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General Physics

by: Talon Thompson

General Physics PHYS 2110

Talon Thompson
Utah State University
GPA 3.6

David Riffe

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David Riffe
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This 26 page Class Notes was uploaded by Talon Thompson on Wednesday October 28, 2015. The Class Notes belongs to PHYS 2110 at Utah State University taught by David Riffe in Fall. Since its upload, it has received 16 views. For similar materials see /class/230473/phys-2110-utah-state-university in Physics 2 at Utah State University.

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Date Created: 10/28/15
Class Notes 4 Phys 2110 Motion and Potential Energy Graphs I INTRODUCTION Often one is concerned with the motion of an object that is acted on by a single conserva tive force For example the object might be a satellite in orbit around the Earth where the single conservative force is the force of gravity on the satellite For such a force we can de ne a potential energy PE and the body speeds up and slows down in such a way that its total mechanical energy ME is conserved ie is constant In these notes we discuss how a graph of the potential energy vs position can be used to discern the motion of the body For con creteness we consider two different conservative forces one due to the gravitational pull of the Earth near its surface and the other due to an ideal spring There are two important concepts to keep in mind when using a potential energy graph to understand an object s motion 1 The negative ofthe slope ofthe graph is equal to the force on the object That is the force is the negative of the spatial derivative of the potential energy function 2 The difference between the total mechanical energy ME of the object which is constant and the potential energy PE equals the kinetic en ergy KE That is KE ME 7 PE II GRAVITY The potential energy associated with a mass m at some vertical height y above the Earth s surface is given by 1 In this expression yo can be any height for ex ample you can choose yo 1 m if you wish Above that point the potential energy increases and is positive below it the potential energy de creases and is negative Don t worry about the sign of the potential energy it s only changes in PE MW 7 yo Enerqy J FlGi 1 Potential energy mgy 7 yo of a 1 kg mass as height y 7 yo Energy 1 yryouu FlGi 2 Potential energy of a 1 kg mass illustrating the decrease in potential energy as the mass moves from point A to point B to point 0 potential energy that are interesting 7 and those can be either positive or negative Everywhere the slope of the graph is mg Thus the gravitational force on the mass is 7mg 777 because the force points down remember we de ned the positive direction to be up That is the gravitational force is given by F9 7th e 210 7mg lt2 Figure 1 shows the gravitational potential en ergy for a 1 kg object near the surface of the Earth Suppose we place such an ogject 075 m above yo point A on Fig 2 and let go We know that gravity will accelerate the object downwards with acceleration 9 As it does so it will lose potential energy PEg and gain kinetic energy KE in such a way that its total mechan ical energy ME remains constant This motion can be deduced by looking at the potential energy graph using the two con cepts outlined above Initially KE 0 and from the graph we see that PE 75 J Therefore ME 75 J However at point A there is a force acting because the slope is not zero The object will thus accelerate in the direction of the force which is negative downwards As long as there are no other forces the objects s PE will decrease as it falls and its KE will increase At point B y 7 yo 025 m for example PEg has dropped to 25 J thus KE has increased to 50 J so that KE PE 75 J the total me chanical energy has remained the same If the object can fall below the y yo mark don t be prejudiced yo doesn t have to be at the surface of the Earth it can be anywhere PEg becomes negative At C y 7 yo 705 m for example PEg equals 75 J so the KE must equal 125 J that is 125 J 7 5 J 75 J Until the Earth gets in the way not pictured the object continues to lose PEg and gain KE III THE SPRING FORCE The potential energy associated with a mass attached to a spring depends on how much the spring is stretched or compressed Let the equi librium length of the spring be L0 and it s cur rent length be L L can be greater stretch or less compression than L0 The potential energy associated with a spring that is either stretched or compressed is 1 PEEP 5k L 7 L02 3 where k is the spring constant 7 a quantity that measures the stiffness of the spring k is measured in Nm The value of k depends on what the spring is made of and how loosely or tightly coiled the spring is We usually replace the difference L 7 L0 by the variable m so that the potential energy is expressed as V 1 PEEP EM 4 Energy J FIGI 3 Potential energy ofa spring 12kx2 VS position x Lower curve C 40 N m upper curve k 200 Nm Energy J m x m FIGI 4 Potential energy of a spring illustrating the Change in potential energy as the mass moves from A to B to 0 As discussed above the negative of the slope derivative of the potential energy is the force From this equation for PESp we thus have 1 1 2 7E gt 7 7km Figure 3 shows two potential energy graphs one for an object attached to a spring with spring constant k 40 Nm lower curve and one for an object attached to a spring with spring con stant k 200 Nm upper curve In contrast to gravity where the force is the same at every position height for a spring the force some times points to the right for negative values of z 7 compressions and sometimes to the left for positive values of z 7 stretches The mag nitude of the force is zero equilibrium when x 0 no compression or stretch and grows as the magnitude of z grows F917 5 Now let s see how motion works for an ob ject attached to a spring The potential energy for an object attached to a spring is pictured in Fig 4 for a value of k 200 Nm We start by stretching the spring by 007 m point A If the object is released from rest it has a to tal mechanical energy of 049 J At point A the potential energy has a positive slope so the object will experience a force in the negative di rection lt accelerates and as it does so it gains KE and loses PEEP At point B m 005 m the objects s PE is 025 J and its KE must be 024 J As the object passes through z 0 the position of no stretch it experiences instan taneously no force the PE slope is zero there but it is moving with a KE of 049 J lt con tinues past z 0 and begins to compress the spring slowing down as it does so Finally at z 7007 m C the object has lost all of its KE and PESp is back to 049 J At that in stant the object is at rest but it feels a force pushing to the right the PE slope is negative so the force points in the positive direction so it begins to accelerate to the right Eventually the object gets to z 007 m again and the Whole process repeats over and over Thus the object oscillates back and forth with kinetic and potential energy constantly being exchanged as it oscillates Class Notes 1 Phys 2110 The Standard Model and the Structure of Matter 1 The Physicist7s Periodic Table The ancient Greeks are credited with the idea that the objects of the natural world are com posed of smaller indivisible particles Indeed the word atom is derived from the Greek word atomos which means undivided This concept of matter being composed of very small elemen tary particles persists today The latest and highly veri ed theory of fundamental particles is known as the Standard Model which is a complex mathematical theory that describes both the elementary particles and the interac tions or forces between those particles The elementary particles that compose mat ter are divided into two classes leptons and quarks There are 6 basic leptons listed in the following table The electron is probably the most familiar to you Except for their masses the muon and tau lepton are identical to the elec tron Associated with each of these three leptons is a particular neutrino Neutrinos are the light est of the elementary particles that have mass more about massless particles later electron e electron neutrino Ve muon a muon neutrino 1m tau 739 tau neutrino 17 There are also 6 basic quarks listed in the next table charm quark ltcgt down quark d bottom quark b Now these twelve elementary particles can be arranged into an elementary particle periodic table of sorts This table consists of three gen erations of four particles each shown below Each generation consists of two leptons one electron like lepton and its associated neutrino and two quarks Generally within each gener ation the neutrinos are the lightest the leptons the next lightest and the quarks the heaviest As one moves from left to right from one gen eration to the next the respective particles are essentially identical except that the respective particles become heavier For example the u c and t quarks are nearly identical for example all have the same electric charge of 23e but the t is heavier than the c which is heavier than the u The neutrinos are an interesting set of par ticles First because they only interact weakly see below with other particles they do not par ticipate in the