Notes for Thermal Physics course, CMI, Autumn 2016 Govind S. Krishnaswami, November 16, 2016

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1 Notes for hermal Physics course, CMI, Autumn 2016 Govind S. Krishnaswami, November 16, 2016 hese lecture notes are very sketchy and are no substitute for books, attendance and taking notes at lectures. More information will be given on the course web site htt:// Please let me know (via of any comments or corrections. Hel from Sonakshi Sachdev in rearing these notes is gratefully acknowledged. Last udated: 22 Dec, Contents 1 hermodynamic systems and states hermodynamic state of a gas and a aramagnet hermodynamic rocesses Work Exact and inexact differentials Ideal gas laws Heat transferred and the first law of thermodynamics Heat transferred and its mechanical equivalent First Law of hermodynamics First law for systems with (,, state sace, three δq equations and secific heats First law for aramagnet Alications of the first law Gay-Lussac and Joule exeriment: irreversible adiabatic exansion Joule-Kelvin (or Joule-homson orous lug exeriment and enthaly Heat caacities of ideal gases Mayer s cycle for mechanical equivalent of heat Adiabatic transformation of an ideal gas Pressure and temerature in the atmoshere Carnot Cycle Second law of thermodynamics Kelvin and Clausius ostulates and their equivalence Consequences of the 2nd law for efficiency of engines Absolute thermodynamic temerature Equality of absolute thermodynamic and gas temeratures Clausius heorem Entroy and some consequences of the 2nd law Absolute temerature as an integrating denominator and the entroy Exactness of ds = δq/ and the energy equations Entroy of an ideal gas Proerties of entroy and examles hermal exansion, comressibility and tension coefficients for an ideal gas Boltzmann s statistical interretation of entroy Heat engines viewed through a S diagram hermodynamic otentials Legendre transform from internal to free energies and enthaly Interretation of Helmholtz free energy Gibbs criteria for thermodynamic equilibrium Maxwell relations Extensivity, Euler equation and Gibbs-Duhem relaton

2 7 Phase transitions General features of 1st and 2nd order transitions Claeyron-Clausius equation from energy equation Condition for hase coexistence and Claeyron s equation from Gibbs free energy van der Waal s gas vdw equation of state, virial exansion, isotherms Critical isotherm and critical oint Maxwell construction Entroy and Caloric condition for vdw gas Remarks on aramagnetic to ferromagnetic hase transition hird law of thermodynamics 53 9 Fourier s equation for heat conduction 55 1 hermodynamic systems and states hermodynamics arose in art from the needs of steam engine builders; the need to understand the ossibility and limits of conversion of mechanical/electrical/chemical energy into heat and vice versa; the attemts to otimize efficiencies of heat engines and refrigerators. An imortant role was layed by the study of the behavior of gases and of the heats released/absorbed in chemical reactions. Another motivation came from the remarkable henomena discovered in hase transitions such as the boiling or freezing of water. Before discussing the state of a thermodynamic system, let us recall the concet of the dynamical state of a mechanical system. he state of a mechanical system of oint articles is secified by giving the ositions and momenta of all the articles at an instant of time. he state variables of a thermodynamic system deend on the sort of system it is. Some examles of thermodynamic systems include gases and liquids (fluids, solids, magnets, chemical solutions, electrolytes, blackbody radiation, stars, black holes etc. hese systems tyically contain a very large number of microscoic constituents and it would be imractical to follow their individual motions. What is more, in most cases, the thermodynamic understanding has receded a detailed understanding of the nature of microscoic degrees of freedom. hus thermodynamics deals with macroscoic roerties of these systems without reference to ositions and momenta of constituents. As oosed to the dynamical state of a gas of N molecules, which would be secified by N ositions and momenta, the thermodynamic state is defined by just a handful of state variables (a number indeendent of N. Perhas the simlest thermodynamic system is a homogeneous gas (or fluid comosed of a single chemical substance (e.g. oxygen or water at rest (no flow. olume, constant ressure and density are obvious mechanical roerties associated with the uniform fluid. In addition to mechanical roerties, thermodynamics deals with the temerature of the gas, which may be measured with, say, a mercury thermometer 1. For the resent we use the same thermometer for all temerature measurements to make them comarable. hus thermodynamics ostulates the existence of temerature as a thermal roerty of 1 Oerationally, we have a column of mercury of height h at some reference temerature (say freezing oint of water at atmosheric ressure. We say that the water freezes at temerature h. When heated, the mercury exands, the temerature is defined simly to be the height of the column. 2

3 a system. It is a arameter of state being defined by the instantaneous state of the system indeendent of its history. It is observed that a fluid when left to itself ( isolated for some time settles into a state of thermal equilibrium, characterized by a common temerature everywhere. hus the zeroth law of thermodynamics ostulates that arts of a system (or different systems are in thermal equilibrium if they have the same temerature. Furthermore, if system A is in thermal equilibrium with system B and B is in equilibrium with C then A is in equilibrium with C: all three have a common temerature. A consequence of this transitivity, is that it allows us to conclude that a bar of iron and a bucket of water are in thermal equilibrium even without bringing them into contact, rovided we check that each is searately in equilibrium with a third system, which lays the role of a thermometer. 1.1 hermodynamic state of a gas and a aramagnet hermodynamic state of a homogeneous fluid. For the uroses of thermodynamics, a uniform fluid has one mechanical degree of freedom, its volume. It is found that thermodynamic roerties (like energy are largely indeendent of the shae of the region occuied by the fluid as long as its volume is large and surface area not too large comared to that of a shere of the same volume. Just as momentum is conjugate to the osition of a article in mechanics, we will see that ressure is the variable conjugate to volume 2. he thermodynamic state of a fixed mass of homogeneous fluid is fixed once the volume, the (uniform ressure and temerature t have been secified. hus the thermodynamic state sace is three dimensional. However, not all triles (,, t reresent equilibrium states. Equilibrium states have the secial roerty of remaining unchanged in time under fixed external conditions. For instance, a flowing gas is not in equilibrium (at the very least, one would need to secify the velocity and ressure at different locations in the gas to determine its state. Equilibrium states lie on a 2D surface determined by a relation f(,, t = 0 called the equation of state. he equation of state allows us to exress one of the state variables, say (, t, as a function of the other two. We will see that for many dilute gases at sufficiently high temeratures and low ressures, (ideal gases, the equation of state is aroximately = nr where the roortionality constant nr is the roduct of the number of moles of gas (amount of substance and the universal gas constant and is the so-called gas thermometer temerature. Following James Watt, the state of such a gas can conveniently be reresented by a oint in the lane (indicator diagram with abscissa along the horizontal axis and ordinate along the vertical axis. An isotherm is a curve along which temerature is constant. A P - diagram is sometimes convenient esecially when studying hase transitions, say between liquid and vaour. he first derivatives of,, with resect to eachother are used to define three material roerties which describe the resonse to a change in one due to another keeing the third 2 Hydrostatic ressure at a oint A is introduced as follows: Consider a small fictitious membrane (flat surface assing through A in a gas/fluid. It is found that the fluid on one side of the membrane exerts a force normal to the membrane and towards the other side. his normal force er unit area is the hydrostatic ressure. he same value of ressure is obtained by considering the normal force er unit area due to the fluid on the other side. In hydrostatics, the two oosing forces balance. Moreover, the value of ressure at a oint is indeendent of the orientation of the membrane through the oint: it is a ositive scalar quantity. 3

4 fixed. hus we have the coefficients of thermal exansion, thermal tension 3 and isothermal comressibility α = 1 (, β = 1 (, κ = 1 (. (1 For instance, comressibility κ measures the fractional change in volume of a gas due to an increase in ressure. he minus sign is because gases are comressed uon the alication of ressure. he recirocal of comressibility is called the bulk modulus. he fact that, and satisfy the equation of state leads to a remarkable relation among the three coefficients α, β and κ. β κ = α. (2 his is obtained by relacing x, y, z by,, in the trile roduct identity we now derive. In general, if x, y, z satisfy a functional relation f(x, y, z = 0 that allows us to locally exress each as a function of the other two z = z(x, y, y = y(z, x and x = x(y, z, then 4 ( z x y ( x y z ( y = 1. (3 z x o see this we write the differentials with subscrits denoting artial derivatives (E.g. z x = ( z x, the y variable that is held fixed in the artial derivative is not indicated exlicitly as it is clear from the context dz = z x dx + z y dy, dx = x y dy + x z dz, and dy = y z dz + y x dx. (4 Now we regard x and y as indeendent variables & write dz in terms of dx & dy in the formula for dx dx = x y dy + x z (z x dx + z y dy. (5 Since dx and dy are indeendent differentials, their coefficients must vanish. he coefficient of dx vanishes rovided x z z x = 1 (i.e., ( ( x z z y x = 1. Since we could equally well have taken any other y air as indeendent variables, we must also have x y y x = y z z y = 1. Now requiring that the coefficient of dy vanish gives x z z y = x y which becomes z x x y y z = 1 uon using the revious result. hermodynamic variables can often be classified as extensive or intensive. Extensive quantities are roortional to the amount of material. For examle, internal energy ( U, the various free energies (e.g. Helmholtz free energy F and entroy ( S are extensive variables, they scale with the volume (ignoring boundary/surface effects, the number N of articles resent or the total mass m. By dividing by N, or m we obtain the corresonding secific energies or entroy. Intensive quantities such as ressure, temerature and chemical otential (to be introduced later are indeendent of the amount of substance. We will see that (,, (µ, N and (t, S are canonically conjugate airs of intensive and extensive variables. Magnets as thermodynamic systems: he analogs of ressure and volume for a magnetic material are the external magnetic field ( H intensive and magnetization ( M by convention diole moment er unit volume, intensive. A ferromagnet such as an iron bar magnet (a magnetic diole has a ermanent magnetization (ointing, say in the z direction. he 3 ension is another word for force and thermal tension refers to the change in ressure (force/area as temerature is varied. 4 o remember this formula, begin with any of the artial derivatives as the first factor. he latter two factors are obtained by cyclically ermuting the variables in the first artial derivative. 4

5 magnetization leads to a magnetic field in the neighbourhood of the magnet, which aligns iron filings. By measuring the magnetic field, one can infer the magnetization. he magnetization of a ferromaget is ermanent (more recisely sontaneous in that it is non-zero even in the absence of an external magnetic field. From a microscoic view oint, the magnetization is the vector sum of molecular magnetic moments er unit volume. A aramagnetic substance has no magnetization of its own. However, in the resence of an external magnetic field H, it acquires a magnetization M that oints along H. Ferromagnets lose their sontaneous magnetization above the Curie temerature t c (770 Celsius for iron and become aramagnets. his is a hase transition which bears some resemblance to the liquid to gas vaourization transition. For simlicity we will assume that H, and therefore M, oints in the z -direction and denote them by their z -comonents H and M. hus the thermodynamic state sace of a uniformly magnetized homogeneous aramagnet is three dimensional with coordinates H, M and t. H and M are conjugate variables like and for a gas. In equilibrium, there is an equation of state that relates M, H and t. For a linear aramagnetic medium M = χh where M is the magnetization er unit volume ( the volume of the magnet is essentially a constant, unlike for a gas. his simly states that the induced magnetization is roortional to the alied magnetic field. he magnetic suscetibility χ deends on temerature. At high temeratures (t > t c the aramagnetic suscetibility is given aroximately by the Curie-Weiss 5 law χ C/(t t c for an aroriate temerature scale. C is called Curie s constant. 1.2 hermodynamic rocesses A thermodynamic rocess is a transformation of a system from an initial state through a continuous succession of intermediate states to a final state. It may be reresented by a curve in the state sace. A transformation in which the system is always infinitesimally close to an equilibrium state is called a reversible or quasi-static transformation. For a homogeneous gas, it may be reresented by a curve that lies entirely on the equilibrium surface in the - -t state sace; it may also be reresented by a curve in a - diagram. Starting from an equilibrium state, a reversible transformation may be realized through a slow variation of external conditions so that the system always has time to adjust itself to the new equilibrium state corresonding to the altered conditions. For examle, the slow exansion of a gas in a container fitted with a horizontal iston, as small weights are slowly removed from the iston and ket aside at the same height. he rocess may be reversed by relacing the weights onto the iston slowly as the gas contracts. If the iston is lowered raidly, the ressure would not be uniform and there would also be velocity currents set u in the gas, making it ass through non-equilibrium states. A transformation assing through equilibrium states from A to B is called reversible since it can be reversed by alying the changes to the external conditions slowly in the oosite order, so that the system asses through the same intermediate states but from B to A. A rocess that is not reversible (not quasi-static is called irreversible. Irreversible rocesses may not be reresentable as curves in the t state sace, since we may need to secify other variables like velocity and there may not be a single ressure that can be assigned to the whole gas. Note that a slow rocess need not be reversible. Examles may be found among dissiative rocesses such as the slow frictional heating of the susension of a swing as it executes damed 5 Named after the Polish-French and French hysicists Marie Curie and Pierre Weiss. 5

6 Figure 1: From K Huang, Introduction to statistical hysics:,, t thermodynamic state sace of a gas of uniform comosition. Reversible rocesses are curves on the equilibrium EOS surface. Irreversible rocesses which can be reresented in such a diagram go off the equilibrium surface. oscillations or the slow heating of a resistor connected to a battery. Even if we try to reverse these rocesses (by reversing the direction of oscillation of the swing or reversing the direction of current heat is still roduced, not absorbed! he sontaneous flow of heat (whether quickly or slowly from a body at a higher temerature to one at a lower temerature is irreversible. Indeed we do not see heat sontaneously flowing from a cold body to a hot one. hermodynamic rocesses in which, or remain constant are called isobaric, isochoric (isoiestic and isothermal. A cycle or cyclic rocess is a transformation in which the initial and final states are the same. On a - diagram, a cycle is reresented by a closed curve (which could have self-intersections. Cyclic rocesses are imortant in ractical alications as they can be reeated, as in a steam engine or refrigerator. Cyclic rocesses are also of great concetual value, as we will learn. 1.3 Work When a gas exands, it does work W on the surroundings; the surroundings do work W on the system. More generally, a system can erform work (mechanical - movement of a iston, electrical/chemical - movement of charges/ions in an electric otential etc. during a thermodynamic rocess. For examle consider a gas enclosed in a cylinder with a movable iston of area A. If is the uniform ressure in the gas, the force exerted on the iston is A. If the iston moves out a distance l, the work done is W = A l. In general, if the volume of a gas increases by d in an infinitesimal exansion, then the work done by the gas is δw = d. Suose the state can be reresented on a - diagram, then a transformation from state ( a, a to ( b, b is reresented by an oriented curve γ a b joining these oints. he work done W (γ = b a d is given by the area under the curve (see figure. Note that the ressure could change during the rocess and that the area can be negative, as in a comression ( b < a. In articular, the work done in a cycle (reresented in a - diagram is the area enclosed by the curve. For a simle closed curve, the area is ositive if traversed clockwise and vice-versa. d and δw both vanish in an infinitesimal isochoric transformation. In electromagnetic theory it is shown that the work done by the external field H in increasing the magnetization of a aramagnet from M to M + dm is HdM. hus the work done by a 6

