8.044 Lecture Notes Chapter 5: Thermodynamcs, Part 2

Transcription

1 8.044 Lecture Notes Chapter 5: hermodynamcs, Part 2 Lecturer: McGreevy 5.1 Entropy is a state function Efficiency of heat engines Refrigerators and heat pumps Legendre transforms, thermodynamic potentials, and Maxwell relations Reading: Adkins, Chapters 4, , 7 (Chapters 8 and 9 are recommended but not required they give examples. Baierlein, Chapters and 10 (But: we ll come back to Chapter 10 when we discuss chemical potential in Chapter 8. Ignore chemical potential for now. 5-1

2 5.1 Entropy is a state function his is a fact we know from the microscopic discussion in Chapter 4. Here we will illustrate its thermodynamic consequences. Example: hree different expansion processes for a hydrostatic system 1 Free 2 Quasistatic Isothermal 3 Quasistatic Adiabatic In each case, expand from e.g. i to f 2 i. What is in each case? 1 Free expansion: d > dq/ since it is not quasistatic. In particular, > 0. For an ideal gas, k B N ( ( ln + 3 ( E N 2 ln + const N Recall: E does not change. Neither does N or. k B N ( ( f ln ln N k B N ln 2 2 (Quasistatic Isothermal expansion: ( i k B N N for free expansion ( ln f i }{{} 2 Isothermal means the temperature doesn t change. ince the temperature didn t change in protocol (1 above, this is another way to get to the same final state. But this time we re getting there quasistatically, so d dq/. 5-2

3 ince is constant and E E(, ame answer: 0 de dq + dw dq dw +P d d P d Nk B d f f d d Nk B Nk B ln k B N ln 2 i i for isothermal expansion his had to be true in order for to be a state function it only depends on the endpoints, not on how we got there. 3 Quasistatic adiabatic expansion: ( f i quasistatic: d dq adiabatic: dq 0 d 0. 0 for quasistatic adiabatic expansion Different final state means we can have a different. k B N ( ( ln + 3 ( E N 2 ln + const N 0 ln + 3 ln E const 2 E 3/2 const o if increases by a factor of 2, E decreases by a factor of 2 2/3. (ince E, so does. Aside: E 3 2 Nk B 3/2 const P Nk B const (P 3/2 P 3/2 5/2 P 5/3 const which is what we found before for the shape of an adiabat. 5-3

4 C P C for a general hydrostatic system. Recall from Chapter 3 that: C P C α }{{} ( P ( U }{{} 0 for ideal gas +P ( he reason α 1 has a name is that it is easy to measure. On the other hand, the P term ( U which was zero for ideal gas is not so easy to measure. (I guess we could imagine doing adiabatic free expansion lots of times into different size containers. We can use the fact that entropy is a state function to find an alternate useful expression for it. he fact that is a state function means we can think of it as a function of any complete set of independent thermodynamic variables. It started its life in our discussion as a function of (E,, but (, would work just as well. d ( d + ( On the other hand, the 1st Law for quasistatic processes also gives an expression for d: Use calculus on U: du ( U d + ( U d 1 ( U du }{{} d + P }{{ d } dq dw d 1 du + P d d to get d + 1 (( U d (1 + P d (2 Now we can equate the coefficients in the two expressions, (1 and (2, for d. ( 1 ( U ( (( 1 U + P ( ( he fact that is a state variable means the mixed partials must be equal. his gives 1 2 U 1 2 U 1 (( U + P + 1 ( P 2 (( ( U P + P 5-4

5 We ve rewritten ( U in terms of things that are easier to measure P and its derivatives. o see that this is something nicely measurable we do one more manipulation: We ve shown that ( ( P C P C α On the other hand, the reciprocity relation implies ( P ( P 1 ( P 1 ( κ 1 α α κ. ince we can write it in terms of things that have historical names, it must be measurable. C P C α2 κ his expression (or its derivation involves basically everything we ve learned so far. A few things we can deduce from it: We know that C P C 0. o consistency with this expression means κ > 0. (If this is not true, the system will explode or collapse. C P C 0 if α 0. hat happens when ( 0. P his happens for water at 4 C. Aside: we ve shown along the way here that ( ( P. his is an example of a Maxwell Relation. (A different kind of Maxwell relations are shown at right. here are actually more of them, and we ll derive them systematically soon in