construction of atoms However their abundance is such that they make up a substantial fraction of the mass of the universe An important fact about this periodic table is that only the particles in the lightest family are stable neutrinos excepted That is aside from the neutrinos only the electron up quark and down quark are stable against decay What this means is that because they do not live very long the heavier quarks and leptons are also not important in the construction of everyday mat ter These heavier particles have only been ob served in high energy collisions in which they are created for a brief amount of time after which they decay into other particles For example if a muon a is created in a high energy collision it will decay with a probability of 099 into an electron a neutrino and an antineutrino within about 2 x 10 6 second Note every particle also has associated with it an antiparticle counter part which is identical except for a change in sign of some of the quantities such as electric charge associated with the particle For exam ple a positron or positively charge electron is an antielectron 2 Fundamental Forces Each of the elementary particles listed above has a set of intrinsic properties mass electrical charge weak charge and color Each of these four intrinsic properties results in an interac tion or force between the particles On the most fundamental level these forces between the particles arise through exchange of other par ticles which are generally known as bosons There are 4 basic types of interactions and thus four types of exchange particles as listed below For example two particles with mass inter act gravitationally by exchanging gravitons77 while two electrically charged particles interact electromagnetically by exchanging photons77 Quarks interact via the strong force by ex changing gluons and since they are electrically charged and have mass they also interact elec tromagnetically and gravitationally lnterestingly while the W and Z bosons have mass gravitons photons and gluons are mass less One important consequence of a particle having no mass is that it always travels at the speed of light In nature ie experimentally no one has ever seen a free quark that is a quark all by itself Quarks either come in quark antiquark pairs that are bound to each other these pairs are generically called mesons or quark triples which are generically known as baryons Be cause the strong force is so strong when one has a quark antiquark pair or triple of quarks the quarks cannot be separated 3 The Atomic Nature of Matter OK so the question is now how is everyday matter constructed out of electrons up quarks and down quarks As you should already know any atom is composed of a nucleus surrounded by some number of electrons Each nucleus is composed of protons and neutrons A proton is composed of 1 d and 2 u quarks a triplel while a neutron is composed of 2 d and 1 u quark another triple The strong force not only holds the quark triples together to form neutrons and protons it also hold the neutrons and protons together to from the nucleus More exotic types of matter which are com posed of different combinations of quarks and antiquarks are again only found shortly after very high energy collisions This is because 1 it takes a tremendous amount of energy to create these other particles and 2 again these heavier particles decay into other stable particles very quickly It is one of the most amazing facts of nature that essentially everything in the world around us is made from fewer than 100 naturally occur ring di erent kinds of atoms In atoms that are electrically neutral the number of electrons equals the number of protons in the nucleus An element is some material that consists of atoms all of which contain the same number of protons Thus atoms with one proton are said to con stitute the element hydrogen atoms with two protons constitute the element helium While atoms of a given element all have the same num ber of protons in their nuclei they may have dif ferent numbers of neutrons Two atoms with the same number of protons but different num ber of neutrons are said to be different isotopes of the same element Different isotopes behave almost identically as far as chemical reactions are concerned because chemical reactions involve atomic electrons only not the atomic nuclei Protons and neutrons both weigh about 2000 times more than electrons So most of the stuff77 of an atom resides in its nucleus Nonetheless atoms are mostly empty space The most common isotope of hydrogen consists of one proton and one electron Suppose we represent the proton in a hydrogen atom by the follow ing dot About how far away from this dot would the electron be on average if this dot were the actual size of the proton Where the period next to the dot is Maybe a centimeter 10 cen timeters A meter No Actually the electron would spend most of its time roughly 100 meters away The average diameter of the electron or bit in hydrogen is about 100000 times the diam eter of the proton ln atoms with more protons the electrons spend more time nearer the nuclei but no matter how many protons and electrons they contain atoms are mostly empty Despite that it is very hard to squeeze the electrons of an atom closer to their nucleus It is also difficult to make the electrons of two atoms interpenetrate If that weren t true it would be impossible for objects to have moreor less permanent shapes and sizes 4 States of Matter The number of atoms in a macroscopic object may well exceed 1023 The interactions of these vast swarms of atoms lead to different states of matter In all materials at all temperatures the constituent atoms are in ceaseless disorganized 7 thermal 7 motion In solids the microscopic agitation of atoms is sufficiently con ned that the atoms typically do not exchange places As a consequence solids have an essentially perma nent shape In uids 7 that is gases and liq uids 7 however atoms can pass by each other This swapping of atomic positions produces the macroscopic phenomenon of ow and the mi croscopic phenomenon of diffusion or atomic mixing Fluids ow around inside solid contain ers and adopt shapes de ned by the containers Fluids don t have a permanent shape At the atomic level solids are characterized by the regular and enduring long range order of their atoms Liquids are characterized by or der only at very short range and by disorder on any length scale longer than a few atomic dis tances Gases are characterized by disorder on any length scale Biological materials typically share features of both the solid and uid states For exam ple biological membranes that surround cells or sub cellular components are basically two dimensional highly ordered structures that also have a large degree of mobility within them Their constituent phospholipid molecules tend to be aligned parallel to each other but can move about within the plane of the membrane quite rapidly by diffusion Such highly ordered but yet uid structures are termed liquid crys tals A second signi cant example is the gel like nature of cellular cytoplasm Gels have some of the properties of solids including a rigidity but can be greatly deformed as well Cytoplasm is a complex material consisting of thousands of different macromolecules including proteins nucleic acids phospholipids polysaccharides as well as smaller organic molecules and salts Un der the control of several lamentous proteins that supply an internal structural rigidity the cytoplasm can be changed back and forth be tween conditions that are either more uid like or more solid like 5 Mass Density and the Sizes of Atoms Mass is a fundamental property of matter For now it is sufficient to think of mass as a measure of the substance of a body The Sl unit of mass is the kilogram kg A kg is roughly the mass of a rock that is the size of a grapefruit A kg weighs about 22 pounds It is possible to determine the masses of individual atoms with a delicate scale called a mass spectrometer Compared with a rock an atom doesn t have very much mass A rule for determining the mass of a par ticular type of atom is to look up the number of atomic mass units per atom77 designated uatom for the atom of interest on a periodic table then multiply by 166 x 10 27 kgu The number of atomic mass units per atom is essen tially the average number of protons plus neu trons in all isotopes of the element in question found on Earth We now show how knowledge of macro scopic properties can be converted into knowl edge about atoms Let s start with the question how many atoms are contained in a 1 kg mass of known composition Suppose for example we are told that the mass is solid gold A periodic table tells us that gold has about 197 uatom So the mass in kg of a gold atom is 197uatom x 166 x 10 27kgu 327 x 10 25kgatom The 1 kg mass of gold is some number of atoms times the mass per atom so if we divide the latter value into 1 kg we nd that 1 kg of gold contains 1kg327 x 10 25kgatom 3 x 1024atoms That s a typical number for solids 1 kg of a solid contains from about 1024 to about 1026 atoms A most interesting question is how far apart are the atoms in a solid or liquid The macro scopic measurement of density allows us to an swer that question The density p is de ned as massvolume If we divide the density of a solid