7 Figure 2: From A Sommerfeld, hermodynamics and Statistical mechanics: diagram for a cycle of a steam engine. Work done is equal to shaded area enclosed and is ositive as the curve runs clockwise. he isobar at the high ressure 1 corresonds to the steam in the cylinder exanding while in contact with the boiler. he isobar at the low ressure 2 corresonds to comression when in contact with the atmoshere or condenser. magnet in the rocess is δw = HdM. U to a sign, this is analogous to δw = d for the work done by a gas. It is very imortant to recognize the emirical fact that the work done during a rocess γ a b in general deends on the rocess and not just on the initial and final states. For examle consider initial and final equilibrium states of water at temeratures t a and t b > t a, but at a common ressure, say atmosheric ressure. he corresonding volumes are determined by the equation of state. Now we can go from state a to b in more ways than one. For instance, we could stir the liquid for some time and heat it u through friction, thereby doing a significant amount of work. Alternatively, we could simly heat the liquid using a flame, in which case hardly any work is done (the liquid does a little work in exanding against atmosheric ressure. his ath-deendence of the work done is reflected in the fact that areas under different curves joining a to b in a diagram are not necessarily the same. In other words, unlike temerature, there is no work function or roerty of state W such that W (b W (a is the work done during the rocess. In articular, the work done in an infinitesimal rocess δw deends on external effects and is not the differential of any state function W. his is why we do not denote it dw. We say that the work done in an infinitesimal rocess is not an exact differential. 1.4 Exact and inexact differentials wo indeendent variables: Given a differentiable function, say of two variables σ(x, y, its differential is dσ = σ σ x dx+ y dy. Under suitable hyotheses (it suffices for σ to have continuous second artial derivatives, the mixed second artials are equal: 2 σ x y = 2 σ y x. his is called Schwarz s or Clairaut s theorem. On the other hand, suose we are given a differential exression (also known as a one-form or Pfaffian differential ω = f(x, ydx + g(x, ydy. We ask whether ω is the differential of some function σ, i.e., is ω = dσ? σ would in a sense be an integral of ω and we would call ω a erfect or exact differential. A necessary condition for this (it is also a sufficient condition in a sufficiently small local region follows from the above remark: f and g must satisfy f y = g x. Denoting the comonents of ω by the vector field on the lane v = (f, g = f ˆx + gŷ, this integrability condition states that v must be curl free v = (g x f y ẑ = 0. In the last section we asserted that the work differential (work done by gas in an infinitesimal 7

8 exansion δw = d is not exact. As another examle, consider the Pfaffian differential ω = Cd + nr d in the variables and. Here n and R are constants while C is indeendent of. he above integrability condition is not satisfied since nr/ is not identically zero. Interestingly, notice that ω/ is an exact differential, the integrability condition is satisfied and we can integrate it to find a function σ such that dσ = ω. Indeed ω/ = Cd/ + nrd/ = d(c log + nr log + σ 0 dσ. Notice that σ is defined only u to an additive constant. is called an integrating factor or integrating denominator for the inexact differential ω. More on this later. Now if ω is an exact differential, then its line integral between a air of oints a, b on the lane is indeendent of the ath γ connecting them. Indeed, let the ath γ : [t a, t b ] R 2 be given by r(t = (x(t, y(t with r(t a = a and r(t b = b. hen the tangent vector to the curve is ṙ(t and the line element dl = ṙ(tdt and γ ω = γ v dl = γ fdx+gdy = tb t a (fẋ+gẏdt = tb t a (σ x ẋ+σ y ẏdt = tb t a dσ(r(t dt = σ(r(t b σ(r(t a. dt (6 he last equality follows from the fundamental theorem of calculus. Path-indeendence follows since the line integral deends only on the values of the function σ at the end oints. In articular, the integral of an exact differential around a closed contour C vanishes dσ = 0. his may also be seen by an alication of Green s theorem (2d version of Stokes theorem fdx + gdy = (g x f y dxdy = 0 or v dl = ( v ẑ dxdy = 0. (7 C S Here S is the two-dimensional region bounded by the closed curve C. Conversely, a differential ω whose line integral around every closed curve vanishes is an exact differential. hree indeendent variables: he above statements have a natural generalization to 3D. he differential ω = fdx + gdy + hdz is locally exact rovided the first artials of f, g and h satisfy f y = g x, g z = h y and h x = f z. (8 Here we use a short-hand notation for artial derivatives: f y = C S ( f y x,z is the artial derivative with resect to y holding x and z fixed. hese three conditions simly state that the coefficient vector field v = (f, g, h, is curl-free: v = (h y g z, f z h x, g x f y = 0. (9 So when v is curl free, we may exress it as the gradient of a scalar v = σ or ω = dσ. Furthermore, the line integral b a ω of the exact differential ω is the difference in values σ(b σ(a indeendent of the ath. he above result can also be obtained from Stokes theorem in three dimensions. Given a sufficiently smooth vector field v in a simly connected region, its line integral around a closed curve C can be re-exressed as the surface integral of its curl over any surface S in the region whose boundary is C : v dl = ( v ds. (10 C S Now if v is curl-free, the RHS vanishes, so that its line integral around any contractible closed ath vanishes. 8

9 1.5 Ideal gas laws It is emirically found that many gases in equilibrium satisfy certain common ( ideal gas laws at temeratures much higher than their condensation oint and at low ressures. At 20 C and atmosheric ressure of one Atm (or 760 mm of Hg or 760 torr, helium, hydrogen, oxygen, carbon dioxide and air to a reasonable aroximation behave ideally while water vaour (steam does not. An ideal gas is what we would get when we exand a real gas to very large volumes. From the molecular standoint, an ideal gas is one whose molecules are oint-like (i.e. occuy negligible volume comared to that of the container and do not interact with each other (exert no forces. Boyle s and Mariotte s Law (1658, 1676: For a given mass of a gas at a fixed temerature, the roduct of ressure and volume occuied by the gas in equilibrium is a constant: P (i (i = Θ (i (t, m. Based on the investigations of Boyle and Mariotte, the constant Θ (i (t, m could deend (and in fact does deend on temerature t, mass m of gas and on the chemical nature of the gas (indicated by the secies label i. Gas temerature scale: By a gas thermometer, we mean the following. We select a secified volume (say 1 cc of a gas (such as He 2 that is far from condensation at a reference temerature (say the freezing oint of water at atmosheric ressure of 1 atm. As noted, gases are observed to exand on heating and contract on cooling. We simly define the gas temerature to be a constant multile of the volume, with the ressure being held fixed. he constant is often fixed by requiring the difference in gas temeratures at the boiling and freezing oints of water to be 100. Note that fixing the constant through such a secification also removes the deendence on the volume of gas we started with, it could have been 100 cc instead of 1 cc. It is a remarkable emirical fact (see also the Charles-Gay-Lussac law below that a wide variety of gas thermometers (with different gases, but all of which have the same volume at the reference temerature and ressure agree on their assignment of gas temeratures as long as the gases are far from condensing. he temerature scale defined this way is called the gas thermometer scale (or absolute gas temerature or the Kelvin scale. Water freezes at 1 atm ressure at = Kelvin. By construction > 0 since the volume of a gas is a ositive number. Charles and Gay-Lussac s Law (1787, 1802: (See N D Hari Dass, he rinciles of thermodynamics All gases, whatever may be their density and the quantity of water which they hold in solution, and all vaours exand equally between the same degrees of heat. Here same degrees of heat means between the same initial and final temeratures (freezing and boiling oints t f, t b of water at atmosheric ressure in Gay-Lussac s exeriments. By exand equally he aarently meant the change in volume as a fraction of original volume was the same for all gases. he exansion of all gases was carried out at the same fixed ressure (atmosheric ressure. If the secies of gas is labelled i, we may summarize the law by (i t b (i t f (i t f = f(c(t f, t b (11 for all gases. In other words, the fractional change in volume is indeendent of secies, but we have allowed for a deendence on the ressure and (through c(t f, t b on the initial and final temeratures. Note that though the volumes (i t b, (i t f deend on the mass of gas resent, the ratio on the left is indeendent of the mass, since it is found that twice the mass of a gas occuies 9

10 twice the volume at a fixed temerature and ressure. So the RHS is also indeendent of mass of gas used. Now alying the Boyle-Mariotte law, we get Θ (i (t b, m Θ (i (t f, m Θ (i (t f, m = f(c(t f, t b. (12 he deendence on ressure cancels between numerator and denominator on the LHS since the exansion is carried out at a fixed ressure. hus the RHS must be indeendent of ressure and we may set f( = 1, absorbing any constant into c(t f, t b. Moreover, the RHS is indeendent of secies. he simlest way for the LHS to be indeendent of secies is for Θ (i (t, m to be secies indeendent. However, this contradicts exeriment: utting this in Boyle s law, we would find that the roduct for mass m of any gas is the same function of temerature Θ(t, m. However, it is found that a common mass m of different gases at the same temerature have different constant values for the roduct. his is also why in setting u the gas thermometer, we did not take equal masses of different gases but rather equal volumes at a common temerature and ressure. he next simlest ossibility is that Θ (i (t, m = ν (i (m Θ(t is the roduct of a secies-deendent function ν (i (m and a secies-indeendent function of temerature. he constant ν (i (m cancels in the quotient on the LHS of (12, ensuring that the RHS is indeendent of both i and m. Putting this in the Boyle-Mariotte law, we deduce that for a fixed mass m of any gas (labeled i at a fixed temerature t, (i (i = ν (i (m Θ(t. (13 Here Θ(t is a universal function of temerature. Now we are free to choose our temerature scale. he law takes its simlest form when we use the gas thermometer scale of temerature, in which case Θ( is a universal constant multile, say R, of (since we defined the gas thermometer temerature to be a multile of the volume of the gas holding, m fixed. hus the simlest way of satisfying the observations of Boyle-Mariotte-Charles-Gay-Lussac is for gases to obey (i (i = ν (i (m R (14 where is the gas thermometer temerature. As we will see, ν (i (m deend on the secies via a characteristic roerty which we now recognize as their molecular masses. his brings us to Avogadro s ostulate. Avogadro s ostulate (1811: Equal volumes of all gases at the same temerature and ressure have the same number of coruscles (molecules, but (in general different masses. However, the molecular constitution of matter took nearly a century to be confirmed. In the interim, it made sense to reformulate Avogadro s ostulate in terms of measurable quantities. Ostwald introduced the concet of moles for this urose. A gram-mole or gram molecular mass µ of a gas was defined as a definite number of grams of a gas. For examle, one gram-mole of hydrogen is 2 grams of the gas (i.e. µ = 2 for H 2. A gram mole of oxygen is 32 grams (i.e. µ = 32 for O 2. hese assignments were based on the roortions in which elements combined to form chemical comounds (the subject of Dalton s and Gay-Lussac s laws of multile roortions 6 and integral 6 Dalton s law of multile roortions: When two elements combine with each other to form more than 10

11 volume ratios 7, but could also be deduced from calibrating gas thermometers with the same mass of different gases with each other. In current language, one gram mole of a comound is as many grams as there are units in the sum of atomic weights of the constituent elements. We often refer to one gram mole of a gas as simly one mole of the gas. Avogadro s law in Ostwald s formulation states that equal volumes of all gases at a common ressure and temerature contain equal numbers of moles of the gas. So suose we consider one gram-mole of gas i (i.e. µ i grams as well as one gram mole of gas j (i.e. µ j grams at the same ressure and temerature. he Avogadro-Ostwald ostulate along with the Boyle-Mariotte-Gay-Lussac-Charles law imlies (i = (j or ν (i (µ i R = ν(j (µ j R or ν (i (µ i = ν (j (µ j i, j. (15 hus the constants ν (i (µ i are indeendent of secies if we consider one gram mole of each secies. We can take this universal constant to be one, by a rescaling of R. hus for one mole of an ideal gas, = R. If we had considered n gram moles instead of one gram mole, the volume occuied at a fixed temerature and ressure is found to be multilied by n, so we arrive at = nr for n gram moles of any sufficiently ideal gas. In other words, ν (i (m = m/µ i. Ideal gas law: Combining Avogadro s ostulate with the Boyle-Mariotte and Charles-Gay- Lussac laws, it is found that the equation of state of an ideal gas takes a articularly simle form when we use the gas thermometer scale. For m grams of a gas with gram molecular mass 8 µ, = (m/µr. (16 Here R = Joules/Kelvin er mole is the universal gas constant. From Ostwald s definition, the quotient n = m/µ is the number of moles of the gas. By Avogadro s law, one mole of all (nearly ideal gases contain the same number of molecules, namely N A = (Avogadro s number, estimated by JJ Loschmidt in 1856 and more accurately measured by J B Perrin in 1909 after Einstein s work on Brownian motion. he ideal gas EOS may also be exressed as a relation among ressure, density ρ = m/ and temerature = ρr/µ. Alternative forms are = nr and = Nk B where N = N A n is the number of molecules and k B = R/N A = J/K is Boltzmann s constant. If we introduce the secific volume v = /m which is the recirocal of density, then the ideal gas law states that v = R/µ. As noted, from a microscoic viewoint, an ideal gas is a collection of molecules of a given chemical secies occuying a volume much larger than the size of the molecules and whose inter-molecular forces can be ignored. he Hamiltonian for such a collection of molecules is the sum (1/2M a (2 a where the various molecules are labelled by a and M is the mass of each molecule. It is not surrising, then, that all gases that can be aroximated by such a treatment behave in the same way excet for the difference in the masses of the molecules. he molecular one comound, the masses of one element that combine with a fixed mass of the other are in a ratio of small whole numbers. E.g. 24g of carbon combine with 32g of oxygen to form 56g of carbon monoxide. 24g of C also combine with 64g of oxygen to give 88g of carbon dioxide. So the ratio of oxygen masses that combine with the same mass of carbon is 1 : 2. 7 Gay-Lussac s law of integral volume ratios: he ratio between the volumes of the reactant gases and the roducts can be exressed in simle whole numbers. E.g. 4g of hydrogen gas combine with 32g of oxygen to give 36g of water vaour. At 0 C and 1 atm ressure, 44.8 l of hydrogen combines with 22.4 l of oxygen to give 44.8 l of water vaour. he ratio of volumes of reactants to roduct is 3 : 2. 8 For examle, the gram molecular mass of H 2 molecule is 2 while that of that of He 4 2 atom is 4. 11

12 mass enters the ideal gas law through µ (M = µ/n A and is the only way in which the gas manifests its chemical identity. For an isothermal transformation of a fixed amount of an ideal gas, is a constant. hus, on a - diagram, an isotherm is reresented by a hyerbola with the and axes as asymtotes. he work done by the gas in a reversible isothermal exansion from volume 1 to 2 is W = 2 1 ( d = nr 2 1 d = nr log( 2/ 1 = nr log( 1 / 2. (17 2 Heat transferred and the first law of thermodynamics 2.1 Heat transferred and its mechanical equivalent Caloric theory of heat: Prior to the mid 1800s heat was regarded as a substance, the caloric which was though to be like a fluid that flowed while its amount was conserved 9. his is in contrast to the current view that heat is a form of energy that can be transformed (artly into (or arise from other forms of energy. he heating of water by a flame was assumed to involve flow of the caloric (heat-fluid from the fire to the water. When a bottle of hot water was brought into contact with a bottle of cold water, it was suosed that the heat fluid flowed from hot to cold water till the temeratures were equalized. he theory was not without merit. Carnot (1824 correctly deduced limits to the conversion of heat into work based on the caloric theory. However, the caloric theory had to be relaced since it contradicted exeriment. he amount of heat fluid was not conserved: heat could be roduced entirely from mechanical or electrical work, as in the heating of a resistor/filament of a light bulb or the melting of two ieces of ice when rubbed together. he caloric theory of heat was relaced by a more mechanical theory of heat. he amount of heat transferred in a rocess may be quantified through temerature measurements. One calorie of heat transferred is defined as the quantity of heat required to raise the temerature of one gram of water from 14 to 15 C at atmosheric ressure. It is imortant to recognise that while we can measure and define the heat exchanged between systems, we have not defined the heat content of a system. Mechanical equivalent of heat Count Rumford (also known as Benjamin homson, 1798 famously demonstrated that water could be made to boil while the shaft of a canon was being mechanically bored. Water could also be heated through stirring using immersed addles that were made to rotate by letting a weight descend under gravity. Several careful exeriments showed that the heat transferred (in calories was roortional to the work done (say in Joules or ergs, through a universal roortionality constant δq = (1/JδW. Precisely, to transfer one calorie of heat, the same amount of work, i.e Joules had to be done. hus Joules of work done is the mechanical equivalent of one calorie of heat transferred. 2.2 First Law of hermodynamics In 1842 Robert Mayer resented a mechanical theory of heat, based on the equivalence of heat and mechanical energy, and formulated the 1st law of thermodynamics. We will state the first 9 An even earlier (now discarded theory of combustion was the Phlogiston theory, develoed from 1667 onwards. 12

