1. CARNOT CYCLE Consider working substance to be an Ideal Gas Note p – v axes i) substance starts at with temperature T2, adiabatic compression to B. Since Work done on substance increases internal energy. ii) substance expands isothermally from and absorbs heat. Since=constant, iii) substance undergoes further expansion (adiabatic) from Again Work done is 1 P dv =

2. iv) substance returns to original pointby undergoing an isothermal compression from D A, and the heat exhausted is equal to Q 2. Since du=0 for this isothermal process, dq = dw: or Since the working substance returns to its original state, the work done by the substance is equal to the net heat absorbed, Q 1 -Q 2 Work done = Q 1 -Q 2 This can be readily verified by summing the individual work terms. This can be easily verified by adding up the individual terms. Efficiency of the carnot cycle is 2 Q: How can efficiency be increased? (area inside the “rectangle”) Only by transferring heat from a warm body to a cold body can heat be converted to work in a cyclic process.

3. For a CARNOT CYCLE Proof: It can be shown that = constant for adiabatic process Adiabatic legs For A B For C D Isothermal legs For B C For D A Combining 4 equations We already know that 3

4. Therefore So 4 maximum (Carnot) efficiency of a power cycle The maximum efficiency of an energy cycle is given by the Carnot efficiency, which can be expressed as: If we consider the temperature difference between the equator and the poles: T pole (T c ) T equator (T h ) We note that T h varies little throughout the year (estimate as ~301 K), compared with T c Winter hemisphere: estimate T c ~ 263 K Summer hemisphere: estimate T c ~ 283 K --> More of the absorbed energy is converted into “work” in the winter hemisphere

5. The hurricane as an energy cycle Consider a hurricane as a Carnot cycle: http://apollo.lsc.vsc.edu/classes/met130/notes/chapter15/vertical_circ.html T sea surface T outflow 1 4 3 2 Analyze each leg of the cycle: 1: absorption of energy from ocean, ~isothermal expansion 2: convection in the eyewall, ~adiabatic expansion 3: upper level outflow, ~ isothermal compression (reject heat via radiation) 4: sinking branch, ~adiabatic compression If we estimate SST~300 K, and outflow T~200 K, U Hawaii class notes: Latent heat released when water vapor condenses in clouds is the key. Hurricanes are giant engines that convert heat into wind energy. Consider a rain rate of 2 inches per day over an area of 300 mi radius: Over a 7 day lifecycle, the energy released is equal to 50,000 1 MT nuclear explosions, equivalent to the total explosive yield of the nuclear arsenals of the US and USSR at the height of the Cold War! www.soest.hawaii.edu/MET/Faculty/businger/.../18cHurricanes1.pdf

6. Our next state function: ENTROPY (consequence of Second Law) There exists a function called entropy S, of the extensive variables of a system, defined for all equilibrium states, such that the values assumed by the extensive variables are those that maximize S (at equilibrium) From the viewpoint of classical thermodynamics, entropy is defined as where in this case heat is added to a substance undergoing a reversible transformation. Clausius (1822-1888) stated that the cyclic integral of δQ/T is always less than or equal to zero: The cyclic integrals of work and of heat are not zero (engines). Entropy is a STATE VARIABLE. (We can make up an internally reversible path from A to B to compute change in entropy from the integral of δQ/T.) Since dq= Tds, the First Law can be written as, Even though we imagined a reversible process to substitute for dq, this equation applies to both reversible and irreversible transitions because they are state variables