body by the mass per atom we obtain the num ber of atoms per unit volume If we take the reciprocal of that we get the volume per atom Now ifwe pretend that each atom is a little cube of side do the volume per atom is d8 Thus do is the cube root of the volume per atom it is also the average distance between adjacent atoms Let s do an example For the element iron the number of atoms per unit volume is these days to use the nanometer 1 x 10 9 m 1 nm Thus 226 x 10 10 m can be written as either 226 A or 0226 nm The average distance between atoms in any elemental solid is roughly the same as for iron Now here is a very familiar result it is ex ceedingly dif cult to increase the density of a solid by squeezing it In other words in a solid the atoms are crammed together about as closely as possible This fact and the fact that the aver age spacing of atoms is about 02 to 03 nm for all elemental solids tells us the very interesting and surprising result that all atoms are about the same size 7 despite the fact that their atomic masses vary by a factor of over 200 You might be tempted to conclude that be cause liquids ow and have no permanent shape that the spacing of atoms in liquids would be a lot larger than in a solid Let s see How about in water Water is a molecular liquid The umolecule for water is about 18 2 for the two hydrogen atoms and 16 for the oxygen atom Since there are three atoms per molecule the average mass per atom is 6 u The density of water is 1000 kgm3 Consequently the aver age molecular spacing in water is about 022 nm 7 more or less the same value as in solids The remarkably different physical properties of solids and liquids arise from only very small differences in how their atoms are spaced What can we say about atomic spacing in 7900kgm3 558uat0m X 166 X 10727kgugases7 The most familiar gas is air a mixture of 853 x 1028atomsm3 The volume per atom in solid iron is then 13 853 x 1028atomsm3 1 116 x 10 29m3atom and the cube root of that do 226 x 1010 m is the average atomic spacing The distance 10 10 m recurs frequently when considering atoms You will sometimes nd 1 x 10 10 m referred to as 1 angstom 1 A though in keep ing with the SI conventions it is more fashionable primarily nitrogen and oxygen molecules Let s say that the average umolecule for air is about 29 Because nitrogen and oxygen molecules con tain two atoms the average uatom for air is about 145 The density of air at room temper ature and at sea level atmospheric pressure is 129 kgm3 a value that is something like 1000 times less than water The average atomic spac ing in air is about 27 nm that is about 10 times greater than in a solid or liquid If we squeeze a quantity of air down to 11000th of its nor mal volume it becomes a liquid the densities of liquid oxygen and liquid nitrogen are almost exactly 1000 times that of air Because biological materials have properties midway between the solid and liquid states the spacing of atoms in them is also about 02 to 03 nm We can use this to assess how many atoms one might nd in a typical biological cell Cells have somewhat different sizes but a typical cell is roughly about 20 x 10 6 m 20 micrometers 20 pm on a side That is a cell has a volume roughly about 8 x 10 15 m3 obtained by cubing 20 pm If a typical atom spacing is 025 nm the volume occupied by an atom is about 025 nm3 15 x 10 29 m3atom Consequently the number of atoms per cell is about 8 x 10 15m3cell 15 x 10 29m3atom 5 x 1014atomscell A cell has lots of stuff in it All cells contain DNA for example Drawings of pieces of DNA in textbooks show it as a long double helix struc ture But just how long is it DNA consists of multiple subunits called base pairs C G77 and A T The number of atoms per pair is 27 Typical animal cells have about 5 x 109 pairs in their DNA which corresponds to about 14 x 1011 atoms Suppose that all of the atoms in the DNA molecule were strung end to end in a linear chain The chain would be about 14 x 1011atoms x 25 x10 10matom 35m longl Of course clumping atoms into base pairs of about 30 atoms each saves space Even so if the pairs were strung out in a linear chain the DNA would still be about 1 m long Obviously DNA in a cell can t be a linear chain because it would burst through the cell membrane It must be stored in a tight coil when not in use77 and only small portions must be pulled apart when transcription or replication occur Similar conclusions can be made about other important ingredients of a cell such as large proteins for example Class Notes 8 Phys 2110 Heat Engines and the Second Law of Thermodynamics I INTRODUCTION The science of thermodynamics was born from the realization that microscopic energy such as the internal kinetic energy in an ideal gas for example can be used to do macroscopic useful work W One of the rst industrial appli cations of this science was the development ofthe steam engine Today we understand that ther modynamics governs the useful extraction of mi croscopic energy in all cases including the work done by an internal combustion engine or the work done by a biological system such as the human body We have already studied the rst law of thermodynamics AUUf7UQ7W 1 which tells us that the change in internal en ergy U of a system is equal to the heat ow Q into the system minus the work W done by the system By considering the rst law you might come to the conclusion that we could con tinually extract useful work W in the following manner First we connect our system to an ex ternal reservoir that is at a temperature that is higher than that of the system1 Because the reservoir is at a higher temperature heat Q ows from the reservoir into the system At the same time we construct the system so that it can con tinually do work W without having its internal energy U change Thus from the rst law Eq 1 we would have W Q which implies that any amount of heat that we extract from the reservoir can be used to do useful work The con servation of energy is satis ed and everything is hunky dory Well in fact no one has ever gured out how to construct such a system In reality the heat extracted from the high temperature reservoir can never be fully utilized to do sustained use ful work There is always a certain amount of heat Q that must be removed from the system to another lower temperature reservoir as we M u T m I PRESSURE M n 2 n 25 n 3 VOLUME mm FIG 1 PV diagram for an ideal monatomic gas Path 1 is along an adiabat path 2 along an isotherm Along paths 3 and 4 the temperature is increasing and heat is owing into the system Along paths 5 and 6 the temper ature is decreasing and heat is owing out of the system shall see in any such system that can do sus tained useful work The combination of 1 a high temperature reservoir 2 low temperature reservoir and 3 the system that does the work on the outside world is generically known as a heat engine The heat that must be re moved from the system is know as waste heat In a biological system such as the human body this waste heat is partially responsible for the increase in body temperature that accompanies physical exertion II HEAT ENGINES To understand the basis of the claim that we cannot convert all heat to sustained useful work lets let the system be an ideal monatomic gas in discussing heat engines the system is also known as the working substance of the heat engine In particular let s study the P V diagram for a monatomic gas as shown in Fig 1 Recall the series of solid lines in the diagram are isotherms while the dotted lines are adiabats For exam ple if we let the gas expand or contract in such a way that it follows one of the solid lines then its temperature remains constant If it moves along one of the dotted lines then there is no heat ow into or out of the system Conversely if the sys tem is allowed to expand or contract so that it crosses the solid lines then its temperature will be changing For example if the system expands along an adiabat as shown by path 1 then its temperature decreases along that path since it is moving from a higher temperature isotherm to a lower temperature isotherm Similarly if the system expands or contracts such that it crosses the adiabats heat will either be owing into or out of the system For example along path 2 heat is owing into the system If the system moves generally from left to right or bottom to top path 3 or 4 eg the temperature increases and the heat ow is into the system Conversely if the system moves generally from right to left or top to bottom path 5 or 6 eg the temper ature decreases and the heat ow is out of the system The other thing that we must keep in mind about the P V diagram is that if we move the system between two points on the diagram then the work W done by the system is the area under the curve between the two points The work is positive if the second point is at a larger volume V than the rst point while the work is negative if the second point is at a lower V Looking at Fig 2 ifwe move from point A to point B along path 1 the system does positive work Note that we could also move from point A to point B along path 2 The system would still do positive work but it would be less than that along path 1 This is because the area under path 2 is less than that under path 1 Let s now think about how we can make a useful heat engine out of our ideal gas Consider