Chapter 18Entropy

3600 seconds, or 1000×3600 joules. Thus at a rate of10 cents per kilowatt hour, the thermal energy of a moleof helium gas is worth only .01 cents.

The value .01 cents does not sound like much, but thatwas the value of the energy in only one mole of helium.Most substances have a greater molar heat capacitythan helium due to the fact that energy is stored ininternal motions of the molecule. Water at roomtemperature, for example, has a molar heat capacitysix times greater than that of helium. Thus we wouldexpect that the thermal energy in a mole of watershould be of the order of 6 times greater than that ofhelium, or worth about .06 cents.

A mole of water is only 18 grams. A kilogram of water,1000 grams of it, is 55 moles, thus the thermal energyin a kilogram or liter of water should have a value in theneighborhood of .06 cents/mole×55 moles = 3.3 cents.Now think about the amount of water in a swimmingpool that is 25 meters long, 10 meters wide, and 2meters deep. This is 500 m3 or 500×103 liters, theirbeing 1000 liters/ m3. Thus the commercial value ofthe heat energy in a swimming pool of water at roomtemperature is 5× 105 × 3.3cents or over $16,000.

The point of this discussion is that there is a lot ofthermal energy in the matter around us, energy thatwould have enormous value if we could get at it. Thequestion is why don’t we use this thermal energy ratherthan getting energy by burning oil and polluting theatmosphere in the process?

CHAPTER 18 ENTROPY

The focus of the last chapter was the thermal energy ofthe atoms and molecules around us. While the thermalenergy of an individual molecule is not large, thethermal energy in a reasonable collection of mol-ecules, like a mole, is a noticeable amount. Suppose,for example, you could extract all the thermal energyin a mole of helium atoms at room temperature. Howmuch would that energy be worth at a rate of 10 centsper kilowatt hour?

This is an easy calculation to do. The atoms in heliumgas at room temperature have an average kineticenergy of 3/2 kT per molecule, thus the energy of a moleis NA 3 23 2KT = 3 23 2RT, where NA is Avagadro’snumber and R = NAK. Since ice melts at 273 K andwater boils at 373 K, a reasonable value for roomtemperature is 300 K, about one quarter of the way upfrom freezing to boiling. Thus the total thermal energyin a mole of room temperature helium gas is

thermal energyin a mole ofroom temperaturehelium gas

= 32RT

= 32 × 8

joulesmole K

× 300K

= 3600 joules

This is enough energy to lift 1 kilogram to a height of360 feet!

To calculate the monetary value of this energy, we notethat a kilowatt hour is 1000 watts of electric power for

Movie Undive

18-2 Entropy

INTRODUCTIONA simple lecture demonstration helps provide insightinto why we cannot easily get at, and use, the $16,000of thermal energy in the swimming pool. In thisdemonstration, illustrated in Figure (1), water is pumpedup through a hose and squirted down onto a flat plateplaced in a small bucket as shown. If you use a vibratorpump, then the water comes out as a series of dropletsrather than a continuous stream.

To make the individual water droplets visible, and toslow down the apparent motion, the water is illumi-nated by a strobe. If the time between strobe flashes isjust a bit longer than the time interval which the dropsare ejected, the drops will appear to move very slowly.This allows you to follow what appears to be anindividual drop as it moves down toward the flat plate.

For our discussion , we want to focus on what happensto the drop as it strikes the plate. As seen in the seriesof pictures in Figure (2), when the drop hits it flattensout, creating a wave that spreads out from where thedrop hits. The wave then moves down the plate into thepool of water in the bucket. [In Figure (1) we see severalwaves flowing down the plate. Each was produced bya separate drop.]

Let us look at this process from the point of view of theenergy involved. Before the drop hits, it has kineticenergy due to falling. When it hits, this kinetic energygoes into the kinetic energy of wave motion. Thewaves then flow into the bucket, eventually dissipate,and all the kinetic energy becomes thermal energy ofthe water molecules in the bucket. This causes a slight,almost undetectable, increase in the temperature of thewater in the bucket (if the water drop had the sametemperature as the water in the bucket, and we neglectcooling from evaporation).

We have selected this demonstration for discussion,because with a slight twist of the knob on the strobe, wecan make the process appear to run backwards. If thetime interval between flashes is just a bit shorter thanthe pulse interval of the pump, the drops appear to risefrom the plate and go back into the hose. The situationlooks funny, but it makes a good ending to the demon-stration. Everyone knows that what they see couldn’tpossibly happen—or could it?

Does this reverse flow violate any laws of physics?Once a drop has left the plate it moves like a ball thrownup in the air. From the point of view of the laws ofphysics, nothing is peculiar about the motion of thedrop from the time it leaves the plate until it enters thehose.

vibratorpump

waterdroplets

Figure 1Water droplets are createdby a vibrator pump. If youilluminate the drops with astrobe light, you can makethem appear to fall or rise.

18-3

Where the situation looks funny is in the launching ofthe drop from the plate. But it turns out that none of thelaws of physics we have discussed so far is violatedthere either. Let us look at this launching from the pointof view of the energy involved. Initially the water in thebucket is a bit warm. This excess thermal energybecomes organized into a wave that flows up the plate.As seen in Figure (3) the wave coalesces into a drop thatis launched up into the air. No violation of the law ofconservation of energy is needed to describe thisprocess.

While the launching of the drop in this reversed picturemay not violate the laws of physics we have studied, itstill looks funny, and we do not see such things happenin the real world. There has to be some reason why wedon’t. The answer lies in the fact that in the reversedprocess we have converted thermal energy, the disor-ganized kinetic energy of individual molecules, intothe organized energy of the waves, and finally into themore concentrated kinetic energy of the upward trav-elling drop. We have converted a disorganized form ofenergy into an organized form in a way that nature doesnot seem to allow.

It is not impossible to convert thermal energy intoorganized kinetic energy or what we call useful work.Steam engines do it all the time. In a modern electricpower plant, steam is heated to a high temperature byburning some kind of fuel, and the steam is sent throughturbines to produce electricity. A certain fraction of thethermal energy obtained from burning the fuel ends upas electrical energy produced by the electric generatorsattached to the turbines. This electric energy can thenbe used to do useful work running motors.

The important point is that power stations cannotsimply suck thermal energy out of a reservoir like theocean and turn it into electrical energy. That wouldcorrespond to our water drop being launched by thethermal energy of the water in the bucket.

Figure 3Rising wave launching water drop.

Figure 2Falling water drop creating wave.

18-4 Entropy

Even more discouraging is the fact that power plants donot even use all the energy they get from burning fuel.A typical high efficiency power plant ends up discard-ing, into the atmosphere or the ocean, over 2/3 of theenergy it gets from burning fuel. Less than 1/3 of theenergy from the fuel is converted into useful electricalenergy. Car engines are even worse. Less than1/5 of the energy from the gasoline burned goes intopowering the car; most of the rest comes out the exhaustpipe.

Why do we tolerate these low efficiency power plantsand even lower efficiency car engines? The answer liesin the problem of converting a disorganized form ofenergy into an organized one. Or to state the problemmore generally, of trying to create order from chaos.

The basic idea is that a disorganized situation does notnaturally organize itself—in nature, things go the otherway. For example, if you have a box of gas, and initiallythe atoms are all nicely localized on one side of the box,a short time later they will be flying around throughoutthe whole volume of the box. On their own, there isalmost no chance that they will all move over to that oneside again. If you want them over on one side, you haveto do some work, like pushing on a piston, to get themover there. It takes work to create order from disorder.

At first, it seems that the concepts of order and disorder,and the related problems of converting thermal energyinto useful work should be a difficult subject to dealwith. If you wished to formulate a physical law, howdo you go about even defining the concepts. What, forexample, should you use as an experimental definitionof disorder? It turns out, surprisingly, that there is aprecise definition of a quantity called entropy whichrepresents the amount of disorder contained in a sys-tem. Even more surprising, the concept of entropy wasdiscovered before the true nature of heat was under-stood.

The basic ideas related to entropy were discovered in1824 by the engineer Sadi Carnot who was trying tofigure out how to improve the efficiency of steamengines. Carnot was aware that heat was wasted in theoperation of a steam engine, and was studying theproblem in an attempt to reduce the waste of heat. In

his studies Carnot found that there was a theoreticallymaximum efficient engine whose efficiency dependedupon the temperature of the boiler relative to thetemperature of the boiler’s surroundings.

To make his analysis, Carnot had to introduce a newassumption not contained in Newton’s law of mechan-ics. Carnot’s assumption is equivalent to the idea thatyou cannot convert thermal energy into useful workin a process involving only one temperature. This iswhy you cannot sell the $16,000 worth of thermalenergy in the swimming pool—you cannot get it out.

This law is known as the Second Law of Thermody-namics. (The first law is the law of conservation ofenergy itself.) The second law can also be expressed interms of entropy which we now know represents thedisorder of a system. The second law states that in anyprocess, the total entropy (disorder) of a system eitherstays the same or increases. Put another way, it statesthat in any process, the total order of a system cannotincrease; it can only stay the same, or the system canbecome more disordered.

To develop his formulas for the maximum efficiency ofengines, Carnot invented the concept known as aCarnot engine, based on what is called the Carnotcycle. The Carnot engine is not a real engine, no one hasever built one. Instead, you should think of it as athought experiment, like the ones we used in Chapter1 to figure out what happened to moving clocks if theprinciple of relativity is correct.

The question we wish to answer is, how efficient canyou make an engine or a power plant, if the second lawof thermodynamics is correct? If you cannot get usefulwork from thermal energy at one temperature, howmuch work can you get if you have more than onetemperature? It turns out that there is a surprisinglysimple answer, but we are going to have to do quite abit of analysis of Carnot’s thought experiment beforewe get the answer. During the discussion of the Carnotengine, one should keep in mind that we are making thiseffort to answer one basic question—what are theconsequences of the second law of thermodynamics—what are the consequences of the idea that order doesnot naturally arise from disorder.

