Chem rd law of thermodynamics., 2018 Uwe Burghaus, Fargo, ND, USA

Transcription

1 Chem rd law of thermodynamics, 2018 Uwe Burghaus, Fargo, ND, USA

2 Class 1-4 Class 5-8 Class 9 Class Chapter 5 Chapter 6 Chapter 7 Chapter 8, 12 tentative schedule 1 st law 2 nd law 3 rd law Gibbs intro, 1 st law, enthalpy, heat capacity, gas expansions energy conservation Entropy Equilibrium stat thermo version of the laws

3 6 in 1: 2 nd law I PChem he entropy of an isolated system which is not in equilibrium will tend to increase over time, approaching a maximum value at equilibrium. No heat engine can run with 100% efficiency dq rev 0 S S S sys S surr qrev Not really fundamental / official versions: q Heat cannot flow from a colder body to a hotter. irr S 1 rev cold hot Clausius inequality 0 for reversible { }process >0 for irreversible A 2 nd class perpetual mobile does not exists.

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5 . Engel / P. Reid, Physical chemistry, chapter 5.8 to ch (1 st Ed. / 2 nd Ed.) Or R. Chang, Physical Chemistry for the Biosciences chapter 4.5 (pages 92-98) I PChem Molecular thermodynamics D.A. McQuarrie, J.D. Simon Chapter 7 P. Atkins / J. Paula, Physical chemistry chapter 4.4 (7 th Ed.) Or I.N. Levine, Physical Chemistry, chapter 5.7 (5 th Ed.) Not everyone likes Wikipedia, but this site is good. Bottom line, use more than one source to study. 2018, U. Burghaus

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7 Plan today 3 rd law of thermodynamics I PChem Example (as a motivation) 3 rd law of thermodynamics More examples Properties of the entropy

8 3 rd law: What about the temperature dependence of the entropy? I PChem

9 Example I PChem What about the temperature dependence of the entropy? Motivation: Calculate the value of S in heating n moles of a monoatomic ideal gas from i to f at constant pressure?

10 S 2 S c p d By the way, where does this equation come from? I PChem ds = dq S 2 = S dq dh=q (for P=const.) dh=cd S 2 = S 1 + C d = S 1+ Cd(ln) = S 1 +Cln( 2 1 )

11 Laws in hermodynamics I PChem 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 he 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. 2 c S2 S1 p d S 1 =??? What is the reference 1 point for the entropy?

12 What is the reference point for the entropy? 2 c S2 S1 p d S 1 ( 1 ) = S 1 ( 1 zero Kelvin) =??? 1 I PChem 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 S 1 ( 1 ) = S 1 (zero Kelvin) = 0 Plank s version: S of a pure, perfect crystal is zero at zero Kelvin.

13 Max Karl Ernst Ludwig Planck ( ) German theoretical physicist Nobel Prize in Physics in 1918 re-naming in 1948 of the Kaiser Wilhelm Society as the Max Planck Society

14 Laws in hermodynamics? I PChem 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 he total Sentropy surr is increasing { in any irreversible process S } 0. 3rd law lim S ( ) 0 K 0 Simple enough Or,??

15 I PChem lim S ( ) 0 K 0 Plank s version: S of a pure, perfect crystal is zero at zero Kelvin. Nernst's heat postulate: Nernst s original postulate from 1907 states that the entropy change in any isothermal process approaches zero as the temperature approaches zero.

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

17 Is there a 4 th law (5 th law)? I PChem he 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. herefore, by extrapolation, there could never be a 4 th law. Lars Onsager's 4 th law. mous-scientists/physicists/lars-onsagerinfo.htm

18 Example I PChem 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. What is the total S? Parameters: 1 = 373 K (25C) and 2 = 100C H vap = 41 kj/mol and c = 75 J/(Kmol) both per mole If you need a numerical example. See whiteboard See e.g. Molecular thermodynamics D.A. McQuarrie, J.D. Simon Chapter 7-3

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20 freezing point boiling point entropy Absolute entropies I PChem S vaporization S fusion 0 0 temperature 3 rd law

21 I PChem Seen in a student exam. Entropy is an abstract concept one cannot actually determine entropies.

22 I PChem C p S > experimental Measure heat capacity (experimentally) qrev p c d solid liquid gas < experimental Extrapolate to 0 Kelvin f b C H C H C f V S S0 d d d 0 f b f S V H solid liquid gas b V B 3 rd law solid liquid Liquid gas H f : fusion (solid liquid) H v : vaporization (liquid gas) f : freezing B : boiling

23 I PChem Seen in a student exam. Entropy is an abstract concept one cannot actually determine entropies. Yes No S C d S and C are closely related