7. Generalized Statement of the Second Law We need to generalize the definition of entropy since real systems are typically spontaneous and irreversible, moving from a state of non-equilibrium to a state of equilibrium. Second law can be formulated as 4 postulates: 1.There exists a STATE VARIABLE for any substance called the ENTROPY. 2.Entropy (S) may change by one of two ways; the substance may contact a thermal reservoir (heat source), or S may change due to “internal” changes in the substance. ds e (externally-forced changes in S) ds i (internally-forced changes in S) Examples: ds e Heat flow into a substance causing molecules to rearrange ds i Mixing a gas, forcing molecules to change position. Total entropy change ds = ds e + ds i 3.Changes in S due to external influences are given by 4.For reversible transformations, For irreversible transformations, So reversible and irreversible processes are distinguished by entropy changes. For reversible processes, there are no entropy changes associated with internal processes, or they have ceased as the system undergoes only reversible changes, say via contact with a heat reservoir. where dq = heat imported or exported, while substance is in contact with a thermal reservoir at temperature

8. Now consider the system + its surroundings ( = “universe”) In a reversible process any heat flow between the system and surroundings must occur with no finite temperature difference between the system and surroundings.  Otherwise the process would be irreversible Thereforefor reversible process. As expected and can be shown via a similar argument, In general Statement of 2 nd Law! UNIVERSE Since Irreversible process

9. Calculations of Entropy Changes For reversible processes, For irreversible processes,is not necessarily equal to However since S is a state function,depends only upon endpoints. Hencecan be found by devising a series of reversible transformations that are equivalent to the irreversible transformations. 1. Reversible adiabatic process Hence, hence 2. Reversible phase change at constant T, p constant 3. Reversible, isothermal process where m=mass, L= latent heat since

10. 4. Reversible change of state for an Ideal Gas From First Law: then Heating termVolume term

11. 5. Constant-pressure heating (irreversible) For constant pressure: Hence for 6. Similarly, for constant-volume heating (irreversible) More entropy calculations at end of notes

12. “Thermodynamic Potentials” The internal energy, when expressed as a function of S and V, is one of the so-called “thermodynamic potentials”. The differential of a thermodynamic potential gives the first and second laws of thermodynamics combined: S and V are the “native variables” for U (although we can express U as a function of any two variables) Remember enthalpy is: So S and P are the “native variables” for H. We define two more thermodynamic potentials that have “native variables”, respectively, of T and V and T and P.  Very useful for equilibrium calcs!

13. Helmholtz & Gibbs Functions Consider the following system/reservoir arrangement, Total work done by the system, whether it be in a reversible or irreversible process is, From principle of entropy, First Law (1) (2) A general expression that entropy must stay constant or increase systemreservoir Q -Q Q is positive if it flows into the system w is positive if done by the system dq=du+dw w=Q-(u 2 -u 1 )

14. For the reservoir,heat flow out of reservoir Therefore we have Note T= constant; reservoir has infinite heat capacity. From (3) we have the following inequality, From First Law or Eq (1) above (substitute for Q) For the irreversible case the work done is less than since some of the heat added to the system can go into changing the ‘internal’ entropy of the system. (3) Reversible process Irreversible process (4)

15. The difference in this energy between 2 equilibrium states at the same temperature is Hence change in F sets an upper limit to the work that can be done in any process between 2 equilibrium states at constant T. F is often referred to as the ‘free energy’ since ∆F represents the maximum amount of work that can be done in a transformation at constant temperature. (5)

16. Returning to the equation This equation is general and can be applied to any system, change of state, phase, or even a chemical reaction. In general, work in processes we are considering consist of pdV work and other possible forms of work, e.g. work required to form a curve interface between 2 phases, frictional dissipation, etc. Let represent pdV type work and represent other forms of work, so we now have If we consider a constant volume process, Hence at constant T and V, Helmholtz energy sets an upper limit to the non- pdV work that can be done in such a transition. (6) (7)

17. For a reversible constant volume process Now consider a situation where no pdV and no work is done, then Hence for a process at constant volume, for which (and T=constant), F can only decrease or remain constant. A transition will not occur if. (8) Wikipedia definition: The Helmholtz free energy is a thermodynamic potential which measures the “useful” work obtainable from a closed thermodynamic system at a constant temperature and volume. For such a system, the negative of the difference in the Helmholtz energy is equal to the maximum amount of work extractable from a thermodynamic process in which temperature and volume are held constant. Under these conditions (constant T and v), it is minimized at equilibrium. Constant v constraint is not always helpful for us …