13 law and later describe the cyclic rocess he used to determine the mechanical equivalent of heat. A similar rocess had been considered by Sadi Carnot somewhat earlier. Hermann Helmholtz (1847 made Mayer s ideas on the first law mathematically more recise. he first law was the work of these and several other scientists. he first law ostulates the existence of the internal energy as a state function of a thermodynamic system. It osits that the heat added to a system is equal to the sum of the increase in its internal energy (denoted U by Rudolf Clausius and the work done by the system. Q = U + W. (18 We may regard the first law as a generalization of the statement of conservation of energy. For an isolated system (thermally and mechanically/electromagnetically etc, the internal energy is conserved. For a thermally insulated system, the work done by the system is equal to the dro in its internal energy W = U. Given the state of a system, we can seak of the internal energy of that state, but not the heat content or work of the system in that state, heat and work are not state functions, they are measures of external effects. If the caloric theory was true, heat would be a state function, equal to the amount of caloric fluid. For a cyclic rocess, the initial and final states are the same so U = 0 and the 1st law becomes Q = W, i.e., the work done is equal to the heat absorbed. For an infinitesimal reversible exansion of system volume d against a ressure, the work done is δw = d. So for infinitesimal reversible rocesses, we write δq = du + d. (19 Unlike the heat and work differentials δq and δw which are not exact, the internal energy differential du is an exact differential. So for a cyclic rocess du = 0. On the other hand, there is no state function Q such that δq is equal to dq. he 1st law allows us to measure heat exchanged in energy units. It is found that 1 calorie of heat exchanged is equivalent to Joules of energy. he first law is sometimes interreted as stating the imossibility of constructing a machine (eretuum mobile or eretual motion machine of the first kind that can create or destroy energy. Rather, energy can be transformed into other forms such as mechanical/electrical work and heat. When the number of articles N in the system can change, the internal energy of a system could change even if no mechanical work is done on the system nor any heat transferred to it. It is conventional to define the chemical otential µ as the energy required to add one article to a thermally and mechanically isolated system. he first law is then generalized to Q = du + W µ N or δq = du + d µdn (20 for an infinitesimal rocess. We may interret this equation as saying that the heat added to a cu of water by a flame is equal to the sum of the increase in its internal energy, the work the water does in exanding against atmosheric ressure and the energy lost to articles exelled from the container. 13

14 Based on the manner they enter the equation for energy conservation, we say that and are conjugate variables as are µ and N. In each air we have one intensive and one extensive variable. Notice that the roduct of conjugate variables has dimensions of energy. 2.3 First law for systems with (,, state sace, three δq equations and secific heats If the state of a thermodynamic system (such as a homogeneous gas can be reresented by a oint in P sace (subject to the equation of state, then we may exress the state function U in terms of any one of the three airs of indeendent variables. For instance, taking (, as indeendent variables, the infinitesimal increase in volume and internal energy are d = ( d + ( d and du = ( U d + ( U d. (21 hen the 1st law δq = du +d imlies the following three δq equations for the infinitesimal heat added to a system with the indicated variables taken as indeendent: ( [( ] U U (, δq = d + + d, [ ( U ( ] [( ( ] U (, δq = + d + + d, ( [ ( U ] U (, δq = d + + d. (22 We define the heat/thermal caacity of system as the ratio C = δq/d of infinitesimal heat added (usually reversibly to the consequent increase in temerature. he heat caacity generally deends on how the system is heated. Of articular significance are the rincial heat caacities C, C at constant volume and ressure. aking (, and (, as indeendent variables in (22 we obtain exressions for C, and C in terms of derivatives of state functions C = ( U and C = ( U + ( = ( H Since 0 and volumes of gases grow with increasing temerature, we see that C C. C exceeds C since at constant ressure, additional heat must be sulied for the gas to do work while exanding. Note that C P can also be written as C P = (U + where the artial derivative is evaluated at constant ressure. It is therefore natural to introduce the new extensive state function enthaly H = U + and exress C as the temerature derivative of H at constant ressure, just as C is the temerature derivative of U at constant volume. he heat caacities are extensive as U and are. It is often convenient to define the corresonding intensive quantities c, c by considering the heat exchanged er unit mass (secific heats or er mole or molecule (molecular heats. Exercise: Use the first law taking, as indeendent variables to show that the difference in heat caacities is given by (( ( U C C = +. (24 14 (23

15 ANS: ake and as indeendent variables and exress the heat added as δq = du + d = (U d + (U d + d. Now divide by d and take the limit at constant ressure. LHS becomes ( ( ( δq C = = (U + (U + (25 d Now notice that the middle term is just C. hus we get the desired result. 2.4 First law for aramagnet Recall that the thermodynamic state sace of a aramagnet is three dimensional with coordinates given by external magnetic field H ( z comonent, z -comonent of magnetization M and temerature. From electromagnetic theory, the work done by the external field in increasing the magnetization from M to M + dm is µ 0 HdM (see below. hus the first law takes the form (Paul Langevin, 1905 δq = du µ 0 HdM (26 Comaring with δq = du + d for a gas, we see that roughly, and µ 0 H lay similar roles while volume and magnetization are analogous. he equation of state for a aramagnet bears some resemblance to the ideal gas law. Indeed, comaring the Curie-Weiss law with the ideal gas law, M = χh = CH ( c and ρ = µ R, (27 we see that Curie s constant C lays the role of µ/r while magnetization (er unit volume is like density and H is like ressure. c for the aramagnet is relaced with the gas thermometer temerature. Let us briefly indicate how the formula for the work done arises. he electromagnetic energy density is given by 1 2 (B H+E D where E, D, B, H are the electric field, electric dislacement field, magnetic flux density and magnetic H field with B = µ 0 (H+M. Restricting attention to the magnetic art, the infinitesimal change in magnetic energy density is δe = 1 2 (H db+b dh. For a linear magnetic medium, M = χh where χ is the suscetibility, so that B = µ 0 (1 + χh. hus δe = 1 [ ] db H db + µ 0 (1 + χh = H db. (28 2 µ 0 (1 + χ We now eliminate db in favour of M and H using db = µ 0 (dh + dm δe = µ 0 H dh + µ 0 H dm = d(h 2 /2µ 0 + µ 0 H dm. (29 he 1st term on the RHS is an exact differential and can be absorbed into the internal energy. he second term is identified as the work done by the external H field in increasing the magnetization by dm. As for a gas, we may define heat caacities and secific heats of a aramagnet at constant magnetic field or constant magnetization. See Haridass or Sommerfeld for further discussion. 15

16 Figure 3: Joule free exansion exeriment and Joule-Kelvin orous lug exeriment (from Sommerfeld 3 Alications of the first law 3.1 Gay-Lussac and Joule exeriment: irreversible adiabatic exansion Gay-Lussac (1807 confined a gas in a thermally insulated cylindrical container fitted with a movable iston. Initially, the gas was in equilibrium at temerature 1, ressure 1 and occuied volume 1. he iston was suddenly moved outwards, increasing the available volume to 2. he gas exanded (through a comlicated flow and eventually reached equilibrium at a lower ressure 2 in volume 2. Remarkably, there wasn t much change in temerature 2 1. his was confirmed by J P Joule (1845 who reeated the exeriment more carefully, this time with two glass jars connected by a tube with a sto-cock. When the cock was released, gas (air, hydrogen from the filled jar flowed into the evacuated jar and a new equilibrium was established, with barely any change in temerature. hough the exeriment re-dates the first law, let us aly the first law to the initial and final states. Since the gas exands into a vacuum (zero ressure, we may assume it does not do any work. Insulation ensures the gas does not exchange any heat. hus the internal energy must be unchanged. aking volume and temerature as indeendent variables, U( 1, 1 = U( 2, 1. Since the gases obey the ideal gas EOS, we conclude that the internal energy of an ideal gas is indeendent of volume, U = U(. 3.2 Joule-Kelvin (or Joule-homson orous lug exeriment and enthaly his was a successor to Gay-Lussac s and Joule s free exansion exeriments. It gave greater confidence to the conclusion that the internal energy of a gas obeying the ideal EOS (air, hydrogen etc. is a function of temerature but not volume. Interestingly, the results of the exeriment could be interreted in terms of enthaly. Gas at a higher ressure 1 is forced slowly from the left chamber through a ie (filled with a orous lug made of cotton wool to the right chamber at lower ressure 2 (the ressures are maintained at the same values throughout. Both chambers were insulated and had the same cross-sectional area A. he ie was made of beechwood, a thermal insulator. Consider a volume 1 of gas between two vertical cross-sections on the left, that emerges on the right and occuies volume 2 between two vertical cross-sections. he distance traversed by the gas on the left is 1 /A and 2 /A on the right. On the left, the force 1 A on the gas does work 1 A 1 /A = 1 1. On the right, the oosing force 2 A does work 2 A 2 /A = 2 2 on the 16

17 gas. hus the total work done on this mass of gas is he rocess may be assumed adiabatic. Denoting the initial and final internal energies of the gas by U 1 and U 2, the first law imlies U 2 U 1 = or U = U (30 he enthaly H = U + of a given mass of gas is conserved as it asses through the insulated lug. Furthermore, it was found that the temerature of the gas was almost unchanged. Assuming the ideal gas law, U 2 U 1 = = nr( (31 hough the volume has changed, U has not, so the internal energy of an ideal gas may be taken indeendent of volume U = U(. Since the gas is ushed down a ressure gradient ( 1 > 2, 2 1 imlies 2 > 1, i.e. the gas exands. Careful measurements indicate that 1 and 2 are not quite the same (under ordinary conditions 2 < 1 for air but 2 > 1 for hydrogen. In fact, many real (as oosed to ideal gases can be cooled as they exand while assing down a ressure gradient. his is the basis of the Joule-homson effect. 3.3 Heat caacities of ideal gases We concluded from Joule s exeriment that the internal energy of a fixed mass of an ideal gas could be taken indeendent of volume. his is sometimes called the caloric condition. It follows that C ( = du d or du = C ( d. his, along with the ideal gas EOS and 1st law leads to the useful relation C C = nr between heat caacities at constant ressure and volume. o see this, we write the first law in form δq = C dt + d and substitute for d from the differential of the EOS = nr : hus we get We may now read off the heat caacity at constant ressure d + d = nrd. (32 δq = (C + nrd d. (33 C = (δq/d = C + nr. (34 his is consistent with our earlier observation that c must exceed c : at constant ressure art of the heat sulied goes into exanding the gas leaving less heat to raise its temerature. he relation is often exressed in terms of molar secific heats c c = R (c = C /n is the heat caacity er mole or molecular heat at constant ressure and is in good agreement with exeriments. While thermodynamics has allowed us to find a relation among heat caacities, it does not give us a way of determining their values. he kinetic theory of gases, which takes into account the molecular structure of matter allows us to estimate c. For monoatomic gases (such as helium and neon and other Noble gases and mercury vaour one finds the molar secific heat c = (3/2R. For diatomic molecules (such as hydrogen, nitrogen and oxygen and air, c = (5/2R. Now a monoatomic secies, aroximated as a oint mass has three translational degrees of 17

18 Figure 4: From Sommerfeld hermodynamics and Statistical Mechanics. equivalent of heat. Mayer s cycle for the mechanical freedom. A diatomic molecule, regarded as a rigid rod of zero thickness has three translational and two rotational degrees of freedom. It is thus lausible that c = (f/2r where f is the number of degrees of freedom. In fact, kinetic theory leads to the law of equiartition of energy, whereby each degree of freedom contributes R/2 to c. A non-collinear olyatomic molecule (such as NO 2 has three rotational and three translational degrees of freedom so that c = 3R. Using the thermodynamic relation c c = R we have c = (1 + f/2r. he ratio of secific heats γ = c /c = C /C ( adiabatic index lays an imortant role in adiabatic transformations to be discussed shortly. hermodynamics and kinetic theory together redict that γ = 1 + 2/f. In articular γ > 1. For mono-, di- and generic olyatomic gases, γ = 5/3, 7/5 and 4/ Mayer s cycle for mechanical equivalent of heat o areciate the historical and hysical significance of the formula for the difference between heat caacities of an ideal gas we study Mayer s cycle (see Fig. 4. his was a cyclic rocess considered by R Mayer in his discovery of the first law of thermodynamics and in calculating the mechanical equivalent of heat. Consider n moles of an ideal gas in state 1 at atmosheric ressure 1 and occuying volume 1 at temerature 1. It is reversibly heated at constant volume to state 2 at temerature 2 > 1 and ressure 2 > 1. In this rocess, the gas does no work, but its internal energy is increased by U 1 = Q 1 = 2 1 C d. (35 Next the gas is allowed to exand isothermally to state 3 with volume 2 thereby reducing its ressure back to 1. By the caloric condition, there is no change in internal energy. Finally, the gas is comressed at constant ressure 1 from volume 2 to 1, thereby returning it to its initial state at temerature 1. In this final stage the work done by the gas is negative 1 ( 1 2 and the heat added to the gas is 1 C d. hus the increase in internal energy in the third rocess is 2 2 U 3 = Q W = ( 2 1 C d. (

19 Being a cycle, there is no net change in internal energy U 1 + U 2 + U 3 = 0 or 2 1 (C C d = 1 ( 2 1. (37 Now suose we make each of the rocesses infinitesimal, so that 2 1 = d and 2 1 = d, then 1 d = (C C d. (38 Looked at in isolation, we may interret this as a relation valid for a rocess at constant ressure 1. However, from the ideal gas law = nr, at constant ressure we must have d = nrd. hus the difference of heat caacities must satisfy C C = nr = (m/mr or c c = R/M. (39 where M is the molecular mass and c, c are the secific heats er unit mass m. Now the secific heats had been measured (and exressed in calories er gram. he gas constant had also been measured (and exressed in ergs er degree er mole or Joules er degree er mole. By equating the two we arrive at the mechanical equivalent of heat: 1 calorie of heat is equivalent to Joules of work. 3.5 Adiabatic transformation of an ideal gas Previously, we argued that a reversible isothermal exansion/contraction of an ideal gas is reresented by a hyerbola = constant on the lane. It is similarly interesting to find the curves on the lane reresenting reversible transformations in which no heat is exchanged, i.e., adiabatic transformations. A gas exands adiabatically if it is enclosed in a thermally insulated container and does work slowly ushing out a movable iston. Since δq = 0 the first law gives du + d = 0. aking, as indeendent variables, Joule s exeriment imlies that du = C d is indeendent of the change in volume. Eliminating = nr/ using the ideal gas law, we get C d + nr d = 0 or C d + (C C d = 0 or d + (γ 1d = 0 (40 where we used C C = nr and γ = C /C. Now assuming γ is indeendent of temerature (as is found to be aroximately the case, we may integrate to get log( γ 1 = constant. hus for adiabatic transformations of an ideal gas we must have γ 1 = constant or γ = constant or γ 1 γ. (41 hus adiabats are defined by curves on the lane along which γ is constant. Since γ > 1 we see that adiabats are steeer than isotherms if and are taken as abscissa and ordinate resectively. 3.6 Pressure and temerature in the atmoshere he ressure and temerature are known to dro with height in the atmoshere. Hot air at the Earth s surface rises while cooler air dros down in convection currents. Since air is a oor 19