18-5

WORK DONE BYAN EXPANDING GASThe Carnot thought experiment is based on an analysisof several processes involving the ideal piston andcylinder we discussed in the last chapter. We willdiscuss each of these processes separately, and then putthem together to complete the thought experiment.

The ideal piston and cylinder is shown in Figure (4). Agas, at a pressure p, is contained in the cylinder by africtionless piston of cross-sectional area A. (Since noone has yet built a piston that can seal the gas inside thecylinder and still move frictionlessly, we are nowalready into the realm of a thought experiment.) Aforce F is applied to the outside of the piston as shownto keep the piston from moving. The gas, at a pressurep, exerts an outward force

p newtonsmeter2

× A meter2 = pA newtons

on the cylinder, thus F must be given by

F = pA (1)

to keep the cylinder from moving.

If we decrease the force F just a bit to allow the gas inthe cylinder to expand, the expanding gas will do workon the piston. This is because the gas is exerting a forcepA on the cylinder, while the cylinder is moving in thedirection of the force exerted by the gas. If the pistonmoves out a distance ∆x as shown in Figure (5), thework ∆W done by the gas is the force pA it exerts timesthe distance ∆x

∆W = pA ∆x (2)

After this expansion, the volume of the gas has in-creased by an amount ∆V = A∆x. Thus Equation 2can be written in the form ∆W = pA∆x or

∆W = p∆V (3)

Equation 3 is more general than our derivation indi-cates. Any time a gas expands its volume by an amount

∆V, the work done by the gas is p∆V no matter whatthe shape of the container. For example, if you heat thegas in a balloon and the balloon expands a bit, the workdone by the gas is p∆V where ∆∆V is the increase inthe volume of the balloon.

Exercise 1In our introduction to the concept of pressure, wedipped a balloon in liquid nitrogen until the air insidebecame a puddle of liquid air (see Figure 17-19). Whenwe took the balloon out of the liquid nitrogen, the airslowly expanded until the balloon returned to its originalsize. During the expansion, the rubber of the balloonwas relatively loose, which means that the air inside theballoon remained at or very near to atmospheric pres-sure during the entire time the balloon was expanding.

(a) If the final radius of the balloon is 30 cm, how muchwork did the gas inside the balloon do as the balloonexpanded? (You may neglect the volume of the liquidair present when the expansion started.) (Answer:

1.1 × 104 joules.)

(b) Where did the gas inside the balloon get the energyrequired to do this work?

pA

∆x

∆V = A∆x

Figure 5The work done by an expanding gas isequal to the force pA it exerts, timesthe distance ∆∆x the piston moves.

gas at a pressure p

piston of area A

F

Figure 4In the ideal piston and cylinder, the pistonconfines the gas and moves frictionlessly.

18-6 Entropy

SPECIFIC HEATS CV AND Cp

In our earlier discussion of specific heat, we dealtexclusively with the “molar” specific heat at constantvolume CV. We always assumed that we kept the gasat constant volume so that all the energy we addedwould go into the internal energy of the gas. If we hadallowed the gas to expand, then some of the energywould have gone into the work the gas did to expand itsvolume, and we would not have had an accuratemeasure of the amount of energy that went into the gasitself.

Sometimes it is convenient to heat a gas while keepingthe gas pressure, rather than volume, constant. This ismore or less the case when we heat the gas in a balloon.The balloon expands, but the pressure does not changevery much if the expansion is small.

Earlier we defined the molar heat capacity CV as theamount of energy required to heat one mole of asubstance one kelvin, if the volume of the substance iskept constant. Let us now define the molar heatcapacity Cp as the amount of energy required to heatone mole of a substance one kelvin if the pressure iskept constant. For gases Cp is always larger than CV.This is because, when we heat the gas at constantpressure, the energy goes both into heating and expand-ing the gas. When we heat the gas at constant volume,the energy goes only into heating the gas. We can writethis out as an equation as follows

energy requiredto heat 1 moleof a gas 1K atconstant pressure

Cp =increase in thermalenergy of the gaswhen the temperatureincreases 1K

+work doneby theexpandinggas

(4)

Noting that since CV is equal to the increase in thermalenergy of the gas, and that the work done is p∆V, weget

Cp = CV + p∆V (5)

In the special case of an ideal gas, we can use the idealgas law pV = nRT, setting n = 1 for 1 mole

pV = RT (1 mole of gas) (6)

If we let the gas expand a bit at constant pressure, we getdifferentiating Equation 6, keeping p constant**

p∆V = R∆T (if p is constant) (7)

** By differentiating the equation (pV = RT), wemean that we wish to equate the change in (pV) to thechange in (RT). To determine the change in (pV), forexample, we let (p) go to (p + ∆p) and (V) go to(V + ∆V), so that the product (pV) becomes

pV → (p + ∆p)(V +∆V)= pV + (∆p)V + p∆V + (∆p)∆V

If we neglect the second order term (∆p)∆V then

pV → pV + (∆p)V + p∆V

Then if we hold the pressure constant (∆p = 0),we seethat the change in (pV) is simply (p∆V).Since R isconstant, the change in RT is simply R∆T.

18-7

If the temperature increase is ∆T = 1 kelvin, thenEquation 7 becomes

p∆V = R (1 mole, p constant, ∆T = 1K) (8)

Using Equation 8 in Equation 5 we get the simple result

Cp = CV + R (9)

The derivation of Equation 9 illustrates the kind ofsteps we have to carry out to calculate what happens tothe heat we add to substances. For example, in goingfrom Equation 6 to Equation 7, we looked at the changein volume when the temperature but not the pressurewas varied. When we make infinitesimal changes ofsome quantities in an equation while holding the quan-tities constant, the process is called partial differentia-tion. In this text we will not go into a formal discussionof the ideas of partial differentiation. When we encoun-ter the process, the steps should be fairly obvious asthey were in Equation 7.

(The general subject that deals with changes producedby adding or removing heat from substances is calledthermodynamics. The full theory of thermodynamicsrelies heavily on the mathematics of partial deriva-tives. For our discussion of Carnot’s thought experi-ments, we need only a small part of thermodynamicstheory.)

Exercise 2(a) Back in Table 2 on page 31 of Chapter 17, we listedthe values of the molar specific heats for a number ofgases. While the experimental values of did not agreein most cases with the values predicted by theequipartition of energy, you can use the experimentalvalues of CV to accurately predict the values of Cp forthese gases. Do that now.

(b) Later in this chapter, in our discussion of what iscalled the adiabatic expansion of a gas (an expansionthat allows no heat to flow in), we will see that the ratioof Cp CVCp CV plays an important role in the theory. It iscommon practice to designate this ratio by the Greekletter γ

γ ≡ Cp CVCp CV (10)

(i) Explain why, for an ideal gas, γ is always greater than1.

(ii) Calculate the value of γ for the gases, listed in Table2 of Chapter 17.

Answers:

gas γ

helium 1.66

argon 1.66

nitrogen 1.40

oxygen 1.40

CO2 1.28

NH4 1.29

18-8 Entropy

ISOTHERMAL EXPANSIONAND PV DIAGRAMSIn the introduction, we pointed out that while a swim-ming pool of water may contain $16,000 of thermalenergy, we could not extract this energy to do usefulwork. To get useful energy, we have to burn fuel to getheat, and convert the heat to useful work. What seemedlike an insult is that even the most efficient power plantsturn only about 1/3 of the heat from the fuel into usefulwork, the rest being thrown away, expelled either intothe atmosphere or the ocean.

We are now going to discuss a process in which heat isconverted to useful work with 100% efficiency. Thisinvolves letting the gas in a piston expand at constanttemperature, in a process called an isothermal expan-sion. (The prefix “iso” is from the Greek meaning“equal,” thus, isothermal means equal or constanttemperature.) This process cannot be used by powerplants to make them 100% efficient, because the pro-cess is not repetitive. Some work is required to get thepiston back so that the expansion can be done overagain.

Suppose we start with a gas in a cylinder of volume V1and let the gas slowly expand to a volume V2 as shownin Figure (6). We control the expansion by adjustingthe force F exerted on the back side of the piston.

While the gas is expanding, it is doing work on thepiston. For each ∆V by which the volume of the gasincreases, the amount of work done by the gas is p∆V .The energy required to do this work must come fromsomewhere. If we did not let any heat into the cylinder,the energy would have to come from thermal energy,and the temperature would drop. (This is one way to getwork out of thermal energy.)

However, we wish to study the process in which the gasexpands at constant temperature. To keep the tempera-ture from dropping, we have to let heat flow into thegas. Since the temperature of the gas is constant, thereis no change in the thermal energy of the gas. Thus allthe heat that flows in goes directly into the work doneby the gas.

To calculate the amount of work done , we have to addup all the p∆V ’s as the gas goes from a volume V1 toa volume V2. If we graph pressure as a function ofvolume, in what is called a pV diagram, we can easilyvisualize these increments of work p∆V as shown inFigure (7).

Suppose the pressure of the gas is initially p1 when thevolume of the cylinder is V1. As the cylinder movesout and the gas expands, its pressure will drop as shownin Figure (7), reaching the lower value p2 when thecylinder volume reaches V2. At each step ∆Vi, whenthe pressure is pi, the amount of work done by the gasis pi∆Vi . The total work, the sum of all the pi∆Vi, isjust the total area under the pressure curve, as seen inFigure (7).

The nice feature of a graph of pressure versus volumelike that shown in Figure (7), is the work done by thegas is always the area under the pressure curve, nomatter what the conditions of the expansion are. If wehad allowed the temperature to change, the shape of thepressure curve would have been different, but the workdone by the gas would still be the area under thepressure curve.

V F2

V F1

Figure 6Isothermal expansion of the gas in a cylinder. Theforce F on the cylinder is continually adjusted sothat the gas expands slowly at constant temperature.

Figure 7The work done by an expanding gas is equal to the sumof all p ∆∆V 's , which is the area under the pressure curve.

