Lecture: 1 Revision of basic properties of gases. Thermodynamics

Transcription

1 Lecture: Revson of basc propertes of gases Thermodynamcs Thermodynamcs s the theory concerned wth systems of large numbers of partcles. The system may be a gas, lqud or sold or mxtures of these, confned to some contaner. The nature of the contaner and the materals nsde defne the type of system under study. The man characterstc of the systems that are dealt wth n thermodynamcs s that ther macroscopc behavor can be descrbed n terms of relatvely smple physcal laws despte of ther complcated nternal mechancs. Ths smplfcaton arses from the statstcal nature of the descrpton that thermodynamcs provdes. The task of thermodynamcs are; Frst, to defne approprate physcal quanttes that (unambguously) characterze the macroscopc state of the system. These parameters are called macroscopc or state varables. Second, we must fnd equatons that relate the state varables by functons that are generally vald (ndependently of what are the elements that consttute the system). In the begnnng we wll need to axomatcally defne certan thermodynamc varables to formulate laws of general valdty. In thermodynamcs we deal wth dfferent types of systems, characterzed by the way they nteract wth the surroundngs: - Isolated systems: no nteractons wth the surroundngs - Closed systems: allows exchange of energy wth the surroundngs - Open systems: allows exchange of partcles and en An solated system s only an dealzaton. In practce these do not exst. A system can also be homogenous f the physcal propertes are dentcal n all ponts of the system or heterogeneous f they change dscontnuously at certan boundares. The dfferent sub-systems enclosed by these boundares are called phases. Once we have defned the system we wll specfy macroscopc varables to descrbe t. These varables should be hosen n such a way that they only depend on the state of the system. The state of the system wll be specfed after a suffcent amount of tme such that the macroscopc parameters no longer change wth tme. Ths wll be called the equlbrum state of the system. We begn wth a descrpton of gases, and ther macroscopc varables. Gas state Gases are the smplest state of mater to descrbe. They are also bound to the early stages of thermodynamcs. We wll often use gases as examples durng the development of ths course. A gas s a homogeneous system that occupes the entre avalable volume. In order to completely specfy the state of a system composed of a gas n a contaner we need to provde 4 varables: -Amount of substance (n number of moles n, n number of molecules, etc.) -Volume (V, n l, or m )

2 -Pressure (p) 4-Temperature (T) Are all these varables ndependent?. o. There s an equaton of state that relates them. Therefore only of them need to be specfed, together wth the equaton of state n order to descrbe the system completely. The problem s that the equaton of state s not known to us n general. We can propose a model to the system, and then wrte an equaton of state that s consstent wth our model. ut the equaton of state s n general a complcated functon of the basc varables defned above. Pressure The pressure of the gas s an mportant varable. It s the force per unt area and t has unts of 5 m. Ths unt receves the name Pascal (Pa). One bar s equvalent to0 Pa. 5 One atm. s equvalent to.05 0 Pa. The standard pressure for specfyng thermodynamc data s defned as bar, and s usually denoted as p =bar. We wll use lower case p to denote the pressure. Pressure s measured wth a barometer or a barometer. The manometer s used to measure the pressure of a contaner contanng some gas, whle the barometer s used to measure the pressure of the atmosphere. A= π r V =A h=h π r lq m lq = ρlqa h or m lq = ρlqhπ r () And the force exerted by the lqud column s F lq = ρlqgπ r h () so that the pressure at the base of the column s Flq ρlqgπr h p= = = ρ gh lq, A π r () whch does not depend on the radus of the column. There are other methods to measure pressure. The student s encouraged to read about these methods. Temperature.

3 Energy can flow from one substance to another when they are n contact. Temperature s the property that tells us the drecton of the energy flow. We denote the temperature by the symbol T. We measure the temperature n arbtrary unts that derve from the apparatus used to measure them, and from some standard temperatures chosen by the scentsts. The Celsus scale s set by the dfference between the temperature of a system of coexstng water ce and water vapor (trple pont) whch s defned as 0C, and the bolng pont of water, whch s defned as 00C. The dfference s dvded nto 00 degrees. If we assume that a substance A s n thermal equlbrum wth a substance. An that the substance s n equlbrum wth substance C, then t s found that C s n equlbrum wth A. Ths expermental fndng s extremely smple, and as far as we can tell, t s always true. We cal ths general law the Zeroth law of thermodynamcs. Thanks to the general valdty of ths law we can buld thermometers. Queston for the student: Identfy how the zeroth law of thermodynamcs needs to be vald n order to be able to use a mercury thermometer. Example: If energy flows from body A to body when they are n contact t s sad that TA > T. If no energy flows, then we say TA = T. Ths stuaton s called thermal equlbrum. We wll descrbe ths n great detal later n ths class. However, thermodynamcs does not tells us anythng about how fast the equlbrum state s acheved. The temperature of a system s obtaned by usng an auxlary system whch presents a unque value of an easly observable state quantty when t s n thermal equlbrum. Ths devce s called thermometer. One example s the volume of a lqud whch s has a well defned value when t s n thermal equlbrum. If the temperature of the thermometer ncreases, the volume of the lqud generally ncreases. y brngng the thermometer nto contact wth a system of unknown temperature, the volume of the lqud wll vary as the system exchanges energy wth the thermometer. Once a suffcent has passed the volume of lqud stops changng and we can measure t. The thermometers based on the expanson of a lqud are not perfect. A more accurate thermometer can be constructed by enclosng a gas n a contaner. y measurng the volume at constant pressure, we can know the temperature. The temperatures reported by the gas thermometer are dfferent from those from the Celsus scale. They are expressed n Kelvn (K) unts. The magntude of one Kelvn unt equals that of one Celsus degree. ut 0C=7K. From ths t s clear that relatng the temperature to the volume change requres that we know about how these two varables are related mathematcally. Ths s the equaton of state.

4 Gases and some ther equatons of state It s found by experments that at low pressures many gases follow a smple relaton between ther volume V, T and the number of moles n. nrt p = (4) V where R s a constant that appears to be the same for many gases ( R = 0.08 dm atm/molk=8.4 J/molK. Ths s an example of an equaton of state, and t s called the Perfect gas or the deal gas equaton Thus, ths constant receved the name Unversal gas constant, or gas constant for short. In the lmt of zero pressure all gases obey qute well ths relaton. Isotherms If we fx the temperature of a gas n a contaner wth a pston that allows us to compress t or to expand t by keepng a constant amount of gas nsde, we wll observe that the pressure ncreases when we decrease the volume. Fgure: Isotherms of an deal gas The curves are hyperbolas, and are called sotherms ( so means same, therm means temperature). We can see that the perfect gas equaton of state predct ths: nrt p= = Constant (5) V V If we have a constant amount of gas, and we fx the temperature, then any two dfferent volumes and ther correspondng pressures must be related by the same constant (nrt), leadng to: pv=pv Also, we can see why at constant pressure, we have 4

5 nrt V V V= = Constant T whch leads to = another well known formula. p T T Intermolecular nteractons: Repulsve nteractons occur at very short dstances, less than one molecular dameter. Attractve forces are of ntermedate range, on the order of several molecular dameters. ecause ther dfferent length scales, the repulsve forces manfest only at very hgh pressures. Attractve forces are effectve when the molecules are close but not at contact, and are neffectve at low pressures when the molecules are relatvely separated. Fgure: The par nteracton potental n a real gas. In the fgure the potental energy U(r) versus the ntermolecular separaton r s shown. Ths potental energy s for a par of molecules. At very large r, they do not experence any force. At ntermedate r they have attractve forces (remember that the force s the (mnus) dervatve of the potental energy functon wth respect to r ). There s a crtcal value of r* at whch the repulsve force balances the attractve component. At ths pont the dervatve s zero and there s no net force. At shorter dstances the slope of the potental energy curve becomes negatve. In ths dstance range, the force ncreases rapdly wth a small decrease n the ntermolecular separaton. We have roughly regons: I) At very hgh pressures: Repulson domnates the forces II) At ntermedate dstances: attracton domnates the force. III)At long dstances the nteracton nearly vanshes. 5

6 Real Gases: The gases n general devate from the perfect gas behavor. Ths s notorous at low temperatures and at hgh pressures. The reason for these devatons s the exstence of ntermolecular forces that domnate the collectve behavor of the gas. The dstance dependence of the molecular nteractons (or ntermolecular forces) s evdent n the factor Z (called the compressblty of the gas). pv pv Z= m V = where Vm = s the molar volume of the gas. nrt RT n For an deal gas ths s just unty at all condtons. For a real gas, a plot of the compressblty versus p Z(p), s very dfferent from the deal gas case: Fgure: The compressblty (Z s also known as compresson factor) of some gases. y now t s clear that the deal gas equaton of state wll not be a good descrpton of the system except at very low pressures. One way to solve ths s to expand the compressblty n a power seres: pv = + p+ C p + D p + f ' = C' = D' =... = 0 we recover the deal gas. nrt Z= ' ' '... 6

7 The coeffcents ', C', D '... can be measured or they can be calculated from statstcal mechancs. From ths compressblty we get an equaton of state called the vral equaton. Isotherms of a real gas We mantan one mole of CO n a contaner enclosed by a pston, at a constant temperature of 0C. Fgure: Isotherms for CO (Scanned from P.W. Atkns, Physcal Chemstry, 6 th ed.) We start at pont A and reduce the volume (by pushng a pston). ear A the gas follows the relaton pv m =Const (oyle s law). And t s approxmately an deal gas. When the volume was reduced to Vm 0.4 l/ molpont, devatons from oyle s law begn to occur. y the tme we reached pont C. for whch we need approxmately 6 MPa for CO, the pressure stays constant even though we are reducng the volume drastcally. What happens here?. Observaton nsde of the contaner reveals that a lqud phase s formed just to the left of pont C. As we further reduce Vm by movng from pont C to pont D and then to pont E the volume of the lqud phase ncreases but the pressure s constant. We call ths pressure the vapor pressure of the lqud at that temperature (n ths case at 0C). 7