the P V diagram in Fig 3 We could start the system at point A in this diagram and let it ex pand along an isotherm as indicated by path 1 In fact along path 1 we do have total conversion of Q let7s call this Qm since this is heat that is owing into the system along the isotherm to W since AU 0 However we have not achieved M u T m message m nus m 15 n 2 n 25 n 3 VOLUME mu FlGi 2 PV diagram for an ideal monatomic gas Along path 1 the work done by the system is greater than the work done along path 2 Because point B is at a greater Volume than point A the work W is positive in both cases very much the gas is sitting at a larger volume In order to get any sustained net work out of the system we must get the system back to a smaller volume state so that it can reexpand and do more work for us We could simply have the sys tem move back along path 1 but the work done by the engine on this reverse path is the nega tive of that just done by the engine and thus the net work done by the heat engine would be zero However we can move the system back to point A along a different path For example we could move it back to point A along paths 2 and 3 In this case the net work in starting out at A and moving along paths 1 2 and 3 back to A is positive Why is this Well remember that the work done in moving between any two points is the area under the curve The area under path 1 when the system is doing positive work is greater that the area under path 3 when the system is doing negative work so the net work done by the system is positive We can now run the engine around this thermodynamic cycle as many times as we want and extract as much useful work as we need One more point since the net work is the area under path 1 minus the PRESSURE m n I n 15 n 2 n 25 n 3 VOLUME mm FIG 3 P V diagram for an ideal monatomic gas Path 1 is along an isotherm Path 2 is along an isochor Path 3 is along an adiabat Heat ow in occurs along path 1 Heat ow out Qout occurs along path 2 The net work in one thermodynamic cycle equals the area inside the closed loop area under path 3 the total net work in one cy cle is simply the area enclose by the cycle This is true for any cycle Of course the system must move around the closed path clockwise 1f the system moves around the cycle counterclockwise then the outside word is doing the net positive work on the system This is the basis of re frigeration which we brie y discuss below Now back to our statement about waste heat which we now designate as Qom since it is heat that leaves the system Notice in Fig 3 that along path 1 heat is owing into the system since the system is moving from left to right across the adiabats remember there are actually an in nite number of adiabats only a few have been drawn The only way that we could get back to point A from point B without throwing away any heat would be to move along an adiabat from point B to A but this is impossible In fact any path that gets the system back to point A must cross the adiabats in the right to left or top to bottom direction which means that some amount of heat Q0 must leave the system2 For the cycle shown the waste heat Q0 is ex tracted along path 2 since path 3 is along an adiabat The other important fact about any cycle where the system does net work is that the waste heat Q0 leaves the system when the system is at least on average at a lower temper ature than when Qin ows into the system This can be seen for the cycle in Fig 3 by noting that Qm occurs while the system is at a constant tem perature along path 1 while Qom occurs as the system is cooling along path 2 Thus in mov ing from point B to point C along path 2 the system must be hooked up to a reservoir that is at a lower temperature than that when the sys tem was moving along path 1 or else Qom would not ow out of the system Summarizing a heat engine is a system that moves along a thermodynamic cycle extracting Qin from a high temperature reservoir expelling waste heat Q0 to a lower temperature reservoir and in the process doing useful net work W Because Uf U around a complete ther modynamic cycle the rst law tells us that W Qm 7 Qom The efficiency 5 of the en gine is de ned as the ratio of W to Qm around a complete cycle That is e W17Qout39 Qm Qm Because Q0 7 0 the efficiency can never be equal to 1 Therefore no heat engine is 100 efficient III SECOND LAW OF THERMODYNAMICS That the efficiency of a heat engine can never be 100 is the essence of the second law of thermodynamics although there are many other different but equivalent statements 0fthis law Lets look at two of the more common state ments of this law One statement of the law is one cannot devise a thermodynamic cycle whereby heat is converted to work at a single temperature However keep ing the system at a single temperature would mean that the system would stay on the same isotherm As we have seen in our discussion of Fig 3 no net work can be done in that case Another statement of the second law is with out external work done on the system heat only flows from a hot to a colder substance This is equivalent to the statement in the last para graph as the following example shows If we could convert heat to work at a single temper ature then we could devise a heat engine that did net useful work with only heat input at that temperature the mechanical work could then be converted to internal energy through friction at some higher temperature location and this in ternal energy could then placed into a higher temperature reservoir The net result would be the ow of heat from the lower temperature heat engine input reservoir to a higher temper ature reservoir without any external work on the system In fact we can move heat from a lower temperature reservoir to a higher temperature reservoir it just takes external work on the sys tem Refrigerators and heat pumps are exam ples of devices that indeed transfer heat from a cold temperature location to a high temperature location 7 but they certainly do not do this spon taneously External work must be performed on the working substance of these devices in order to perform this heat transfer IV ISOCHOR 7 ISOBAR HEAT ENGINE Any particular type of heat engine is de ned by the paths that compose its thermodynamic cycle For example a fairly simple heat engine to think about is the isochor isobar heat engine illustrated in Fig 4 The net work done for this cycle is easy to calculate because it is a rectan gle W AP AV where AP P1 7 P2 and AV V2 7 V1 are the lengths of the vertical and horizontal sides of the rectangle respectively 3 By noting which direction the paths cross the adiabats we deduce that along paths 1 and 4 heat ows into the system and along paths 2 and 3 heat ows out of the system Although it is a bit tricky to show the average temperature during the Qm paths is larger than the average temperature during the Q0 paths Let s see what we can say about the efficiency of an isochor isobar engine To do this we need M u T Nquot I2 25 u u voumumnz V FIG 4 P V diagram for an ideal monatomic gas show ing an isochor isobar thermodynamic cycle Paths 1 and 3 are along isobars Paths 2 and 4 are along isochors Heat ow in occurs along paths 1 and 4 Heat ow out Qom occurs along paths 2 and 3 The net work W in one thermodynamic cycle equals the area inside the closed loop APAV to know two of the following three items W Qin and Qom Along the isochors W 0 while along the isobars the work is simply P time the change in volume which is AV along path 1 and 7AV along path 3 As we discussed in class along an isobar Q 52NkB AT Combining this with PAV NkB lATl yields lQl 52 PAV Similarly along an iso chor Q 32NkB AT Combining this with VAP NkB lATl yields lQl 32 VAP The results for all four paths are summarized in the table below Using these expressions the efficiency can now be calculated from Eq 2 as W e7 Qm AP AV 5P AV 3 3 E 1 V1AP The right hand side can be rearranged to express the efficiency as 4 This equation actually tells us something very interesting First the ef ciency can be made as large as possible if 1 V1 is as small as possible and V2 as large as possible so that VlAV z 0 and 2 P2 is made as small as possible and P1 as large as possible so that P1 z AP In these limits we come close to the maximum ef ciency of 25 40 for an isochor isobar heat engine That is a lot of waste heat at least 60 of the input heat but we could probably live with that the heat engine can still do a considerable amount of useful work V ISOTHERMADIABAT HEAT ENGINE Perhaps the most famous heat engine is the isotherm adiabat heat engine also known as the Carnot cycle heat engine named after the rst person to write down the second law of ther modynamics Sadi Carnot The Carnot cycle is illustrated in Fig 5 Starting at A the system is rst expanded isothermally at a high temper ature TH path 1 the Qm phase and then adia batically path 2 It is then compressed isother mally at a low temperature To path 3 the Q0 phase and then adiabatically path 4 back to point A While it is dif cult to build a real Carnot engine due to its oblique P V diagram it is conceptually simple since there is heat ow only along the isotherms Thus both the high temperature and low temperature reservoirs can each be held at a xed temperature the hot reservoir just above TH and the cold reservoir just below To Furthermore this engine