20 Figure 5: From K. Huang Introduction to Statistical Physics. Adiabat is steeer than an isotherm for an ideal gas. conductor of heat, its motion can be taken to be adiabatic (i.e. ρ γ or γ 1 = constant. We will not attemt to model this flow, but instead assume a steady state where the ressure and density are nevertheless related by the adiabatic relation. o get a crude estimate of the temerature and ressure variation with height, we consider a thin horizontal layer of gas of height dz and area A in mechanical equilibrium. he weight of the air in the layer is A ρ g dz. In equilibrium, the ressure at the bottom of the layer exceeds that on the to of the layer by the weight er unit area: d = ρgdz. he negative sign is because ressure dros with increasing z. It turns out that the temerature variation with height is simler (essentially linear than that of ressure or density. So we eliminate density in favour of temerature using the EOS = ρr/µ where µ is the mean molecular mass of air (between 28 and 32 for N 2 and O 2. hus d = µg dz (42 R We may now eliminate ressure in favour of temerature by use of the adiabatic condition γ 1 γ = constant. Differentiating, we get d = γ d γ 1. (43 hus we arrive at a constant rate of change of temerature with height ( d γ 1 dz = µg. (44 R γ Find the numerical value of the temerature gradient. Also find an equation for d/dz and solve it. 3.7 Carnot Cycle A reservoir or source of heat or heat bath at a given temerature is an idealized body that can exchange heat with other thermodynamic systems without suffering a change in temerature and without erforming work. A large body of water at a common temerature throughout behaves as a reservoir. Its temerature and volume may be taken to be roughly constant as it comes into contact with other small systems. he engine of a car rovides an examle of a cyclic heat engine. It absorbs heat at a high temerature t 2 (generated from combustion of fuel, does some work in moving the car, exels waste heat via the exhaust to the low temerature reservoir (surroundings and returns to its 20

21 Figure 6: From K Huang: Carnot cycle on a diagram. initial state. Here t denotes some emirical temerature scale, such as given by a mercury thermometer. A reversible engine is one that oerates around a reversible cycle, this is of course an idealization since no real engine is truly reversible. he Carnot engine is a reversible cyclic heat engine that converts heat Q 2 absorbed at a high temerature (t 2 reservoir to work W while also exelling some waste heat Q 1 at a low temerature (t 1 reservoir. It is of great imortance in understanding the limits of conversion of heat into work. Consider a fluid (e.g. an ideal gas whose state can be reresented by a oint in the lane. A Carnot cycle is a clockwise oriented closed curve ABCDA consisting of two isotherms and two adiabatics (recall that adiabats are steeer than isotherms for an ideal gas. A simle examle of a Carnot engine consists of a gas enclosed in a cylindrical container whose lateral walls are thermally insulated. A thermally insulated movable iston on the to of the cylinder allows work to be done by/on the gas while the diathermic base of the cylinder allows heat to be exchanged. One cycle of the engine consists of the following four stages. o begin with, the gas in the container is laced on the heat source at t 2 and brought to equilibrium at temerature t 2 while occuying a volume A. AB : he container is laced on a heat source (reservoir at t 2 and the iston is raised slowly causing the gas to exand isothermally from A B and absorb heat Q 2. BC : he container is laced on an insulator and the gas is allowed to exand adiabatically (reversibly from volume B to C while its temerature dros from t 2 to t 1. CD: he container is laced on a heat reservoir at temerature t 1 isothermally from C to D while exelling heat Q 1. and slowly comressed DA: he container is laced on an insulator and the gas is slowly comressed from D back to the initial volume A, raising its temerature from t 1 to t 2. Let us aly the 1st law to one cycle of the Carnot engine. Since it has returned to its initial state, there is no change in internal energy (we do not assume here that the fluid is an ideal gas. hus the work W done by the gas must equal the net heat absorbed: W = Q 2 Q 1. It is an emirical observation (that we will rove later using the 2nd law of thermodynamics that if W > 0, Q 1 cannot be zero or negative, i.e., some waste heat must be surrendered at 21

22 the low temerature reservoir ( exhaust. he work done is also given by the area W = d enclosed by the closed curve ABCDA in the lane. he efficiency of the Carnot cycle is defined as the ratio of work done W = Q 2 Q 1 to heat Q 2 absorbed at the high temerature reservoir η = W Q 2 = Q 2 Q 1 Q 2 = 1 Q 1 Q 2. (45 No real engine with η = 1 has been constructed. We will show later that the 2nd law imlies η < 1. Since the Carnot cycle is reversible, we can run it in reverse ADCBA to get a refrigerator. In this case the signs of Q 2, Q 1 and W are all reversed. Since the curve is traversed counterclockwise d is also negative. In other words, work W is erformed on the refrigerator, allowing it to absorb heat Q 1 from the low temerature reservoir (at 1, the interior of the fridge and deosit heat Q 2 = W + Q 1 at the high temerature reservoir (at 2, the environment outside the fridge. If E is a Carnot engine, then we will denote the corresonding refrigerator by Ē. 4 Second law of thermodynamics he second law is a ostulate built on the observation that heat is not conducted from lower to higher temeratures. his leads to Clausius formulation. Interestingly there is an equivalent formulation of the second law due to Kelvin, that uts limits on the conversion of heat into work. Recall that while the first law recludes creation of energy, it does not lace restrictions on the conversion of heat into work or vice versa. Emirically, there is no restriction on the conversion of work comletely into heat. Indeed, a body at any temerature can be heated by friction, thereby converting mechanical work entirely into heat (electrical energy can similarly be converted into heat using a resistor. However, there are limits to the conversion of heat into work. If not, it would be ossible to roduce essentially an unlimited amount of energy using a device that extracts heat from our surroundings. he second law states the imossibility of creating such a eretuum mobile of the second kind. 4.1 Kelvin and Clausius ostulates and their equivalence Clausius ostulate: here is no thermodynamic rocess whose only final result is to transfer heat from a reservoir at a lower temerature to a reservoir at a higher temerature. Kelvin ostulate: here does not exist a thermodynamic rocess whose only final result is to extract heat from a reservoir which is at the same temerature throughout and convert it entirely into work. A useful re-statement of the Kelvin ostulate is this: If the only final result of a thermodynamic rocess is to extract heat Q from a reservoir at temerature t and convert it entirely into work W then W = Q 0. In other words, work was done on the system and converted into an equal amount of heat. he Kelvin ostulate recludes the construction of an engine that converts heat from one reservoir at a fixed temerature into work. Loosely, a Carnot engine is the next best thing, it does work by oerating between reservoirs at two different temeratures. 22

23 We will now show the equivalence of the Clausius and Kelvin ostulates. Convince yourself that to show A B it is enough to show that the falsehood of B imlies the falsehood of A. Clausius Kelvin: Suose the Kelvin ostulate is false. It would then be ossible to extract heat Q 1 from a reservoir at temerature t 1 and convert it to entirely into work W = Q 1. Now we could use this work W to deliver an equal amount of heat (e.g. via friction to a reservoir at any temerature, say one at temerature t 2 > t 1. We would thus have transferred heat from low to high temerature with no other effect, in violation of Clausius s ostulate. Kelvin Clausius: Conversely, suose the Clausius ostulate is false and we have a device that delivers heat Q 2 from a reservoir at t 1 to a reservoir at t 2 with t 2 > t 1. hen we could run a Carnot engine between the reservoirs at t 2 and t 1 to extract heat Q 2 from the reservoir at t 2 and deliver Q 1 < Q 2 to the reservoir at t 1 while erforming work Q 2 Q 1 > 0. In effect, we would have a machine whose only final effect is to extract heat Q 2 Q 1 from a reservoir at t 1 and convert it entirely into work, in violation of Kelvin s ostulate. Cautionary remarks: (1 In alying the second law, care must be taken in checking the hyotheses. In articular, the second law does not rohibit comlete conversion of heat into work, if that is not all that haens. For instance, in isothermal exansion of an ideal gas, the heat sulied to the gas is equal to the work done by the gas. here is no change in internal energy of the gas. However, at the end of the rocess, the gas has exanded and occuies a larger volume. his violates the assumtion in Kelvin s ostulate that there is no other final effect... So isothermal exansion of an ideal gas does not contradict the second law. On the other hand, a Carnot engine, being cyclic, is able to do work without any net change in state of the gas! But this clever device comes at a cost: it needs a second reservoir to accet waste heat! (2 Similarly, consider isobaric exansion of an ideal gas. Here the heat sulied to the gas goes artly into work done and artly into increasing the temerature (and internal energy of the gas (note that = nr imlies that must increase if increases at constant, n. Since U(, = U(, the internal energy must go u in an isobaric exansion. hus in isobaric exansion, not only do we not have comlete conversion of heat into work, we also see other effects like increase in volume of the gas and its temerature and internal energy. he second law does not rohibit isobaric exansion. 4.2 Consequences of the 2nd law for efficiency of engines A cyclic engine cannot absorb heat at two reservoirs and do work. If a cyclic heat engine (not necessarily a Carnot engine nor even a reversible one oerating between temeratures t 2 > t 1 does work W > 0 by absorbing heat Q 2 > 0 at t 2 and exelling heat Q 1 at t 1, then Q 1 must both be strictly ositive. his seemingly obvious but non-trivial statement is a consequence of the 2nd law. In articular, it means a cyclic engine cannot absorb heat at both reservoirs and do work. Proof: o see why, suose Q 1 0 so that the engine absorbs heat Q 1 at t 1. If Q 1 = 0 then we already have a contradiction with Kelvin s ostulate, so we may assume Q 1 > 0. hen we could bring the two reservoirs into thermal contact and allow heat of magnitude Q 1 to be conducted from t 2 to t 1 so that the low temerature reservoir would suffer no net heat gain/loss. After one cycle, the engine would have taken heat Q 2 Q 1 from the reservoir at t 2 and converted it comletely into work, contradicting the Kelvin ostulate. (An alternate argument is to convert art of the work done into heat amounting to Q 1 and 23

24 deliver it to the low temerature reservoir, so that the latter suffers no net heat exchange. Again we arrive at a contradiction with the Kelvin ostulate. Fundamental theorem on heat engine efficiencies: Now consider two cyclic engines ( unrimed and rimed working between the same temeratures t 2 > t 1. Let the amounts of heat absorbed and exelled at t 2 and t 1 and the work done be denoted Q 2, Q 1, W and Q 2, Q 1, W resectively. aking W, W > 0 all these Q s are ositive by the revious result. hen the second law imlies that if the first (unrimed engine is reversible, then Q 2 Q 1 Q 2 Q 1 or η η. (46 In other words, the efficiency η = 1 Q 1 /Q 2 of any cyclic engine is bounded above by the efficiency η = 1 Q 1 /Q 2 of the reversible cyclic engine between the same air of temeratures. We will rove this shortly. Corollary: If both engines are reversible, then Q 2 Q 1 = Q 2 Q 1 or η = η. (47 his is a consequence of the fundamental theorem. Indeed, if the rimed engine is reversible we have Q 2 /Q 1 Q 2/Q 1. Combining with Q 2 /Q 1 Q 2 /Q 1 on account of the reversibility of the first engine, the result follows. In other words, all reversible cyclic engines oerating between the same air of temeratures have the same universal efficiency (though they might absorb/exel different amounts of heat. he Carnot engine is a secial cyclic reversible engine that uses an ideal gas as its working substance, but its efficiency is universal. Proof of the fundamental theorem in the secial case Q 2 = Q 2 : We will run the reversible engine in reverse as a refrigerator and suly it with some of the work outut Q 2 Q 1 from the rimed heat engine (the rest of the work can be used to clean the refrigerator!. hus the total work done by the cyclic rocess defined by combining the engines is W tot = Q 2 Q 1 (Q 2 Q 1. (48 For simlicity, let us assume that Q 2 = Q 2 so that there is no net heat exchanged with the high temerature reservoir t 2. It follows that the combined engine absorbs heat Q 1 Q 1 from reservoir t 1 and does an equal amount of work W tot = Q 1 Q 1. his would violate the second law unless Q 1 Q 1 so that the combined engine simly converts works into heat. Dividing by Q 2 = Q 2 we get Q 1 Q 2 Q 1 Q 2 or Q 2 Q 1 Q 2 Q 1 or η η. (49 he above roof may be generalized to treat the case Q 2 Q Absolute thermodynamic temerature We concluded above that all reversible cyclic engines working between the same air of temeratures t 2 > t 1 have the same Q 2 /Q 1 ratio (or efficiency η = 1 Q 1 /Q 2, irresective of the 24

25 heats absorbed/exelled or work done. hus for a reversible cycle, Q 2 /Q 1 is some universal function of the temeratures Q 2 = f(t 1, t 2. (50 Q 1 Note that f(t 1, t 2 > 0 since Q 1,2 > 0 assuming the engine does ositive work. We will show now that the function f(t 1, t 2 satisfies the reroducing/multilicative roerty f(t 0, t 1 f(t 1, t 2 = f(t 0, t 2 (51 where t 0 is any temerature. o see this, suose C 1 (t 1, t 0 and C 2 (t 2, t 0 are two reversible cycles working between the indicated temeratures. For i = 1, 2, C i absorbs heat Q i at t i and exels heat Q 0 at t 0. hen we must have Q 2 Q 0 = f(t 0, t 2 and Q 1 Q 0 = f(t 0, t 1. (52 Now consider the combined reversible cycle C consisting of one cycle each of C 2 and the reverse of C 1. C absorbs heat Q 2 at t 2 and exels Q 1 at t 1 with no net heat exchanged at t 0. hus we must have Q 2 Q 1 = f(t 1, t 2. (53 Combining, we get the advertised multilicative roerty Q 2 = f(t 1, t 2 = f(t 0, t 2 Q 1 f(t 0, t 1. (54 Since t 0 is an arbitrary temerature, we may treat it as a fixed reference temerature and regard f(t 0, t i as a function θ of t i alone. hus we write f(t 0, t = θ(t > 0. (55 It follows that f(t 1, t 2 can be exressed in terms of the single function of one variable θ: Q 2 = f(t 1, t 2 = θ(t 2 Q 1 θ(t 1. (56 Note that θ(t is not uniquely defined: we may redefine θ(t by a ositive multilicative constant without affecting Q 2 /Q 1. It now makes sense to switch from the emirical temerature t (which is defined using the secial roerties of a thermometric material, say mercury, used in a thermometer to the socalled absolute thermodynamic temerature θ(t. It is conventional to choose the multilicative constant in θ so as to ensure that the boiling and freezing oints of water at atmosheric ressure differ by 100: θ(t b θ(t f = 100. he existence of absolute thermodynamic temerature is a consequence of the 2nd law. It is indeendent of the roerties of the working substance in the heat engine, as it is defined in terms of the universal efficiency of reversible cycles. We may now write this universal efficiency as η rev = 1 Q 1 Q 2 = θ 2 θ 1 θ 2. (57 25