8 When we reach pont E all the gas has dsappeared and the pston now rests aganst the surface of the lqud. After pont E f we try to reduce the volume t takes consderable pressure. P rses sharply wth a small reducton of V. m Crtcal constants: We can see two types of sotherms n the fgure. I) the ones that have a horzontal (p=constant) stretch and, II) the ones that do not have ths. The length of the horzontal secton where gas and lqud coexst decreases f the temperature s ncreased. Evdently, there s gong to be a temperature for whch the length of the length of ths porton of the sotherm wll reduce to one pont. Ths temperature receves the name of Crtcal temperature T C. If the experment s done at any temperature above TC the sotherms do not exhbt the p=constant feature. Furthermore, at no pressure t s possble to see the coexstence of lqud and gas f the system s mantaned at any temperature above T C. The sotherm at TC s the crtcal sotherm, and the crtcal pont s characterzed by the crtcal pressure and molar volume p C, VT. C, C Gas p /MPa V / cm / mol T / K C C C CO He Table crtcal parameters for common gases Other equatons of state The deal gas law had no gas-dependent parameter t s vald for all gases provded that the pressure s low enough. The vral equaton has lots of parameters, that must be determned for each gas, and they are a functon of the temperature. Clearly t s not a very useful equaton. There are other equaton of states that have less parameters, and are better that the deal gas, but fal to accurately descrbe the real gases. The most mportant of them s the Van der Waals equaton of state. We wll construct t as follows. egnnng from the deal gas law, pv=nrt where molecules are pont objects wth no sze, we account for the repulsve nteracton by assertng that the molecules behave as small but mpenetrable spheres. Thus, f a gas s n a contaner, of volume V t does not have the entre volume avalable. Instead t has a volume V-nb, where n s the number of moles, and b s called the covolume and t s the volume taken out by n moles of sphercal molecules. The perfect gas should be corrected to be p(v-nb)=nrt. Ths s by tself another equaton of state, and t s called the rgd spheres gas. ext we account for the attractve forces, whch wll tend to decrease the frequency of collsons aganst the walls of the contaner as well as the momentum of the molecules before they ht the walls. Ths double effect s due to the fact that the molecules near the wall are beng pulled back by the bulk of the gas n the contaner. The bgger the number 8

9 densty of molecules n the contaner n, the larger ths effect wll be. Then the pressure V s reduced because of ths double effect, and so t wll be n proporton to the square of n so we have V nrt n n p = a ths s rewrtten as p + a ( V - nb) = nrt ( V - nb) V real gas pressure V corrected volume correcton to the pressure corrected volume corrected pressure (6) Ths s the van der Waals equaton of state, a andb are two parameters that are specfc for each gas. Gas atm l a b mol - 0 l mol Ar.45. CO Table : van der Waals parameter for gases The behavor of the van der Waals (vdw) equaton of state The vdw equaton s a cubc functon of v: n an abn p + a ( V - nb) = nrt pv + nbp = nrt V V V multplyng by the volume sq. pv + Van nbpv abn = V nrt (7) pv V n( bp + RT ) + Van abn = 0 (8) the compressblty s pv an nbp an + = nrt VnRT nrt V nrt Z vdw an nbp an = + + Z VnRT nrt V nrt g (9) The vdw equaton can be wrtten more compactly n terms of the molar volume: 9

10 n nrt n V p+ a ( V - nb) = nrt p= a and Vm = V V nb V n RT a RT a p = = V b V V b V n n m m (0) () RT RT a RT ) f Vm >>> band T s very hgh, then p >>> p= Vm b Vm Vm Vm recover the deal gas behavor. and we a RT ) When both Vm b and V V b m m we encounter the lqud-gas coexstence zone. ) For T < TC the sotherms oscllate. Ths s unphyscal (decrease V and P goes down! unstable system) For T = TC the oscllaton reduces to a horzontal nflecton pont. We can obtan the value of the crtcal constants for a vdw gas n terms of a andb : RT a p = and both frst and second dervatves must vansh at the horzontal Vm b Vm nflecton pont: dp dv m d p = 0 = 0 at T = TC. dv m Ths gves us two equatons from whch pc, VC, TCcan be obtaned. dp dv m RTC a = + = 0 ( V b) V m m () RT = 6 = 0 d p C a 4 dvm ( Vm b) Vm () from the frst dervatve Eqn. () RTC a RTC = and C C C C V b V V V b ( m ) ( m ) m a = 6 ( ) V 4 m m from the second dervatve Eqn. (): 0

11 RT C a = 6 C ( V ) V 4 m b m equatng these results we get RTC RTC = = and rearrangng C C C C C ( Vm b) Vm ( Vm b) Vm b Vm C C C Vm= ( Vm b) Vm= b (4) RTC a = from the frst dervatve C C V b V Reeplacng ths result n ( m ) ( m ) RT C = a RT a a = RTC =8 7b 7b C ( b b) ( b) b ( ) =8 a TC 7 Rb and we obtan the crtcal pressure fro the equaton of state so that (5) p C R 8a a 4a a a 4 a = = = = so that b7rb 9b 7b 9b 9b 7b a pc = (6) 7b ote that the crtcal compressblty for the van der Waals gases s C pv C m a b ZC = = = 0.75 (7) RT 7 8a C R b 8 7Rb for all gases. Thus we can check f ths s true: Gas C pc / atm V / cm / mol TC / K Z C m Ar CO Table : crtcal compresson factor for common gases. ote that the result s reasonably constant, but lower than 0.75 that the vdw equaton predcts. In addton, CO devates more than the other two gases.

12 Lecture: Thermodynamc equlbrum -Thermal equlbrum If a system and all ts nternal parts A,,C satsfy that TA = T = Tc =... t s sad that t s n thermal equlbrum. -Mechancal equlbrum When all the forces n the system balance we say that the system s un mechancal equlbrum. -Chemcal equlbrum If a chemcal reacton or a phase transton takes place nsde of the system, the number of partcles or molecules of dfferent knd or ther numbers present n the dfferent phases can change. We have seen ths occurrng when we dscussed the sotherms of CO. Increasng the force F causes the CO (g) molecules to dsappear and to become part of the lqud. When the force s changed and after a suffcent tme, the number of molecules n the gas and lqud phase reman constant n tme. If we look closely we see that there s a contnuous exchange of molecules between the phases. ut from a macroscopc pont of vew, the average number of molecules n the gas and lqud phases remans constant Fgure Ths stuaton s called Chemcal Equlbrum We wll gve more detaled descrpton of the chemcal equlbrum later on n ths curse. A system that s n mechancal, thermal and chemcal equlbrum at the same tme s sad to be n thermodynamc equlbrum. Functons of state When a system s n thermodynamc equlbrum ts propertes do not depend on tme. ut f we change the pressure, for example, the system accommodates to the new value of the pressure by changng some other of ts propertes (perhaps the volume). These propertes (p, T, V etc) are called thermodynamc varables or coordnates of the system. However, once the system s at equlbrum the thermodynamc varables adopt values that are the same regardless of how we prepared the system. For example, for a pure gas wth no chemcal reactons or phase changes, T = f( p, V) where f s a functon (equaton of state). If p s adjusted frst, and then the volume s adjusted, the temperature adopts a new value. If nstead the same volume adjustment s done frst, and then the pressure

13 adjustment, the same fnal temperature s obtaned. Thus t s sad that T s a functon of state. If n general we have a functon of state G=g(x,y) where x,y could be p and V for nstance. A small change n the functon of state G can be wrtten as g g dg = dx + dy x y y x A whch we rewrte as dg = A( x, y) dx + ( x, y) dy () usually functons of state are smooth and contnuous wth well defned dervatves at any pont (x,y). For ths type of functons we can calculate the second dervatve, g g = () x y y x y x y x Ths last equaton s only satsfed for certan functons that we call analytcal. Eqn. can be combned wth Eqn to get ( xy, ) Axy (, ) = x y y whch has two consequences: x () ) dg = A( x, y) dx + ( x, y) dy can be ntegrated to gve a functon of state (or an equaton of state). ) dg s an exact dfferental whch only depends on the dfference n the functon of state between two states but not on the path between them dg = A( x, y) dx + ( x, y) dy can be consdered to be a dot product of G = A ˆ ( x, y) + ˆj( x, y) and dr = dx ˆ + ˆjdy. We now can calculate the lne ntegral of Gdr as: Gdr = dscurlg ( ) (4) wth our defnton of the vector (4 ) s the z-component of the curl: xy (, ) Axy (, ) ( curl( G) ) = 0 z = (5) x y y x but Eqn. (5) s equal to zero because the crossed dervatves (Eqn. ()) are equal (whch s our assumpton here). Then t wll be true that Gdr = 0 (6) If we separate ths ntegral over a closes curve nto two portons or paths, we get

14 Gdr = Gdr + Gdr = 0 r r path path whch mples Gdr = Gdr + Gdr = 0 r r path path Fgure Concluson: We can choose dfferent paths to go from state r = ( x, y ) to state r = ( x, y ) gx (, y) gx (, y). And the change n G s ndependent of the path chosen, but t only depends on r = ( x, y ) and r = ( x, y ). Then the functon of state can always be defned by ntegratng r gxy (, ) = Gdr (7) r However, to have a unque value that only depends on r = ( x, y) we must specfy a reference value r = ( x, y ), and gxys (, ) now relatve to gx (, y ). Fortunately we are gong to be nterested only n the dfferences between the value of the functon G. For a small change n the value of G and usng (7) r+ dr r r dr dg g( x dx, y dy) g( x, y) G dr G dr + = + + = = G dr (8) r r r ths means that dg only depends on r+ dr and r, and not on our choce of r whch allows us to defne a standard state wthout the rsk of loosng generalty of the conclusons that we arrve at. Extensve use of ths s done n the feld of thermochemstry whch we shall use as an example of applcaton of thermodynamcs next class. Some thermodynamc functons do not satsfy these propertes, (.e. they are not analytcal, or they are not state functons). The dfferental form of such functons (n the 4

15 form gven by Eqn. () wll not be an exact dfferental n the sense that Eqn. () wll not be satsfed for them (the crossed second dervatves wll not be equal). The frst law of thermodynamcs In mechancs the energy of a system s gven by the Hamltonan functon, whch s constant for a conservatve system. If we nclude all the varables that descrbe the processes, all systems are conservatve. It follows that the energy s always conserved. We consder a system A surrounded by the rest of the unverse, and we say that the system has a certan amount of energy U n t. Then f energy s taken from the unverse and put nto the system, the total Utotal=Uunverse + U(A) s constant. A well known fact s that energy can be deposted n system A by dong work on t or by supplyng heat. Ether way we do ths we arrve at system A regardless of how do we depost the energy n t. Ths s the statement of the frst law of thermodynamcs Δ U = q+ W (9) where q s the amount of heat transferred to the system and W s the work done on t. If we solate a system such as kg of water, so no heat can flow trough the walls of the contaner Fgure If we perform mechancal work on ths system ntally at temperature T by lettng the weght of mass m get lower n the lab, such that the propellers str the water causng t to heat up (the nternal energy of the water ncreases. We could measure the fnal temperature T after a certan amount of mechancal work has been done on the system W = mgh mechancal 5