has the property that QWtQm TcTH so that the ef ciency can written as T70 TH E 7 5 M u T Nquot GAS PRESSURE m nus n1 n1 n2 n2 n3 VOLUMEmquot3 FIG 5 P V diagram for an ideal monatomic gas show ing an isothermadiabat Carnot thermodynamic cycle Paths 1 and 3 are along isotherms Paths 2 and 4 are along adiabats Heat ow in occurs along path 1 Heat ow out Qom occurs along path 3 The net work W in one thermodynamic cycle equals the area inside the closed loop Keep in mind that these two temperatures are absolute temperatures In principle if the cold reservoir were at absolute zero then the ef ciency would indeed be 100 However one can never reach absolute zero this is essentially the third law of thermodynamics so that no heat en gine is 100 ef cient However Eq 5 clearly shows that the smaller the ratio of To to TH the more ef cient will be the engine VI REFIGERATORS AND HEAT A refrigerator or heat pump is simply a heat engine that is run in reverse reverse meaning that the thermodynamic cycle is traversed coun terclockwise Consider the cycle in Fig 3 If we run the system around this cycle counterclock wise then the outside word does net work on the system Furthermore path 2 is now the Qm path and path 1 is the Q0 path Remember path 2 is at a lower average temperature than path 1 so we have heat owing in from a low temperature reservoir and heat owing out to heat pump a high temperature reservoir 7 a refrigerator or 1 We assume that the reservoir is so large that the a slightly lower temperature than the systemi heat ow from or to it does not change its temper 2 The standard convention for heat ow is that Q aturei However we do not need to assume that is positive for heat ow in and negative for heat the reservoir is at a constant temperature As we ow out However when discussing heat engines will nd out when we discuss entropy to make the we let both and Q01 be positive but keep in engine as ef cient as possible we want the tem mind that always ows in and Q01 always perature of the reservoir to be only slightly dif ows out ferent in temperature from the systemi For heat 3 Keep in mind that throughout this discussion ow in the reservoir will be at a slightly higher temperature than the working substance For heat ow out Qout the reservoir will be at both AP and AV as de ned are positive Class Notes 3 Phys 2110 Drag Forces I THE IMPORTANCE OF DRAG FORCES In introductory discussions of an object mov ing through the Earth s atmosphere the resis tance of the air on the motion of the object is usually neglected The main reason for this ne glect is that the equation of motion for the object Newton s second law is then fairly sim p187 m6 7mg 1 V This equation of motion includes only the gravitational force magnitude mg Because of the simplicity of Eq 1 one can exactly solve for the position Ft and velocity 17t of the ob ject As it turns out the path of an object sub ject to this equation of motion is a parabola as illustrated in Fig 1 The solid curve in Fig 1 is the path that a baseball would take if it left Sammy Sosa s bat with an initial speed of 150 mihr at an initial angle of 45 degrees with re spect to the horizontal Notice that the ball is calculated to travel a distance of 1600 feet Even Sammy would have a difficult time actually hit ting a baseball such a distance What s wrong with the calculation As you have probably guessed the neglect of air resis tance results in a signi cant overestimation of the distance the baseball travels If the appro priate force due to air resistance is included in the calculation of the ight of the baseball then the dashed curve in Fig 1 results The result is still quite a long home run or long foul ball but we now have a realistic distance 500 ft for one of Sammy s hits Drag forces of which air re sistance is one example are generally important when a solid object moves through a uid either a liquid or gas Other examples include a bicy clist riding a bike a boat moving through water or the analytical technique of electrophoresis Here we discuss two equations that are often used to describe drag forces Keep in mind that the equations for these forces are not exact in the sense that for example Newton s law of gravi tational force is This is because they describe with only a few parameters the interaction of atoms at the rather complicated interface be tween the solid and the uid Still the equations are quite useful in a variety of circumstances II LOW VELOCITY HIGH VISCOSITY Let s now consider an object moving at relatively slow velocity slow will be de ned later through a fairly Viscous think sticky medium A good example might be a small metallic sphere such as a BB moving through molasses For the moment we will neglect grav ity and assume that the object is subject only to the drag force A good approximation to the drag force of the viscous medium on the solid object is 131 7K 4m 0 2 Here K1 is a positive constant that depends upon both the object and the uid through which it moves What does this equation tell us It simply says that the drag force F1 is proportional to the velocity and that the force points in the direction opposite to the velocity so that the force always opposes the motion of the object Recall that 19 is a unit vector in the direction of the velocity This equation for the force is most ideally suited to describing an object with circular cross section moving through the medium For such an object the constant K1 is equal to K1 Cl RT where 01 is another constant B is the radius of the object perpendicular to the direction of mo tion and 7 is the viscosity of the medium The units of viscosity are kg m 1 s 1 or equivalently N s m For a sphere 01 6 Equation 3 explicitly shows how the drag force depends on the size of the object and the medium through 400 E g 200 0 39 I I I I 200 400 600 800 1000 1200 1400 1600 DISTANCE Without Air Resistance 39 39 39 39 With Air Resistance F GI 1 Flights of baseballs hit by Sammy Sosa7 rst without air resistance solid line and with air resistance dotted line which it moves For objects that are not circular parameter S is the cross sectional area perpen in cross section Eq 3 can still be used to good dicular to the direction of travel of the object approximation However7 the parameter R will units m27 and p is the density kgm3 of instead be some distance roughly half the size of the medium For an object with circular cross the object section such as a sphere Eq 5 can be written as III HIGH VELOCITY7 LOW VISCOSITY K2 CD n39RZp 6 Another situation commonly encountered is an Object moving fairly fast through a 10W where a is the radius of the cross section viscosity medium The drag force then takes a Equation 4 With K2 giVen by Eq 6 was form different than Eq39 An example Of such used to describe the drag force in the more re motion is a baseball moving through the air In alistic calculation of the baseball ight shown in this case the drag force is better represented as Fig39 1 a 1 2 I IV SO WHICH DRAG FORCE SHOULD F2 iiKg U U7 I USE Where K2 is a pOSitive COHStant that again de Because all normal uids have both density pends Upon prOpertles Of the SOhd ObJeCt and and viscosity both types of drag force act on the uid This equation says that the drag force I 7 I I I I I I I I any object mov1ng through a uid However7 in Is agam Opposme to the dlrecuon Of the V8100 many cases we only have to use one of the forces ity7 but this time the force is proportional to the I I I I I to describe the drag To see how this works we square of the velocity7 as opposed to being linear have made several graphs Of the magnitude Of in the velocity In this case the constant K2 is each force VS the speed U Of an Object shown in usually eXpressed as Fig 2 Notice that for very low velocities part a of the gure 7 check the m axis scale the 5 linear v drag force solid line is much larger than the quadratic I force dotted line7 but at very The parameter CD is known as the drag co large velocities the 122 force becomes signi cantly ef cient and is typically close to 057 but it does larger than linear v force7 as shown in part depend upon the exact shape of the object The K2 CDSp However7 there is always one speed where the which is known as the crossover speed If we use Eqs 3 and 6 for K1 and K2 in a 7 then for a sphere we can express the crossover speed as 12 03905 Uc39ross So if the speed of the object is much greater than 120059 then F2 can be used to describe the drag if the speed is much less than 1207055 then F1 can be used However if the velocity v is close to 1207059 then the sum F1 F2 must be used to accurately describe the drag on the object For one of your homework problems you will show for a baseball moving through air that the crossover speed is much less than any typical speed important to the game of baseball Thus the 122 force can be used to describe the drag on the home run balls that come off of Sammy s bat MAGNITUDE OF FORCE V GRAVITY AND DRAG FORCES Let s now consider the case as in the lab c 39 with the falling coffee lter where an object is 1000 l39 moving only in the vertical direction subject to 39 both the force of gravity and an appropriate drag force If we release the object from rest then ini 39 tially the only force on it is that due to gravity 500 39 quot However as the object speeds up the drag force 39 increases