V2V1

V1ppressure

area =

∆V

1 p1

V2p2p2

pi

pi

i

∆Vi

volume

18-9

Isothermal CompressionIf we shoved the piston back in, from a volume V2 toa volume V1 in Figure (7), we would have to do workon the gas. If we kept the temperature constant, then thepressure would increase along the curve shown inFigure (7) and the work we did would be preciselyequal to the area under the curve. In this case work isdone on the gas (we could say that during the compres-sion the gas does negative work). When work is doneon the gas, the temperature of the gas will rise unless welet heat flow out of the cylinder. Thus if we have anisothermal compression, where there is no increase inthe thermal energy of the gas, then we have the pureconversion of useful work into the heat expelled by thepiston. This is the opposite of what we want for a powerplant.

Isothermal Expansion of an Ideal GasIf we have one mole of an ideal gas in our cylinder, andkeep the temperature constant at a temperature T1 , thenthe gas will obey the ideal gas equation.

pV = RT1 = constant (11)

Thus the equation for the pressure of an ideal gas duringan isothermal expansion is

p = constantV

(11a)

and we see that the pressure decreases as 1/V. Thisdecrease is shown in the pV diagram of Figure (8).

ADIABATIC EXPANSIONWe have seen that we can get useful work from heatduring an isothermal expansion of a gas in a cylinder.As the gas expands, it does work, getting energy for thework from heat that flows into the cylinder. Thisrepresents the conversion of heat energy at one tem-perature into useful work. The problem is that there isa limited amount of work we can get this way. If weshove the piston back in so that we can repeat theprocess and get more work, it takes just as much workto shove the piston back as the amount of work we gotout during the expansion. The end result is that we havegotten nowhere. We need something besides isother-mal expansions and compressions if we are to end upwith a net conversion of heat into work.

Another kind of expansion is to let the cylinder expandwithout letting any heat in. This is called an adiabaticexpansion, where adiabatic is from the Greek (a-not +dia-through + bainein-to go). If the gas does workduring the expansion, and we let no heat energy in, thenall the work must come from the thermal energy of thegas. The result is that the gas will cool during theexpansion. In an adiabatic expansion, we are convert-ing the heat energy contained in the gas into usefulwork. If we could keep this expansion going we couldsuck all the thermal energy out of the gas and turn it intouseful work. The problem, of course, getting the pistonback to start the process over again.

Figure 8In the isothermal expansion of an ideal gas, we have pV = constant. Thus the pressure decreases as 1/V.

V2V1

V1ppressure

area undercurve = workdone by gas

pV = constant(p = const / V)

1 p1

V2p2p2

volume

isothermal expansion

18-10 Entropy

It is instructive to compare an isothermal expansion toan adiabatic expansion of a gas. In either case thepressure drops. But in the adiabatic expansion, thepressure drops faster because the gas cools. In Figure(9), we compare the isothermal and adiabatic expan-sion curves for an ideal gas. Because the adiabaticcurve drops faster in the pV diagram, there is less areaunder the adiabatic curve, and the gas does less work.This is not too surprising, because less energy wasavailable for the adiabatic expansion since no heatflowed in.

For an ideal gas, the equation for an adiabatic expan-sion is

pVγ = constant; γ =

Cp

CV(12)

a result we derive in the appendix. (You calculated thevalue of γ for various gases in Exercise 2.) Theimportant point now is not so much this formula, as thefact that the adiabatic curve drops faster than theisothermal curve.

If we compress a gas adiabatically, all the work we dogoes into the thermal energy of the gas, and thetemperature rises. Thus with an adiabatic expansionwe can lower the temperature of the gas, and with anadiabatic compression raise it.

Exercise 3In the next section, we will discuss a way of connectingadiabatic and isothermal expansions and compres-sions in such a way that we form a complete cycle (getback to the starting point), and get a net amount of workout of the process. Before reading the next section, it isa good exercise to see if you can do this on your own.

In order to see whether or not you are getting work outor putting it in, it is useful to graph the process in a pVdiagram, where the work is simply the area under thecurve.

To get you started in this exercise, suppose you beginwith an ideal gas at a pressure p1, volume V1, andtemperature T1 , and expand it isothermally to p2, V2, T1

as shown in Figure (10a). The work you get out is thearea under the curve.

If you then compressed the gas isothermally back to p1, V1, T1 , this would complete the cycle (get you back to

where you started), but it would take just as much workto compress the gas as you got from the expansion.Thus there is no net work gained from this cycle. A morecomplex cycle is needed to get work out.

If we add an adiabatic expansion to the isothermalexpansion as shown in Figure (10b) we have the start ofsomething more complex. See if you can complete thiscycle, i.e., get back to p1, V1, T1 , using adiabatic andisothermal expansions or compressions, and get somenet work in the process. See if you can get the answerbefore we give it to you in the next section. Also showgraphically, on Figure (10b) how much work you do getout.

Figure 10apV diagram for an isothermal expansionfrom volume V1 to volume V2 .

V2V1

V1ppressure

pV = constant

pV = constant (for γ = 1.667)

1p1

p2

p3

volume

adiabatic expansion

isothermal expansion

γ

V1

pressure

V2volume

V1T1p1

V2T1p2

isothermal expansion

Figure 9Comparison of isothermal and adiabaticexpansions. In an adiabatic expansion the gascools, and thus the pressure drops faster.

18-11

THE CARNOT CYCLEWith the isothermal and adiabatic expansion and com-pression of an ideal gas in a frictionless cylinder, wenow have the pieces necessary to construct a Carnotcycle, the key part of our thought experiment to studythe second law of thermodynamics.

The goal is to construct a device that continuallyconverts heat energy into work. Such a device is calledan engine. Both the isothermal and adiabatic expan-sions of the gas converted heat energy into work, butthe expansions alone could not be used as an enginebecause the piston was left expanded. Carnot’s re-quirement for an engine was that after a complete cycleall the working parts had to be back in their originalcondition ready for another cycle. Somehow the gas inthe cylinder has to be compressed again to get the pistonback to its original position. And the compressioncannot use up all the work we got from the expansion,in order that we get some net useful work from thecycle.

The idea for Carnot’s cycle that does give a net amountof useful work is the following. Start off with the gasin the cylinder at a high temperature and let the gasexpand isothermally. We will get a certain amount ofwork from the gas. Then rather than trying to compressthe hot gas, which would use up all the work we got,cool the gas to reduce its pressure. Then isothermallycompress the cool gas. It should take less work tocompress the low pressure cool gas than the work wegot from the high pressure hot gas. Then finish the

Figure 10bpV diagram for an isothermal expansionfollowed by an adiabatic expansion.

V1

pressure

V2 V3volume

V1T1p1

V2T1p2

V3T3p3

adiabatic expansion

isothermal expansion

cycle by heating the cool gas back up to its originaltemperature. In this way you get back to the originalvolume and temperature (and therefore pressure) of thecylinder; you have a complete cycle, and hopefully youhave gotten some useful work from the cycle.

To cool the gas, and then later heat it up again, Carnotused an adiabatic expansion and then an adiabaticcompression. We can follow the steps of the Carnotcycle on the pV diagram shown in Figure (11). The gasstarts out at the upper left hand corner at a hightemperature T1 , volume V1, pressure p1. It then goesthrough an isothermal expansion from a volume V1 toa volume V2, remaining at the initial temperature T1 .The hot gas is then cooled down to a low temperature

T3 by an adiabatic expansion to a volume V3. The cool,low pressure gas is then compressed isothermally to avolume V4, where it is then heated back to a highertemperature T1 by an adiabatic compression. Thevolume V4 is chosen just so that the adiabatic compres-sion will bring the temperature back to T1 when thevolume gets back to V1.

Figure 11The Carnot Cycle. The gas first expands at a hightemperature T1. It is then cooled to a lowertemperature T3 by an adiabatic expansion. Then it iscompressed at this lower temperature, and finallyheated back to the original temperature T1 by anadiabatic compression. We get a net amount ofwork from the process because it takes less work tocompress the cool low pressure gas than we gotfrom the expansion of the hot high pressure gas.

V1T1

T1

T3

T3

V1 V4 V2 V3

pressure

p1

V2p2

V3p3

V4p4

volume

compression

isothermal

adiabaticcom

pression

adiabatic expansion

isothermal expansion

p3

p4

p2

p1

18-12 Entropy

In this set of 4 processes, we get work out of the twoexpansions, but put work back in during the twocompressions. Did we really get some net work out?We can get the answer immediately from the pVdiagram. In Figure (12a), we see the amount of workwe got out of the two expansions. It is the total areaunder the expansion curves. In Figure (12b) we seehow much work went back in during the two compres-sions. It is the total area under the two compressioncurves. Since there is more area under the expansioncurves than the compression curves, we got a netamount of work out. The net work out is, in fact, justequal to the 4 sided area between the curves, seen inFigure (13).

Thermal Efficiencyof the Carnot CycleThe net effect of the Carnot cycle is the following.During the isothermal expansion while the cylinder isat the high temperature TH, a certain amount of thermalor heat energy, call it Q H, flows into the cylinder. Q Hmust be equal to the work the gas is doing during theisothermal expansion since the gas’ own thermal en-ergy does not change at the constant temperature.(Here, all the heat in becomes useful work.)

During the isothermal compression, while the cylinderis at the lower temperature TL , (the gas having beencooled by the adiabatic expansion), an amount of heat

Q Lis expelled from the cylinder. Heat must be ex-pelled because we are doing work on the gas bycompressing it, and none of the energy we supply cango into the thermal energy of the gas because itstemperature is constant. (Here all the work done be-comes expelled heat.)

Since no heat enters or leaves the cylinder during theadiabatic expansion or compression, all flows of heathave to take place during the isothermal processes.Thus the net effect of the process is that an amount ofthermal energy or heat Q H flows into the cylinder at thehigh temperature TH, and an amount of heat Q L flowsout at the low temperature TL, and we get a net amountof useful work W out equal to the 4-sided area seen inFigure (13). By the law of conservation of energy, the

V1 1

T1

T3

T

V1 V3

pressure

p1

V2p2

V3p3

volume

Figure 12aThe work we get out of the two expansions isequal to the area under the expansion curves.