18. Next, consider a process that occurs at constant pressure. In this case the pdV work is Then returning to the equation: We have: (8) Now define the Gibbs Free Energy as, Then for a transition between 2 states at the same T,p: and substituting into (8) we have ∆G represents the total non- pdV work for a constant T,P process. This is of fundamental importance for cloud physics. (dG = -SdT + VdP)

19. During nucleation, some energy is released when molecules move from the vapor to the “liquid” phase is the work (energy required) necessary to form the surface of the small liquid (or ice) embryo from the vapor phase: If So G can only decrease in a transition occurring at constant T, P. If the hypothetical nucleus is too small (known as an unstable nucleus or "embryo"), the energy that would be released by forming its volume is not enough to create its surface quick look ahead to one application we’ll be discussing:

20. Thermodynamic Potential Given Consider a closed system which is one where no mass is allowed to cross the boundaries of the system. Taking total differentials, Using First Law in the form: We can substitute for (1) and (2) the dU term (1) (2)

21. Consider the following series of equations, It is useful to identify the coefficients of the equations on the left hand side with partial derivatives of the variables on the right hand side. For: Therefore we have Since Therefore

22. Since And Since And Since P, V, T and S can be expressed as partial derivatives of U, F, G and H, the latter variables are referred to as Thermodynamic Potentials.

23. Maxwell Relations The “Gibbs relations” are the following equations that we have already seen: We recognize that they are exact differentials and have the form We can deduce the “Maxwell Relations”: Can get unmeasurables in terms of measurables

24. Stable and Unstable Equilibrium Stable Equilibrium: a state in which all irreversible processes have ceased, and the system is in a state of maximum entropy. Reversible transformations are possible (for which ). For state equilibrium, This equilibrium condition is for an isolated system (no heat transfer, no mass transfer) A typical system we will deal with is NOT in thermal isolation, but rather in contact with a heat reservoir (e.g., as approximated by the atmosphere). a)For a transition at constant volume, in thermal contact with a reservoir at temperature T, the stable state is defined by, b) For a transition at constant pressure and temperature, (in contact with a thermal reservoir), stable state is, MAX ENTROPY Helmholtz free energy is a minimum Gibbs free energy is a minimum

25. Applying equilibrium criteria We can now add moisture to the air parcel we’ve computed changes for –We could have added water to the “dry air mixture” and computed a new average molecular weight, etc., BUT we know that simple approach, assuming water stays as a vapor during the adiabatic ascent, is only valid over small regions –That is, we’ll need to consider phase changes for the water First, we’ll figure out where vapor, liquid and ice are in equilibrium, and the energy changes associated with the transitions –Can calculate changes such as heat released so we can treat the energy changes in our parcel –Can also figure out if the change is favored (e.g., if the transition is allowed to occur thermodynamically, given specified conditions)

26. LATENT HEATS: During a phase change, the pressure is constant and equal to the saturation pressure (which is a function of T only). Adding heat to a substance at constant pressure and resulting in a change in physical state, i.e. change in phase, is described by, where L is the latent heat (J kg -1 ) for the particular transition and m is mass transferred between phases. The heat absorbed goes into changing the molecular configuration/structure. For water substance: Heat of fusion Heat of condensation Heat of sublimation SOLIDVAPORLIQUID LfLf LcLc LSLS 26 L is an energy change to go from one phase to another Starting with First Law in the form dq = dh when dp = 0

27. Phase transitions: what are the equilibrium conditions for a mixed-phase system, e.g. a mixture of water and water vapor? Consider a system consisting of liquid water and H 2 O (v) undergoing a change in state, Thermodynamics of Moist Air vapor liquid vapor liquid ″ denotes liquid phase ″′ denotes vapor phase