from zero until its magnitude equals 39 the force due to gravity Because the gravita 39 tional and drag forces point in opposite direc quot39 tions the net force on the object is zero Thus 0 20 40 the object no longer accelerates its velocity re SPEEWWS mains constant and it falls at a constant rate This is the condition of terminal velocity FIG 21 1717 SOlid line and 1727 0th line 7 5 Setting the force of gravity equal to the drag speed 2 Here K1 1N sm and K2 1N sm2 A A force in the general case where we must include The two forces are equal at the crossover speed vmoss Note the di erent xaxis scales in the di erent graphs bOth klnds Of drag forces glves us an equatlon that determines the terminal velocity Uterm of a freely falling object two forces are equal From Eqs 2 and 4 it can be shown the two forces are equal at a speed 1 of mg Klvterm l 5K2vtzerm39 9 If the object is a sphere then we can use Eqs K1 3 and 6 and substitute for K1 and K2 in Eq 2 727 7 g which yields 7T 2 CDPR39Ute39rm2 mg 67rntherm 10 As discussed above however we often only need to include one of the drag force terms For example if the linear v drag force is the important drag force then the second term on the right hand side of Eq 9 can be neglected Solving Eq 9 for the terminal velocity then results in m Ute39rm If the v2 drag force is the dominant drag force then the rst term on the right hand side of Eq 9 can be neglected resulting in 2mg Ute39rm VI ELECTROPHORESIS The technique of electrophoresis is often used to identify proteins or DNA sequences The ba sis for this technique lies in the difference in drag force that different molecules undergo in a uid In the technique the sample under investigation a particular protein say is placed at one end of the electrophoresis cell which contains the medium that provides the drag force An elec tric eld is applied that quickly accelerates the charged molecules to terminal velocity For a molecule with charge q unit Coulomb C the electric force on the molecule is given by 13 where E is the electric eld units N C l So instead of gravity and the drag force balancing it is now the electric force and the drag force that balance each other The terminal veloc ity depends on the size and thus the molecular weight of the molecule The molecules are al lowed to drift for a certain amount of time From the distance that they travel during that time their molecular weight can then be identi ed Class Notes 97 Phys 2110 Entropy and the Second Law I THERMODYNAMIC DEFINITION OF ENTROPY Entropy is a term that crops up from time to time in everyday conversations This is usu ally due to its intimate association with disorder Less commonly known is the association of an increase in entropy with a lessened ability to do work We will discuss both of these ideas here To begin we consider the relationship of entropy to our previous discussions of the rst and sec ond laws of thermodynamics and heat engines First of all7 entropy is a function of state or state function of the system7 similar to the internal energy U of the system For example7 if we know the volume V7 number of particles N7 and temperature T of an ideal monatomic gas7 then we know the internal energy7 which can be expressed as U 32NkBT Similarly7 if we know V7 N7 and T of the ideal monatomic gas we also know the entropy7 which can be written as U 3 V N N 0 Here C is a constant that depends upon the mass of the atoms that make up the gas This is all well and good7 but you are prob ably wondering how does entropy relate to the laws of thermodynamics The basic thermody namic de nition of entropy is as follows Sup pose the system which could be the working substance of a heat engine is at a temperature T and we add a small amount of heat AQ to it The entropy S of the system changes as a result of this heat ow and is given by S NkB 1n 1 V M2 7 T lt2 Note that if AQ is negative heat ow out then the entropy decreases Let s consider what Eq 2 means in terms of the P V diagram for an ideal gas Figure 1 shows the familiar P V di agram with isotherms and adiabats indicated as AS M u T m PRESSURE M I I I n 2 n 25 n 3 VOLUME mm FIG 1 PV diagram for an ideal monatomic gas Path 1 is along an isotherm Path 2 is along an isochor Path 3 is along an adiabat7 which has a constant Value of entropy Because entropy is a state function each point on the diagram has a particular Value of entropy solid and dotted lines7 respectively Recall7 along an adiabat the heat ow AQ is zero Equation 2 tells us then that the entropy S of the system does not change along an adiabat That is7 an adiabat is a constant entropy curve in the same way that an isotherm is a constant temperature curve The fact that the entropy is de ned by Eq 2 and that it is a state function given by Eq means that heat ow which is not a state function is7 in fact7 directly related to the change of a state function of the system 7 the change in entropy Practically speaking7 this now puts heat ow on the same level as the in ternal energy We can now think about the P V diagram with this understanding of entropy Keep in mind that because entropy is a state function7 each point on the P V diagram has a certain value of entropy S associated with it just as each point also has a certain value of U7 T7 P7 and V Let s now suppose that we start with the system at point A in Fig 1 If the gas is expanded along an adiabat along path 3 from point A the entropy of the system remains con stant but the temperature decreases because the internal energy is being converted into work during the expansion If however the system expands along an isotherm along path 1 from point A the temperature remains constant but the entropy increases because heat ows into the system If the system completes a thermo dynamic cycle for example starting at A and following paths 1 2 and 3 back to A then the entropy of the system is the same as when it left point A precisely because the entropy is a state function Another way to say this is that because S is a state function for any thermody namic cycle starting at A and ending back at A S 0 Notice however for a thermodynamic cycle Q Qm 7 Qom is not necessarily zero because the net work is not necessarily zero re member W Q for a close path so heat ow is not a state function Now even though we have a simple relation ship between heat added and the change in en tropy 7 Eq 2 it is not necessarily trivial to use Eq 2 to calculate AS for any given path on the P V diagram This because for most paths the temperature is constantly changing along the path In such a case we must then resort to in tegral calculus and use the in nitesimal version of Eq 2 1S dQT However there are two types of paths where Eq 2 can be directly applied isotherms where T is constant and adiabats where S is constant II ENTROPY AND THE CARNOT CYCLE With this in mind let s think about the Carnot engine which consists of paths that lie along only either an isotherm or an adiabat Re ferring to Fig 2 because AS 0 around the whole path from A to A say and because AS is zero on paths 2 and 4 it must be that ASpathl iASpath 3 Thus from Eq 2 we have that PRESSURE m nus n1 m n 2 n 25 n 3 VOLUME mm FIG 2 P V diagram for an ideal monatomic gas show ing an isothermadiabat Carnot thermodynamic cycle Paths 1 and 3 are along isotherms Along path 1 the tem perature is TH along path 3 the temperature is To Heat ows into the system along path 1 QM heat ows out along path 3 Qom Paths 2 and 4 are along adiabats constant entropy curves Qin th TH To 3 where TH is the temperature along path 1 and To is the temperature along path 3 Rearranging Eq 3 we thus have th E Qin TH7 which is why the efficiency 1 7 QWtQm for the Carnot heat engine can be written as 17 TCTH 4 III THE ENTROPY FORM OF THE SECOND LAW OF THERMODYNAMICS When a Carnot engine runs around a com plete cycle no net entropy change occurs to the system working substance because the nal state equals the initial state and entropy only depends on the state of the system What about the entropy change of the system s surround ings In each cycle the high temperature reser voir loses some heat 762139 thus its entropy decreases Similarly in each cycle the entropy of the low temperature reservoir increases due to the heat ow Qom into that reservoir Using Eq 2 we can easily calculate the entropy change of the two reservoirs during a complete cycle as suming that the temperature of each reservoir is constant Let s assume that the hot reservoir temperature is constant at THE which must be larger than the system temperature TH along path 1 in Fig 2 else heat would not ow from the reservoir to the system and that the cold reservoir temperature is TOR Using Eq 2 we can thus write down the total entropy change of the uniuerse system surroundings for the complete cycle as AStotal Assystem l Assurroandings 7 th 7 Qout TH TO THE TOR Rearranging the right hand side of Eq 5 we can write the change in the entropy of the uni verse as 1 1 1 1 AStotal Qm mgtQ mt E i 6 This equation tells us something very impor tant Because THR gt TH and To gt TOR we must have that AStoml gt 0 That is the total entropy change of the universe increases for a complete thermodynamic cycle Whoal In fact we do not have to think about a com plete thermodynamic cycle to come to a simi lar conclusion Let s say we have two objects