26 We may also exress the ratio of work done to heat exelled at the low temerature reservoir: W Q 1 = Q 2 Q 1 Q 1 = θ 2 θ 1 θ 1 = θ 2 θ 1 1. (58 Now running the engine in reverse, the signs of Q 1, Q 2 and W are reversed. W/Q 1 is then the work required to extract a unit amount of heat from the interior of the (reversible refrigerator. θ 2 is tyically room temerature, and is fixed. As exected, the work (sulied through electrical energy required grows as the temerature of the interior θ 1 falls. 4.4 Equality of absolute thermodynamic and gas temeratures he absolute thermodynamic temerature θ coincides with the ideal gas thermometer temerature defined earlier. o see this, we consider a Carnot cycle, i.e. a reversible engine with an ideal gas as working substance comosed of the following four stages. (A - B isothermal exansion absorbing heat Q 2 from reservoir at gas temerature 2 ; (B - C adiabatic exansion (C- D isothermal comression at gas temerature 1 exelling heat Q 1 and (C - A adiabatic comression. We will now show that Q 2 /Q 1 = 2 / 1. Proof: he heat Q 2 absorbed during the isothermal exansion is equal the work done W AB since the internal energy of the ideal gas is unchanged (by Joule s exeriment, internal energy of an ideal gas deends on temerature an not volume, so U A = U B. hus from (17 Q 2 = W AB = R 2 log B A. (59 Similarly, the heat Q 1 exelled during the isothermal comression is hus Q 1 = W CD = R 1 log C D (60 Q 2 Q 1 = 2 1 log( B / A log( C / D. (61 It turns out that the ratios of volumes are equal, B / A = C / D. o see this, recall that in an adiabatic rocess, γ 1 is constant where γ = c /c. hus 2 γ 1 B = 1 γ 1 C and 2 γ 1 A = 1 γ 1 D. (62 Dividing one by the other, we get B / A = C / D. hus the ratio of heat absorbed to heat exelled is the ratio of gas thermometer temeratures Q 2 Q 1 = 2 1. (63 Previously, we found that Q 2 /Q 1 = θ 2 /θ 1 is the ratio of absolute thermodynamic temeratures. hus and θ can differ at most by a multilicative constant. However we must have = θ since both temerature scales have been chosen so that the boiling and freezing oints of water are 100 degrees aart at atmosheric ressure. We will denote this common absolute temerature scale by, the Kelvin scale. In articular, the efficiency of a Carnot engine is η = (64 26

27 4.5 Clausius heorem Consider a system S that undergoes a cyclic rocess P in which it comes into contact with reservoirs at temeratures 1,, n from which it absorbs heat in the amounts Q 1,..., Q n and erforms work W = i Q i. By the 2nd law, some of the Q is must be negative (heat exelled, otherwise, heat would have been comletely converted into work without other net effect. Clausius theorem states that Q i 0 (65 i i with equality for a reversible cycle. It is imortant here that be the absolute thermodynamic temerature. By taking a limit as n we may arrive at a cycle in which the system absorbs heat δq( from a distribution of sources with temeratures. Clausius theorem then becomes δq 0 (66 with equality for a reversible cycle. Note here that is the temerature of the articular reservoir from which the system S receives heat δq, it is not necessarily the temerature of S while it receives that heat. After all, in general heat cannot naturally be conducted from a cold to hot body, so the temerature of S must be bounded above by that of the reservoir if it is to receive heat (and must be greater than or equal to if it is to give u heat. If the heat is received reversibly, then the temeratures of the system and reservoir coincide. Proof: Clausius theorem is a consequence of the 2nd law and makes elegant use of Carnot cycles. In addition to the n reservoirs at i (i = 1,..., n, it is convenient to introduce a reference reservoir at an arbitrary temerature 0 > 0 and a sequence of Carnot engines C i oerating between temeratures 0 and i. C i is chosen to absorb heat Q 0 i at 0 and exel heat Q i to the reservoir at i. Since S absorbs Q i at i during the rocess P, there is no net heat exchanged at the n reservoirs in the combined cycle consisting of P, C 1, C n. In this combined cycle, an amount of heat Q 0 = i Q 0 i = 0 i Q i i (67 is absorbed at 0. Now the work done by S in rocess P is W = i Q i while the work done in C i is Q 0 i Q i. hus the total work done in the combined cycle is W total = i Q i + i (Q 0 i Q i = Q 0. (68 hus, the only final result of the combined cycle is to convert heat Q 0 from 0 entirely into work. his would violate the 2nd law unless Q 0 0, i.e., 0 Clausius inequality follows since 0 is a fixed ositive temerature. i Q i i 0. (69 Now if P is a reversible rocess, we can run it as well as each of the C i backwards. In the reversed combined cycle, the signs of all the Q s would be reversed leading to the reversed 27

28 Clausius inequality i Q i i 0. (70 Combining the inequalities for the forward and backward rocess we conclude that i Q i/ i = 0 for a reversible rocess. 5 Entroy and some consequences of the 2nd law he most imortant consequence of the second law is the existence of a new state function called entroy. 5.1 Absolute temerature as an integrating denominator and the entroy Clausius theorem states that for any reversible cyclic rocess, δq = 0. (71 rev hus the line integral of the differential δq/ vanishes around any closed curve lying on the sace of equilibrium states. Now suose P and P are two reversible rocesses between the same air A, B of initial and final equilibrium states, and let us denote by P the reverse of rocess P. hen P P is a reversible cyclic rocess (A B A to which Clausius theorem alies δq δq 0 = P P P = δq P. (72 In other words, the line integral of δq/ is indeendent of the (reversible ath chosen. On the other hand, we have observed that the heat change P δq is ath-deendent (even for reversible rocesses. his means that though the infinitesimal heat change δq is not an exact differential on the sace of equilibrium states, dividing it by the integrating factor turns it into the exact differential δq/. Entroy of equilibrium states: he second law of thermodynamics has allowed us to define an absolute temerature which is an integrating denominator for the heat δq added in an infinitesimal reversible rocess. he exact differential δq/ = ds defines an extensive state function (entroy uto an additive constant. hus, with resect to a reference equilibrium state A, the entroy of any equilibrium state B that can be reached from A by a reversible rocess is given by δq S(B = S(A + R(A B. (73 R(A B is any reversible ath from A to B. By the exactness of δq/, S(B is indeendent of the choice of reversible ath. A different choice of reference state will add a common constant to the entroy of all equilibrium states. Note that we have not defined the entroy of states that are not in equilibrium, we will address this later. hus, loosely we may say that the zeroth, first and second laws of thermodynamics each ostulates the existence of a new state function: temerature, internal energy and entroy. 28

29 For an infinitesimal reversible rocess 10, we may write the heat added to the system as δq = ds. hus a reversibly adiabatic rocess is the same as an isentroic rocess. In articular, the heat caacities may be exressed as C = ( S and C = For infinitesimal rocesses we have a new form of the 1st law: ( S. (74 ds = du + d or du = ds d. (75 In this form, S and are the natural variables that the internal energy deends on. In articular, if the function U(S, is known, then we may comute the temerature and ressure, as functions of S and : ( ( U U = and =. (76 S S However, the internal energy is often not directly measurable and we seldom have an exlicit formula for U as a function of S and (e.g. for an ideal gas it is easier to exress U as a function of, U = C. Remarkably, we can eliminate the internal energy from (76. Indeed, taking one more artial derivative and equating mixed artials first of Maxwell s four relations: ( = S 2 U S = 2 U S we arrive at the (. (77 S hree more Maxwell relations follow by use of the trile roduct identity: ( ( ( ( ( S =, = = S S, ( S. (78 Later, we will re-derive these remaining Maxwell relations using the three other thermodynamic otentials (Helmholtz free energy, enthaly and Gibbs free energy in lace of the internal energy. 5.2 Exactness of ds = δq/ and the energy equations Energy equation: he fact that ds is an exact differential imlies that its comonents (regarded as a vector must have zero curl. Choosing and as indeendent variables we have ds = 1 ( U 1 d + U + d (79 hus we must have ( 1 U = ( 1 U + or 1 2 U = 1 U U (80 10 For irreversible infinitesimal rocesses, δq is well-defined, but we have not defined ds. 29

30 he mixed artials cancel by Schwarz s theorem leaving the so-called -form of energy equation ( ( U =. (81 he energy equation exresses the volume derivative of internal energy at constant (which is difficult to measure in terms of the more directly measured ressure and temerature. For a fluid satisfying the ideal gas EOS = nr the energy equation becomes ( U = (nr/ = 0. (82 his is consistent with our earlier assumtion (based on Joule s exeriment, that the internal energy of an ideal gas is a function of temerature and not volume. It now seems as if we can dro this assumtion and instead derive the volume-indeendence of U using the energy equation. However, this is not quite true, because the assumtion that U is indeendent of volume has already cret into the derivation of the energy equation. Strictly seaking, the temerature that aears in the energy equation (or as an integrating factor for δq is the absolute thermodynamic temerature. o show the equality of gas and thermodynamic temeratures (see 4.4, we assumed that the internal energy of an ideal gas is a function of temerature alone, so that it does not change during an isothermal comression in a Carnot cycle. hus the volumeindeendence of the internal energy of an ideal gas and the functional form of U( (i.e. that C is a constant are indeendent ostulates on the nature of an ideal gas, that do not follow from the alication of the first and second laws of thermodynamics to the EOS = nr. We will aly the energy equation later on to understand the sloe of hase boundaries in a diagram for first order hase transitions (Claeyron equation. wo other forms of the energy equation may be obtained as above by choosing (, and (, as indeendent variables. In the former case the 1st law ds = du/ + (/ d becomes (subscrits denote artials ( 1 ds = U + ( 1 d + U + d. (83 he condition for vanishing curl is ( 1 U + ( 1 = U +. (84 he mixed second artials cancel and we get the form of the energy equation ( ( ( U =. (85 For a gas satisfying the ideal EOS, the RHS vanishes, which is consistent with the assumtion that U(, is indeendent of ressure. Similarly, taking and as indeendent, ds = ( 1 U + 30 d + 1 U d. (86

31 he resulting (, form of the energy equation is ( ( U ( = + ( U ( For a gas obeying the ideal EOS, this equation is satisfied if both we have assumed based on Joule s exeriment. ( U. (87 and ( U vanish, as 5.3 Entroy of an ideal gas he second law guarantees that is an integrating denominator for the infinitesimal heat δq absorbed in a reversible rocess. Let us demonstrate this for an ideal gas and thereby identify the state function (entroy S such that δq/ = ds. For n moles of an ideal gas = nr/ and du = C d with C indeendent of (this is called the caloric condition, U(, = U( is indeendent of volume. So the first law δq = du + d imlies ds = δq o carry out exlicit integration, we assume that C good aroximation. It follows that hus, the entroy of an ideal gas is = C d + nrd. (88 is indeendent of temerature, which is a ds = d ( log C + log nr = d log ( C nr. (89 S(, ideal gas = C log + nr log + S 0 = log ( C nr + S 0 (90 where S 0 is an additive constant of integration. he dimensional constants in S 0 must ensure that the argument of the logarithm is dimensionless and that S is extensive (the term nr log entails logarithmic violation of extensivity under n λn, λ. It turns out that the third law of thermodynamics constrains the value of S 0. For a monoatomic ideal gas C = (3/2nR from kinetic theroy, so S(, monoatomic = nr log( 3/2 + S 0. (91 Notice that if an ideal monatomic gas undergoes an isentroic rocess, then 3/2 must be constant. More generally using nr = C P C we find that C C P C must be a constant. aking the C th root this means γ 1 must be constant. Entroy may also be exressed in terms of and or and by using the EOS = nr/ and the relation C = C + nr: S(, = C log nr log(/nr + S 0 and S(, = C log(/nr + C log + S 0. (92 Entroy rise in free Joule exansion of an ideal gas. In free Joule exansion of an ideal gas from a container of volume to one of volume, no work is done and no heat is added to the gas. Its final temerature and internal energy are the same as their initial values. hough the rocess is irreversible, the initial and final states are equilibrium states, so we may evaluate their entroies. he constant S 0 cancels out and we obtain a simle formula for entroy rise in free exansion of an ideal gas: S(, S(, = nr log( /. (93 31

32 5.4 Proerties of entroy and examles Clausius inequality: Recall that we defined the entroy of an equilibrium state B with reference to an equilibrium state A as the line integral of the exact differential δq/ along any reversible ath R from A to B. Suose in addition to the reversible ath R(A B, the equilibrium states A and B are also connected by a (not necessarily reversible ath P. hen the closed ath P R reresents a cyclic rocess to which we may aly Clausius theorem 11 P R δq 0 P δq R δq = S(B S(A. (94 hus the line integral of δq/ along any ath P joining equilibrium states A and B is bounded above by the entroy difference. Equality holds if (and only if P is reversible. Since S is a state function, the entroy change in any rocess P between equilibrium states A and B is always given by S(B S(A. It may also be exressed as the line integral P δq/ only if P is reversible. Law of non-decreasing entroy for insulated systems: For a thermally insulated system, the heat added δq during any rocess must vanish. hus we must have S(B S(A: the final entroy of a thermally insulated system must be at least as large as its initial value. his is often loosely stated as the law of increase of entroy of an isolated system. On the other hand, the entroy of systems that can exchange heat with their surroundings could increase or decrease. For examle, we might reduce the entroy of a body at temerature h by letting heat Q be conducted from it to another body at a lower temerature l < h. (1 Heat conduction: Suose an isolated system consists of two large bodies at temeratures h and l with h > l. A small amount of heat Q > 0 is conducted from the hotter one to the colder one. From Clausius theorem, the changes in entroy satisfy the following inequalities ( S h δq Q h < 0 and ( S l δq Q l > 0. (95 Here we have assumed that the small heat transferred does not significantly change the temeratures of either of the big bodies. hus the change in entroy of the system is given by ( 1 S = ( S h + ( S l Q 1 > 0 as h > l. (96 l h We see that entroy is roduced in heat conduction. he entroy of the colder body has increased. he entroy of the hotter body tyically decreases, though our inequality allows it too to increase. (2 Heating through friction: he initial energy in a swing is dissiated as it executes damed oscillations, heating u the susension through friction. In this rocess mechanical work W done on the swing is comletely converted into heat Q = W and the temerature of the susension is raised from i to f > i. he change in entroy must satisfy S f i δq f > δq = 1 f δq = Q > 0. (97 i f f i f 11 Note that in general we may not consider R P since P may not be reversible! 32