6 5.2 Efficiency of heat engines Recall: 1. A heat engine takes a system (some assembly of some substance around a closed cycle over and over. It returns to the initial state after one cycle. 2. Heat is transferred into and out of the substance. ome part of the cycle involves a cold sink. Conventions: Q H is the heat taken in from the hot reservoir. Q C is the heat spat out to the cold sink. W OU is the work done by the system. 3. Work is performed. 4. Efficiency is a subjective thing, in that it is defined as the ratio of (what we want to (what we use up it depends on our goals. For heat engines, efficiency is defined by η work out heat in from hot source W out Q H 1 Q C Q H hot source Q engine Q H C cold sink W OU If all of Q H enters from a reservoir at a single H, and if all of Q C exits to a reservoir at a single C, and if the engine operates quasistatically then this is called a Carnot engine. 1 hen the following is true: Carnot: H Q H H C Q C C where H is the change in entropy of the system during the heating stage of the cycle and H is the change in entropy of the system during the cooling stage of the cycle. When we heat the system, we increase its entropy; when we extract heat, we decrease its entropy. 1 Below we will consider more complicated things, e.g. more isothermal legs and more adiabatic legs. hey will compare unfavorably with Carnot. 5-6

7 is a state function, and we are going in a closed cycle, so during one whole cycle: 0 O. Finally, if we assume quasistatic operation, then the adiabatic legs involve no change in entropy: 0 dq d for adiabatic steps. o: 0 O Q C C Q H H Q C C Q H H η 1 C H. here can be many implementations of such a Carnot engine: e.g. one where the substance in question was an ideal gas, or one involving a paramagnet. hey all look the same on a - diagram: H Q in 1 H C 4 Q C out 3 he P- diagram and the H-M diagram would look quite different from each other: 5-7

8 If everything is done quasistatically d }{{} area enclosed in diagram dq Q H Q C 1st Law W OU On the other hand, is a state function means that 0 d quasistatic any closed cycle If the cycle is not traversed quasistatically, dq d > dq dq < 0 dq 0 Clausius heorem and the inequality is saturated for quasistatic cyclic processes ( Carnot engines. 5-8

9 Carnot is the best wo arguments that Carnot efficiency is the maximum possible: Consider a cyclic but not quasistatic engine which takes heat from some reservoir at H and dumping into a cold sink at C, just like the engine we just studied. Q H hot source, H Q H H Q Q H H other engine W OU Carnot, running backwards Q C Q C Q Q C C cold sink, C C Clausius tatement of 2nd Law: Q H Q H 0 Q H Q H. (imilarly for Q C Q C. W OU Q H W OU Q H η other engine η carnot engine. 5-9

10 Carnot is the best, part 2 Q: Can we do better if we construct some complicated protocol involving reservoirs at different temperatures? Consider the cycle of an arbitrary, quasistatic engine in the plane: Q IN η upper path from 1 to 2 Q OU d + lower path from 2 to 1 hese facts combine to imply that W Q IN Q IN Q OU Q IN 1 Q OU Q IN. 2 d max d max ( lower path from 1 to 2 η 1 Q OU Q IN 1 min ( 2 1 max ( min max η Carnot 2 d min d min ( 2 1. It is less than the efficiency of a Carnot engine with just two reservoirs, one at min and one at max. A: No. 1 [End of Lecture 13.] 5-10

11 5.3 Refrigerators and heat pumps Refrigerator: run the cycle backwards, extract heat at the cold end and dump heat into the hot reservoir. Accomplishing this requires that we do work on the system, Q H W + Q C. Assume that this is a Carnot refrigerator, i.e. everything is reversible, and there s just two temperatures involved H, C. We know: W Q H 1 C H, Q C Q H C H but this isn t what we care about in judging whether this is a good refrigerator. H Q Q H refrigerator C C W η refrig what we want what we use up heat extracted from cold end work done on system Q C W Q C Q H Q C C H C For H C 1+ a little, η refrig 1 easy to cool without doing a lot of work, low power. For C 0, η refrig 0. It becomes increasingly difficult to cool something as 0. 0 is the point at which no further heat can be extracted. Reaching 0 requires an infinite amount of work. Heat Pump ame setup, but with a different goal. uppose we want to heat the hot end, not cool the cold end. η heat pump Q H W Q H if reversible Q H Q C H H C Now we have completed the rigorous version of Chapter 1. Go back and read Chapter 1 again. One more item in thermodynamics, however: 5-11

12 5.4 Legendre transforms, thermodynamic potentials, and Maxwell relations A while back, we saw that we could start with 2 E E(, in terms of which the 1st Law for quasistatic processes is: de d P d and from this construct a new state variable: H E + P enthalpy dh de + P d + dp dh d + dp (3 Note then that the enthalpy H is most naturally H(, P. It is most useful in analyzing processes at constant pressure, in which case (3 reduces to dh dq (this is true even if the process is not quasistatic. E and H are two examples of thermodynamic potentials. he step of going from E(, to H(, P is an example of a Legendre ransform. A place where you might have seen this operation is going between Lagrangian and Hamiltonian descriptions in classical mechanics. As there, the same physics can be described using E or H. ometimes one is more convenient than the other. For example: C ( E C P complicated expression with derivs of E ( H C complicated expression with derivs of H C P P 2 We wrote it as U U(, ; if you like, this is the relation (U,, rearranged by solving for the internal energy. 5-12