in thermal contact one at temperature Tim and one at temperature Todd and we let a small amount of heat AQ ow between them The entropy change of the universe is then 1 1 AStotal M i Again this must be greater than zero since Tim gt Tcold Thus for any spontaneous heat flow process between two objects the entropy of the universe increases This is one consequence of the entropy form of the second law of thermodynamics which can be stated as follows for any process the entropy of the uniuerse neuer decreases Processes where AS 0 are known as reversible because the universe can be put back into the same state by reversing the process as it was before the process occured Processes where AS gt 0 are known as irreversible because the the universe cannot be put back into its original state In other words let s say that an irreversible pro The entropy of the universe will necessarily have increased Because entropy is a state function the universe will obviously not be in the same state as it was before the process The second law then states that we cannot get back to the initial state of the universe because AS cannot be negative For the thermodynamic cycle situation de scribed by Eq 6 the only way that the process could be reversible so that AS 0 would be if THR were only in nitesimally in the mathe matical sense larger than TH and T0 were in nitesimally larger than TOR in other words if THR TH 1 1T and To TOR dT Then we could ignore the difference between TH R and TH and the difference between To and TOR in Eq 6 and the total entropy change of the universe would be zero However in reality reversibil ity is only a theoretical limit to what practi cally happens in any situation In all real world cases thermodynamic processes are irreversible and the entropy of the universe increases Note that the entropy form of the second law directly implies the heat ow form of the second law heat always spontaneously flows from a hotter to colder substance which we discussed in our previous heat engine handout Keep in mind that entropy can decrease in a particular region of the universe For the Carnot cycle as heat ows out of the system its entropy is decreasing However the increase in entropy of the low temperature reservoir is greater than the decrease in entropy of the system resulting is a net increase in entropy for the universe Similarly sometimes one hears that life ex ists in the face of the second law of thermody namics77 meaning presumably that the orderly arrangement of molecules required to make liv cess OCCUIS ing matter is a low state of entropy and that this somehow violates the second law Nothing could be further from the truth Consider for exam ple a beaker of water that has been supercooled to a temperature below 0 C That can be done if there are no dust particles and the beaker isn t vibrated The liquid water wants to become ice and will do so if given the slightest encour agement such as a gentle tap When the ice forms the atoms are arrayed in closely packed crystalline structure a beautifully ordered array Certainly the ice is in a lower state of entropy than when it was a liquid with its atoms free to move about helter skelter How did this happen if the second law says entropy has to increase The answer is that when the water freezes it re leases energy called latent heat and this heat ow to the surroundings increases the entropy of the surroundings If the net change in entropy of the universe is positive the process has the possi bility to occur Thus crystals and all other low entropy states are perfectly acceptable to the second law as long as their appearance increases the entropy of something else So is the entropy of the entire universe con tinually increasing Because thermodynamics was born from observations about systems that are on the length scale of a few meters interact ing with surroundings that are also typically a few meters in size it does make sense to ask if we look at a much larger system such as our galaxy for example does the second law hold From one perspective the question about the entropy of the entire universe is probably im possible to even ask because the entropy of the whole universe is probably not de nable neither is the total energyl However if we de ne the surroundings to be the region of space that is interacting with the system then it is probably safe to say that if the entropy of the system de creases even if the system is a galaxy or cluster of galaxies or something even larger then the entropy of the surroundings must increase by a greater amount Related to this discussion about the entropy of the universe it is certainly true that gravity tends to reduce entropy 7 by forming planets stars solar systems galaxies clusters of galax ies etc So if this order is appearing what part of the universe is paying for the order by becom ing disordered The answer at least partially lies in the gravitational eld itself As matter is organized into planets stars and so forth en ergy and entropy are produced in the gravita tional eld in the form of gravitational waves At present most gravitational theorists believe that the entropy increase partially present in the gravitational eld is greater than the decrease in the organized matter satisfying the second law IV ENTROPY AND ENERGY AVAILABLE TO DO WORK A consequence of this new form of the sec ond law is that a heat engine that utilizes irre versible steps that is any real engine must be even less ef cient than a heat engine that oper ates reversibly which already cannot be 100 ef cient This can be seen by thinking about the Carnot cycle illustrated in Fig 2 Let s as sume that we have a hot reservoir at THE and a cold reservoir at TOR which are both xed temperatures The most ef cient engine that can utilize these two reservoirs is the reversible engine where TH is just in nitesimally below THR and To is just above TOR Since the ef ciency of the engine is give by e 1 7 THTc the reversible engine has an ef ciency equal to e 1 7 THETOR Any other engine that can utilize these two reservoirs must have TH lt TH R and To gt TOR Therefore TcTH is larger for any irreversible engine operating between these two reservoirs and the ef ciency is smaller than for the reversible engine This is a generic result whenever entropy in creases the amount of microscopic energy avail In other words an increase of entropy degrades the usefulness of microscopic energy A very instructive example of this result is the so called free expansion77 of a gas Sup pose an ideal gas is enclosed in a container with rigid thick insulating walls Initially the gas is in equilibrium and is con ned to half the con tainer say by a removable partition The other half of the container is evacuated Suddenly the able to do work is decreased partition is removed and the gas rushes freely into the vacuum Of course this process leads to turbulent swirling of the gas during which pres sure density and temperature rapidly change from placeto place in the container That is during this turbulent phase the thermodynamic properties of the gas are not well de ned so that its changing state does not even exist on the P V diagraml Eventually however the magic of internal equilibrium asserts itself and the gas reestablishes a state of thermodynamic equilib rium with well de ned T P V etc The rapid expansion of the gas is highly irreversible Cer tainly merely reintroducing the partition will not cause the gas to spontaneously compress back to its original state Nonetheless the gas has well de ned initial and nal states so Uf 7 U and Sf 7 S can be calculated by connecting the initial and nal states with any process including a reversible process We rst argue that Uf 7 Ui is zero for free expansion This is because during the ac tual intervening events the gas moves no walls so no work is done and the walls are insulating so no heat ows We assume that the parti tion is so light that removing it doesn t cost any energy If the gas is ideal its internal energy depends on temperature only so the tempera ture of the nal state must equal the tempera ture of the initial state Using PV ngT we can state that if PiT and V are the gas s ini tial thermodynamic properties then Pf Pi2 Tf Ti and Vf 2 are its nal ones There are in nitely many processes that could connect these two states The most direct perhaps is a reversible isothermal expansion By looking at the P V diagram we know that isothermal expansion is accompanied by an increase in en tropy Thus the entropy of the freely expanded gas must have increased We can use Eq 1 to calculate this increase in entropy Using lnAB lnA lnB we can rewrite Eq 1 as S NkB mm 111 so that gas ready to do work T gas will never do work T FlGi 3 Before and after removal of the partition that separates two halves of the container The gas is initially in the lefthand side of the container AS NkB 1nVfV NkB ln2 9 which is positive So how is this example of free expansion con nected with the degradation of the usefulness of energy Remember the initial and nal states have exactly the same internal energy But as we will now argue the nal states isn t as use ful as the initial state s Here s the story Go back to the initial unexpanded state Insert a paddle wheel contraption in the evacuated por tion of the chamber as shown in Fig 3 Pull the partition down halfway