16 If now repeat the experment by keepng the weght m fxed, so no mechancal work s done, and we start at temperature T agan, but now we pass a current on the resstor that act as a heater. The amount of electrcal work done on the resstor swelec = I Rt whch t s completely converted nto heat at the system, we see that the fnal temperature T can be reached after some tme. Once the system has reached T t s ndstngushable from the system at T obtaned by lowerng the mass m through a dfference n heght of h. We can also supply some mechancal work W and some heat q and produce the very same temperature change from T to T. Whch s exactly what the frst law says (Eqn. (9)). We adopt the conventon that W > 0 f the energy of the system ncreases. Durng a process the nternal energy of the system ncreases f the process that caused the change can be produced by lowerng a weght somewhere n the surroundng of the system and usng ths mechancal work to cause the change. If the system does work on the surroundngs, then t wll be equvalent to dsplace a mass aganst a force F. The total work done s zf W = Fdz = F dz = F ( Z Z ) z f Z f Z zf z f < then W > 0 we can always create an magnary process that does the work on the system by lowerng a weght h= Z f Z and F = mg. Work of compresson and expanson: The system s enclosed n a contaner that s covered by a frctonless pston of zero mass, and area A under an external pressure p ext. The magntude of the external force s F = pext A. We shall assume that the moton of the pston s quasstatc so no turbulences n the system or the surroundngs are generated durng the process such that no heat spreadng phenomena can occur wthout us knowng about them. If the system expands and rses the pston by dz the work done s Fgure 4 Aganst the surroundnngs FΔ z = pext A dz =δw ote that t s the external pressure what matters and not the nternal pressure what matters. ut A dz = dv or δ W = pextdv For a compresson δ W > 0 an for a expanson δ W < 0 6

17 In general work can be dfferent from expanson-compresson. ut the general deal s the same. The system varables can be of two dfferent types: I) the varables lke the volume or the energy of the system depend on how bg the system s. These varables are called extensve. II)The varables lke the temperature or the nternal pressure do not depend on the sze of the system, and these are called ntensve varables. y carefully lookng at the equaton of the work of expanson-compresson we can see that t s the product of the form Work=(change n an extensve varable)x(ntensve varable) Ths s the general form of any type of work; the factor (change n an extensve varable) s called the generalzed dsplacement the factor (ntensve varable) s called the generalzed force. Type of work defnton Generalzed force Generalzed dsplacement Expanson/compresson p dv pressure Volume Surface expanson electrcal ext γ dσ Surface tenson Surface area φdq Electrc potental charge Expanson aganst a constant pressure In ths case the system does work aganst a constant force so that we can take t out of the ntegral: zf Vf Vf z V ext ext V ext f W = Fdz = p dv = p dv = p ( V V ) We can represent ths n a p-v dagram and the shaded area represents the magntude of the work done: Reversble expansons: Fgure 5 7

18 To acheve a reversble process we must be able to revert ts drecton by applyng an nfntesmal change n the force that drves the process. For an expanson ths can be done by a small change n the external pressure so that t can became smaller or larger than the nternal pressure by the applcaton of such nfntesmal change. ut ths s to say that the external pressure only dffers by an nfntesmal amount from the nternal pressure. Therefore they must be essentally dentcal at all tmes!. Ths means that the system s at equlbrum durng the process. Ths s not a trval statement because a process s what occurs spontaneously to take a system from whatever state to an equlbrum state. For an expanson/compresson work, f we do t reversbly then W Vf Vf = p dv pdv V ext = V ow n order to calculate the amount of work we need to know the value of p at all tmes. Snce the pressure depends on the volume of the system, we can use the equaton of state: Vf W = p( V) dv V If we have an deal gas n the cylnder enclosed by the pston n Fgure and we let the gas expand reversbly, the system wll do an amount of work Vf Vf nrt V f W = p( V) dv = dv nrtln V = V V V assumng that the temperature and n are constant, we obtan the work of reversble expanson/compresson, whch s negatve f the gas expands and postve f we compress t. Also note that f the system expands at a hgher temperature the amount of work t does on the surroundngs s hgher. As before, we can represent ths n a p-v dagram and the shaded area represents the magntude of the work done: Fgure 5. Reversble expanson. The grey shaded rectangle represents the work done durng the correspondng rreversble process. 8

19 If we were to compress the gas from Vf to V, we need to do less work than we would need f we were to do ths rreversbly. The dfference equals the dfference In the grey shaded rectangle and the area under the sotherm T. Durng a reversble expanson the gas does more work (area under the sotherm T for work done aganst a varable pressure) than durng the correspondng rreversble expanson (lower rectangle area, expanson aganst pf). In general the frst law of thermodynamcs can nclude other forms of work; Δ U = q+ W = q+ W + W + W + exp-comp elec magn... The correspondng terms depend on the system that we are tryng to descrbe. ote that the frst law does not say what would happen, or n whch drecton a process wll occur. Despte of ts mundane smplcty, a great deal can be done wth t. efore we move on to the second law of thermodynamcs, we wll dscuss an mportant applcaton of thermodynamcs; calormetry. 9

20 Lecture Calormetry Measurng Δ U : Adabatc omb Calormeter *A change of state n the thermodynamc sense- s ntated nsde a rgd contaner of volume V. *The bomb s mmersed n a strred water bath. *The water bath s n an nsulated contaner. THIS ETIRE SYSTEM IS THE : Adabatc omb Calormeter ecause the bomb can do no work, (V=constant) all that can happen s that the bomb gets hotter or cooler. Then the change n the nternal energy s gven by: Δ U = q(seen by the system whch s nsde the bomb). Δ Usystem = q= Δ Ucalormeter = C ΔTcalormeter because the entre system + calormeter s adabatcally enclosed Then the temperature change n the water bath s proportonal to the amount of heat q exchanged wth the bomb: qsystem = C Δ Tcalormeter and thus, Δ Usystem = C Δ Tcalormeter C s a postve parameter that measures how much heat needs to be transferred to the calormeter n order to rse ts temperature by one degree, and t s called the calormeter constant. From now on we drop the subndex calormeter n Δ T. In order to measure ΔU we must know the calormeter constant. To measure C we need to do a separate experment; we nsert an electrc heater (merely a resstor) n the water bath and apply some electrcal work. Ths causes the temperature to rse. ecause we can measure the amount of work that we supply and we can measure Δ T ; Δ U = q+ welec = welec > 0 welec welec = I V t C = ΔT but Δ T > 0 and the calormeter and the bomb are absorbng energy (the electrcal work s all converted to heat by the joule effect at the resstor). Clearly s C > 0. All we have to do s to measure I, V, ΔT and the tme t durng whch the electrc current s passed and we get C. Example: A of current passes trough a heatng resstor for 000 seconds, from a V battery. The temperature changes by +5.5K. welec 6000J welec = I V t=a 000s V=6000J C = = = 6.5 kj / K ΔT 5.5K Enthalpy ormally we do not do experments at constant volume. It s vastly more convenent to let the system expand or contract under the (supposedly) constant pressure of the atmosphere. Then the system pushes the boundares of the contaner dong work n the envronment. ow the change n the nternal energy s no longer equal to the energy 0

21 suppled as heat; some of the nternal energy s converted to mechancal work. So we have Δ U < q.under these condtons the heat exchanged s equal to another thermodynamc property of the system. The enthalpy (whch we shall descrbe n greater detal later) s defned as H = U + pv for any system. The change of enthalpy n a thermodynamc process, s n dfferental form: dh = du + d( pv ) = du + Vdp + pdv we can now replace our defnton for the change n nternal energy du = δ q + δwextra + δwexpanson-compresson = δq + δwextra pextdv so that dh = du + Vdp + pdv = δ q + δ wextra pextdv + Vdp + pdv ecause we are measurng a state functon t only has a meanngful value once we have attaned thermodynamc equlbrum. Then the external pressure wll be equal to the nternal pressure. Thus, dh = δ q + δ wextra + Vdp If no other type of work other than expanson-compresson occurs, dh = δ q + Vdp and at constant pressure dh = δ q. Then when the system s not constraned to stay at constant volume but t s at constant pressure, the measurement of the heat released gves us the change n the enthalpy of the system. We can now ntegrate the dfferental form to get an overall enthalpy change whch we would measure f we start at an equlbrum state and let the system evolve to a new thermodynamc equlbrum state: f Δ H = δ q where we note that the heat s not an exact dfferental. Ths change ΔH s for a change of state At constant pressure, and when no addtonal types of work are done on or by the system. Processes n whch Δ H > 0 are sad to be endothermc, and processes wth Δ H < 0 are called exothermc. Example: A phase transton between two crystallne forms of tn (Sn AW=8.69 g/mol) s observed at 000kPa and at 98K Sngrey Snwhte Δ U=+04.4kJ/mol The denstes of the two forms of tn are: ρ = 5.75 g/ cm Sngrey ρsn =7. g/ cm whte What s the nternal enthalpy change Δ H? Δ H = HSn H ( ) whte Sn = U grey Sn + pv whte Sn U whte Sn + pv grey Sngrey at 000kPa we have Δ H =Δ U + p( V V ) =Δ U + pδ V Snwhte Sngrey

22 8.69g For.0 mol of tn, we have 8.68g so that V = ρsn Then, we get V = 6.7 cm / mol Snwhte VSn = 0.55 cm / mol grey These are the volumes occuped by one mol of the respectve forms, and are called molar volumes. For one mol we have: 6 cm 6 m pδ V = 0 Pa( ) 0 = 4.8 J / mol mol cm and Δ H =Δ U + pδ V =, ( 4.8) J / mol =,00. J / mol We notce that the term pδ V s neglgble. In general f no gases are nvolved n the transformaton, we could safely gnore pδ V. For systems n whch gases are formed or consumed n a process pδv s usually consderable. ut we can approxmate the value of ths term by assumng deal gas behavor: H = U + pv = U + nrt now we can dfferencate ths dh = du + d( nrt ) = du + RTdn For a process that takes the system from one equlbrum stuaton to another we wll have an overall change of enthalpy f Δ H =Δ U + RTΔ n= U U + RT ( n n ) assumng constant temperature and f gass gass deal gas behavor for all the gasses n the system. Example: f The reacton H( g) + O( g) HO( l) Δ n= ( ngass ngass ) = 0 = And for ths process we have Δ H =Δ U + RTΔ n =ΔU RT Some useful defntons n thermochemstry: Standard Enthalpy Changes Changes n the enthalpy of a system undergong a process (physcal or chemcal) are normally reported under a specal set of condtons whch we refer to as standard condtons. The change n enthalpy for a process n whch the ntal and fnal substances are n ther standard state s called Standard Enthalpy Change ΔH. The standard state of a 5 substance at a specfed Temperature s ts pure form at.0 ar or.0 0 Pa.

