The Story of Spontaneity and Energy Dispersal. You never get what you want: 100% return on investment

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Rhoda Whitehead
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1 The Story of Spontaneity and Energy Dispersal You never get what you want: 100% return on investment

2 Spontaneity Spontaneous process are those that occur naturally. Hot body cools A gas expands to fill the available volume A spontaneous direction of change is where the direction of change does not require work to bring it about.

3 Spontaneity The reverse of a spontaneous process is a nonspontaneous process Confining a gas in a smaller volume Cooling an already cool object Nonspontaneous processes require energy in order to realize them.

4 Spontaneity Note: Spontaneity is often interpreted as a natural tendency of a process to take place, but it does not necessarily mean that it can be realized in practice. Some spontaneous processes have rates sooo slow that the tendency is never realized in practice, while some are painfully obvious.

5 Spontaneity The conversion of diamond to graphite is spontaneous, but it is joyfully slow. The expansion of gas into a vacuum is spontaneous and also instantaneous.

6 2 ND LAW OF THERMODYNAMICS

7 Physical Statements of the 2 nd Law of Thermodynamics Kelvin Statements No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work

8 It is impossible for a system to undergo a cyclic process whose sole effects are the flow of an amount of heat from the surroundings to the system and the performance of an equal amount of work on the surroundings. It is impossible for a system to undergo a cyclic process that turns heat completely into work done on the surroundings.

9 Clausius statement It is impossible for a process to occur that has the sole effect of removing a quantity of heat from an object at a lower temperature and transferring this quantity of heat to an object at a higher temperature. Heat cannot flow spontaneously from a cooler to a hotter object if nothing else happens

10 The 2 nd Law of Thermodynamics The 2 nd Law of Thermodynamics recognizes the two classes of processes, the spontaneous and nonspontaneous processes.

11 Implications of the 2 nd Law No heat engine can have an efficiency as great as unity No macroscopic process can decrease the entropy of the universe

12 Are you kidding me? Thermodynamic Cat

13 Hot Reservoir Heat Engine Work I approve! Cold Reservoir Heat

15 What determines the direction of spontaneous change? The total internal energy of a system does NOT determine whether a process is spontaneous or not. Per the First Law, energy is conserved in any process involving an isolated system.

16 What determines the direction of spontaneous change? Instead, it is important to note that the direction of change is related to the distribution of energy. Spontaneous changes are always accompanied by a dispersal of energy.

17 Energy Dispersal Superheroes with energy blasts and similar powers as well as the Super Saiyans are impossible characters. They seem to violate the Second Law of Thermodynamics!

18 Power Genki dama

19 Energy Dispersal A ball on a warm floor can never be observed to spontaneously bounce as a result of the energy from the warm floor

20 Energy Dispersal In order for this to happen, the thermal energy represented by the random motion and vibrations of the floor atoms would have to be spontaneously diverted to accumulate into the ball.

Energy Dispersal It will also require the random thermal motion to be redirected to move in a single direction in order for the ball to jump upwards.

21 Energy Dispersal It will also require the random thermal motion to be redirected to move in a single direction in order for the ball to jump upwards. This redirection or localization of random, disorderly thermal motion into a concerted, ordered motion is so unlikely as to be virtually impossible.

22 Energy Dispersal and Spontaneity INDEED! Spontaneous change can now be interpreted as the direction of change that leads to the dispersal of the total energy of an isolated system!

23 Entropy A state function, denoted by S. While the First Law can be associated with U, the Second Law may be expressed in terms of the S

24 Entropy and the Second Law The Second Law can be expressed in terms of the entropy: The entropy of an isolated system increases over the course of a spontaneous change: ΔS tot > 0 Where S tot is the total entropy of the system and its surroundings.

25 Entropy A simple definition of entropy is that it is a measure of the energy dispersed in a process. For the thermodynamic definition, it is based on the expression:

26 Entropy as a State Function To prove entropy is a state function we must show that ds is path independent Sufficient to show that the integral around a cycle is zero or ds dq 0 T Sadi Carnot (1824) devised cycle to represent idealized engine w 3 w 4 Hot Reservoir q c q h Engine T h Cold Reservoir T c -w 1 -w 2 Step 1: Isothermal reversible T h Step 2:Adiabatic expansion T h to T c Step 3:Isothermal reversible T c (sign of q negative) Step 4: Adiabatic compression T c to T h

27 Carnot Engine How is that possible?

28 Carnot Cycle

29 Carnot Cycle Step 1: ΔU=0 Step 2: ΔU=w Step 3: ΔU=0 Step 4: ΔU=-w

30 Efficiency of Heat Engines Efficiency is the ratio of the work done by an engine in comparison to the energy invested in the form of heat for all reversible engines e or η = w q h = q h qc q h = T h Tc T h = 1 T c T h All reversible engines have the same efficiency irrespective of their construction.