Figure 12bThe work required to compress the gasback to its original volume is equal tothe area under the compression curves.

V1 1

T3

T

V1 V3

pressure

p1

V3p3

volume

T3V4p4

TH H

L

TL

pressure

heat in at high temperature T

heat out at low temperature T

net work done during cycle

volume

Figure 13The net work we get out of one complete cycle is equalto the area bounded by the four sided shape that liesbetween the expansion and compression curves.

18-13

work W must be equal to the difference between Q Hin and Q L out

W = QH – QL (13)

We see that the Carnot engine suffers from the sameproblem experienced by power plants and automobileengines. They take in heat QH at a high temperature(produced by burning fuel) and do some useful workW, but they expel heat QL out into the environment. Tobe 100% efficient, the engine should use all of QH toproduce work, and not expel any heat QL. But theCarnot cycle does not appear to work that way.

One of the advantages of the Carnot cycle is that we cancalculate QH and QL, and see just how efficient thecycle is. It takes a couple of pages of calculations,which we do in the appendix, but we obtain a remark-ably simple result. The ratio of the heat in, QH, to theheat out, QL, is simply equal to the ratio of the hightemperature TH to the low temperature TL .

QH

QL=

TH

TL

for a carnotcycle based onan ideal gas

(14)

One suspects that if you do a lot of calculation involv-ing integration, logarithms, and quantities like thespecific heat ratio, and almost everything cancels toleave such a simple result as Equation 14, then theremight be a deeper significance to the result than ex-pected. Equation 14 was derived for a Carnot cycleoperating with an ideal gas. It turns out that the resultis far more general and has broad applications.

Exercise 4A particular Carnot engine has an efficiency of 26.8%.That means that only 26.8% of QH comes out as usefulwork W and the rest, 73.2% is expelled at the lowtemperature TL . The difference between the high andlow temperature is 100 K ( TH –TL = 100 K). What arethe values of TH and TL? First express your answer inkelvins, then in degrees centigrade. (The answershould be familiar temperatures.)

Exercise 5

If you have a 100% efficient Carnot engine, what canyou say about TH and TL?

Reversible EnginesIn our discussion of the principle of relativity, it wasimmediately clear why we developed the light pulseclock thought experiment. You could immediately seethat moving clocks should run slow, and why that wasa consequence of the principle of relativity.

We now have a new thought experiment, the Carnotengine, which is about as idealized as our light pulseclock. We have been able to calculate the efficiency ofa Carnot engine, but it is not yet obvious what that hasto do either with real engines, or more importantly withthe second law of thermodynamics which we arestudying. It is not obvious because we have not yetdiscussed one crucial feature of the Carnot engine.

The Carnot engine is explicitly designed to be revers-ible. As shown in Figure (14) , we could start at point1 and go to point 4 by an adiabatic expansion of the gas.During this expansion the gas would do work but noheat is allowed to flow in. Thus the work energy wouldcome from thermal energy and the gas would cool from

TH to TL .

pressure

volume

expansion

isothermal

adiabaticexpansion

adiabatic compression

isothermal com

pression

1

2

T

3

4

H

TL

Figure 14The Carnot cycle run backward.

18-14 Entropy

The next step of the reverse Carnot cycle is an isother-mal expansion from a volume V4 to a volume V3.During this expansion, the gas does an amount of workequal to the area under the curve as shown in Figure(15a). Since there is no change in the internal energy ofa gas when the temperature of the gas remains constant,the heat flowing in equals the work done by the gas.This is the same amount of heat QL that flowed outwhen the engine ran forward.

In going from point 3 to point 2, we adiabaticallycompress the gas to heat it from the lower temperature

TLto the higher temperature TH . Since the compres-sion is adiabatic, no heat flows in or out.

In the final step from point 2 to point 1, we isothermallycompress the gas back to its original volume V1. Sincethe gas temperature remains constant at TH, there is nochange in thermal energy and all the work we do,shown as the area under the curve in Figure (15b), mustbe expelled in the form of heat flowing out of thecylinder. The amount of heat expelled is just QH, theamount that previously flowed in when the engine wasrun forward.

We have gone through the reverse cycle in detail toemphasize the fact that the engine should run equallywell both ways. In the forward direction the enginetakes in a larger amount of heat QH at the hightemperature TH, expels a smaller amount of heat QLat the lower temperature TL , and produces an amountof useful work W equal to the difference QH – QL.

In the reverse process, the engine takes in a smalleramount of heat QLat the low temperature TL, andexpels a larger amount QH at the higher temperature

TH. Since more heat energy is expelled than taken in,an amount of work W = QH – QLmust now be sup-plied to run the engine. When we have to supply workto pump out heat, we do not usually call the device anengine. The common name is a refrigerator. In arefrigerator, the refrigerator motor supplies the workW, a heat QLis sucked out of the freezer box, and a totalamount of energy QL + W = QH is expelled into thehigher room temperature of the kitchen. If we have aCarnot refrigerator running on an ideal gas, then theheats QLand QH are still given by Equation 14

QL

QH=

TL

TH(14 repeated)

where TLand TH are the temperatures on a scalestarting from absolute zero such as in the kelvin scale.

Exercise 6

How much work must a Carnot refrigerator do to remove1000 joules of energy from its ice chest at 0° C and expelthe heat into a kitchen at 27° C?

Figure 15Heat flow when the Carnot cycle runs backward.Since more heat flows out than in, some work W isrequired for the cycle. The net effect is that the workW pumps heat out of the gas, giving us a refrigerator.

V V21

TH

TL

pressure

volume

1

4

3

2

heat expelled equals work done on the gas

b) A lot more work is required, and a lotmore heat is expelled when we compressthe hot gas isothermally.

a) During the isothermal expansion, someheat flows into the gas to supply the energyneeded for the work done by the gas.

V V34

TH

TL

pressure

heat in equals work done by the gas

volume

1

4

3

2

18-15

ENERGY FLOW DIAGRAMSBecause of energy conservation, we can view the flowof energy in much the same way as the flow of somekind of a fluid. In particular we can construct flowdiagrams for energy that look much like plumbingdiagrams for water. Figure (16) is the energy flowdiagram for a Carnot engine running forward. At thetop and the bottom are what are called thermal reser-voirs—large sources of heat at constant temperature(like swimming pools full of water). At the top is athermal reservoir at the high temperature TH (it couldbe kept at the high temperature by burning fuel) and atthe bottom is a thermal reservoir at the low temperature

TL . For power plants, the low temperature reservoir isoften the ocean or the cool water in a river. Or it maybe the cooling towers like the ones pictured in photo-graphs of the nuclear power plants at Three Mile Island.

In the energy flow diagram for the forward runningCarnot engine, an amount of heat QH flows out of thehigh temperature reservoir, a smaller amount QLisexpelled into the low temperature reservoir, and thedifference comes out as useful work W. If the Carnotengine is run on an ideal gas, QH and QLare alwaysrelated by QH/QL = TH/TL .

Figure (17) is the energy flow diagram for a Carnotrefrigerator. A heat QLis sucked out of the lowtemperature reservoir, an amount of work W is sup-plied (by some motor), and the total energy

QH = QL + W is expelled into the high temperaturereservoir.

Maximally Efficient EnginesWe are now ready to relate our discussion of the Carnotcycles to the second law of thermodynamics. Thestatement of the second law we will use is that youcannot extract useful work from thermal energy at onetemperature. (The colloquial statement of the first lawof thermodynamics — conservation of energy—is thatyou can’t get something for nothing. The second lawsays that you can’t break even.)

Up until now we have had to point out that our formulafor the efficiency of a Carnot engine was based on theassumption that we had an ideal gas in the cylinder. Ifwe use the second law of thermodynamics, we canshow that it is impossible to construct any engine, byany means, that is more efficient than the Carnot enginewe have been discussing. This will be the main resultof our thought experiment.

WQL

QH

high temperature reservoir

low temperature reservoir

TH

TL

Carnotrefrigerator

Figure 16Energy flow diagram for a Carnot engine. Since energyis conserved, we can construct a flow diagram forenergy that resembles a plumbing diagram for water. Ina Carnot cycle, QH flows out of a “thermal reservoir”at a temperature TH . Some of this energy goes out asuseful work W and the rest, QL , flows into the lowtemperature thermal reservoir at a temperature TL .

WQL

QH

high temperature reservoir

low temperature reservoir

TH

TL

Carnotengine

Figure 17The Carnot refrigerator is a Carnot engine runbackwards. The work W plus the heat QL equals the heat

QH pumped up into the high temperature reservoir.

18-16 Entropy

Let us suppose that you have constructed a Superengine that takes in more heat QH

*from the hightemperature reservoir, and does more work W* , whilerejecting the same amount of heat QL as a Carnotengine. In the comparison of the two engines in Figure(18) you can immediately see that your Super engine ismore efficient than the Carnot engine because you getmore work out for the same amount of heat lost to thelow temperature reservoir.

Now let us run the Carnot engine backwards as arefrigerator as shown in Figure (19). The Carnotrefrigerator requires an amount of work W to suck theheat QL out of the low temperature reservoir and expelthe total energy QH = W + QL into the high tempera-ture reservoir.

You do not have to look at Figure (19) too long beforeyou see that you can use some of the work W* that yourSuper engine produces to run the Carnot refrigerator.Since your engine is more efficient than the Carnotcycle, W* > W and you have some work left over.

The next thing you notice is that you do not need the lowtemperature reservoir. All the heat expelled by yourSuper engine is taken in by the Carnot refrigerator. Thelow temperature reservoir can be replaced by a pipe andthe new plumbing diagram for the combined Superengine and Carnot refrigerator is shown in Figure (20).The overall result of combining the super engine andCarnot refrigerator is that a net amount of work

Wnet = W*–W is extracted from the high temperaturereservoir. The net effect of this combination is toproduce useful work from thermal energy at a singletemperature, which is a violation of the second law ofthermodynamics.