33 So heating through friction is an entroy raising rocess. Remark: Suose equilibrium states A and B of a thermally insulated system are connected by a reversible rocess (which must necessarily be adiabatic δq = 0. hen S(B = S(A. On the other hand, if they are connected by an irreversible rocess (again with no heat exchange then S(B > S(A. Evidently, the same air of equilibrium states of a thermally insulated system cannot be connected by both a reversible and irreversible rocess with no heat change. E.g. Free Joule exansion of an ideal gas from equilibrium state (, 1 to equilibrium state (, 2 with no heat exchange is an irreversible rocess between two states of an ideal gas that differ in volume 2 > 1 but have the same temerature. It involves an entroy increase of nr log( 2 / 1. Can we find a reversible way of adiabatically exanding the same ideal gas between the same equilibrium states? his is not ossible. he conditions of being a reversibly adiabatic exansion γ = const as well as a reversibly isothermal exansion P = const cannot be simultaneously satisfied for an ideal gas with γ 1. here are reversible ways of exanding such a gas isothermally, but they involve work and hence heat exchange and therefore cannot be adiabatic. E.g. let the gas exand while comressing a sring connecting the iston to a fixed wall, the rocess can be reversed by exanding the sring while comressing the gas. Entroy of comosite systems Suose a system s is comosed of two homogeneous arts s 1 and s 2, each in equilibrium with its own entroy. hen it is often (though not always the case that we can define the entroy of s to be the sum S = S 1 + S 2. For instance, suose the internal energies are additive, U = U 1 + U 2 and the work done by s is the sum of the works done W = W 1 + W 2. hen the heat added to s would also be additive δq = δq 1 + δq 2. If is the temerature of the reservoir from which heat δq is received, then the entroy S(B (defined as B A δq/ of the state B of the combined system (relative to state A satisfies the additive roerty: S(B = B A δq = B1 A 1 δq 1 + B2 A 2 δq 2 = S 1(B 1 + S 2 (B 2. (98 Here A 1 and B 1 are the states of subsystem s 1 when the total system is in states A and B, the same alies to A 2 and B 2. he additivity of internal energies may fail for instance if the energy of the interface (common boundary surface of the two systems is comarable to that of either s 1 or s 2 in isolation. When the above additivity roerty holds, we may use it to define the entroy of nonhomogeneous systems that are not in equilibrium. We subdivide the system s into homogeneous arts s = i s i each of which is in equilibrium (say at temerature i and ressure i and define the entroy of s to be the sum of entroies. In this manner we may try to define the entroy of a flowing fluid, which is evidently not in equilibrium. Infinitesimal version of Clausius inequality: Secializing the Eq. (94 to an infinitesimal rocess P we may write S(B S(A ds and P δq/ δq/. So for an infinitesimal rocess in which a system absorbs heat δq from a reservoir at temerature, ds δq. (99 If the rocess is irreversible, then ds > δq. Equality holds for a reversible rocess. We had initially assumed that the rocess P begins and ends at equilibrium states. his assumtion may be relaxed in situations where we can define the entroy for non-equilibrium states. Entroy and loss: he entroy increase in an irreversible rocess of an isolated system may be interreted as a loss in the ability to do useful work. o see this, we consider two rocesses: 33

34 (a free Joule exansion of n moles of a thermally insulated ideal gas from (, 1 to (, 2 with 2 > 1 and (b slow (reversible isothermal exansion of the same ideal gas between the same initial and final states while comressing a sring. In free exansion, no work is done, but the entroy of the gas rises by S = nr log( 2 / 1. In the slow isothermal exansion, there is no change in U as does not change, so the work done W = nr log( 2 / 1 must equal the heat Q absorbed. hus the increase in entroy of the gas in this isothermal exansion is S = δq/ = nr log( 2 / 1. We see that in both free and reversible exansion, there is an identical increase in the entroy of the gas. his had to be the case as both rocesses have the same initial and final states, and entroy is a state function. How do the rocesses differ? (a Free exansion is irreversible, the entroy of the system as a whole (gas + insulated container increases by S. In (b, the entroy of the surroundings of the gas (the sring and reservoir decrease, as they give u heat Q at a fixed temerature. In fact, since the gas + surroundings form an isolated system undergoing a reversible rocess, the total entroy is unchanged. Indeed, the rocess can be reversed by isothermally comressing the gas slowly by letting the sring exand. hus we may interret the increase in entroy in free exansion as a loss in useful work that the gas could do. In (b there is no increase in entroy for the system as a whole, and useful work done by the gas is stored in the sring. Increasing entroy and stability: An isolated system (thermally and mechanically isolated must have a fixed internal energy (as δq = δw = 0. We showed above that its entroy is a non-decreasing function of time. So if an isolated system is in a state of maximal entroy consistent with its energy then it can undergo only those transformations to other states with the same energy and maximal value of entroy. In ractice all macroscoic rocesses are irreversible (rolling of balls, running of engines etc.. In other words in real isolated hysical systems, entroy is strictly increasing with time. So even if there is another state with the same maximum entroy, a real system cannot access it. hus, a maximum entroy state of an isolated system is a stable state: once in such a state, the system remains in it if isolated. 5.5 hermal exansion, comressibility and tension coefficients for an ideal gas Some measurable coefficients we have defined for a gas/fluid are (here H = U+ is enthaly: ( ( ( ( δq U δq H heat caacity at constant volume C = = and ressure C = = d d comressibilities: isothermal κ = 1 ( and isentroic κ S = 1 ( S coefficient of thermal exansion α = 1 ( and tension β = 1 (. (100 For an ideal gas we may determine the coefficients of thermal exansion α and tension β as well as the isothermal comressibility κ directly from the EOS = nr : α = 1 ( = nr = 1, β = 1 ( = nr = 1 and κ = 1 ( = nr = 1 2. (101 We see that an ideal gas is more comressible at low ressures. At very high ressures, gases liquify and may often be treated as incomressible. he coefficients of thermal exansion and tension are both equal to 1/. At high temeratures, there is hardly any fractional change in volume or ressure of an ideal gas when the temerature is slightly increased. 34

35 o find the adiabatic comressibility κ S = 1 ( S of an ideal gas, we use the formula S(, = C log + C log C log nr + S 0. (102 ( An adiabatic reversible rocess is isentroic, so we have ds = 0 and in articular S = 0, which becomes, uon dividing by C 1 γ + 1 ( = 0 or κ S = 1 S γ. (103 his formula for κ S deends on the heat caacity ratio γ = C /C which is not determined by thermodynamic rinciles alone. In fact, C and C deend on the microscoic nature of the gas (mono/diatomic etc. hermodynamics, however, does tell us that C C v = nr. An alternative aroach to comuting κ S is to recall that γ is constant for reversible isentroic transformations of an ideal gas. aking the differential of this, we get κ s = 1/γ. 5.6 Boltzmann s statistical interretation of entroy hermodynamics rovides a macroscoic descrition of systems indeendent of their microscoic structure (atomic/molecular structure of matter. Statistical mechanics (SM is a framework based on the dynamics of microscoic constituents. Boltzmann gave a statistical interretation for entroy and the law of its increase for isolated systems. SM begins with the microscoic dynamical state of a system (say a gas, secified by ositions and momenta of a large number of molecules. On the other hand, the thermodynamic state of the same system is defined by the values of a few macroscoic variables such as,,. here are tyically very many microstates that corresond to a given macrostate, since for examle, different molecular motions can result in the same total energy and ressure in a fixed volume. Suose π is the fraction of microscoic states corresonding to a given macrostate, it is interreted as the statistical robability of the given thermodynamic state. Boltzmann related the robability π of a macrostate to its entroy: S = k B log π where k B = R/N A is Boltzmann s constant. Based on this relation, we see that the entroy of an isolated system is non-decreasing rovided transformations of the system to states of lower robability are forbidden. o make Boltzmann s relation lausible, let us show that if there is a functional relation S = f(π between entroy and robability, then it must be of the logarithmic form given by Boltzmann. Following Fermi, consider a thermodynamic system s in a articular state with entroy S = f(π where π is the robability of the state. Suose s is comosed of two searate arts s 1 and s 2 with entroies S 1 = f(π 1 and S 2 = f(π 2 where π 1,2 are the robabilities of the corresonding states. hen under the conditions for extensivity, S = S 1 +S 2 while π = π 1 π 2 so that f(π 1 π 2 = f(π 1 + f(π 2. his functional relation must hold for any robabilities π 1,2 = x, y, i.e. f(xy = f(x + f(y for all 0 x, y 1. o determine f we assume it is differentiable and suose that y = y 0 + ɛ for small ɛ. hen aylor exanding, f(x(y 0 + ɛ = f(x + f(y 0 + ɛ f(xy 0 + ɛxf (xy 0 = f(x + f(y 0 + ɛf (y 0. (104 By the functional relation f(xy 0 = f(x + f(y 0 so we have xf (y 0 x = f (y 0 or y 0 xf (y 0 x = y 0 f (y 0 = const. = k. (105 35

36 Figure 7: S diagram of a Carnot cycle. From K Huang, Introduction to statistical hysics. Now denoting y 0 x as χ and integrating we have χf (χ = k or f(χ = k log χ + const. (106 As desired, we find that S = f(π = k log π uto an undetermined additive constant. 5.7 Heat engines viewed through a S diagram A convenient feature of diagrams is that the work done in a rocess is given by the area d under the curve reresenting the rocess. he heat added, however, does not have a simle geometric interretation in a diagram. In view of the second law, ds is the heat reversibly added. hus, if we choose the conjugate variables and S as indeendent variables, then the heat added in a reversible rocess is once again given by the area ds under the curve in a S diagram. Isentroic (reversible adiabatic and isothermal rocesses are reresented by vertical and horizontal lines in a S diagram. A Carnot cycle becomes a rectangle with edges along the isotherms = 2 and = 1 and along the adiabats S = S 1 and S = S 2 corresonding to isothermal exansion and comression and adiabatic comression and exansion (see figure. he heat absorbed at 2 is 2 (S 2 S 1 = A + B while the heat exelled at the low temerature is 1 (S 2 S 1 = B. he net heat absorbed is the area A enclosed, which must also equal the work done in the cycle. hus the efficiency is η = A/(A + B. 6 hermodynamic otentials 6.1 Legendre transform from internal to free energies and enthaly Combining the first and second laws of thermodynamics for infinitesimal reversible transformations of a gas, the increase in internal energy is du = ds d. hus the internal energy, which is our first examle of a thermodynamic otential, is naturally a function of entroy and volume 12. he other two conjugate variables are given by its artial derivatives = ( U S and = ( U. he change in internal energy is articularly simle for adiabatic (ds = 0 or isochoric (d = 0 rocesses. However, many rocesses take lace at constant temerature (room S temerature or at constant ressure (atmosheric ressure. hus, it would be convenient to have thermodynamic otentials with other indeendent variables. 12 Of course, we may view U as a function of and as we did in deriving the ds equations. What we mean here are the indeendent differentials that aear in du directly as a consequence of the 1st and 2nd laws. 36

37 he Legendre transform (L, familiar from Lagrangian and Hamiltonian mechanics 13 allows us to change indeendent variables. For instance, to switch from (S, to (, as indeendent variables, we define the Helmholtz free energy F = U S. It follows that df = du ds Sd = d Sd. hus F is naturally a function of (, with the other two given by = ( F and S = ( F. (107 We will see that the free energy may be interreted in terms of the work that a system can do at constant temerature. o obtain F = U(S, S as a function of, we must exress S in terms of and using the relation = U S. Since this is just the condition for U S to be extremal, we may write F (, = ext S (U(S, S (108 Enthaly is the thermodynamic otential that deends on (S,. It is defined via the relation H = U + so that dh = du + d + d = ds + d and ( ( H H = and =. (109 S It is a Legendre transform of the internal energy S H(S, = ext (U(S, + (110 Gibbs free energy is the last thermodynamic otential, with indeendent variables (,. It may be obtained from U(S, by a succession of two Legendre transforms G = U S + = F + = H S (111 Evidently dg = df + d + d = Sd + d so that ( ( G G S = and = Gibbs free energy may be obtained as a Legendre transform of U, F or H :. (112 G(, = ext,s (U(S, S + = ext (F (, + = ext S (H(S, S (113 In summary, we have defined three new state functions (thermodynamic otentials via Legendre transforms from the internal energy: F = U S, H = U + and G = U S +. (114 Note that we cannot obtain thermodynamic otentials with (, or (, S as natural indeendent variables via Legendre transformation from U(S,. his is because the L only allows us to relace a variable by its conjugate, i.e., or S. Note also that U and H are 13 he Lagrangian is a function of coordinates q and velocities q, while the Hamiltonian is a function of coordinates and momenta. H = q L with = L or H(q, = ext q q( q L(q, q. 37

38 defined u to additive constants. On the other hand, the Helmholtz and Gibbs free energies are defined u to a linear function of temerature, since entroy has been defined only u to an additive constant. We will return to this issue when we discuss the third law of thermodynamics. he first and second laws of thermodynamics may be exressed in terms of any one of the four thermodynamic otentials: du(s,, N = ds d + µdn, df (,, N = Sd d + µdn, dh(s,, N = ds + d + µdn and dg(,, N = Sd + d + µdn. (115 For an ideal gas, the Helmholtz free energy exressed as a function of its natural variables, is F ideal gas = U S = C (C log + nr log + U 0 S 0. (116 It is easily verified that = ( F leads to the ideal gas equation of state while its entroy is recovered by comuting ( F. Similarly, the Gibbs free energy G(, of a ideal gas is G = U S + = C + U 0 (C log + nr log + S 0 + nr = C + U 0 (C log nr log(/nr + S 0. (117 Notice that both free energies are undetermined uto a linear function of temerature due to additive constants in S and U. Why are F and G called free energies? Let us find out. 6.2 Interretation of Helmholtz free energy For a mechanical system, the law of conservation of energy says that the work done is equal to the decrease in internal energy. For a thermodynamic system, the work done W = U + Q may be less/more than the decrease in internal energy deending on whether heat is absorbed or given u by the system. he second law allows us to use the free energy to get an uer bound on the work that a system in contact with a reservoir at constant temerature can do. Consider a transformation of a system from state A to B while in contact with a heat reservoir at temerature. By the second law B A δq S(B S(A. (118 Since the temerature of the reservoir is fixed, we have an uer bound on the heat received by the system Q = B A δq (S(B S(A. (119 Note that the temerature of the system need not equal, in other words the rocess need not be isothermal. By the first law W = Q U, so we get an uer bound on the work done W U(B + U(A + (S(B S(A. (120 If we assume that the initial and final temeratures of the system are equal to, then the RHS is just the difference in Helmholtz free energies W F (A F (B = F. (121 38

39 hus the maximum work that a system can do while in contact with a reservoir at constant temerature is the decrease in free energy. In this sense, F is the energy freely available to do work, hence the name. If the rocess is reversible, then the inequality is saturated and F (A F (B is the actual amount of work done. Note that to be reversible, the system would have to have the same temerature as the reservoir throughout. If the rocess is irreversible, we have a strict uer bound on the work done. 6.3 Gibbs criteria for thermodynamic equilibrium We observed that the entroy of an isolated system is a non-decreasing function of time. What is more the change in entroy of an isolated system S > 0 if the rocess is irreversible, with equality for reversible rocesses. All macroscoic natural rocesses are found to be irreversible and are observed to roceed towards increasing entroy. What is more, any state of an isolated system with maximal entroy consistent with its conserved internal energy is articularly stable since there is no thermodynamic rocess that can further increase its entroy. he foregoing considerations lead us to Gibbs criterion for thermodynamic equilibrium: an isolated system with fixed energy and number of articles is in equilibrium if its entroy is maximal. Gibbs also gave another criterion for thermodynamic equilibrium that traces its origins to mechanical equilibrium. We are familiar with the fact that a article moving in a otential is in equilibrium if it is at a minimum of the otential. Gibbs second criterion states that if the entroy of an isolated system is held fixed, then it is in equilibrium when its energy is minimized. Neither entroy nor internal energy is an easily measured roerty of thermodynamic systems. Moreover, U is naturally a function of S and, but entroy is not easily controlled in the lab. Proerties that are more easily controlled are temerature, ressure and volume. his is where the other thermodynamic otentials come in handy. Recall that the natural variables of Helmholtz free energy, and Gibbs free energy are (, and (, resectively. hus it is useful to reformulate the condition for thermodynamic equilibrium in terms of F and G. Now in an infinitesimal rocess, ds δq/ (irresective of whether the system is isolated or not, with equality if it is reversible. So by the first law, ds du + d. If the system is mechanically isolated, then no work is done and so ds du or du ds 0. But this is just the condition df (, 0 for a system at constant temerature. hus we see that the Helmholtz free energy is a decreasing (recisely non-increasing function of time for a mechanically isolated system at constant temerature. he same alies to finite transformations of a mechanically isolated system at constant temerature: from the revious section, W = 0 F (A F (B. As a consequence, a system at constant temerature (equal to that of its environment undergoing isochore transformations (mechanically isolated, e.g. if is constant and d = 0 is in equilibrium when its free energy is a minimum. Recall that a mechanical system is in equilibrium when the otential is a minimum. F being a thermodynamic analogue, is sometimes called the thermodynamic otential at constant volume. We may also view the rincile of minimization of Helmholtz free energy as a constrained minimization roblem. Recall Gibbs second criterion for equilibrium: minimize internal energy holding entroy fixed. his is equivalent to minimizing U S where is a Lagrange multilier enforcing the constraint on entroy. Similarly, let us consider a system at constant ressure and temerature. his alies to chemical reactions haening at room temerature and at atmosheric ressure. Now G = 39