23 Examples: Standard vaporzaton enthalpy change ΔH vap of water at 7K: HOl ( ) HOg ( ) Δ H vap(7 K) = kj/ mol The Standard of reacton enthalpy change ΔH vap of water at 98K: Cs ( ) + H( g) CH( g) Δ H (98 K) =+ 7 kj/ mol Reac ote that the frst process nvolves one mole of water. If we multply the reacton by (both sdes) clearly the enthalpy change wll be twce as much. Ths s true n general. Enthalpy of phase transtons: A( s) A( l) ΔH fus( T) Fuson A( l) A( s) Δ H ( T) = ΔHfus ( T) A( l) A( g) ΔH vap ( T) Vaporzaton A( g) A( l) Δ H ( T) = ΔHvap ( T) Δ sub Sublmaton Δ = Δ sub A( s) A( g) H ( T) Changng the drecton of a process changes the sgn of A( g) A( s) H ( T) H ( T) ΔH. There are a number of processes that are of nterest. We wll gve a few examples here. Enthalpy of Ionzaton In ths process an electron detaches from an atom or a molecule n gas phase: + Eg ( ) E( g) + e( g) ΔH( T) Δ n =+ Snce Δ n =+ we have gas gas Δ H ( T) =Δ U ( T) + RTΔ n=δ U + RT Ionzaton energy: The nternal energy change n ths process s called the onzaton energy, f the process s carred out at the temperature of 0K. Δ U ( T = 0 K) = E. Δ H ( T) =Δ U ( T) + RT = E + RT The value of RT at room temperature s.5kj/mol, whch s much smaller than the typcal onzaton energes E 50 kj / molor even bgger. Then, we can gnore the RT term and assume that ΔH ( T) E at almost any temperature realzable n the laboratory. At,000K s ~7.5 kj/mol and at 0,000K RT ~75kJ/mol. Electron Gan enthalpy Ths s the process n whch an electron bnds to an atom or a molecule n gas phase: Eg ( ) + e( g) E( g) + ΔH ( T) Δ n = ea gas

24 Δ Hea ( T) =ΔUea ( T) RT When the electron detaches at T=0K, the energy change was tradtonally called electron affnty E ea. However, the value of E ea has been taken (also tradtonally) wth the opposte sgn of the nternal energy change assocated wth the process of bndng an electron: Eea = Δ Uea ( T = 0 K) And we have Δ H ( T) =ΔU ( T) RT = E RT ea ea ea Enthalpy of bond dssocaton and bond formaton Consder a molecule, (a group of atoms connected by chemcal bonds). The standard reacton enthalpy change for the reacton n whch one mole of chemcal bonds are dssocated s: A ( g) A( g) + ( g) ΔH A-( T) s called the (standard) bond dssocaton enthalpy (change). The words n parenthess are usually omtted. It s mportant to realze that ths standard quantty s specfed for one mole of broken bonds n gas phase. One specal case s a reacton n whch ALL the chemcal bonds n a molecule are dssocated. Such process receves the specal name of enthalpy of atomzaton, and clearly t wll have an enthalpy change equvalent to the sum of all the bond dssocaton enthalpes n the molecule. For example: HOg ( ) H( g) + Og ( ) + H( g) ΔH a ( T) We can see that Δ Ha( T) =Δ HH OH( T) +Δ HO H( T) = 499 kj / mol+ 48 kj / mol The molecule adjusts after loosng the frst H and the second bond dssocaton s slghtly smaller. ote that there are two moles of bonds per mole of water. Also, we must know the chemcal structure n order to know how many bonds per molecule are there n t. Evaporaton of a sold can also be nterpreted as breakng all the bonds n the sold, so we have for example: a( s) a( g) Δ H (98 K ) =Δ H (98 K) = 07 kj / mol Enthalpy of formaton a In order to calculate enthalpy changes for a varety of transformatons t became useful to make a table of enthalpy changes. Ths table s n the CRC Handbook of Physcal Chemstry. One of the quanttes that you wll fnd s the Enthalpy of Formaton of a substance relatve to ts elements n the standard form. Ths s the enthalpy change of the process of formng OE mole of the substance n ts standard state, DIRECTLY from the elements n ther standard states. However, t s clear that we need to choose a standard state. Ths seems to reduce the generalty of the enthalpy of formaton. However, ths s not the case, because H and U are state functons that can not be known absolutely, and we can only measure ther sub 4

25 changes. Then, t was defned (arbtrarly) that the standard state of any substance s ts stable form at one bar. The standard state of water at 98K s lqud water snce at that temperature the stable for of water s the lqud. The standard form of carbon at 98 K s graphte, snce that s the stable form at one bar and 98K (damond s not the stable form, but ts converson to graphte s very slow). The standard form of ntrogen 500K s the datomc gas, snce ths s the stable form at 500K and bar. The standard enthalpy change of formaton of the gas ammona s: ( g) + H( g) H( g) Δ H f (98 K) = 46. kj / mol Another example s Cs ( ) graphte + O( g) + H CHOHl ( ) Δ H f (98 K) = 8.66 kj/ mol The standard form of methanol at one bar and 98K s the pure lqud form. Mg( s) + O( g) MgOs ( ) Δ H f (98 K) = kj/ mol Mg( s) Mgs ( ) Δ H f (98 K) = 0.0 kj/ mol Ths last process takes Mg(s) and forms the exact same Mg(s). Then t s clearly a process that must have zero enthalpy change. Ths s why The enthalpy of formaton of any element n ther standard form s ZERO Enthalpy of combuston A partcularly mportant reacton enthalpy change s the one assocated wth complete combuston of one mole of the substance n oxygen. Mostly ths s done for organc compounds. The standard enthalpy change of combuston ΔH c for some reactons: 7 CH CH( g) + O( g) CO( g) + H O( l) Δ Hc (98 K) =,560 kj / mol CH OH ( l) + O( g) CO( g) + H O( l) Δ Hc (98 K) = 76 kj / mol ote that the product of combuston of all organc compounds sco( g) + H O( l). For compounds that contan ntrogen, the combuston forms CO( g) + HO( l) + ( g) ecause the products are always the same for combustng dfferent organc compounds, knowng the combuston enthalpes allows us to calculate the enthalpy change for reactons that transform varous organc compounds. For ths reason the standard combuston enthalpes of many organc compounds are documented. For example let us calculate the enthalpy change for the transformaton CH CH( g) + O( g) + HO( l) CH OH ( l) Ethane Methanol Δ H (98 K) =? c 5

26 We use the fact that the enthalpy s a state functon, whch means that we can take any alternatve path from ethane to methanol, and the enthalpy change wll be the same. We can do ths by wrtng the followng ΔHRe acton CH CH( g) + O( g) + HO( l) CH OH ( l) ΔH c(ethane) ΔH c(methanol) CO ( g) + H O( l) If we follow the paths ndcated by the arrows we go from ethane to methanol n to alternatve ways, but they should have the same value for the enthalpy change. Ths trangular dagram s called a cycle. In order to be able to calculate the desred Δ H R We must frst balance the reactons: 7 CH CH( g) + O( g) CO( g) + H O( l) Δ Hcethane ( )(98 K) =,560 kj / mol CO( g) + H O( l) CH OH ( l) + O ( g) Δ H = Δ Hc( methanol) (98 K) = 76 kj / mol The next step s to add these two reactons n such a way that CO( g) and H O( l) cancel out. In order to do so, we must frst multply the second reacton by : 7 CH CH( g) + O( g) CO( g) + HO( l) ΔH + CO ( g) + 4 H O( l) CH OH ( l) + O ( g) - ΔH 7 CH CH( g) + O( g) + CO( g) + 4 H O( l) CO ( g) + H O( l) + CH OH ( l) + O ( g) ΔH Whch after smplfcaton gves C ( ethane) C( methanol ) R ( ethane methanol ) CH CH ( g) + O ( g) + H O( l) CH OH ( l) Δ H =Δ H + ( - ΔH ) R ( ethane methanol ) C ( ethane) C ( methanol ) In summary: Reactons can be multpled by a factor, and the value of ΔH R s multpled by the same factor. (a factor or - correspond to nvertng the reacton). Reactons can be added, and the value of ΔHR ( total) for the sum of reacton s the sum of the correspondng ΔH R of the reactons beng added. 6

27 orn-haber cycle Ths s the generalzaton of the case descrbed for the combuston reactons. In general you may want to calculate the value of ΔH R for a partcular reacton. Then all that s needed s to wrte an alternatve path from the reactant sde to the product that we want. Ths alternatve path can have several steps (reactons) for whch ΔH s known. Consder for example the formaton of sodum chlorde from chlorde and sodum ons n the gas phase, to form the sold acl. Ths type of process and ts reverse process are mportant because they are used n laser ablaton, magnetron sputterng and pulsed laser deposton. If we were to produce laser ablaton on a sodum chlorde target, the products + wll be the ons a ( g) + Cl ( g) When these recombne on a cold surface, they form a flm composed of crystals of acl: + a ( g) + Cl ( g) acl( s) Δ H (98 K) =? What s the enthalpy change nvolved n the process of formaton of the crystal? The ons n the gas phase are solated, whle the sold s an ordered lattce, because of ths, the value of such ΔH s called the lattce enthalpy of sodum chlorde. R 7

28 As we dd before we try to obtan the same product by an alternatve way. In the fgure we see that the values of ΔH R for the steps -5 are all known. Then ther sum should be equal to the lattce enthalpy. In other words, gong around the complete cycle, the net enthalpy change should be ZERO. ased on ths dagram, we wrte: formaton a( s) + Cl( g) acls ( ) Δ HR =Δ H f ( acl) = 4.5 kj/ mol + sublmaton a( s) ag ( ) Δ HR =Δ Hsub( a) =+ 07. kj/ mol + bond dssocaton Cl( g) Cl( g) Δ H R = Δ H( Cl Cl) = +.68 kj / mol + onzaton + - a( g) a( g) + e Δ HR4 =Δ H( a( g)) = kj/ mol + electron affnty Cl( g) + e Cl ( g) Δ H R5 =Δ Hea( Cl( g)) = 5. kj / mol + + Lattce Enthalpy a ( g) + Cl ( g) acl( s) Δ H R6 =ΔHlattce formaton(acl) 0 =Δ H +ΔH +Δ H +Δ H +Δ H +ΔH f ( acl) sub( a) ( Cl Cl ) ( a( g )) ea( Cl ( g )) lattce formaton(acl) and then we solve for the OLY one ΔH that s unknown to us: Δ H = ( Δ H +Δ H +Δ H +Δ H +ΔH ) lattce formaton(acl) f ( acl ) sub( a) ( Cl Cl ) ( a( g )) ea( Cl( g )) from whch we get Δ Hlattce formaton(acl) = kj / mol Enthalpy of formaton of ons n soluton A smlar cycle can be wrtten to obtan the enthalpy of hydraton. The hydraton s the process that begns wth the ons n gas phase and takes them nto water. If a solvent other than water s used the process s called solvaton. Reacton s replaced by lots of water! a( s) + Cl( g) acl( aq) Δ H R' =Δ H f ( acl( aq)) = kj / mol Wth ths we can get the sum of the hydraton enthalpes for a + ( g) + Cl ( g). It s also possble to calculate the values for the two processes + + a ( g) + H O( l) a ( aq) ΔH + hyd ( a ) Cl ( g) + H O( l) Cl ( aq) ΔH hyd ( Cl ) 8