28. Define Hence Since both states are stable equilibrium states, G 1 =G 2, and therefore But since the system is closed Therefore

29. At equilibrium, the specific Gibbs free energy is the same for both phases At the triple point, Lets look at g between 2 phases in more detail: From earlier discussions, Where s′″ is the specific entropy of the vapor phase. Hence the slope of vs. T gives the specific entropy. Likewise for the liquid phase Note the slopes are <0, since larger values of g correspond to lower T’s. The difference between entropies is Since liquid a d f vapor g T c b,e vapor liquid

30. Clausius-Clapeyron Equation This equationis very important since it describes the variation in pressure with temperature, for a system consisting of two phases in equilibrium. For the case of a liquid/vapor or vapor/solid system, this “pressure” is the saturation vapor pressure. For a system in equilibrium, Suppose the system undergoes a small change to a new pressure and temperature, p+dp, T+dT such that the new state is also a stable state. Each phase changes by amounts. Then for each phase: Since the new state is also an equilibrium state,

31. Or, Since For vapor-liquid system For vapor-ice system Geometrical Interpretation TT PP VAPOR SOLID LIQUID  Triple point CLAUSIUS-CLAPEYRON EQUATION s, v are specific quantities! For vapor-liquid system Critical point S-L L-V S-V gives the slope of the equilibrium line between the 2 phases involved. (on P-V-T diagram.) =latent heat for vapor-liquid interface

32. Looking again at the Clausius-Clapeyron Equation we have three interfaces, Sincealways, the slope of the equilibrium line between the 2 phases involved is determined by the difference in specific volumes. 1. vapor  liquid solid  vapor liquid  solid vapor  liquid vapor  solid ] ALWAYS For all materials!

33. 2. For substance that expands upon freezing (like water). Hence substance contracts upon freezing. MOST SUBSTANCES! TT PP V S L Triple

34. Integrate the Clausius-Clapeyron Eq n/ to get e s as a function of T. For a vapor/liquid system, Assume Therefore GAS LAW after integration

35. More entropy calculations 35

36. 7. Irreversible change of state of an Ideal Gas Let n moles of an ideal gas of irreversibly change state to This irreversible process can be broken into 2 reversible processes: 1.) A slow expansion to at constant -put gas in a frictionless piston-cylinder, place in constant temperature bath; slowly move piston out 2.) Hold gas at, heat gas to temperature and pressure Hence Constant-volume heatingIsothermal expansion Rapid expansion

37. 8. Irreversible phase change Let 10g of supercooled water at -10°C change to ice at -10°C. Consider constant pressure. This process is IRREVERSIBLE e.g. if the ice is slowly heated, it will not become water at -10°C. Water at -10°C Water at 0°C 1 Ice at 0°C Ice at -10°C 3 2 IRREVERSIBLE Find equivalent reversible path and compute entropy change

38. 1.) Isobaric heating of water 2.) Phase change 3.) Isobaric cooling of ice

39. More entropy calculations Calculate the change in entropy when 5g of water at 0°C are raised to 100°C and converted into vapor (steam) at that temperature. 1. Raise temperature of H 2 O (l) from 0°C to 100°C Let Specific heat of water 2. Convert to vapor (steam) at 100°C

40. Find for the melting of 5g of ice at 0°C at 1 atm. This process is reversible since both ice and liquid states are possible at 0°C. (Equilibrium states!) For freezing of liquid water, and Why the specific signs for melting vs. freezing?

41. Let n moles of a perfect gas undergo an adiabatic free expansion into a vacuum (this is Joule’s experiment). –Express ∆S in terms of the initial and final temperature and volume. –Calculate ∆S if The initial state isand the final state is Although the process is adiabatic, (dq=0), entropy change is non-zero because this process is irreversible. Entropy of course changes (and increases) since the molecules changed their configuration expanding into the vacuum. Hence With This is the so-called Entropy of Mixing.