31 Refrigeration/ Heat pump

32 Refrigeration

33 Coefficient of performance (COP or β or c) COP = q c w = q c q h q c = T c T h Tc COP describes the q c in this case as the minimum energy to be supplied to a refrigeration-like system in order to generate the required entropy to make the system work.

34 SUPERENGINE?

35 Carnot Cycle - Thermodynamic Temperature Scale The efficiency of a heat engine is the ratio of the work performed to the heat of the hot reservoir e= w /q h The greater the work the greater the efficiency Work is the difference between the heat supplied to the engine and the heat returned to the cold reservoir q h Hot Reservoir Cold Reservoir Heat Engine Work Heat -q c w w = q h -(-q c ) = q h + q c Therefore, e = w /q h = ( q h + q c )/q h = 1 + (q c /q h )

Carnot Cycle - Thermodynamic Temperature Scale William Thomson (Lord Kelvin) defined a substance-independent temperature scale based on the heat transferred between two Carnot cycles sharing an

36 Carnot Cycle - Thermodynamic Temperature Scale William Thomson (Lord Kelvin) defined a substance-independent temperature scale based on the heat transferred between two Carnot cycles sharing an isotherm He defined a temperature scale such that q c /- q h = T c /T h e = 1 - (T c /T h ) Zero point on the scale is that temperature where e = 1 Or as T c approaches 0 e approaches 1 Efficiency can be used as a measure of temperature regardless of the working fluid Applies directly to the power required to maintain a low temperature in refrigerators q h Hot Reservoir Cold Reservoir Heat Engine Work Heat -q c Efficiency is maximized w Greater temperature difference between reservoirs The lower Tc, the greater the efficiency

37 Entropy For a measurable change between two states, In order to calculate the difference in entropy between two states, we find a reversible pathway between them and integrate the energy supplied as heat at each stage, divided by the temperature.

38 Example

39 Practice

40 Reversible temperature changes S = T 1 T 2 C p dt T The specific heat of water is J/g K.

41 Change in entropy of the surroundings: ΔS sur If we consider a transfer of heat dq sur to the surroundings, which can be assumed to be a reservoir of constant volume. The energy transferred can be identified with the change in internal energy du sur is independent of how change brought about (U is state function Can assume process is reversible, du sur = du sur,rev Since du sur = dq sur and du sur = du sur,rev, dq sur must equal dq sur,rev That is, regardless of how the change is brought about in the system, reversibly or irreversibly, we can calculate the change of entropy of the surroundings by dividing the heat transferred by the temperature at which the transfer takes place.

42 Change in entropy of the surroundings: ΔS sur For adiabatic change, q sur = 0, so DS sur = 0

43 Entropy changes: Expansion Entropy changes in a system are independent of the path taken by the process ΔS = nr ln V 2 V 1 Total change in entropy however depends on the path: Reversible process: ΔS tot = 0 Irreversible process: ΔS tot > 0

44 Irreversible processes

45 Entropy changes: Phase Transitions Δ trans S = Δ transh T trans Trouton s rule: An empirical observation about a wide range of liquids providing approximately the same standard entropy of vaporization, around 85/88/90 J/mol K. Δ vap S = 10.5 R

48 Entropy of gas mixing

49 Exercise

50 w = 0 ΔU = q = 312 J ΔS = 1.00 J/K

51 Third Law of Thermodynamics At T = 0, all energy of thermal motion has been quenched, and in a perfect crystal all the atoms or ions are in a regular, uniform array. The localization of matter and the absence of thermal motion suggest that such materials also have zero entropy. This conclusion is consistent with the molecular interpretation of entropy, because S = 0 if there is only one way of arranging the molecules and only one microstate is accessible (the ground state).

52 Third Law of Thermodynamics The entropy of all perfect crystalline substances is zero at T = 0.

53 Nernst heat theorem The entropy change accompanying any physical or chemical transformation approaches zero as the temperature approaches zero: ΔS 0 as T 0 provided all the substances involved are perfectly crystalline.

54 Lewis statement If the entropy of each element in some crystalline state be taken as zero at the absolute zero of temperature, every substance has a finite positive entropy but at the absolute zero of temperature the entropy may become zero, and does so become in the case of perfect crystalline substances.

Unattainable absolute zero Giauque s adiabatic demagnetization has led to temperatures of less than 0.000001 K (1 μk) in the nuclear spins of a magnetizable system.