Exercise 7

Suppose you build a Super engine that takes the sameamount of heat from the high temperature reservoir asa Carnot engine, but rejects less heat QL

*< QL than aCarnot engine into the low temperature. Using energyflow diagrams show what would happen if this Superengine were connected to a Carnot refrigerator. (Youwould still be getting useful work from thermal energy atsome temperature. From what temperature reservoirwould you be getting this work?)

W

QL

QL

high temperature reservoir THCarnot refrigeratorSuper engine

QL

QHnet

Figure 19Now run the Carnot enginebackward as a refrigerator.

high temperature reservoir

low temperature reservoir

TH

TL

Carnot refrigeratorSuper engine

WQL

QHWQL

QH* *

Figure 18Comparison of the Super enginewith the Carnot engine.

WQL

QH WQL

QH

high temperature reservoir

low temperature reservoir

TH

TL

Carnot engineSuper engine

* *

Figure 20If you connect the Super engine to the Carnotrefrigerator, you can eliminate the low temperaturereservoir and still get some work Wnet out. Thismachine extracts work from a single temperature, inviolation of the second law of thermodynamics.

18-17

ReversibilityWe have just derived the rather sweeping result that ifthe second law of thermodynamics is correct, youcannot construct an engine that is more efficient than aCarnot engine based on an ideal gas. You may wonderwhy the cycles based on an ideal gas are so special. Itturns out that they are not special. What was specialabout the Carnot engine is that it was reversible, that itcould be run backwards as a refrigerator. You can useprecisely the same kind of arguments we just used toshow that all reversible engines must have preciselythe same efficiency as a Carnot engine. It is arequirement of the second law of thermodynamics.

There were two reasons we went through the detailedsteps of constructing a Carnot engine using an ideal gasin a frictionless piston. The first was to provide oneexample of how an engine can be constructed. It is nota very practical example, commercial engines arebased on different kinds of cycles. But the Carnotengine illustrates the basic features of all engines. In allengines the process must be repetitive, at least twotemperatures must be involved, and only some of theheat extracted from the high temperature reservoir canbe converted to useful work. Some heat must beexpelled at a lower temperature.

While all reversible engines have the same efficiency,we have to work out at least one example to find outwhat that efficiency is. You might as well choose thesimplest possible example, and the Carnot cycle usingan ideal gas is about as simple as they get. Because ofthe second law of thermodynamics, you know thateven though you are working out a very special ex-ample, the answer QH/QL = TH/TL applies to all re-versible engines operating between two temperaturereservoirs. This is quite a powerful result from the fewpages of calculations in the appendix.

APPLICATIONS OFTHE SECOND LAWDuring the oil embargo in the middle 1970s, there wasa sudden appreciation of the consequences of thesecond law of thermodynamics, for it finally becameclear that we had to use energy efficiently. Since thattime there has been a growing awareness that there isa cost to producing energy that considerably exceedswhat we pay for it. Burning oil and coal depletesnatural limited resources and adds carbon dioxide tothe atmosphere which may contribute to global climatechanges. Nuclear reactors, which were so promising inthe 1950s, pose unexpected safety problems, both nowas in the example of Chernobyl, and in the very distantfuture when we try to deal with the storage of spentreactor parts. Hydroelectric power floods land thatmay have other important uses, and can damage theagricultural resources of an area as in the case of theAswan Dam on the Nile River. More efficient use ofenergy from the sun is a promising idea, but technologyhas not evolved to the point where solar energy cansupply much of our needs. What we have learned isthat, for now, the first step is to use energy as efficientlyas possible, and in doing this, the second law ofthermodynamics has to be our guide.

During the 1950s and 60s, one of the buzz words formodern living was the all electric house. These houseswere heated electrically, electric heaters being easy andinexpensive to install and convenient to use. And it alsorepresents one of the most stupid ways possible to useenergy. In terms of a heat cycle, it represents the 100%conversion of work energy into thermal energy, whatwe would have called in the last section, a 0% efficientengine. There are better ways of using electric powerthan converting it all into heat.

You can see where the waste of energy comes in whenyou think of the processes involved in producingelectric power. In an electric power plant, the first stepis to heat some liquid or gas to a high temperature byburning fuel. In a common type of coal or oil firedpower plant, mercury vapor is heated to temperaturesof 600 to 700 degrees centigrade. The mercury vaporis then used to run a mercury vapor turbine which coolsthe mercury vapor to around 200° C. This coolermercury vapor then heats steam which goes through asteam turbine to a steam condenser at temperatures

18-18 Entropy

around 100° C. In a nuclear reactor, the first step isoften to heat liquid sodium by having it flow throughpipes that pass through the reactor. The hot sodium canthen be used to heat mercury vapor which runs turbinessimilar to those in a coal fired plant. The turbines areattached to generators which produce the electric power.

Even though there are many stages, and dangerous andexotic materials used in power stations, we can esti-mate the maximum possible efficiency of a power plantsimply by knowing the highest temperature TH of theboiler, and the lowest temperature TLof the condenser.If the power plant were a reversible cycle runningbetween these two temperatures, it would take in anamount of heat QH at the high temperature and rejectan amount of heat QL at the low temperature, where

QH and QL are related by QH/QL = TH/TL (Eq. 14).The work we got out would be

W = QH – QL

amount of work froma reversible cycle (15)

We would naturally define the efficiency of the cycleas the ratio of the work out to the heat energy in

efficiency = W

QH=

QH – QL

QH(16)

If we solve Equation 14 for QL

QL = QHTL

TH

and use this in Equation 16, we get

efficiency =

QH – QL

QH=

QH 1 – TL/TH

QH

efficiency =

TH – TL

TH

efficiency of areversiblecycle

(17)

Since by the second law of thermodynamics no processcan be more efficient than a reversible cycle, Equation17 represents the maximum possible efficiency of apower plant.

The important thing to remember about Equation 17 isthat the temperatures TH and TL start from absolutezero. The only way we could get a completely efficientengine or power plant would be to have the low

temperature at absolute zero, which is not only impos-sible to achieve but even difficult to approach. You cansee from this equation why many power plants arelocated on the shore of an ocean or on the bank of a largeriver. These bodies are capable of soaking up largequantities of heat at relatively low temperatures. If anocean or river is not available, the power plant will havelarge cooling towers to condense steam. Condensingsteam at atmospheric pressure provides a low tempera-ture of TL = 100° C or 373 K.

Equation 15 also tells you why power plants run theirboilers as hot as possible, using exotic substances likemercury vapor or liquid sodium. Here one of thelimiting factors is how high a temperature turbineblades can handle without weakening. Temperaturesas high as 450° C or around 720 K are about the limitof current technology. Thus we can estimate themaximum efficiency of power plants simply by know-ing how high a temperature turbine blades can with-stand, and that the plant uses water for cooling. You donot have to know the details of what kind of fuel is used,what kind of exotic materials are involved, or howturbines and electric generators work, as long as theyare efficient. Using the numbers TH = 720 k ,

TL = 373 k we find that the maximum efficiency isabout

maximumefficiency

=TH – TL

TH=

720 – 373

720

= .48(18)

Thus about 50% represents a theoretical upper limit tothe efficiency of power plants using current technol-ogy. In practice, well designed power plants reach onlyabout 33% efficiency due to small inefficiencies in themany steps involved.

You can now see why the all electric house was such abad idea. An electric power plant consumes three timesas much fuel energy as it produces electric energy.Then the electric heater in the all electric house turnsthis electric energy back into thermal energy. If thehouse had a modern oil furnace, somewhere in theorder of 85% of the full energy can go into heating thehouse and hot water. This is far better than the 33%efficiency from heating directly by electricity.

18-19

Electric CarsOne of the hot items in the news these days is theelectric car. It is often touted as the pollution freesolution to our transportation problems. There areadvantages to electric cars, but not as great an advan-tage as some new stories indicate.

When you plug in your electric car to charge batteries,you are not eliminating the pollution associated withproducing useful energy. You are just moving thepollution from the car to the power plant, which may,however, be a good thing to do. A gasoline car mayproduce more harmful pollutants than a power station,and car pollutants tend to concentrate in places wherepeople live creating smog in most major cities on theearth. (Some power plants also create obnoxioussmog, like the coal fired plants near the Grand Canyonthat are harming some of the most beautiful scenery inthe country.)

In addition to moving and perhaps improving thenature of pollution, power plants have an additionaladvantage over car engines—they are more efficient.Car engines cannot handle as high a temperature as apower plant, and the temperature of the exhaust from acar is not as low as condensing steam or ocean water.Car engines seldom have an efficiency as high as 20%;in general, they are less than half as efficient as a powerplant. Thus there will be a gain in efficiency in the useof fuel when electric cars come into more common use.

(One way electric cars have for increasing their effi-ciency is to replace brakes with generators. Whengoing down a hill, instead of breaking and dissipatingenergy by heating the brake shoes, the gravitationalpotential energy being released is turned into electricenergy by the generators attached to the wheels. Thisenergy is then stored as chemical energy in the batteriesas the batteries are recharged.)

The Heat PumpThere is an intelligent way to heat a house electrically,and that is by using a heat pump. The idea is to use theelectric energy to pump heat from the colder outsidetemperature to the warmer inside temperature. Pump-ing heat from a cooler temperature to a warmer tem-perature is precisely what a refrigerator does, whiletaking heat from the freezer chest and exhausting it intothe kitchen. The heat pump takes heat from the cooleroutside and exhausts it into the house.

As we saw in our discussion, it takes work to pump heatfrom a cooler to a higher temperature. The ratio of theheat QL taken in at the low temperature, to the heat QHexpelled at the higher temperature, is QH/QL = TH/TLfor a maximally efficient refrigerator. The amount ofwork W required is W = QH – QL. The efficiency ofthis process is the ratio of the amount of heat deliveredto the work required.

efficiency ofheat pump

=QH

W=

QH

QH – QL

=TH

TH – TL

(19)

where the last step in Equation 19 used QL = QHTL/TH.

Exercise 8

Derive the last formula in Equation 19.

When the temperature difference TH – TL is small, wecan get very high efficiencies, i.e., we can pump a lotof heat using little work. In the worst case, where

TL = 0 and we are trying to suck heat from absolutezero, the efficiency of a heat pump is 1—heat deliveredequals the work put in—and the heat pump is actinglike a resistance heater.