13 A mathematical interlude on Legendre ransform Consider a function y(x whose second derivative y (x is nowhere zero. (o it is everywhere concave, or everywhere convex. [ hink of E E( at fixed : E 3 2 Nk B, 3/2 const E(, 2/3. ] Geometric interpretation: Consider the point (x, y (x, y(x along the curve y(x; draw the tangent line through this point. It has slope y y (x. Its y-axisintercept is at I y xy. We can represent the information about the same curve as either y(x or as I(y. Here s the protocol for doing the latter: Pick a y. Draw a line with that slope. lide it up and down until it has intercept I(y. Repeat for however many values of y you want. his collection of straight lines (determined by I(y is the envelope of tangents of the curve of interest y(x. y(x I(y Contain the same information, determine the same curve. 5-13

14 In the definition of I y xy, you recognize the formula for Legendre transformation. Rewrite in terms of thermo letters: y(x E }{{} think of this as y (, }{{} think of this as x and is just a spectator. ( E y P. }{{} I ( y y xy }{{} E + P H(, P. H P o E(, and H(, P describe the same physics. 3 Now we can Legendre transform back: In the plot I have shown the curve I(y. 4 Also indicated are the point (y, I and the vertical-axis-intercept of the tangent through that point. Its height is I y I y Ĩ. 3 As long as E is a convex (or concave function of. his is in fact a requirement for stability. 4 I m using the I(y for the function I plotted above, which was y(x (x his means y (x 2(x 2 ; I solved this for x(y (4 + y /2 and plugged that into I(y y(x xy 2 2y (y 2 /

15 We can represent the data in I(y just as well by I Ĩ(. But y I y x Ĩ I + xy y hat is, Ĩ y(x. o the Legendre transform squares to one, i.e. undoes itself. Returning to physics variables again, start with H(, P and dh d + dp. Construct E H P. de d P d o this is E(,. o to go back and forth between E and H we Legendre transform, either way. Maxwell Relations ( E de d P d dh d + dp ( ( E H P P ( H P E is a state variable. H is a state variable. equate mixed partials of E: equate mixed partials of H: ( ( ( ( P P o: Maxwell Relations follow by equating mixed second derivatives of thermodynamic potentials. he second one is the one we encountered previously. P 5-15

16 More thermodynamic potentials more Maxwell relations o far has just gone along for the ride in our Legendre transforms: we haven t touched the d bit. We can Legendre transform that bit, too: (Mnemonic: F is for free. Define F E (Helmholtz Free Energy o F F (,. df de d d df d P d [Preview: in the next Chapter, we will formulate statistical mechanics for a system which is not isolated, and in particular is held at constant temperature by a heat reservoir. F will play a key role.] From this expression, we deduce: ( F ( F P Equating crossed derivatives gives our 3rd Maxwell relation: ( ( P 5-16

17 he 4th Maxwell relation, by the Method of the Missing Box E E(,, ( de d ( P d, P H E + P F E H H(, P, ( dh ( d + dp P P F F (,, ( df d ( P d, P G H G F + P G G(, P, ( dg ( d + dp, P G is for Gibbs Free Energy. he 4th and last thermodynamic potential (for a hydrostatic system: E, F, G, H. You should be able to quickly recreate this derivation of the four Maxwell relations. DO NO MEMORIZE HI UFF. Note that you can do this operation for any pair of conjugate variables. In systems with more thermodynamic variables, there are more thermodynamic potentials. For example in a system with a magnetization, E E(,, M so de d P d + HdM and we can construct a new thermodynamic potential whose name is not standard: for which I I(,, H E HM di d P d MdH. 5-17

18 Conditions for equilibrium (he point of F and G, and why they are called free energies. 1. Consider a system in contact with a bath at temperature. uppose the system is equilibrating at constant. As it equilibrates, the 2nd Law tells us: d dq. bath ince is constant, d 0 dw 0 so the 1st Law is: de dq: df 0 d de bath d (E bath 0 at constant, bath o: at constant,, the system evolves so as to minimize F. 2. uppose instead that work can be done, d 0, as the system equilibrates, but P is held constant. d dq de + P d bath bath d ( E bath + }{{} P assumed constant dg 0 at constant P, bath o: at constant P,, the system evolves so as to minimize G. Note also that at fixed P, dw is the work done by the system. d (E P d df + dw df dw During relaxation to equilibrium at constant and P, the work done by the system is less than or equal to the (magnitude of the change in F, as F decreases. his is why F is called the Free Energy or available energy. his is all the thermodynamics we re going to need, except for Chapter 8: Chemical Potential. Next: the version of tat Mech that we actually use. 5-18