29 For ths we need to obtan a value ndependently: From spectroscopc measurements, t s known that: + ( ) HOl () + H g H ( aq ) ΔH,090 kj / mol + Hydraton( H ) If we use the reacton: () ( ) HOl + HCl g H ( aq) + Cl ( aq) Δ H Hydraton( HCl) = 67.0 kj / mol we could get the value of Δ H. hyd ( Cl ) Problems: - A mass of 0.77g of the sugar D-rbose ( ) CH O are completely burned n excess of pure oxygen nsde a calormetrc bomb, causng the temperature of the calormeter to rse by 0.90K. (The products are only carbon doxde and water). A second experment burnng benzoc acd n excess of oxygen and usng the same calormeter produced a temperature rse of.940k. The nternal energy of combuston for benzoc acd ( ) 7 6 CHO s Δ U =,5 kj / mol. a) Wrte and balance the combuston chemcal reacton. b) Calculate the standard nternal energy of combuston of D-rbose. Answer: b) Δ U =,0 kj / mol -Usng a thermodynamc cycle fnd the hydraton enthalpy of Data (standard condtons, at 98K): Sublmaton enthalpy of Mg=67. kj/mol Frst onzaton energy of Mg=7.646eV Ionzaton energy of Mg + =5.05eV ond dssocaton enthalpy of Cl-Cl =4.6 kj/mol Electron affnty of Cl=.78 ev Enthalpy of dssoluton of MgCl =-50.5kJ/mol (n water) Enthalpy of hydraton of Cl =-8.7kJ/mol Mg +. Answer: -,89. kj/mol - Desgn a orn-haber cycle to calculate Δ H. hyd ( Cl ) Usng data from tables, evaluate ths enthalpy change. 9

30 Adabatc compresson-expanson We would lke to fnd the relatonshp between the temperature and the volume for a perfect gas when t s compressed adabatcally, (q=0). Accordng to the frst law of thermodynamcs, du = δ q + δw = p dv ext for a reversble process du = δwrev = pdv If we compress the gas by dv U ncreases. Ths means that the temperature ncreases. We have for an nfntesmal volume change; du = CvdT = pdv by nsertng the deal gas law, pv = nrt we have nrt Cv dv Cv dt = pdv = dv dt = V nrt V ntegratng C dv V = = ln nrt V V T V v dt T 0 V0 0 Assumng that Cv s not a strong functon of temperature, C T v dt V Cv T V ln and then ln ln nr = = T0 T V nrt T V We can elmnate the logarthms and wrte the expresson: Cv nr T V0 = T0 V whch s vald n general. Usng the smple form of heat capacty that s found from elementary knetc theory of the gases, Cv = nrwe get T V0 = T0 V p0v0 We can elmnate the temperature n ths equaton usng the relatont0 = and a nr smlar expresson for temperature T, to obtan: 0

31 pv nr V p V0 = = nr p V V p V or p V V = p V V whch gves p 0 V0 p0 V0 V0 = p V and we can then express the pressure as = = p V V The curves p versus v gven by the last equaton s called adabates. pv 0 0 pv ote that the sotherms were obtaned fromt = 0 constant nr = nr = p0 V0 = p V The frst law s vald for reversble or rreversble processes: du = δ q + δw = δq + δw rev rev rrev rrev However, the work performed by the system s larger f the process s carred out reversbly. The work that needs to be done on the system n an rreversble compresson process s ALWAYS larger than the correspondng reversble work to carry out the reversble compresson. δwrev < δwrrev and therefore, δw > δw = pdv rrev rev Analogously, because δ qrev + δwrev = δqrrev + δwrrev we conclude that δ qrev + δwrrev > δqrrev + δwrrev = δqrev + δwrev t follows that δ qrev > δ qrrev Carnot Cycle: We carry out successvely the followng processes on a fxed number of moles of a perfect gas:

32 -Reversble sothermal expanson from V to V (n ths step T s constant = T h ) For ths step t holds that V p = because T s constant, then U s also constant, so Δ U = 0 = W+ Q V p Q V V = W = pdv = nrt dv h V V V V ln Q = W = nrth and snce V > V Q > 0 V Heat goes from the surroundngs nto the system. Work s done by transformng heat from the envronment. -Reversble adabatc expanson of the solated gas fromv to V ; ow because the gas does work and we do not allow heat to enter, the system must do work at the expense of ts nternal energy. Therefore t cools down from Th to T c We have found that for ths type of expanson process t holds that T h = Tc V V because Q = 0 and therefore,

33 W =Δ U ( ) 0 = ncvδ T = ncv Tc Th < snce Th > Tc -Isothermal reversble compresson of the solated gas at the temperaturet c from V to V 4 ; Ths process satsfes V4 p V4 = and we also have that Δ U = 0 = W+ Q whch mples Q = W = nrtc ln V p4 V Snce V > V4 so that Q < 0 whch means that heat leaves the system and t goes nto the surroundngs. 4-Reversble adabatc compresson of the solated gas fromv 4 to V, takng the gas to the ntal temperature T h ; Ths process follows the relatonshp T c = Th V V 4 and Q 4 = 0 because the process s adabatc. Then we have, W4 =Δ U ( ) 0 4 = ncvδ T = ncv Th Tc > Th > Tc Δ U4 > 0 For the entre cycle, we calculate the change n the nternal energy of the gas: Δ Utotal = Q+ W+ W + Q+ W+ W4 = 0 and we also have that W = W4 = 0 = 0 V ln V ln 4 Q = nrth and Q = nrtc V V Q T V ln V = nr and also 4 h Q T c V = nr ln V for these two adabatc processes we had that thus, T h = Tc V V and T c = Th V V 4 V4 V V V V 4 = = = V V V V4 V

34 We now replace ths n Q V4 V Q = nrtcln nrtcln V = V then Q V = nrt ln c Tc V Q V Q Q Usng the expresson for = nr ln we get that = whch then leads to T V T T Q T + = 0 h Q T c h h c Ths equaton s vald not only for ths Carnot cycle but for any reversble cyclc process, as far as the experments tell us. Ths means that regardless of the path that we take, Q T s the same. It must be concluded that Q T s a state functon. δ q δ qrev If we do the cycle n nfntesmal steps, = 0 or n an ntegral form 0 T = T Therefore, cycle cycle δq rev T A δqrev + = 0 T δ qrev Ths mples that s ndeed a state functon, and we gve t a specal name: T δ qrev ETROPY=. T In fact any arbtrary cyclc process can be approxmated to any desred accuracy by a large set of Carnot cycles for whch our dervaton s vald: 4

35 δ qrev For each Carnot cycle n the set t must hold that = 0 and therefore for the T Carnot cycle δ qrev entre reversble arbtrary cycle = 0 T Entropy Total cycle δ qrev We showed that δ qrev > δ qrrev but =ds from whch we get TdS = δ qrev and t must T be true thattds > δ qrrev. For a perfectly solated system, 0 = δ qrev then ds = 0 : The entropy s an extremum n any reversble adabatc process. Snce δ qrev > δ qrrev we have that for any rreversble adabatc processtds > δ q rrev = 0 whch mples ds > 0 The entropy ALWAYS ncreases for rreversble processes. In general (reverble or rreversble ) ds 0 The condton of extremum must then be a mnmum, snce the entropy ncreases or stays constant, but t never decreases. These last statements are extremely mportant conclusons, and they have profound physcal consequences. Thermodynamc Temperature: We ntroduced the work of expanson-compressonδ W = pextdv, and another example s the lnear dsplacement δ W = Fdx 5

36 For a Magnetc feld and a system wth magnetc moment μ, In addton to expanson-compresson work we would have δw = pextdv dμ where dv and dμ are the generalzed dsplacements, and and P are the generalzed forces. Heat, lke work s not a state functon. However n a quasstatc process we can have du = δ q (no work). What would happen f we try to wrte δ q as the product of some generalzed force tmes some dsplacement?. δ q= δ q where we could have some dfferent types of heat smlar to what we do wth the dfferent types of work. n δ q= yk( x, x, x... xn) dx ot at exact dfferental k k = force dsplacement And we know that n q δ q dxk ot at exact dfferental k = xk However t s sometmes possble to fnd a functon λ( x, x, x... x n ) such that when we multply λ tmes δ q we obtan a dfferental that s exact: λ( x, x, x... x ) δq = dφ ( x, x, x... x ) = λy dx n n k k = n Φ now, dφ ( x, x, x... xn) = dxk holds! = xk Ths means that whle δ q s not an exact dfferental, dφ s an exact dfferental n If ths s true, then Φ Φ ( λ yk ) λ yk = and = and also x x x x k j k j Φ ( λ y j ) = x x x k j k Assumng that λ and λ yk exst, and further assumng that y j Φ x x j k s contnuous we wll Φ Φ ( λ y j ) ( λ yk ) have = and therefore = j, k =,... n x j xk xk xj xk xj Ths s a necessary and suffcent condton for dφ be an exact dfferental. λ( x, x, x... x n ) s called the ntegraton factor and t always exst for n. n > there may or may not exst an ntegraton factor. Implcatons n thermodynamcs: 6

37 If δ q s exchanged reversbly (quasstatcally) there exsts an ntegraton factor λ = θ where θ s called the thermodynamc temperature or λ = and δ q = ds where S s T T what we call the entropy. S s a state functon because we constructed t to satsfy the defnton of an exact dfferental. The ntegraton factor s not unque: f λ s an ntegraton factor for δ q, then λ f ( Φ )( f s an arbtrary functon of Φ ) Is also an ntegraton factor for δ q!! λ f ( Φ ) δq= f( Φ ) λδq= f( Φ) dφ = dϕ and ϕ = f ( Φ) dφ Ths non unqueness allows for many ways of defnng an equvalent of the entropy, and temperature. 7