18-20 Entropy

To illustrate the use of a heat pump let us assume thatit is freezing outside ( TL = 0° C = 273K ) and youwant the inside temperature to be 27°C = 300K . Thena heat pump could have an efficiency of

efficiency of heatpump running from0°C to 27°C

=TH

TH – TL

=300 k

300 k – 273 k

= 11.1

In other words, as far as the second law of thermody-namics is concerned, we should be able to pump eleventimes as much heat into a house, when it is just freezingoutside, as the amount of electrical energy required topump the heat. Even if the electrical energy is producedat only 30% efficiency, we should still get

.30 × 11.1 = 3.3 times as much heat into the house asby burning the fuel in the house at 100% conversion offuel energy into heat.

Exercise 9This so-called “heat of fusion” of water is 333kJ/kg.What that means is that when a kilogram (1 liter) of waterfreezes (going from 0° C water to 0° ice), 333 kilojoulesof heat are released. Thus to freeze a liter of 0° C waterin your refrigerator, the refrigerator motor has to pump

333 × 103 joules of heat energy out of the refrigeratorinto the kitchen. The point of the problem is to estimatehow powerful a refrigerator motor is required if you wantto be able to freeze a liter of water in 10 minutes.

Assume that the heat is being removed at a temperatureof 0°C and being expelled into a kitchen whose tem-perature is 30°C, and that the refrigerator equipment is100% efficient. (We will account for a lack of efficiencyat the end of this problem.)

In the United Sates, the power of motors is generallygiven in “horsepower”, a familiar but archaic unit. Theconversion factor is 1 horsepower = 746 watts, and apower of 1 watt is 1 joule per second.

Calculate the horsepower required, then double theanswer to account for lack of efficiency. (Answer:0.16 horsepower.)

Exercise 10

Here is a problem that should give you some practicewith the concepts of efficiency. You have the choice ofbuying a furnace that converts heat energy of oil intoheat in the house with 85% efficiency. I.e., 85% of theheat energy of the oil goes into the house, and 15% goesup the chimney. Or you can buy a heat pump which ishalf as efficient as a Carnot refrigerator. (This is a morerealistic estimate of the current technology of refrigera-tion equipment.) At very low temperatures outside, heatpumps are not as efficient, and burning oil in your ownfurnace is more efficient. But if it does not get too coldoutside, heat pumps are more efficient. At what outsidetemperature will the heat pump and the oil furnace havethe same efficiency? Assume that the electric energyyou use is produced by a power plant that is 30%efficient. (Answer: - 26° C.)

18-21

While the piston goes back down, the valves are set sothat a mixture of air and fuel are sucked into the piston.When the cylinder is at the bottom of the piston, wehave a cool, low pressure fuel air mixture filling the fullvolume V4. We are now at the position labeled (4) inFigure (21). It took two strokes (up and down) of thepiston to go from position 3 to position 4.

In the final stroke, the valves are shut and the risingpiston adiabatically compresses the gas back to thestarting point p1 , V1, T1 . While the increase in tem-perature during this compression is what is needed toignite the diesel fuel, you do not want the temperatureto rise enough to ignite the air gasoline mixture in agasoline engine. This can sometimes happen in agasoline engine, causing a knock in the engine, orsometimes allowing the engine to run for a while afteryou have shut off the ignition key and stopped the sparkplug from functioning.

The Internal Combustion EngineWe finish this section on practical applications with abrief discussion of the internal combustion engine. Themain point is to give an example of an engine that runson a cycle that is different from a Carnot cycle. It ismore difficult to apply the second law of thermody-namics to an internal combustion engine because itdoes not take heat in or expel heat at constant tempera-tures like the Carnot engine, but we can still analyze thework we get out using a pV diagram.

The pV diagram for an internal combustion engine isshown in Figure (21). At position 1, a fuel and airmixture have been compressed to a small volume V1by the piston which is at the top of the cylinder. If it isa gasoline engine, the fuel air mixture is ignited by aspark from a sparkplug. If it is a diesel engine, themixture of diesel fuel and air have been heated to thepoint of combustion by the adiabatic compression frompoint 4 to point 1 that has just taken place. One of theadvantages of a diesel engine is that an electricalsystem to produce the spark is not needed. This isparticularly important for boat engines where electricsystems give all sorts of problems. (We said this wasa section on practical applications.)

After ignition, the pressure and temperature of the gasrise rapidly to p2 , T2 before the piston has had a chanceto move. Thus the volume remains at V1 and the pVcurve goes straight up to point 2. The heated gas thenexpands adiabatically, and cools some, driving thepiston down to the bottom of the cylinder. This is thestroke from which we get work from the engine.

We now have a cylinder full of hot burned exhaustgases. In a 4 cycle engine, a valve at the top of thecylinder is opened, and a piston is allowed to rise,pushing the hot exhaust gases out into the exhaustpipes. Not much work is required to do this. This is thepart of the cycle where (relatively) low temperaturethermal energy is exhausted to the environment.

V4V1

V1

1

2

3

4

pressure

spark

exhaust

compression

power stroke

volume

adiabatic

adiabatic

V4

Figure 21PV diagram for an internal combustionengine. When the piston is all the way up inthe cylinder the volume is V1 . When it is allthe way down, the volume has increased to V2 .

18-22 Entropy

We see that the inefficient engine expelled more Q/Tthan it took in. The inefficient, non reversible, enginecreates Q/T while reversible engines do not.

As we have done throughout the course, whenever weencounter a quantity that is conserved, or sometimesconserved, we give it a name. We did this for linearmomentum, angular momentum, and energy. Now wehave a quantity Q/T that is unchanged by reversibleengines, but created or increased by irreversible ineffi-cient ones. We are on the verge of defining the quantityphysicists call entropy. We say on the verge of definingentropy, because Q/T is not entropy itself; it representsthe change in entropy. We can say that when the gasexpanded, the entropy of the gas increased by QH/TH .And when the gas was compressed, the entropy de-creased by QL/TL.

For a reversible engine there is no net change in entropyas we go around the cycle. But for an irreversible,inefficient engine, more entropy comes out than goes induring each cycle. The net effect of an inefficientengine is to create entropy.

What is this thing called entropy that is created byinefficient irreversible engine? Consider the mostinefficient process we can imagine—the electric heaterwhich converts useful work in the form of electricalenergy into heat. From one point of view, the devicedoes nothing but create entropy. If the heater is at atemperature T, and the electric power into the heater isW watts, all this energy is converted to heat and entropyis produced at a rate of W/T in units of entropy persecond. (Surprisingly, there is no standard name for aunit of entropy. The units of Q/T are of course joules/kelvin, the same as Boltzman’s constant.)

The process of converting energy in the form of usefulwork into the random thermal energy of molecules canbe viewed as the process of turning order into disorder.Creating entropy seems to be related to creating disor-der. But the surprising thing is that we have an explicitformula Q/T for changes in entropy. How could it bepossible to measure disorder, to have an explicit for-mula for changes in disorder? This question baffledphysicists for many generations.

ENTROPYThe second law of thermodynamics provided us withthe remarkable result that the efficiency of all revers-ible engines is the same. Detailed calculation of thisefficiency using a Carnot engine based on an ideal gasgave us a surprisingly simple formula for this effi-ciency, namely QH/QL = TH/TL . Our preceding ex-amples involving car engines, power plants, refrigera-tors and heat pumps illustrate how important thissimple relationship is to mankind.

When you do a calculation and a lot of stuff cancels out,it suggests that your result may have a simpler interpre-tation than you originally expected. This turns out to betrue for our calculations of the heat flow in a Carnotengine. To get a new perspective on our equation forheat flow, let us write the equation in the form

QH

TH=

QL

TL(20)

In this form the equation for heat flow is beginning tolook like a conservation law for the quantity Q/T.During the isothermal expansion, an amount QH/THof this quantity flowed into the piston. During theisothermal compression, QL/TL flowed out. We findthat if the engine is reversible, the amount of Q/T thatflowed in is equal to the amount of Q/T that flowed out.The net effect is that there was no change in Q/T duringthe cycle.

To get a better insight into what this quantity Q/T maybe, consider a nonreversible engine operating between

TH and TL , an engine that would be less efficient thanthe Carnot engine. Assume that the less efficientengine and the Carnot engine both take in the sameamount of heat QH at the high temperature TH. Thenthe less efficient engine must do less work and expelmore heat at the low temperature TL. Thus QL for theless efficient engine is bigger than QL for the Carnotengine. Since QL/TL = QH/TH for the Carnot engine,

QL/TL must be greater than QH/TH for the less effi-cient engine

QL

TL>

QH

TH

for an engine thatis less efficient thana Carnot engine

(21)

18-23

Ludwig Boltzman proposed that entropy was related tothe number of ways that a system could be arranged.Suppose, for example, you go into a woodworkingshop and there are a lot of nails on the wall with toolshanging on them. In one particular woodworking shopyou find that the carpenter has drawn an outline of thetool on the wall behind the nail. You enter her shop youfind that the tool hanging from each nail exactlymatches the outline behind it. Here we have perfectorder, every tool is in its place and there is one and onlyone way the tools can be arranged. We would say that,as far as locating tools is concerned the shop is in perfectorder, it has no disorder or entropy.

On closer inspection, we find that the carpenter has twosaws with identical outlines, a crosscut saw and a ripsaw. We also see that the nails are numbered, and seethe cross cut saw on nail 23 and the rip saw on nail 24.A week later when we come back, the cross cut is onnail 24 and the rip saw on 23. Thus we find that hersystem is not completely orderly, for there are twodifferent ways the saws can be placed. This way oforganizing tools has some entropy.

A month later we come back to the shop and find thatanother carpenter has taken over and painted the walls.We find that there are still 25 nails and 25 tools, but nowthere is no way to tell which tool belongs on which nail.Now there are many, many ways to hang up the toolsand the system is quite disordered. We have the feelingthat this organization, or lack of organization, of thetools has quite high entropy.