40 U S +, so at constant,, dg = du ds Sd + d + d = du ds + d 0 (122 as ds δq. hus Gibbs free energy decreases for infinitesimal transformations of a system at constant and. We could arrive at the same result by considering finite transformations. Suose our system undergoes a thermodynamic transformation A B at constant temerature and ressure. he work done is simly W = B A d = ( (B (A, which must be bounded above by the dro in free energy: W F (A F (B. So F (B + (B F (A + (A or G(B G(A. (123 In articular, a system at constant and is in equilibrium if G is a minimum. 6.4 Maxwell relations he first law for infinitesimal reversible rocesses du = ds d imlies ( U S = and =. he equality of second mixed artials of U leads to the first Maxwell identity: ( U S ( ( = S S he differential of Helmholtz free energy df = Sd d imlies ( F. hus we get a second Maxwell relation ( ( S = = S and ( F (124 = (125 Enthaly is naturally a function of S and P with dh = ds + dp. hus ( H S = and = which imlies the Maxwell relation ( H S Finally, for the Gibbs free energy, dg = d Sd, so ( ( =. (126 S S ( = ( G = and ( G = S and ( S. (127 We may obtain additional Maxwell relations by allowing the number of articles to vary by including a chemical otential term in the first law du = ds d + µdn. he Maxwell relations rovide surrising connections among thermodynamic coefficients. We will see an alication in the context of the third law of thermodynamics. 40

41 6.5 Extensivity, Euler equation and Gibbs-Duhem relaton We have mentioned before that if surface effects can be ignored, then the internal energy, work done, heat absorbed and entroy may be taken to be extensive. What this means is that if the volume (and consequently the number of articles is doubled then U, S etc are also doubled. More recisely, extensivity of internal energy is the condition U(λS, λ, λn = λu(s,, N for any real λ > 0. (128 Such a function is called homogeneous of degree one. Euler s theorem on homogeneous functions: More generally, f(x, y, is a homogeneous function of degree n, if for all x, y, for which f is finite, it satisfies f(λx, λy, = λ n f(x, y, λ. (129 Euler s theorem on homogeneous functions gives an alternate characterization of homogeneous functions, assuming differentiability. For definiteness, consider two indeendent variables. Examles of homogeneous functions include f(x, y = 2x 2 7y 2 + xy (degree two and g(x, y = 4x 7y or g = x 2 /y (degree one. Now define the scaling or dilatation/dilation oerator D = x x + y y. Notice that Df = 2f and Dg = g. Euler s theorem states that a homogeneous function is an eigenfunction of the scaling oerator with eigenvalue equal to the degree of homogeneity. Df = ( x x + y f(x, y = nf(x, y. (130 y Let us sketch a roof for one variable, essentially the same alies to any number of variables. So suose f(λx = λ n f(x for all λ, x. Let us consider an infinitesimal scaling by taking λ = 1 + ɛ where ɛ is small. hen we must have f(x + ɛx = (1 + ɛ n f(x (1 + nɛf(x (131 Rearranging, ( f(x + ɛx f(x f(x + ɛx f(x = nɛf(x or x = nf(x. (132 ɛx aking ɛ 0 holding x 0 fixed, the LHS tends xf (x. hus we get Df(x = nf(x as advertised. Alternatively we may use the aylor exansion on the LHS to conclude that xf = nf. f(x + ɛxf (x + O(ɛ 2 = f(x + nɛf(x + O(ɛ 2 (133 Now extensivity imlies that the internal energy is homogeneous of degree one, so it must satisfy ( S S + + N U = U. (134 N he above three artial derivatives of U may be obtained from the first law: du = ds d + µdn U S =, 41 U =, U N = µ. (135

42 hus U = S + µn. (136 Now recall that we had defined the Helmholtz free energy, enthaly and Gibbs free energy via successive Legendre transforms of the internal energy: F = U S, H = U +, G = F + = U S +. (137 As a consequence, when energy is extensive F = µn, H = S µn and G = µn (138 In other words, the chemical otential is simly the Gibbs free energy er article. he above formulae for thermodynamic otentials (in articular G = µn leads to the Gibbs- Duhem relations. Indeed, equating dg = µdn + d Sd with dg = µdn + Ndµ we get the Gibbs Duhem relation in the energy reresentation dµ = N d S d = vd sd (139 N Here v = /N is the secific volume and s = S/N is the entroy er article. We could also obtain this relation by equating du comuted from U = S + µn with the exression given by the first law. 7 Phase transitions 7.1 General features of 1st and 2nd order transitions When water freezes or boils, it undergoes a hase transition. We have also encountered the transition from the ferro- to aramagnetic hases of iron. hese are examles of so-called first and second-order hase transitions. A first order transition is characterized by a nonzero latent heat. For instance, suose we start with water at 30 C and heat it at constant atmosheric ressure, its temerature rises till we hit 100 C. he water starts to boil, with water being converted to vaour. However, the temerature remains constant till all the water has been converted to vaour. he amount of heat required to vaourize unit mass of water at 100 C is the latent heat of vaorization. he ice to water melting transition also involves a latent heat. he boiling/condensation transition can also be examined at constant temerature rather than ressure (See Fig 8(b. For instance, suose we begin with water vaour at = 105 C and atmosheric ressure. As we isothermally comress the vaour (decrease the volume of the container, the ressure increases until it reaches the so-called saturated vaour ressure, when the vaour starts condensing. Further comression does not lead to an increase in ressure. Instead, more and more of the vaour condenses isothermally at the fixed vaour ressure. At this stage, the vaour and liquid hases coexist in equilibrium. Phase coexistence is another characteristic feature of first order transitions. When all the vaour has condensed, further comression results in an abrut and stee increase in ressure, since liquids are not as comressible as gases. Fig 8 shows the equation of state surface in the thermodynamic state sace showing both the melting and boiling transitions. Note that the gas and liquid hases is each described 42

43 by an equation of state, though the EOS function f(,, is different for the two hases. Isotherms for a liquid to vaour transition are shown in a diagram. We will see that for two hases to coexist in equilibrium, their (Gibbs free energies must be equal. hus, free energies are continuous across a first order hase transition. On the other hand the densities and secific entroies of vaour and liquid hases that coexist at a common temerature and ressure are vastly different. (secific entroy of a gas is significantly more than that of the liquid (or solid, there are many more microstates corresonding to a given macrostate of a gas, roughly, gases are more disordered than liquids or solids. Since secific volume and secific entroy are derivatives of the secific Gibbs free energy with resect to ressure and temerature, we infer that the first derivatives of the free energy are discontinuous across a first order hase transition. his motivates Ehrenfest s classification of hase transitions. A hase transition is of order n = 1, 2,... if an n th artial derivative of free energy is the first one to be discontinuous. (a (b (c Figure 8: (a From K Huang: (b From E Fermi: Isotherms for liquid to vaour transition in a diagram (c From K Huang: Phase boundaries for solid-liquid-gas system in a diagram. he line of first order transitions ends at the critical oint reresenting a second order transition. At the trile oint, all three hases coexist. Notice from the and diagrams that the range of volumes over which the liquid and vaour coexist shrinks as we move to isotherms at higher temeratures. At the critical temerature c (374 C for water the coexistence region of the isotherm shrinks to a oint with the critical saturated vaour ressure c (217.7 atm for water. At this critical oint, vaour and liquid coexist with the same density and secific entroy and the latent heat vanishes. he transition changes from first to second order. his may be seen in a diagram (Fig 8(c showing coexistence curves that demarcate the boundaries between the various hases. he hase boundaries corresond to first order hase transitions. It is seen that the line of first order liquid-gas hase transitions ends at the critical oint. A line of first order hase 43

44 transitions often ends at a second order transition. At temeratures higher than c, water exists in a single hase with continuously varying roerties. Note that the melting transition is always first order, there is no critical oint for the solid to liquid transition. Instead we have a trile oint on the diagram (rojection of the trile-line isotherm in sace where solid, liquid and vaour can coexist at a first order transition. At temeratures less than the trile oint temerature, there is no liquid hase and vaour directly condenses to a solid via a first order transition. Secific heats tyically diverge at a second order hase transition ( = c. Recall that c = ( s where s is secific entroy and that entroy S = ( F. hus secific heats are given by second derivatives of the free energy. In a second order transition, the free energy and its first derivative are continuous across the transition but its second derivative suffers an (infinite discontinuity. Interesting henomena in the neighbourhood of the critical oint of water were discovered in building and working with steam engines (detailed measurements were collected in steam tables, which are like star charts in celestial mechanics. One such henomenon is critical oalescence, steam at the critical temerature was found to be oaque to light of ractically any wave length. Usually, a medium is transarent to light of wavelengths different from the tyical length scales of disturbances in the medium. Critical oalescence is interreted as a demonstration of the existence of fluctuations on all length scales in steam at c. It turns out that secific heats measure fluctuations in the internal energy, so diverging secific heats go hand in hand with fluctuations on all length scales. We say that 2nd order hase transitions dislay scale-invariance. he ferromagnetic to aramagnetic transition as is increased beyond the Curie-Weiss temerature is another examle of a second order hase transition. In this case, the Helmholtz free energy and its first derivative (the sontaneous magnetization are continuous, but the second derivatives of free energy, (the magnetic suscetibility and secific heat diverge at the critical temerature. 7.2 Claeyron-Clausius equation from energy equation he Claeyron equation gives a formula for the variation of saturation vaour ressure with temerature when liquid and saturated vaour coexist at the boiling oint (in terms of the latent heat of vaorization associated to the first order transition. It determines the shae of the hase boundary between vaour and liquid hases in a diagram. he key idea is to use the energy equation to exress U at constant volume in terms of at constant temerature. Following Fermi, we consider the region (below the critical oint of the diagram where vaour and liquid coexist. he isotherms are horizontal, so the ressure (saturated vaour ressure ( is a constant, deending only on temerature (and not volume. Similarly, the densities (or secific volumes v g and v l of the vaour and liquid are functions only of temerature. As the mixture is comressed isothermally, vaour condenses without changing the density of either hase. Similarly, let u g and u l be the secific internal energies. At a articular volume along the horizontal isotherm, suose m g and m l are the masses, so that the total mass m = m g + m l. he total volume and internal energy are given by = m g v g ( + m l v l ( and U = m g u g ( + m l u l (. (140 Now if the system is exanded, the ressure and temerature do not change but a mass dm of 44

45 the liquid is converted to vaour. o aly the energy equation to this transformation, we find the change in volume and internal energies: + d = (m l dmv l + (m g + dmv g = + (v g v l dm d = (v g v l dm. (141 Similarly du = (u g u l dm. Since both these are under isothermal conditions, By the energy equation, ( U ( = 1 = u g u l v g v l. (142 ( U +. (143 In our case the ressure is indeendent of volume so we may write d/d in the energy equation. It is ossible to exress ( U in terms of the latent heat. he heat needed to vaourize a mass dm of liquid that coexists with saturated vaour at temerature is δq = du + d = (u g u l + (v g v l dm (144 he heat of vaourization er unit mass is called the latent heat λ. hus λ( = δq ( U dm = u g( u l ( + ( (v g ( v l ( or = λ. (145 v g v l Combining with the energy equation, we arrive at the Claeyron equation d d = λ( (v g v l. (146 he latent heat and v g v l are ositive, so the sloe of the liquid-gas hase boundary is ositive. he Claeyron equation may also be alied to the solid to liquid transition, with λ denoting the latent heat of melting and v g v l relaced with v l v s. For most materials the sloe is again ositive. Water is an excetion, it exands on freezing, so that the sloe is negative in this case. In the liquid to gas case, the Claeyron equation may be simlified. We assume, following Clausius, that the density of the gas is much less than that of the liquid v g v l and treat the vaour as an ideal gas satisfying v g = R/M where M is the molecular mass of the vaour. he resulting Clausius-Claeyron equation is d d Mλ( R 2 or d log dt Mλ( R 2. (147 If we further assume that the latent heat does not vary much with temerature, we may integrate to get an aroximate formula for the saturated vaour ressure as a function of temerature ( 0 e λm/r. (148 45

46 7.3 Condition for hase coexistence and Claeyron s equation from Gibbs free energy We will aly the rincile that the Gibbs free energy is an extremum in equilibrium to rederive the Claeyron equation. As before we have a liquid and its saturated vaour coexisting at a constant temerature and vaour ressure (. Let u l, u g, s l, s g, v l, v g, g l, g g (all functions of be the secific internal energies, entroies, volumes and Gibbs otentials of the liquid and gas er unit mass. If m l, m g are the masses of the two hases, then the total Gibbs otential is the sum G = m l g l + m g g g. If the volume is slightly increased isothermally, the ressure is unchanged but a mass δm of liquid is converted to gas. he change in Gibbs otential δm(g g g l must vanish since the mixture was in equilibrium. hus the secific Gibbs otentials of the two hases must be equal in equilibrium or the Gibbs otential is continuous across a first order hase transition 14. his is the condition for hase coexistence. In other words g g g l = (u g u l (s g s l + (v g v l = 0. (149 Since we seek a formula for d/d, we differentiate with resect to temerature: d d (u g u l d d (s g s l (s g s l + d d (v g v l + d d (v g v l = 0. (150 By the first law, ds = du + dv, so dividing by d three of the terms cancel leaving d d (v g v l = (s g s l or d d = (s g s l (v g v l. (151 However, (s g s l, the heat required to vaourize unit mass of liquid is simly the latent heat λ, so we arrive at Claeyron s equation d d = λ (v g v l. ( van der Waal s gas vdw equation of state, virial exansion, isotherms Figure 9: Interatomic otential energy (R between two identical atoms as a function of searation R between their nuclei. he ideal gas EOS works well at high temeratures and low ressures. However, significant deviations are observed as the gas gets closer to condensation. van der Waals roosed the 14 Note that the derivatives of the secific Gibbs otential with resect to temerature and ressure (negative secific entroy and secific volume are discontinuous across the hase boundary for a first order hase transition. 46