38 The effcency of the Carnot Engne We have establshed that for a Carnot cycle operatng between heat reservors (system and surroundngs) at a temperature Th and Tc wth Th > Tc Q Q + = 0 Th Tc How much work does ths cycle perform? V V 4 Wtotal = W+ W + W+ W 4 = W+ W = nrthln + nrtcln V nc V = v( Tc Th) = ncv( Th Tc) ( W = W ) Only the work done durng the sothermal steps contrbute to the result W total. We also V4 V establshed that = V V from whch we get V = V so that V V4 V V4 Wtotal = nrth ln nrtc ln V4 V V4 V4 Wtotal = nrth ln nrtc ln V V W V = nrln ( T T ) and snce V 4 V4< V we have that ln < 0 V 4 total h c V > 0 Then, Wtotal < 0 whch means that the system undergong a Carnot cycle does work on the surroundngs!. We can make use of the Carnot cycle, whch s often called a Carnot Engne. The effcency of the cycle s the rato of the work done and the heat taken from the reservor at T = Th. How much of that heat s converted to work? Ths heat wasq. Then we take the absolute value of the work done and dvde t by Q. Wtotal W+ W η = = Q Q ote that Q > 0 for the cycle operatng n the drecton we ntally descrbed t. ut, Q+ W = 0 and Q + W = 0 so Q = W and Q = W whch mples that Q+ Q = ( W+ W) so we replace ths n the effcency η : W+ W Q+ Q Q η = = = + Q Q Q ote that Q > Q and Q > 0 so that Q+ Q > 0 We have seen that for ths reversble cycle Q Q Tc Q + = 0 or = T T T Q h c h 8

39 wth ths the effcencyη becomes a functon of the temperatures of the reservors: Wtotal Q Tc η = = + = Q Q Th SnceT h > T c Tc T < and η > 0 and η < h In summary: The hgher Th and the lower T c the hgher the effcency of a Carnot engne. Only f Th or f Tc 0 we wll have 00% effcency of converson of heat nto work. ow consder what happens for an rreversble cycle. Is t possble to have ηrrev > ηrev? To answer ths we consder two heat reservors at temperatures Th and Tc wth Th > Tc that can be put n contact: In ths case we know that - Energy wll flow from the reservor at T h to the reservor at T c. It EVER happens n the opposte drecton. - o work s done at all. Then let us assume that between these two reservors we nstall two cycles that can operate between them: 9

40 Ths engne s REVERSILE and t takes all the work produced by the rreversble engne and t transfers an amount of heat Q from the cold reservor to the hot reservor. It can do ths by usng ALL the work done by the rreversble engne. Ths IRREVERSILE engne takes an amount of heat Q ' from the hot reservor and t produces an amount W of work. The effcences are Wrev Wrrev η rev = and η rrev =. Q Q ' If ηrrev > ηrev then t wll be Q' < Q because all the work done n the rreversble cycle s consumed by the reversble engne. So Q' < Q. Then we are transferrng a net amount of heat Q = Q Q' nto the hot reservor wthout dong any net work. Thus, the hot reservor would get hotter and hotter wthout the need of supplyng any work. 40

41 Thus we have to conclude that t s mpossble that any sort of rreversble cycle would be more effcent that a reversble Carnot cycle, and ηrrev < ηrev ALWAYS. Q Q' ηrev = + > + = ηrrev Q Q' Tc Q Moreover, recall that for the reversble cycle = Th Q Then Tc Q Q' = > or Th Q Q' Tc Q' T > h Q' from whch we see that for the rreversble engne Q ' Q ' T > h T or Q ' Q ' + < 0 If we were able to subdvde the rreversble cycle nto c Th Tc small peces as we dd for the reversble cycle we wll have nfntesmal terms δ q' δ q' + < 0 T T and therefore we wll conclude that for any engne or cycle operatng between a hot and a cold reservor (f the cycle s not reversble) δ q δ qrrev < 0 < 0 T T 4

42 ow consder the path n p-v space: S δq rrev δqrev T T A A SA < = ds =ΔS ( A ) δ qrrev δqrrev = lm δqrrev ds A = < A T T A Only n ths lmt we wll have V VA + δv wth δv 0 p pa + δ p wth δ p 0 whch wll mply that T TA whch justfes that we took /T outsde the ntegral. δ qrrev s the total heat exchanged n the rreversble process gong from A A do any process n an adabatc contaner, then δ q rrev = 0 and then we get that 0 = rrev T δ q < ds or A ds > 0. If we Ths last equaton s the expresson of the second law of thermodynamcs. In words, The entropy of the unverse ncreases for all processes or at best, t stays constant for reversble processes The entropy ncreases untl equlbrum s reached, and ds = 0 after the equlbrum has been reached. As the system evolves n tme S ncreases or t remans constant, but there s no turnng back. It s then sometmes sad that the entropy s what determnes the drecton of the arrow of tme. 4

43 Propertes of the entropy We had that du = δ q + δw + δwother where other means other forms of work than expanson and compresson. For a reversble process n the absence of other types of work, and for a constant composton (number of partcles n the system or number of moles of each component) du = δ q + δw = TdS pdv (*) rev rev s a combned form of the frst and second law of thermodynamcs. ote that p and T do not depend on the sze of the system. If we have a system n equlbrum, the n we dvde t n two sub-systems, the temperature wll be the same on each of the two parts. These varables that do not depend on the sze of the system are called ntensve varables. On the other hand, V s an extensve varable. If you dvde a system n equlbrum n two sub-systems they wll have volumes tat are clearly smaller that that for the ntal subsystem. It appears that n the expresson for the nternal energy, (*) each term s the product of an ntensve varable tmes the dfferental of an extensve varable. Ths s correct, and the entropy s extensve. When the sze of the system changes (number of partcles) clearly the nternal energy U changes. Therefore we need another term that accounts for the ncrease n the nternal energy when the number of partcles changes: du = δ q + δw = TdS pdv + μd rev rev The last term, μd s also the product of an ntensve varable μ whch s called The Chemcal Potental tmes d a change n the number of partcles. We wll dscuss the chemcal potental n greater detal later on. From the expresson combnng the frst and second law, we can derve the followng denttes: du = TdS pdv + μd Combned frst and second laws U S V, = T Temperature U V S, = p Pressure U SV, = μ Chemcal potental The last equatons are totally general equatons of state. For example you can compare the deal gas law after solvng t for the pressure wth the above equaton for pressure. The equaton du = TdS pdv + μd s wrtten from the energy pont of vew, but we can also see the system from the entropy pont of vew: 4

44 du = TdS pdv + μd p μ ds = du + dv d T T T whch leads to: Equlbrum condtons Once we know the entropy as a functon of ts varables SUV (,, ) we can completely specfy the state of the system. It also allows us to antcpate what are the condtons for the system to be at equlbrum. How does S tell us that we have reached equlbrum? Example thermal equlbrum: Consder a system represented as n We now dvde ths system by a rgd, fxed nsulatng and mpermeable wall. ow we have Ths has constraned the system n a number of ways. We now have U = U + U but more mportantly, S = S( U, V ) + S( U, V). If we now remove the nsulatng propertes of the wall, leavng the system dvded by a rgd, fxed and mpermeable wall, heat can now flow from one subsystem to the other. ow that we have removed one constran the system wll evolve to a new equlbrum stuaton. What s the new equlbrum condton? U = U+ U and energy can flow freely from () to (). So that U = U U then, 0 = du = du+ du or du = du we also have that S = S+ S ds = ds+ ds changes n S as the system equlbrates we can replace these dfferental relatons n 44

45 ds ds ds = du+ du = du+ du = du du du V T T T T V ut ds = 0 for small changes du. Then n order to reach ds = 0 for any small du 0 we must have that = 0. Ths surely mples that T = T and we arrve at the T T condton of thermal equlbrum A smlar argument but extendng ths to a movable wall by vrtue of the fxed total volume wll yeld du = du and dv = dvfrom whch we get ds ds ds ds ds = du + dv + du + dv du dv du dv V U V U p p p p ds = du + dv + du + dv = du + dv T T T T T T T T where we see that n order for the equlbrum we wll need T = T (thermal equlbrum) and p = p(mechancal equlbrum) As an exercse: analyze the case n whch the wall becomes permeable to partcles so that at the begnnng. Fnd the equlbrum condton Answer: T = T, p = p = (chemcal equlbrum). Propertes of the entropy The entropy tells us when equlbrum has been reached. In some sense t s smlar to a reachng the mnmum potental energy n a mechancal system. Indeed, the entropy s referred to as a thermodynamc potental and t has propertes that we shall dscuss n order to understand ts mplcatons. To begn, we note that S s an ncreasng functon of U; p μ ds = du + dv d snce T > 0. And we know that the entropy s a maxmum at T T T S S equlbrum. ut s t > 0 or s < 0? In a pctoral way, whch of the two U U followng fgures s the correct behavor of the entropy? 45

46 ? Consder a contaner wth a gas. We can magne that we dvde t n exactly two EQUAL HALVES. o physcal wall, just an magnary dvder: Snce V and are equal and fxed on both sdes, then U0 s also fxed. In addton, Stotal = S+ S = S( U0, V, ) = S( U0, V, ) + S( U0, V, ) left sde lets consder the case that the curvature s postve: rght sde SU ( ) < SU ( +Δ U) + SU ( ΔU) both sdes ths sde has hs sde has wth the same ΔU excess ΔU less energy energy energy 46

47 but clearly the entropy has ncreased as we know t s always true. However, the nternal energy has done somethng unusual: S ncreases at the expense of one sde gettng hotter and the other gettng cooler!! (no thermal equlbrum s possble). Ths s aganst all our expermental evdence, and therefore the entropy can not have postve curvature. Thus, t S can not be that > 0. Let us check tat f the entropy has a negatve curvature wth U respect to the energy we CA have thermal equlbrum. S S(U 0 ) [S(U0 +ΔU)+S(U 0 -ΔU)] S(U 0 -ΔU) S(U) S(U 0 +ΔU) U 0 -ΔU U 0 U 0 +ΔU U In ths case we have that SU ( 0) > SU ( 0 +Δ U) + SU ( 0 ΔU) both sdes ths sde has hs sde has wth the same ΔU excess ΔU less energy energy energy now clearly f S ncreases t has to do so by equalzng the energy on both sdes of the magnary wall. Thus, thermal equlbrum s reached when the entropy ncreases. S Therefore t s that < 0. In other words S s a concave functon on the nternal U energy. Extensvty of the Entropy. Another far reachng property of the entropy arses from the fact that t s an extensve varable. Consder a system, whch we dvde n two halves as before: U,V, U,V, for ths we have seen that S S 47