24 Extrapolate to 0 Kelvin I PChem See e.g. Molecular thermodynamics D.A. McQuarrie, J.D. Simon Chapter 7-4

25 Museum Boerhaave C v /3Nk Dulong-Petit law Dulong & Petit: specific heat capacity is independent of the substance when a factor correcting for differences in the atomic weight is used k/e 3R 3R c p M C p is 3R per mole Pierre Louis Dulong ( ) French Physicist Public Domain Photo Alexis hérèse Petit ( ) French Physicist No Photo available c p e Einstein model Albert Einstein ( ) German-American physicist Low temperature limit c p Debye model 3 Peter Debye ( ) Dutch-American physicist R: gas constant M: molar mass : temperature

26 relatively low temperatures small vibrations within harmonic approximation independent atoms (no coupling of vibrations) pair wise Lenard-Jones potentials atoms vibrate at equilibrium position one frequency leads to qualitatively, but not quantitatively correct results n Debye used a continuous frequency distribution n Lattice vibrations can be described as a gas of noninteracting phonons. Phonons in a box.

27 I PChem Lattice contributions to C v at low : For < 5 K, contribution of electron gas: C F V ) ( ) ( D Debye V C Einstein V E e C / R C F D V ) ( 2 ) ( 5 12 C 0 for 0

28 ln(x) 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) P0 then S (Why? S = nrln(v 2 /V 1 ) V 1 : reference volume V 2 for P 0 ) S 0 (Why: = 0K w = 1 (perfect crystal) S = k B ln(w) = 0, note that w >= 1 and S increases with temperature ) f b H f S increases with temperature HV 0 0 f f b B S diamond < S graphite (Why?: High degree of order reduced S) C C C S S d d d x

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30 Some properties of the entropy (S) S is a state function (S does not depend on the path chosen: initial final state) S is extensive (S tot = S 1 + S 2 + ) vap S >0 and fus S > 0 I PChem S gas > S liquid > S solid (per mole) S ~ size of a molecule (Why?: # of degrees of freedom increase with number of atoms) P0 then S (Why? S = nrln(v 2 /V 1 ) V 1 : reference volume V 2 for P 0 ) S 0 (Why: = 0K w = 1 (perfect crystal) S = k B ln(w) = 0, w > 1) S increases with temperature f b H f HV C C C S S0 d d d S diamond < S graphite (Why?: High degree of order reduced S) 0 f f b B

31 Some properties of the entropy (S) S diamond < S graphite (Why?: High degree of order reduced S) I PChem 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 = ns(products) - ns(reactants)

32 ake Home Message I PChem Is there a manual to science?

33 Laws in hermodynamics - history I PChem 0 th law Definition of the temperature considering the equilibrium of systems. Arnold Sommerfeld Ralph H. Fowler 1 st law U q w 2 nd law 0 for reversible S Ssys Ssurr { } > 0 for irreversible process 3rd law lim S ( ) 0 K 0 3 rd law, Planck s version: S of a pure, perfect crystal is zero at zero Kelvin.

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35 Supplemental I PChem Nernst s original postulate

36 lim ( S) 0 0K I PChem Plank s version: S of a pure, perfect crystal is zero at zero Kelvin. Nernst's heat postulate: Nernst s original postulate from 1907 states that the entropy change in any isothermal process approaches zero as the temperature approaches zero. Unattainability of absolute zero temperature. Nernst's heat postulate V2: It is impossible for any process, no matter how idealized, to reduce the temperature of a system to its absolute-zero value in a finite number of operations. verification : can absolute zero be reached?

37 Supplemental I PChem Really zero?

38 2 c S2 S1 p d S 1 ( 1 ) =??? 1 I PChem 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 S 1 ( 1 ) = S 1 (zero Kelvin) = 0 Plank s version: S of a pure, perfect crystal is zero at zero Kelvin.

39 Plank s version: S of a pure, perfect crystal is zero at zero Kelvin. Experimental result: Deviations of experimental and calculated heat capacities for 0 K are known. Example wo different orientations S residual =??

40 S residual =?? wo different orientations weight 2 N N: # of molecules : total # of arrangements res N S k ln( ) k ln(2 ) Nk ln(2) nrln(2) n: # of moles EXAMPLE: 1 mol CO S res 1Rln(2) ~ 5 EXAMPLE: for a single molecule J Kmol EXAMPLE: 3 or 5 different orientations S S res res N nk ln(3 ) N nk ln(5 ) res S 1kln(2) ~ 0

41 I PChem Calorimetric f b C H C H C S S0 d d d q rev c d p f V 0 f b f B spectroscopically Canonical ensemble ln( Q) S k ln( Q) k N, V Entropy from partition functions Partition functions are available spectroscopically See e.g. Molecular thermodynamics D.A. McQuarrie, J.D. Simon Chapter 7-6

42 lim ( S) 0 0K I PChem Plank s version: S of a pure, perfect crystal is zero at zero Kelvin. Nernst's heat postulate: Nernst s original postulate from 1907 states that the entropy change in any isothermal process approaches zero as the temperature approaches zero. Unattainability of absolute zero temperature. Nernst's heat postulate V2: It is impossible for any process, no matter how idealized, to reduce the temperature of a system to its absolute-zero value in a finite number of operations. verification : can absolute zero be reached?