As the gas rushes into the vacuum it can turn the paddle wheel If the paddle wheel is connected to an electric gen erator for example its motion can produce an electric current That is the rushing gas can do useful work Of course in this case after it fully expands it won t be in the same nal state as before because some of its internal energy will have been transferred to the surroundings via work The gas will be cooler On the other hand suppose we let the gas freely expand as before Then insert into it a half pulled down partition and a paddle wheel as in the second panel of the gure In the ex panded state the gas has the same internal en ergy as in the initial compressed state but now it will never turn the paddle wheel The increase in entropy of the gas has resulted in a loss of usefulness of the internal energy V MICROSCOPIC DEFINITION OF ENTROPY As we discussed in the Section I the purely thermodynamic or macroscopic de nition of entropy states that 1 entropy is a state func tion and that 2 its change can be calculated via AS AQT Entropy can also be de ned from a microscopic viewpoint In fact as we see below entropy connects what we know macro scopically about the system to what we do not know microscopically about the system The microscopic de nition of entropy due to Gibbs is de ned as S k3 lnZ 10 where Z is the number of microstates consis tent with a given macrostate of the system What does this mean Well for a system in thermodynamic equilibrium the macrostate is de ned as the thermodynamic state de ned by values of P V N T U and S for example Of course for any given macrostate there are typi cally an incredibility large number of microstates that could result in that particular macrostate as we now show To understand Eq 10 let s consider the simulation discussed in class where we had 200 sites each starting out with an equal amount of energy illustrated in Fig 4a with one en ergy unit on each site Recall we can de ne the energy distribution function for this system which is a histogram of the number of sites with a given amount of energy In the simulation the initial microscopic state of the system is one where all of the sites have one energy unit so the energy distribution function shown in Fig 4b is a spike at 1 energy unit that is 200 sites high Let s also recall what happens when we ENERGY vs SITE NUMBER 6 a a 4 E Lu 2 0 0 50 1 50 200 SI39IE NUMBER ENERGY DISTRIBUTION FUNCTION 200 m E m a o E E 100 0 0 2 8 10 4 ENERGY FIGI 4 Initial conditions of energy exchange simulation a Initial energy 1 energy unit at each site b En ergy distribution function histogram of sites VS energy turn on the simulation and we let the sites ex change energy After a while there is a much broader distribution function with some sites having no energy some having one energy unit some with two and so forth A plot of the sys tem and distribution function at some later time are shown in Fig 5 For this system P and V aren t really im portant and N is xed at 200 so the only interesting thermodynamic variables are T U and S You might think that U isn t too in teresting since it is xed at 200 energy units but energy is very interesting because it is con stantly being transferred among the individual sites OK so what is the macrostate and what is the microstate for the simulated system The macrostate is de ned by the distribution func tion parts b of Figs 4 and 5 while the mi crostate is de ned by the exact placement of the energy bits parts a of Figs 4 and 5 ENERGY Vs SITE NUMBER gt E a Z a SITE NUMBER ENERGY DISTRIBUTION FUNCTION m E m 6 4 m n 2 p Z ENERGY FIGI 5 Energy exchange simulation after 1700 time steps a Energy at each site b Energy distribution function histogram of sites VS energy The green curve is the theoretical calculation for the longtime average of the distribution function Here is how the entropy enters into all of this In thermodynamics we only ever know the macrostate of the system7 not the exact mi crostate To calculate the entropy in the sim ulation we take the given distribution function macrostate and ask7 how many exact partition ings of energy among the sites microstates are consistent with the macrostate That number is Z in Eq 10 In general Z is an incredibility large num ber For example7 for the distribution function shown in Fig 5b7 Z is approximately 10119 I And this is for a system with only 200 sites and one energy unit per site For a system with 1023 atoms the number of energy con gurations is even more staggering The simulation also illustrates very nicely the ENTROPY vs TIME ENTROPY kB 0 500 1000 TIME STEP 1500 2000 FIGI 6 Entropy S in units of k3 VS time for the simu lated system Initially S 0 It increases to its thermo dynamic Value after approximately 1000 time steps principle that any spontaneous process results in an increase in entropy In the initial7 very spe cial state7 the distribution function is a spike at one energy unit A little thought reveals that there is only one microstate corresponding to this macrostate7 that shown in Fig 4a7 where each sites has exactly one energy unit Thus7 in Eq 10 Z 17 which implies lnZ O7 and thus the entropy is initially zero As the system evolves in time we end up with a distri bution function that has a maximal number of microstates associated with it7 the distribution function shown in Fig 5b At this point the entropy has increased to it maximum value The graph in Fig 6 shows the time evolution of the entropy for the simulated system One last point associated with this simula tion The simulated system has well de ned val ues of N7 U7 and S which is initially increasing during all stages of evolution to the maximum entropy macrostate state7 but it is only after the entropy has reached its maximum value that the temperature T well de ned As we discussed earlier in class7 the temperature is related to the slope of the latetime distribution function Thus the second law also is consistent with the evolution of an isolated system to a well de ned set of thermodynamic variables This is exactly what is meant by thermodynamic equilibrium Referring to the language of dynamical systems and chaos we can say that the attractor state of the system is a macroscopic state with well de ned thermodynamic variables We point out that the evolution to thermo dynamic equilibrium was already discussed with regard to the freely expanding gas Initially the gas is in thermodynamic equilibrium at T1 Vi and Pi When the partition is initially removed and the gas expands into the evacuated space it cannot be characterized by these simple thermo dynamic variables However if we wait a while we will nd out that again it has well de ned values of T V and P thermodynamic equilib rium is again achieved as the entropy increases to its maximal thermodynamic value in the larger volume space VI CHILDREN OF ENTROPY As we discussed above low entropy states such as life are certainly allowed if that state was paid for by an increase in entropy elsewhere lnterestingly life is not only compatible with the second law it may in fact be actually help ing it out There are many examples of phys ical systems with remarkably orderly behavior in space and time that exist because they facil itate the making of entropy elsewhere These are generically called dissipative structures because they help dissipate organization of en ergy One example occurs when a liquid in a con tainer is heated from below and cooled from above When the temperature difference be tween bottom and top is maintained at a small value the most efficient way of creating an en tropy increase is via random atomic collisions 7 heat conduction diffusion of heat In con duction there is no net motion of the uid atoms remain in incoherent motion only When the temperature difference is increased however convection begins ln convection large blobs of the liquid rise because of buoyancy they are less dense than their surroundings to the top where they cool and fall back down Throughout the liquid atoms are still moving incoherently but in the blobs there is coherent motion as well Thus under the right conditions the overall increase in entropy leads to coherent motion When the temperature difference is made a little bigger it is possible for convective motion to organize even more into cylindrical cell like structures In these cells uid moves up the center and falls on the outsides These beautiful cells they have hexagonal cross sections facilitate the produc tion of entropy in the universe and are thus are a favorite trick of nature Dissipation of organi zation can make wonderfully ordered structures as long as those structures help make more en tropy elsewhere Ultimately when the tempera ture difference is made too large turbulent roil ing of the liquid destroys the convection cells At too high a temperature difference convection cells just don t do as well in making entropy else where as does turbulence Consider now our planet There is heat en ergy coming in from a hot source the sun and heat transmitted out to a cold sink the rest of the universe So the interesting question is are the organized structures we see on our planet ie life the result of the overall increase in en tropy that is occurring all around us That is are we indeed children of entropy


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