55 Unattainable absolute zero Giauque s adiabatic demagnetization has led to temperatures of less than K (1 μk) in the nuclear spins of a magnetizable system. William Francis Giauque, , was an American chemist who discovered that ordinary oxygen consists of three isotopes. He received the 1949 Nobel Prize in chemistry for pioneering the process of adiabatic demagnetization to attain low temperatures. Opposing laser beams that effectively stop the translational motion of atoms have acheved an effective temperature of K (3 nk) (Saubamea and friends)

56 Third-Law entropies or Absolute Entropies These are entropies reported on the basis that S(0) = 0.

58 Measurement of Entropy (or molar entropy) for heating

59 Measurement of Entropy (or molar entropy) The terms in the previous equation can be calculated or determined experimentally The difficult part is assessing heat capacities near T = 0. Such heat capacities can be evaluated via the Debye extrapolation

61 Measurement of Entropy (or molar entropy) In the Debye extrapolation, the expression below is assumed to be valid down to T=0. C p, m = at 3 C v, m = at 3 + bt

62 Exercises

63 Statistical Entropy: A molecular look Boltzmann formula: S = k ln W S st = k B ln /W thermodynamic probability Reflects the number of microstates, or the ways in which the molecules of the system can be arranged.

64 Entropy is a reflection of the microstates, the ways in which the molecules of a system can be arranged while keeping the total energy constant. Statistical entropy is a measure of the lack of information about the mechanical state of a system. Example: statistical entropy of a deck of cards

65 General equations for entropy during a heating process S as a function of T and V, at constant P ΔS = ncv ln T f T i + nr ln V f V i S as a function of T and P, at constant V ΔS = ncp ln T f T i nr ln P f P i

66 HELMHOLTZ AND GIBBS ENERGIES

67 Clausius inequality ds dq T The Clausius inequality implies that ds 0. In an isolated system, the entropy cannot decrease when a spontaneous change takes place.

68 Criteria for spontaneity ds dq T 0 In a system in thermal equilibrium with its surroundings at a temperature T, there is a transfer of energy as heat when a change in the system occurs and the Clausius inequality will read as above:

69 Criteria for spontaneity When energy is transferred as heat at constant volume: ds dq T 0 *dq = du TdS du At either constant U or constant S: Which leads to ds U, V 0 du S, V 0 du TdS 0

70 Criteria for spontaneity When energy is transferred as heat at constant pressure, the work done is only expansion work and we can obtain TdS dh At either constant H or constant S: Which leads to ds H, p 0 dh S, p 0 dh TdS 0

71 Criteria for spontaneity We can introduce new thermodynamic quantities in order to more simply express du TdS 0 and dh TdS 0

72 Helmholtz and Gibbs energy Helmholtz energy, A: A = U - TS Gibbs energy, G: G = H - TS da = du TdS dg = dh TdS da T,V 0 dg T,p 0

73 Helmholtz energy A change in a system at constant temperature and volume is spontaneous if it corresponds to a decrease in the Helmholtz energy. Aside from an indicator of spontaneity, the change in the Helmholtz function is equal to the maximum work accompanying a process.

74 Helmholtz energy

77 , useful

79 Variation of the Gibbs free energy with temperature

80 Variation of the Gibbs free energy with pressure

81 Variation of the Gibbs free energy with pressure

82 Homework 1. When mol C 6 H 12 O 6 (glucose) is oxidized to carbon dioxide and water at 25 C according to the equation C 6 H 12 O 6 (s) + 6 O 2 (g) 6 CO 2 (g) + 6 H 2 O(l), calorimetric measurements give Δ r H θ = kj mol -1 and Δ r S θ = J K -1 mol -1 at 25 C. How much of this energy change can be extracted as (a) heat at constant pressure, (b) work? 2. How much energy is available for sustaining muscular and nervous activity from the combustion of 1.00 mol of glucose molecules under standard conditions at 37 C (blood temperature)? The standard entropy of reaction is J K -1 mol Calculate the standard reaction Gibbs energies of the following reactions given the Gibbs energies of formation of their components a) Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) b) C 12 H 22 O 11 (s) + 12 O 2 (g) 12 CO 2 (s) + 11 H 2 O(l)

86 One for the road Life requires the assembly of a large number of simple molecules into more complex but very ordered macromolecules. Does life violate the Second Law of Thermodynamics? Why or why not?

Chapter 3. The Second Law Fall Semester Physical Chemistry 1 (CHM2201)

Chapter 3. The Second Law Fall Semester Physical Chemistry 1 (CHM2201) Chapter 3. The Second Law 2011 Fall Semester Physical Chemistry 1 (CHM2201) Contents The direction of spontaneous change 3.1 The dispersal of energy 3.2 The entropy 3.3 Entropy changes accompanying specific