48 Stotal = S + S = S ( U, V, ) + S ( U, V, ) left sde rght sde ut we can vew the system as OE system: U V SUV (,, ) = S(,, ) = S( U, V, ) or SUV (,, ) = S( U, V, ) In general rather than a factor of ½ we can have any multplyng factor lambda: S( λu, λvλ) = λs( U, V, ) In words, The entropy s a frst order homogeneous functon of the extensve varables. In partcular for λ = we have SUV (,, ) = S( U, V, ) = SU ( m, Vm,) = SU ( m, Vm) The energy per partcle(or per mol) and volume per partcle(or per mol). and SUV (,, ) = SU ( m, Vm ) = Sm s the molar entropy or the entropy per partcle. A smlar reasonng also shows that the nternal energy s also a frst order homogeneous functon on the extensve varables: U( λs, λv, λ) = λu( S, V, ) In fact f we see the nternal energy as a functon of the entropy, we have: du = TdS pdv + μd U( S, V, ) = T S V, so when we dfferencate U( λs, λv, λ) = λu( S, V, ) wth respect to S U( S, V, ) U( λs, λv, λ) d( λs) λ = S S ds V, λv, λ T( S, V, ) T( λs, λv, λ) λt( S, V, ) = λt( λs, λv, λ) whch of course says that T( S, V, ) = T( λs, λv, λ). In the temperature s a ZERO order homogeneous functon of the extensve varables. The same s true for all the ntensve varables: T( S, V, ) = T( S, V, ) = T( Sm, Vm) = λ 48

49 p( SV,, ) = p( S, V, ) = ps ( m, Vm) μ( SV,, ) = μ( S, V, ) = μ( Sm, Vm) Clealry we have varables that are functons of the same two molar varables. Ths mples that there s a relatonshp between T p,and μ. Ths relaton s called the Gbbs-Duhem relaton. We fnd such relaton now. We start from λu( S, V, ) = U( λs, λv, λ) and dfferencate t wth respect to λ : U( S, V, ) = du ( λs, λv, λ) d( λs) du ( λs, λv, λ) d( λv ) du ( λs, λv, λ) d( λ) + + dλs dλ dλv dλ dλ dλ λvλ λs, λ λs, λv usng the fact that the varables are homogeneous : U( S, V, ) = T( S, V, ) S P( S, V, ) V + μ( S, V, ) or smply U = TS PV + μ Ths relaton s called Euler relaton. It s ont a trval result, because t mples that du = TdS pdv + μd can be ntegrated drectly somethng that from the mathematcal pont of vew s not drectly apparent. From the Euler relaton we get by dfferencatng: du = TdS + SdT pdv Vdp + μd + dμ now we subtract du = TdS pdv + μd member by member, and we get 0 = SdT Vdp + d μ Gbbs-Duhem The Gbbs-Duhem relaton tell us that we can not vary T p,and μ ndependently. Once we set the value of any two of them, the thrd one s determned by the Gbbs-Duhem relaton. If know how p depends on ( SV,, ), and μ depends on ( SV,, ) we can calculate TSV (,, ) How do we measure the entropy? In prncple we know that for a reversble process ST ( ) ST ( ) f Tf = T δ q T rev 49

50 At constant pressure we saw that δ qrev = CpdT so that Tf Tf δ q CdT rev p ST ( f) = ST ( ) + = ST ( ) + T T T T It appears that all we need to know s how the heat capacty depends on the temperature. In fact, ths equaton also says that f we know what the heat capacty s at the absolute zero temperature, we can know the ASOLUTE value of the entropy. Ths s perhaps unusual, that we can now the absolute value of a state functon. There are however dffcultes to get such absolute value. In prncple one can measure Cp ( T ) whch gves curves wth some dscontnutes where some phase transtons occur: C p /T lq sold gas S ΔS vap T ΔS fus T f T bol T The entropy changes assocated wth the phase transtons at constant pressure are ds ds fus δ qrev ( fusson) dh ( fusson) ΔH ( fusson) = = Δ S fus = T T T vaporzaton fusson fusson fusson δ q (vap) dh (vap) ΔH (vap) = = Δ S = rev vap Tbol Tbol Tf bol 50

51 Whch we can measure n calormetrc experments. ut t s hard to measure Cp near T=0K. However ernst proposed that at T=)K all the moton s quenched, so f the materal forms a perfectly crystallne sold, the entropy wll be zero. Ths s called ernst heat theorem (also called the Thrd law of thermodynamcs). In the lmt of T=0K, ΔS 0 Therefore, f the entropy at T=0K s zero, we can n prncple know the absolute entropy. Here t s far to say that ths thrd law of thermodynamcs does not have the generalty that the frst laws have. It says that for perfectly crystallne solds Ths rules out mperfect solds, or the presence of defects. There certanly are many mportant examples of such materals. How do we know f at the absolute ZERO temperature there s a resdual entropy?. The answer to ths queston wll be apparent when we dscuss the entropy n terms of statstcal mechancs. For now lets verfy that the Thrd law does work for perfectly crystallne materals: Example: There are two crystallne forms for sulphur, alpha and beta sulphur. T= 69K CdT p S( αs)( T = 69 K) = = S( αs)(0 K) + 7 J/K mol T T = 0 S( α )(69 K) = S( α )(0 K) + 7 J/K mol S S T= 69K CdT p S( βs)( T = 69 K) = = S( βs)(0 K) + 8 J/K mol T T = 0 S( β )(69 K) = S( β )(0 K) + 8 J/K mol S S ΔS( β α )( T = 0 K) = S( α )(0 K) S( β )(0 K) = S S S S = S( α )(69 K) S( β )(69 K) + 8J/K mol- 7J/K mol S S From the enthalpy change measured for the transton at 69K we have ΔH( β ) 40 / S α J mol S ΔS( βs αs)( T = 69 K) = = =.09 J / Kmol T 69K transton replacng ths n the prevous result: ΔS( β α )( T = 0 K) =ΔS( β α )(69 K) + J/K mol =-.09 J/mol+ J/mol 0J/mol S S S S Thus, for sulphur there seems to be a zero entropy at T=0K. 5

52 Thermodynamc potentals We have seen that we can characterze a system by specfyng SUV (,, ) or U( S, V, ) du = TdS pdv + μd p μ ds = du + dv d T T T () These two equatons are wrtten n a dfferental form, and descrbe how does the entropy or the energy change when we vary the parameters that appear as dfferentals. These varables that appear as dfferentals are called the natural varables. For that reason we wrte SUV (,, ) or U( S, V, ). In an experment, t wll be common to fx some of the varables, and observe the varaton of a smaller number of free varables whch are allowed to vary as the system evolves to equlbrum. Once the equlbrum has been reached, we note the changes n these free state functons. We could fx the volume and the number of moles (or partcles) and then we can follow changes n U and S. Or we could fx S and vary V to see how U changes. We may encounter stuatons where t s more convenent to consder T as held constant rather than S. Exercse: Dscuss possble ways of carryng an experment at constant entropy. It s far easer to control the temperature, the pressure, and the number of moles than the entropy. The present formulaton s n terms of the extensve varables. How can we reformulate what we obtaned so far and keep the nformaton that the equatons () contan?. We can do that by defnng new functons that contan the same amount of physcal nformaton. Then we use that functon to predct the equlbrum condtons. We note that n () T s not a natural varable. We would lke to have dt appear n the equaton. Another example s U = p we would lke to have dp n () so that the varable of nterest s p rather V S, than V.To obtan the functons we are lookng for, the varable that appears as a dervatve should appear as a parameter, whle the ones that are parameters should be n the dervatves. Lets see how to do ths n general. If we have a functon f ( x ) f(x) x 5

53 df We would lke to fnd a new functon g of the dervatve of. Explctly, we want g dx wth the condton that g contans the same amount of nformaton about the thermodynamc system as f ( x) does. To stress that ths s not a trval matter, lets try to do t. df To fnd g we could wrte p = whch wll be the ndependent varable of ( ) dx g = g p. Then t appears that f we solve for x we get ( ) g( p) = f x( p) df Whch gves us g n terms of x. It requres that we solve p = = f '( x) for x. So dx from p = f '( x) we obtan x( p) = f ' ( p) (the nverse functon of f '( x ) ) Lets us do ths wth an example df f ( x) = ax + bx+ c p= = ax+ b dx now we use the nverse functon x( p) = f ' ( p) to get x( p ) x p and then we can have ( ) x = and p b a p b g( p) = f ( x( p) ) = f a p b p b ( p b) b( p b) g( p) = a + b + c= + + c= a a 4a a pb b p pb b pb b p pb + b + pb + b ( p pb + b ) c = c = + c 4a a a 4a 4a 4a a a 4a p b g( p) = + c 4a Is that t?. ot really. If nstead of havng f ( x) we had a dfferent functon f ( x ) shfted along the axs f ( x) = a( x x ) + b( x x ) + c 0 0 f( x) = a( x x0) + b( x x0) + c = ax + ax0 axx0 + bx bx0 + c by callng a' = a b' = ax + b c' = ax bx + c

54 we can rewrte f x a x b x c ( ) = ' + ' + ' Clearly f ( x) f( x) because t has dfferent coeffcents b and c. However, f we obtan g ( p ) where df ' ( ) p = = f x we get the same result: dx ( ) p b g ( ) p = + c= g p 4a see ths: p b' p b' p ( ax0 + b) g( p) = a' + b' + c' = + ( c+ ax0 bx0) = a' a' 4a 4a p (4a x0 + b 4 abx0) p b + c + ax0 bx0 = ax0 + bx0 + c + ax0 bx0 = 4a 4a 4a 4a p b 4a + c Ths means that the g( p) obtaned by ths method does not contan the same nformaton that f ( x ) has. Clearly an nfnte number of functons that are shfted along the x axs (.e. dfferent functons) gve the same g( p ) whch means that f we tred to go back from g( p) to f ( x) we can not do t because the transformaton s not unque. We have lost some nformaton n the process of gettng g( p) and now we can not go back to f ( x ). g( p) does not contan enough nformaton to produce f ( x ). What dd we do wrong? We only used the slope of the functon f ( x ). A general approach must produce a dfferent g( p) for each dfferent f ( x ). Fortunately there s an equvalence between the pont geometry and the lne geometry. Ths allows us to represent a curve equally well by ether of the two followng defntons: - A curve s the locus of ponts satsfyng the relaton y = f( x) - A curve s the envelope of a famly of tangent lnes to t. 54

55 Y(x) Y(x) x x Therefore the equaton that allows us to reconstruct the famly f tangents wll allows us to exactly descrbe the functon f(x). ote that for each tangent we need the slope and the ntercept. efore we only used the slope, and ths s why we faled. Any pont n the plane can be descrbed by two numbers x,y so every straght lne n the plane s descrbed by numbers p and ψ. p s the slope and ψ s the ntercept on the y axs. The relaton f(x) selects a subset of ponts n the plane (x,y), and the relaton ψ ( p) selects a subset of lnes n the plane ( p, ψ ) We need to fnd ψ ( p) and ths wll be equvalent to havng f(x). Snce p s the dervatve (slope) t wll play the role of the varable n our thermodynamc functons. df ψ( p) ψ = = ψ ( f '( x) ) we wll fnd ψ graphcally: dx y(x) y 0 (x 0 ) (x 0,y 0 ) ψ(x 0 ) The tangent lne s T( x) = y( x ) + y'( x )( x x ) but the ntercept s T( x) = T( x= 0) = T(0) and ths s ntercept slope x 0 x 55