47 simlest equation of state that qualitatively accounts for the first order vaour to liquid condensation transition. Condensation is due to intermolecular forces. Recall that the molecules of an ideal gas are oint-like and do not interact with each other. Real gas molecules strongly reel when they get closer than about a tenth of a nonometer ( hard-shere core. he intermolecular otential is attractive ( cohesive forces at slightly larger searations and goes to zero asymtotically. Consider n moles of a gas in a container of volume. Suose nb is the total volume occuied by the hard cores of the molecules. he molecules therefore have only an effective volume eff = nb to move around. he ressure of the gas on an outer wall (or a membrane in the gas is due to the force exerted by molecules as they collide with the wall. he attractive interaction between airs of molecules could be exected to reduce this ressure comared to its ideal gas value ideal. van der Waals argued that the reduction in ressure would thus be roortional to the number of airs of articles, i.e., n 2. However, since ressure is intensive, we must have = ideal n2 a 2 (153 where a is a (dimensional constant deending on the inter-molecular force. We may call n 2 a/ 2 the cohesion ressure. van der Waals ostulated the equation of state ( ideal eff = nr or + n2 a 2 ( nb = nr. (154 In other words, the ideal EOS holds rovided we use the ideal ressure and effective volume. When a and b vanish, the effects of intermolecular forces and molecular size are not accounted for and we return to the ideal EOS. irial exansion of vdw EOS: At low density n/ the vdw EOS should aroach that of the ideal gas. he virial exansion quantifies the deviations from ideal behavior through an exansion in owers of n/. o obtain it we rewrite the vdw EOS ( + n2 a 2 ( nb = nr as = nr a nb n2 2 or nr = ( 1 nb 1 na R. (155 Exanding in owers of the small dimensionless quantity nb/ we get nr = 1 + nb ( 1 a ( nb 2 ( nb (156 br When the molecular volume nb we recover the ideal gas law. he first deviation is encoded in the virial coefficient b a/r, which can be measured. Its temerature deendence contains information on the molecular arameters a and b. van der Waals isotherms: he vdw isotherms are lotted on the P lane in Fig 10. For large they aroach the hyerbolae = constant of the ideal gas. It is convenient to multily (154 by 2 and regard the vdw EOS as a cubic equation for for any given, n and : ( 2 + n 2 a( nb = nr 2 or 3 (nb + nr 2 + n 2 a n 3 ab = 0 (157 For high temeratures, there is a unique volume for any given ressure. As the temerature is decreased, this continues to be true till we aroach a critical temerature c and a critical 47

48 Figure 10: From K Huang: vdw isotherms in a P diagram. Figure 11: van der Waals isotherm on a P diagram and Maxwell construction from E Fermi, hermodynamics. he saturated vaour ressure for the temerature of the indicated vdw isotherm GF EDCBA is the one for which the areas above and below the horizontal coexistence line F IDHB are equal. isotherm with a horizontal oint of inflection C. As the temerature is further lowered, there are three volumes corresonding to a given ressure. Comaring with the isotherms of a real gas shown in Fig. (8 we see that they are qualitatively similar to the vdw isotherms for c. For < c the vdw isotherms have a local maximum and minimum ressure while the real isotherms have a constant saturated vaour ressure corresonding to an inhomogeneous stable hase with vaour and liquid coexisting. If an unsaturated gas is comressed isothermally, till the saturated ressure is reached, condensation normally sets in and the gas hase searates into a liquid-vaour mixture. Under further comression, the ressure remains constant with more and more of the vaour condensing until the system consists of ure liquid. It is only on further isothermal comression that the ressure again starts to rise. What then is the nature of the hysical states along vdw isotherms for < c? If a saturated vaour is free of imurities 15 and is carefully comressed, it is ossible for higher ressures to be reached without condensation setting in. his corresonds to a homogeneous but labile (somewhat unstable suersaturated gas which would condense if erturbed (the ortion FE in Fig. 11. he corresonding homogeneous labile state aroached from the liquid end of the vdw isotherm is called undercooled (the ortion BC in Fig. 11. It may be achieved by heating a liquid in a container that is free of vibration to a temerature a little higher than the boiling oint. he vdw isotherm thus corresonds to homogeneous but somewhat unstable states. Remarkably, it is ossible to aly Maxwell s construction to the vdw isotherms to obtain the stable horizontal isotherms. Before discussing this, let us analytically find the critical values of, and. 15 Imurities like dust would serve as nucleation sites for condensation. Suersaturated vaour is used in a Wilson cloud chamber to detect tracks of elementary articles, the articles serve as nucleation sites for condensation. 48

49 7.4.2 Critical isotherm and critical oint he isotherms for n moles of a vdw gas are curves of constant on the lane defined by the EOS ( + n2 a ( nb = nr. (158 2 It is convenient to multily by 2 and regard this as a cubic equation for for any given and : ( 2 + n 2 a( nb = nr 2 or 3 n(b + R 2 + n 2 a n 3 ab = 0 (159 For sufficiently high the grah of the isotherm ( must reduce to hyerbola = nr/ characteristic of an ideal gas. So for large the isotherm has no local extrema and there is a unique volume corresonding to any given ressure (see Fig. 10. For sufficiently low, an isotherm has both a local minimum ressure min and a local maximum ressure max and three volumes 1,2,3 corresonding to any ressure in between ( min < < max. he critical isotherm (at the critical temerature c is the one where the three roots 1,2,3 coalesce at a critical volume c, in other words, the critical isotherm has an inflection oint: = 0, and 2 = 0. (160 2 he corresonding ressure is c and temerature c. o find these values, we ut and Combining we get = nr ( nb 2 + 2n2 a 3 = 0 or 2 2 = 2nR ( nb 3 6n2 a 4 = 0 or R ( nb 2 = 2na 3 (161 R ( nb 3 = 3na 4. (162 2na 3 ( nb = 3na 4 or 2 nb = 3 c = 3nb. (163 Putting this in the first condition we get R c = 8a 27b and finally from the vdw EOS, c = a/2nb 2. hus the critical ressure, volume and temerature for n moles of a vdw gas are c = a 27b 2, c = 3nb and c = 8a 27Rb. (164 he critical volume is thrice the excluded volume nb, so the gas/liquid is quite dense at the critical oint. Universal form of vdw EOS: If the ressure, volume and temerature are exressed in units of the their critical values P = / c, = / c and = / c, the vdw EOS takes a universal form common to all vdw gases irresective of their values of a and b: ( P + 3 ( 2 1 = (165 49

50 States of different vdw gases (different a, b corresonding to the same P,, are called corresonding states. o see this we substitute for,, in terms of P,, in the EOS: ( Pa 27b 2 + n2 a 2 9n 2 b 2 (3nb nb = nr 8a a ( P 27Rb 9b n 8a 2 (3 1 nb = 27b. (166 Cancelling factors we get the universal EOS Maxwell construction As discussed, the vdw isotherms describe unstable homogeneous states (suersaturated vaour for instance in the region between volumes B and F (see Fig 11 where the vdw isotherms dislay a local minimum and a local maximum. In this region, the stable equilibrium state is a liquid-vaour mixture at the constant saturation vaour ressure. he Maxwell construction gives an elegant method of identifying the constant saturated vaour ressure that corresonds to a given vdw isotherm. he height of the horizontal coexistence isotherm BF is determined by the condition that the areas DEF D and BDCB bounded by the horizontal and the the vdw isotherm must be equal. o show this, we follow Maxwell (see Fermi and consider the reversible isothermal cycle BCDEF IDHB. In other words, we first exand the liquid from B along the vdw isotherm BCDEF assing through homogeneous states ending with saturated vaour at F. hen we return from F to B by comressing the vaour at constant saturated vaour ressure via liquid-vaour mixtures F IDHB. Note that the oint D corresonds to two different states: a liquid-vaour mixture when aroached horizontally and a labile homogeneous state when aroached via the vdw isotherm 16. Being a reversible cyclic rocess δq/ = 0 by Clausius theorem. Since it is isothermal as well, is a constant, and the heat added δq must be zero. By the first law for a cycle, the work done is also zero. Since DEF ID is traversed clockwise while BCDHB us traversed anti-clockwise, the total work done is zero rovided the magnitudes of the areas are equal. In real gases, these areas are quite small. Note that the Maxwell construction alies to any equation of state, not just the vdw gas Entroy and Caloric condition for vdw gas Recall the caloric condition ( U = 0 for an ideal gas that followed from the Joule free exansion exeriment. We may view this as a necessary condition for δq/ to be an exact differential. Indeed, taking, as indeendent variables for an ideal gas, we have δq du + d = = 1 ( du 1 d d + U + nr d. (167 he integrability condition ( 1 U = ( 1 U + nr Similarly, let us derive the caloric condition for a vdw gas: δq du + d = = 1 U d + 1 ( U U = 1 U U nr nb an he closed curves DHBCD and DIF ED do not reresent reversible cyclic rocesses. or ( U = 0. (168 d (169 50

51 As before, we get 1 2 U = 1 2 U U + an2 2 2 or ( U = an2 2. (170 = an2 2 So for entroy to exist as a state function for a vdw gas, the caloric condition ( U must be satisfied. In articular, the EOS does not comletely determine the thermodynamic nature of a vdw gas, it must be sulemented by the caloric condition. he above caloric condition imlies that the internal energy of a vdw gas is not indeendent of volume, unlike for an ideal gas (a 0. However, we may show that the heat caacity at constant volume is indeendent of volume C = C (, just as for an ideal gas! Indeed, differentiating the caloric condition w.r.to, 2 U = ( U C = 0 or = 0. (171 Entroy of vdw gas: Assuming the caloric condition is satisfied, the entroy of a vdw gas is determined by ds = δq = 1 ( U d + nr nb d = C ( d + nr d. (172 nb As in the case of an ideal gas, the temerature deendence of C is not fixed by thermodynamics. If it is assumed roughly constant, then we may integrate from 0, 0 to, to obtain ( ( nb S(, C log + nr log. (173 0 nb Remarks on aramagnetic to ferromagnetic hase transition Consider a magnetic material like iron. Recall that the magnetization M is the total diole moment of the samle er unit volume. he thermodynamic state sace of a magnet is arametrized by three variables H, M,, the alied external magnetic field, magnetization and temerature. A aramagnet is one which has zero magnetization in the absence of an external magnetic field H, but which develos a induced magnetization when an external H is turned on. For small H, the induced M is linear in H and satisfies the Curie-Weiss law M = χh where χ is the temerature-deendent suscetibility. In other words, the external field tends to align 17 the atomic dioles in the direction of H. his linear relation ceases to hold for large H, indeed there is a non-linear saturation (see Fig. 12. he magnetization cannot increase indefinitely, it has a maximal value, which is attained when all the atomic dioles are aligned. Now χ is a function of temerature. If the temerature of the aramagnet is increased χ decreases: thermal agitation decreases the alignment brought about by the external field. For convenience let us suose H = Hẑ and M = Mẑ are in the z -direction. hen the isotherm M(H for a higher temerature lies below the earlier one, with saturation occurring at a higher H (see Fig.12. On the other hand, if the temerature is decreased, the suscetibility (sloe of M(H at H = 0 grows and in fact diverges at a critical Curie temerature c (see Fig. 17 A diamagnet is one where the induced magnetization oints in the direction oosite to H. 51

52 13. At c the samle ossesses a residual sontaneous magnetization M s ( c even when the external field H 0 +. he sontaneous magnetization grows as the temerature is decreased below c. Below c the material is a ferromagnet, is has a magnetization even in the absence of an external field, like the familiar bar magnet. We have just described a second order hase transition from a aramagnet to ferromagnet. he free energy, internal energy, secific entroy and magnetization are continuous across the transition 18, but the suscetibility and secific heat diverge at the critical oint. It is found that in the neighbourhood of c, there are fluctuations on a whole variety of length scales in the samle and even far searated atomic sins are strongly correlated - this is the analogue of critical oalescence at the critical oint in the vaour to liquid transition in water. Moreover, the sontaneous magnetization, secific heat and suscetibility have ower law behaviors for c. For instance: M(, H = 0 ( c β, C(, H = 0 ( c α, and χ(, H = 0 ( c γ. (174 he exonents α, β, γ are called critical exonents (there are similar exonents for > c. hese exonents have been measured and found to dislay universality - they are the same for a wide variety of materials undergoing the second order hase transition. A remarkably successful theory of second order hase transitions (which redicts the values of critical exonents and much more, based on the method of the renormalization grou has been develoed beginning in the 1960s (Nobel rize to Kenneth Wilson in A simle microscoic model for a magnet was roosed by Heisenberg 19. In the Heisenberg model we have a crystal with atomic magnetic diole moments m i located at the sites of the crystal lattice; i labels the sites. he magnetic moments arise from the electronic sin (and orbital motion. he Hamiltonian is H = i H i m i i,j n.n J m i m j. (175 he first term reresents the magnetic diole energy due to the external magnetic field. Here H i is the external magnetic field at the location of the i th diole. It is minimized when the dioles are aligned with the external field, as we exect for a aramagnet (reversing the sign of this term would be relevant to a diamagnet. he second term reresents the energy of interaction between nearest neighbour atomic dioles (denoted n.n or ij. J is an interaction strength. We have assumed that the diolar interactions decay raidly with searation, so that we may restrict attention to nearest neighbours. Notice that for J > 0, the interaction energy is minimized when all the dioles oint in the same direction, as we exect for a ferromagnet (if J < 0 we have an anti-ferromagnet, whose energy is lowest when neighbouring dioles oint in oosite directions. Since the interaction energy involves dot roducts of dioles, it is rotation invariant (the dot roduct of two vectors does not deend on the orientation of the frame, but only on the angle between them and their lengths. he direction of the external magnetic field H breaks this rotation symmetry and determines the direction in which the dioles align. 18 Second order hase transitions are often referred to as continuous hase transitions. here is no latent heat. 19 A related simlified model is called the Ising model, in which the magnetic moment on each site can only take the values ±1 in aroriate units, these are the z -comonents of the magnetic moment. 52

53 Figure 12: Plot of magnetization vs alied magnetic field for various temeratures in a magnetic domain. For > c (aramagnetic hase M = χh for small H (linear resonse regime while M saturates for large H. For c (ferromagnetic hase there is a residual (sontaneous magnetization even when H 0 ±. he memory effect (hysteresis loo is also shown. 8 hird law of thermodynamics Our definition of the entroy of a state B deended on the choice of a reference state A: S(B = B A δq/ where the integral is along any reversible rocess. he arbitrariness in the choice of reference state leads to an undetermined constant S 0 in the entroy. his does not matter as long as we are only concerned with differences in entroies. On the other hand, the free energies F = U S and G = U S + are undetermined u to a linear function of temerature, reducing their utility in dealing with states at different temeratures. We also noticed that an aroriate choice of S 0 (deending on some dimensional hysical quantities would be necessary to make the formula for the entroy of an ideal gas dimensionally consistent as well as extensive. here are other situations where the value of S 0 is imortant (E.g. the equations for chemical and gaseous equilibria. he third law was originally formulated by Walther Nernst (1912 in terms of the maximum available work or Helmholtz free energy. He discovered it in the context of electrochemical reactions and exerimentally verified its consequences for the behavior of secific heats. he third law of thermodynamics, as reformulated by Planck, states that as the temerature of a system aroaches absolute zero, its entroy aroaches a constant value indeendent of ressure/volume/density, state of aggregation etc. his constant value may be taken to be zero. In articular, all states of the system at = 0 have S = 0. We consider some simle consequences of the third law. Since S aroaches a constant indeendent of volume and ressure as 0, its derivatives with resect to, must vanish. We may use the Maxwell relations to relate derivatives of entroy to certain measurable coefficients. For examle ( S = ( = α and ( S = ( = β. (176 hus the coefficients of thermal exansion α and tension β must vanish at absolute zero (assuming and do not. Secific heats aroach zero at absolute zero. Consider the secific heats at constant 53