To put a numerical value on how disorganized thecarpenter shop is, we go to a mathematician who tellsus that there are N! ways to hang N tools on N nails.Thus there are 25! = 1.55 × 1025 different ways the 25tools can be hung on the 25 nails. We could use thisnumber as a measure of the disorder of the system, butthe number is very large and increases very rapidly withthe number of tools. If, for example, there were 50 toolshung on 50 nails, there would be 3.04 × 1064 differentways of hanging them. Such large numbers are notconvenient to work with.

When working with large numbers, it is easier to dealwith the logarithm of the number than the numberitself. There are approximately 1051 protons in theearth. The log to the base 10 of this number is 51 andthe natural logarithm, ln 1051 , is 2.3 times bigger or117. In discussing the number of protons in the earth,the number 117 is easier to work with than 1051 ,particularly if you have to write out all the zeros.

If we describe the disorder of our tool hanging systemin terms of the logarithm of the number of ways thetools can be hung, we get a much more reasonable setof numbers, as shown in Table 1.

Setup Number of ways Logarithm of theto arrange tools number of ways

all tools haveunique positions 1 0

two saws canbe interchanged 2 .7

25 tools onunmarked nails 1.5 × 1025 58

50 tools onunmarked nails 3.0 × 1065 148

TABLE 1

The table starts off well. If there is a unique arrange-ment of the tools, only one way to arrange them, thelogarithm of the number of ways is 0. This is consistentwith our idea that there is no disorder. As the numberof tools on unmarked nails increases, the number ofways they can be arranged increases at an enormouspace, but the logarithm increases at a reasonable rate,approximately as fast as the number of tools and nails.This logarithm provides a reasonable measure of thedisorder of the system.

18-24 Entropy

We could define the entropy of the tool hanging systemas the logarithm of the number of ways the tools couldbe hung. One problem, however, is that this definitionof entropy would have different dimensions than thedefinition introduced earlier in our discussions of en-gines. There changes in entropy, for example QH/TH ,had dimensions of joules/kelvin, while our logarithm isdimensionless. However, this problem could be fixedby multiplying our dimensionless logarithm by somefundamental constant that has the dimensions of joules/kelvin. That constant, of course, is Boltzman’s con-stant k, where k = 1.38 × 10–23joules/kelvin. Wecould therefore take as the formula for the entropy (callit S) of our tool hanging system as

S = k ln n entropy of our tool

hanging system (22)

where n is the number of ways the tools can be hung.Multiplying our logarithm by k gives us the correctdimensions, but very small values when applied to asfew items as 25 or 50 tools.

Equation 22 appears on Boltzman’s tombstone as amemorial to his main accomplishment in life. Boltzmanbelieved that Equation 22 should be true in general.That, for example, it should apply to the atoms of thegas inside the cylinder of our heat engine. When heatflows into the cylinder and the entropy increases by anamount QH/TH , the number of ways that the atomscould be arranged should also increase, by an amountwe can easily calculate using Equation 22. Explicitly,if before the heat flowed in there were nold ways theatoms could be arranged, and after the heat flowed in

nnew ways, then Equation 22 gives

change inentropy = k ln nnew – k ln nold =

QH

TH

Since ln nnew – ln nold = ln nnew/nold , we get

ln

nnew

nold=

QH

kTH(23)

Taking the exponent of both sides of Equation 16, usingthe fact that eln x = x , we get

nnewnold

= eQH/kTH (24)

Thus Boltzman’s equation gives us an explicit formulafor the fractional increase in the number of ways theatoms in the gas atoms in the cylinder can be arranged.

Boltzman committed suicide in 1906, despondent overthe lack of acceptance of his work on the statisticaltheory of matter, of which Equation 22 is the corner-stone. And in 1906 it is not too surprising that physi-cists would have difficulty dealing with Boltzman’sequation. What is the meaning of the number of waysyou can arrange gas atoms in a cylinder? From aNewtonian perspective, there are an infinite number ofways to place just one atom in a cylinder. You cancount them by moving the atoms an infinitesimaldistance in any direction. So how could it be that 1024

atoms in a cylinder have only a finite way in which theycan be arranged?

This question could not be satisfactorily answered in1906, the answer did not come until 1925 with thediscovery of quantum mechanics. In a quantum pic-ture, an atom in a cylinder has only certain energylevels, an idea we will discuss later in Chapter 35. Evenwhen you have 1024 atoms in the cylinder, the wholesystem has only certain allowed energy levels. At lowtemperatures the gas does not have enough thermalenergy to occupy very many of the levels. As a resultthe number of ways the atoms can be arranged islimited and the entropy is low. As the temperature isincreased, the gas atoms can occupy more levels, canbe arranged in a greater number of ways, and thereforehave a greater entropy.

The concept of entropy provides a new definition ofabsolute zero. A system of particles is at absolute zerowhen it has zero entropy, when it has one uniquelydefined state. We mentioned earlier that quantummechanics requires that a confined particle has somekinetic energy. All the kinetic energy cannot beremoved by cooling. This gives rise to the so-calledzero point energy that keeps helium a liquid even atabsolute zero. However, a bucket of liquid helium canbe at absolute zero as long as it is in a single uniquequantum state, even though the atoms have zero pointkinetic energy.

18-25

For a while there was a debate among physicists as towhether the second law of thermodynamics was theonly law in nature that could be used to distinguishbetween time running forward and time running back-ward.

When you study processes like the decay of one kind ofelementary particle into another, the situation is sosimple that the concepts of entropy and the second lawof thermodynamics do not enter into the analysis. Inthese cases you can truly study whether nature issymmetric with the respect to the reversal of time. Ifyou take a moving picture of a particle decay, and runthe movie backwards, will you see a process that canactually happen? For example if a muon decays into anelectron and a neutrino, as happened in our muonlifetime experiment, running that moving picture back-ward would have neutrinos coming in, colliding withan electron, creating a muon. Thus, if the basic laws ofphysics are truly symmetric to the reversal of time, itshould be possible for a neutrino and an electron tocollide and create a muon. This process is observed.

In 1964 Val Fitch and James Cronin discovered anelementary particle process which indicated that naturewas not symmetric in time. Fitch found a violation ofthis symmetry in the decay of a particle called theneutral k meson. For this discovery, Fitch was awardedthe Nobel prize in 1980. Since the so-called “weakinteraction” is responsible for the decay of k mesons,the weak interaction is not fully symmetric to thereversal of time. The second law of thermodynamicsis not the only law of physics that knows which waytime goes.

In our discussion of temperature in the last chapter, weused the ideal gas thermometer for our experimentaldefinition of temperature. We pointed out, however,that this definition would begin to fail as we ap-proached very low temperatures near absolute zero. Atthese temperatures we need a new definition whichagrees with the ideal gas thermometer definition athigher temperatures. The new definition which is usedby the physics community, is based on the efficiency ofa reversible engine or heat cycle. You can measure theratio of two temperatures TH and TC, by measuring theheats QH and QC that enter and leave the cycle, and usethe formula TH/TC = QH/QC . Since this formula isbased on the idea that a reversible cycle creates noentropy ( QH/TH = QC/TC ), we can see that the con-cept of entropy forms the basis for the definition oftemperature.

The Direction of TimeWe began the chapter with a discussion of a demonstra-tion that looked funny. We set the strobe so that thewater drops appeared to rise from the plate in the bucketand enter the hose. Before our discussion of the secondlaw of thermodynamics, we could not find any law ofphysics that this backward process appeared to violate.Now we can see that the launching of the drops from theplate is a direct contradiction of the second law. In thatprocess, heat energy in the bucket converts itself at onetemperature into pure useful work that launches thedrop.

When you run a moving picture of some action back-wards, effectively reversing the direction of time, inmost cases the only law of physics that is violated is thesecond law of thermodynamics. The only thing thatappears to go wrong is that disordered systems appearto organize themselves on their own. Scrambled eggsturn into an egg with a whole yoke just by the flick ofa fork. Divers pop out of swimming pools propelled,like the drops in our demonstration, by the heat energyin the pool (see movie). All these funny looking thingsrequire remarkable coincidences which in real life donot happen.

MovieTime reversed motion picture of dive

18-26 Entropy

APPENDIX: CALCULATION OF THEEFFICIENCY OF A CARNOT CYCLEThe second law of thermodynamics tells us that theefficiency of all reversible heat engines is the same.Thus if we can calculate the efficiency of any oneengine, we have the results for all. Since we have basedso much of our discussion on the Carnot engine runningon an ideal gas, we will calculate the efficiency of thatengine.

To calculate the efficiency of the ideal gas Carnotengine, we need to calculate the amount of work we getout of (or put into) isothermal and adiabatic expan-sions. With these results, we can then calculate the netamount of work we get out of one cycle and then theefficiency of the engine. To simplify the formulas, wewill assume that our engine is running on one mole ofan ideal gas.

Isothermal ExpansionSuppose we have a gas at an initial volume V1, pressure

p1 , temperature T, and expand it isothermally to avolume V2, pressure p2 , and of course the sametemperature T.

The P-V diagram for the process is shown in Figure (A-1). The curve is determined by the ideal gas law,which for 1 mole of an ideal gas is

pV = RT (A-1)

The work we get out of the expansion is the shaded areaunder the curve, which is the integral of the pressurecurve from V1 to V2.

Using Equation 1, we get

W = pdV

V1

V2

= RTV dV

V1

V2

= RT dVV

V1

V2

= RTlnVV1

V2

= RT ln V2 – ln V1

= RT lnV2V1

(A-2)

Thus the work we get out is RT times the logarithm ofthe ratio of the volumes.

Adiabatic ExpansionIt is a bit trickier to calculate the amount of work we getout of an adiabatic expansion. If we start with a moleof ideal gas at a volume V1, pressure p1 , and tempera-ture TH, the gas will cool as it expands because the gasdoes work and we are not letting any heat in. Thuswhen the gas gets to the volume V2, at a pressure p2 ,its temperature TC will be cooler than its initial tem-perature TH.

