy change of a reaction aa + bb + cc + dd + r S = S(products) - ns(reactants)

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