The pV diagram for the adiabatic expansion is shownin Figure (A-2). To get an equation for the adiabaticexpansion curve shown, let us assume that we changethe volume of the gas by an infinitesimal amount ∆V .With this volume change, there will be an infinitesimalpressure drop ∆p , and an infinitesimal temperature

Figure A-1Isothermal expansion.

Figure A-2Adiabatic expansion.

p1

p2

V1 V2

TH

TL

pressure

adiabaticvolume

p1

p2

V

T

1 V2

pressure

isothermal

volume

18-27

drop ∆T. We can find the relationship between thesesmall changes by differentiating the ideal gas equa-tions. Starting with

pv = RT

and differentiating we get

p∆V + ∆p V = R∆T (A-3)

We now have to introduce the idea that the expansionis taking place adiabatically, i.e., that no heat is enter-ing. That means that the work p∆V done by the gasduring the infinitesimal expansion ∆V must all havecome from thermal energy. But the decrease in thermalenergy is CV∆T. Thus we have from conservation ofenergy

p∆V + CV∆T = 0

or

∆T =– p∆V

CV(A-4)

The – (minus) sign tells us that the temperature dropsas work energy is removed.

Using Equation A-4 for ∆T in Equation A-3 gives

p∆V + ∆pV = R

– p∆VCV

Combining the p∆V terms gives

p∆V 1 + RCV

+ ∆pV = 0

p∆V

CV + RCV

+ ∆pV = 0 (A-5)

Earlier in the chapter, in Equation 9, we found that foran ideal gas, CV and Cp were related by Cp = CV + R.Thus Equation A-5 simplifies to

p∆V

Cp

CV+ ∆pV = 0 (A-6)

It is standard notation to define the ratio of specificheats by the constant γ

Cp

CV≡ γ (A-7)

thus Equation A-6 can be written in the more compactform

γp∆V + ∆pV = 0 (A-8)

The next few steps will look like they were extractedfrom a calculus text. They may or may not be toofamiliar, but you should be able to follow them step-by-step.

First we will replace ∆V and ∆p by dV and dp toindicate that we are working with calculus differen-tials. Then dividing through by the product pV gives

γ dVV +

dpp = 0 (A-9)

Doing an indefinite integration of this equation gives

γ ln V + ln p = const (A-10)

The γ can be taken inside the logarithm to give

ln Vγ + ln p = const (A-11)

Next exponentiate both sides of Equation A-11 to get

eln Vγ + ln p = econst = another const (A-12)

where econst is itself a constant. Now use the fact that ea + b = eaeb to get

eln Vγeln p = const (A-13)

Finally use eln(x) = x to get the final result

pVγ = const

adiabaticexpansion

(A-14)

18-28 Entropy

Equation A-14 is the formula for the adiabatic curveseen in Figure (A-2). During an isothermal expansion,we have pV= RT where T is a constant. Thus if wecompare the formulas for isothermal and adiabaticexpansions, we have for any ideal gas

pV = const isothermal expansion

pVγ = const adiabatic expansion

γ = Cp /CV ratio of specific heats

Cp = CV + R

(A-15)

The Carnot CycleWe now have the pieces in place to calculate theefficiency of a Carnot cycle running on one mole of anideal gas. The cycle is shown in Figure (11) repeatedhere as Figure (A-3).

During the isothermal expansion from point 1 to 2, theamount of heat that flows into our mole of gas is equalto the work one by the gas. By Equation A-2, this workis

W12 = QH = RTH ln V2/V1 (A-16)

The heat QL expelled at the low temperature TL isequal to the work we do compressing the gas isother-mally in going from point 3 to 4. This work is

W34 = QL = RTC ln V4/V3 (A-17)

Taking the ratio of Equations A-16 to A-17 we get

QH

QL=

THln V2/V1

TCln V4/V3

(A-18)

The next step is to calculate the ratio of the logarithmsof the volumes using the adiabatic expansion formula

pVγ = constant .

In going adiabatically from 2 to 3 we have

p2V2γ = p3V3

γ (A-19)

and in going from 4 to 1 adiabatically we have

p4V4γ = p1V1

γ (A-20)

Finally, use the ideal gas law pV = RT to express thepressure p in terms of volume and temperature inEquations A-19 and A-20. Explicitly use

p1 = RTH / V1 ; p2 = RTH / V2

p3 = RTH / V3 ; p4 = RTH / V4

(A-21)

to get for Equation (19)

RTH

V2V2

γ =RTC

V3V3

γ

or

THV2γ – 1 = TCV3

γ – 1 (A-22)

and similarly for Equation (20)

THV1γ – 1 = TCV4

γ – 1 (A-23)

as you can check for yourself.

V1T1

T1

T3

T3

V1 V4 V2 V3

pressure

p1

V2p2

V3p3

V4p4

volume

compression

isothermal

adiabaticcom

pression

adiabatic expansion

isothermal expansion

p3

p4

p2

p1

Figure A-3The Carnot cycle.

18-29

If we divide Equation A-22 by A-23 the temperatures TH and TC cancel, and we get

V2γ – 1

V1γ – 1 =

V3γ – 1

V4γ – 1

or

V2V1

γ – 1

=V3V4

γ – 1

(A-24)

Taking the ( γ – 1)th root of both sides of Equation A-24 gives simply

V2V1

=V3V4

(A-25)

Since V2 /V1 = V3 /V4 , the logarithms in Equation A-18 cancel, and we are left with the surprisingly simpleresult

QHQL

=THTL

(14 repeated)

which is our Equation 14 for the efficiency of a Carnotcycle. As we mentioned, when you are doing acalculation and a lot of stuff cancels to give a simpleresult, there is a chance that your result is more general,or has more significance than you expected. In thiscase, Equation 14 is the formula for the efficiency ofany reversible engine, no matter how it is constructed.We happened to get at this formula by calculating theefficiency of a Carnot engine running on one mole ofan ideal gas.

18-30 Entropy

IndexAAdiabatic expansion

Calculation of work 18-26In Carnot cycle 18-11Introduction to 18-9

Applications of the second law of thermodynamics 18-17

BBoltzman

Formula for entropy 18-24

CCarnot cycle

As thought experiment 18-4Efficiency of

Calculation of 18-28Discussion 18-12Formula for 18-13, 18-29Reversible engines 18-18

Energy flow diagrams 18-15Entropy 18-22Introduction to 18-11Maximally efficient engines 18-15Refrigerator, energy flow diagrams 18-15Reversible engines 18-13Reversiblility 18-17Second law of thermodynamics 18-4

Cp = Cv + R, specific heats 18-7Cp and Cv, specific heats 18-6Cycle, Carnot 18-11

DDiagrams, PV (pressure, volume) 18-8Differential equation

For adiabatic expansion 18-27Direction of time

Dive movie 18-1Entropy 18-25Neutral K meson 18-25Rising water droplets 18-3

DisorderDirection of time 18-25Entropy and the second law of thermodynamics 18-

4Formula for entropy 18-24

Dive movie, time reversed 18-1

EEfficiency

Of Carnot cycle, calculation of 18-26

Of electric cars 18-19Of heat pump 18-19Of reversible engines 18-18

Electric cars, efficiency of 18-19Energy flow diagrams for reversible engines 18-15Engines

Internal combustion 18-21Maximally efficient 18-15Reversible, efficiency of 18-18

EntropyBoltzman's formula for 18-24Definition of 18-22Number of ways to hang tools 18-23Second law of thermodynamics 18-1

Expanding gas, work done by 18-5Expansion

AdiabaticCarnot cycle 18-11Equation for 18-26PV Diagrams 18-9Reversible engines 18-13

IsothermalCarnot cycle 18-11Equation for 18-26PV Diagrams 18-8Reversible engines 18-13

FFitch, Val, K mesons and the direction of time 18-27

GGamma = Cp/Cv, specific heats 18-7Gas, expanding, work done by 18-5

HHays, Tobias

Dive movie, time reversed 18-1Heat pump, efficiency of 18-19

IInternal combustion engine 18-21Isothermal expansion

Calculation of work 18-26PV Diagrams 18-8

MMaximally efficient engines 18-15Movie

Time reversed dive, second law of thermodynam-ics 18-1

OOrder and disorder

Direction of time 18-25Entropy and the second law of thermodynamics 18-

4Formula for entropy 18-24

18-31

PPower

1 horsepower = 746 watts 18-20Efficiency of a power plant 18-18

PressurePV diagrams 18-8

Adiabatic expansion 18-9, 18-26Internal combustion engine 18-21Isothermal expansion 18-8, 18-26Reversible engines 18-13The Carnot cycle 18-11, 18-26, 18-28

PV diagramsAdiabatic expansion 18-9Carnot cycle 18-11Isothermal expansion 18-8

RReversible engines

As thought experiment 18-13Carnot cycle 18-17Efficiency of 18-18

SSecond law of thermodynamics. See also Carnot

cycle; Thermal energyApplications of 18-17Chapter on 18-1Statement of 18-4Time reversed movie 18-1

Specific heatCp and Cv 18-6Gamma = Cp/Cv 18-7

TThermal efficiency of Carnot cycle. See Carnot cycle,

efficiency ofThermal energy

Dollar value of 18-1Time reversed movie of dive 18-1

Thermal expansionAdiabatic expansion 18-9

Derivation of formula 18-26Formula for 18-10

Isothermal expansion 18-8Derivation of formula for 18-26

Work done by an expanding gas 18-5Thought experiments

Carnot cycle 18-4Time

Direction ofEntropy 18-26Neutral K meson 18-26

Time reversalDive movie 18-1Water droplets 18-2

WWater droplets, time reversal 18-2Work

Calculation of in adiabatic expansion 18-26Calculation of in isothermal expansion 18-26Done by an expanding gas 18-5

XX-Ch18

Exercise 1 18-5Exercise 2 18-7Exercise 3 18-10Exercise 4 18-13Exercise 5 18-13Exercise 6 18-14Exercise 7 18-16Exercise 8 18-19Exercise 9 18-20Exercise 10 18-20