56 T(0) = ψ ( y'( x0 )) then ψ ( y '( x0)) = y( x0) + y'( x0)( x x0) = y( x0) + y'( x0)( x0) we now replace the slope by for x= 0 p : ψ ( y '( x0)) = y( x0) p( x0) x0 Ths s a functon of x 0, we can wrte t smply as a functon of x, so we drop the sub ndex and wrte t as ψ ( y'( x)) = y( x) p( x) xor more smply: ψ ( x) = yx ( ) xpx ( ) ψ ( x) = yx ( ) xpand ψ ( x) s called The Legendre transform of y( x ) dy ote thatψ = y x and that ( ) dx ψ x s the correspondng value of the tangent to y( x) at pont x, ( x, yx ( )). If everythng s correct ψ ( x) should only be a functon of the dervatve dy = p( x) = y'( x) We start from ψ = y xp dx dψ = dy xdp pdx but dy dy = dx = p( x) dx replacng ths n dx dψ = pdx xdp pdx = xdp Thus, dψ = xdpa functon of p as we requred. To obtan ψ ( p) explctly, we have to get rd of x: ψ = y xp ( x) yx ( ) xpx ( ) yx ( ) xy'( x) ψ = = because p = y'( x) ths wll only be possble f we can solve p = y'( x) for x. I other words, x = y' ( p) must exst. Replacng ths n ψ ( x) = yx ( ) xpx ( ) we get ( ) ψ y ' ( p) = y( y' ( p)) y' ( p) p( y' ( p)) a functon of p a functon of p p so we obtan ( ) ψ p = yy ( ' ( p)) py ' ( p) an explct functon of p. Example: yx ( ) = x we have ths functon of x but we want a functon of y '( x) = x p= x dy dx 56

57 and we can solve the dervatve for x = x( p). Ths s x = ψ ( x) = yx ( ) xpx ( ): p p ψ = p p= p p = p ψ ( p) = p p. w we replace n We can always fnd the Legendre transform provded that the dervatve of our functon s strctly monotonc. Ths wll allow us to solve x = x( p) from the dervatve. Example: Show that for y( x) = x the Legendre transform does not exst. Soluton: y '( x) = = pnot solvable for x. ψ ( x) = y( x) xy'( x) = z px= x x= 0so ths ψ ( x) does not contan all the nformaton that yx ( ) = xhas. From ψ ( x) we can completely reconstruct y( x) n a unque way ψ ( p) = y( p) x( p) y( p) = ψ ( p) + xp (**) dψ but dψ = xdp x= = ψ '( p) dp because y '( x) s strctly monotonc, then x = y' ( p) s also strctly monotonc (bjectve) then x = y' ( p) can also be solved for p and p can then be nserted n (**) to yeld: = ψ ' ( ) ( ) = ( ψ ' ( )) or ( ) ( ) p x y p y x etter to see ths n an example: We start fro the legendre transform of p ψ ( p) = (see example above) 4 Then dψ ( p) = pdp 4 or dψ ( p) = pdp dψ = dp p = x ( ) p = x p x = x yx ( ) ( ) ( ) y ψ x ψ ψ x xψ x ( ' ) = ' + ( ' ) = x whch we have found to be because dy dy dx = p ( x ) dx dp dy dx and dψ = dy xdp pdx dψ = pdx xdp pdx dψ dψ = xdp = x dp 57

58 ote that d ψ = x s strctly monotonc. SO we can replace p ( x ) = xn dp ( ) ( ) ( ) ( ) ψ ( ) ψ p = y p x p y p = p + xp or n ths example: y( x) = ( x) + x x= 4x x = x yx ( ) = x Summary: Gven a functon y( x) whch has a strctly monotonc frst dervatve p = y'( x) there exsts a Legendre transform defned by ψ = y xp. In summary to get a functon that replaces a varable by ts dervatve the method s: dy take your functon y(x), and buld a new functon ψ by subtractng the product x dx from t In other words take your ntal functon and subtract from t the product of the varable that you want to elmnate tmes the dervatve of your ntal functon. We can now extend ths to functons of several varables: For f ( xy, ) a functon wth df = p( x, y) dx + q( x, y) dy f f Where pxy (, ) = and qxy (, ) = xy y ψ = f ( xy, ) xp x If we want to replace x by p then we do Ths s t. To convnce yourself, note that ψ has a total dfferental dψ = df pdx xdp Insertng df = pdx + qdy n that dψ = pdx qdy pdx xdp as you can see you get dψ = qdy xdp ow x that was n the dfferental s a parameter, and p that was a parameter s now n the dfferental. Ths s what s called a partal Legendre transform wth respect to x. If we bult ψ = f ( xy, ) xp yqwe would get the Legendre transform wth respect to x and wth respect to y. Enough math. ow lets put ths formal deas nto thermodynamcs. The use of the Legendre transform on the functon U( S, V, ) gves a number of new functons called thermodynamc potentals 58

59 Lets assume that we do not want to control the volume n an experment, but rather we want to control the pressure. Lets see what we get f start from U( S, V, ) and transform t wth respect to V: The total dfferental s du = TdS pdv + μd and we want du ψ ( S, p, ) = U V = U V( p) = U( S, V, ) + pv dv S, p We mmedately recognze ψ ( S, p, ) as the Enthalpy. So H = U + pv, and we have dscussed ts use n calormetry. ow you can see why t was defned n that partcular way. H( S, p, ) The total dfferental of the Enthalpy s dh = du + pdv + Vdp = TdS pdv + μd + pdv + Vdp = dh = TdS + Vdp + μd ow we want to get rd of the entropy n U( S, V, ): du = TdS pdv + μd we construct a new functon F du F = U S = U TS ds V, T The total dfferental of the Helmholtz potental s df = du TdS SdT = TdS pdv + μd TdS SdT df = SdT pdv + μd If we combne the Helmholtz potental wth Euler relaton(u = TS pv + μ ), we get F = U TS = pv + μ F contans exactly the same nformaton bout the system as U( S, V, ) does, but t s a functon of FTV (,, ). Ths new thermodynamc potental s called the Free energy or Helmholtz potental It plays a role n processes at constant volume and constant temperature. We can do a double Legendre transform of U( S, V, ) wth respect to V and S to obtan a functon that elmnates V and S: Ths s the Gbbs potental or Gbbs free enthalpy GT (, p, ) du du G = U S U V = U TS V( p) = U TS+ pv ds V, dv S, T p The total dfferental of ths functon s: 59

60 dg = du TdS SdT + pdv + Vdp = TdS pdv + μd TdS SdT + pdv + Vdp dg = SdT + Vdp + μd A functon GT (, p, ). Combnng G wth Euler relaton we get: G = U TS + pv = TS pv + μ TS + pv = μ Thus, for a system that has only one component, the chemcal potental s smply molar G Gbbs potental: μ =. The Gbbs potental s extremely mportant n chemstry because t s the most convenent potental to descrbe systems at constant pressure and at constant temperature, condtons that are easy to mantan n the laboratory or n an ndustral process. 60

61 Prncple of mnmum Energy The second law of thermodynamcs asserts that there exsts a functon S of the energy of the system SUV (,, ) that s a monotonc ncreasng functon of the nternal energy U. Ths functon reaches an extremum (maxmum) at equlbrum. Ths would be f we use p μ ds = du + dv d as our fundamental relaton. From ths we clearly see that T T T ds = 0 for constant nternal energy once we reach the equlbrum. To see ths consder the fgure: At constant U equlbrum s acheved when S s a maxmum wth respect to another varable X j. If we start wth a system for whch the varable X j s not at the value correspondng to equlbrum, the system constraned to be at constant U wll evolve, and S wll ncrease and X j wll vary durng ths process. The process stops when S acheved a maxmum: ds = 0 Equlbrum. dx j U ds ear the equlbrum pont we have < 0 dx j U Another way of lookng at ths s by takng the fundamental relaton du = TdS pdv + μd. In ths case, we constran the system to have a constant entropy. Ths s represented n the followng fgure: 6

62 At constant entropy, equlbrum s acheved when U s a mnmum wth respect to other (unconstraned) varables X j. Let alone, the system at constant entropy wll evolve towards mnmzng the energy by adjustng X j. The use of thermodynamc potentals We have seen how we can use the entropy to fnd the equlbrum condton. All the thermodynamc potentals contan the same nformaton about the system, and therefore we should be able to predct the equlbrum condtons wth any of them. The choce of the potental to use depends on the system and on what are the varables that we can control. Usng the Free energy (Helmholtz Potental) We know that a system wll evolve to an equlbrum stuaton n whch S s a maxmum, and ds = 0 at equlbrum. F can be used to fnd the equlbrum condton: 6

63 Lets consder a system at constant T : Heath ath T T V F.. System The system plus the reservor are consdered to be solated from the rest of the unverse. For the system plus the reservor t must always be: dstotal = dssystem + ds bath 0 (the equal sgn s at equlbrum) The system and the bath are n thermal contact and heat can be exchanged between them. In addton, the volume of the system s not fxed. Ths means that the system can do work on the bath. If δ qsystem s the heat exchanged from the system to the bath. δwsystem s the work done by the system the frst law for ths case s du = δq + δw dubath = δqbath + δwbath ecause the entre thng s solated, system system system δwsystem = δwbath δqsystem = δqbath From the second law we have TdS = δ qrev δ qrrev δwrev δwrrev then du system TdSsystem = δwsystem (rev) δwsystem(rrev) At constant Temperature we can wrte ths n the form: du ( TS ) = df = δw (rev) δw (rrev) system system system system system 6

64 It follows that the change n F s the work reversbly done by or on the system under constant temperature.(sothermal reversble work). As we can see ths reversble work s more negatve than the correspondng rreversble work. (f the system does work on the bath, t does more work reversbly than when t does t rreversbly. Conversely, f the bath does work on the system, the work s smaller (less of a postve work) when done reversbly. Ths means that we obtan the maxmum useful work f we carry out the process reversbly. In ths case the equal sgn of dstotal = dssystem + ds bath 0 holds, from whch we get: dstotal = 0 = dssystem + ds bath δ q (rev) system ds bath = dssystem = = ( du system δwsystem (rev)) T T TdStotal = TdSsystem + TdSbath = 0 ow we replace (*) for ds n the last expresson, bath (*) TdStotal = TdSsystem du system + δwsystem (rev) = 0 dfsystem TdStotal = df (rev) 0 system + δwsystem = (Isothermal, reversble) TdStotal = df (rrev) 0 system + δwsystem or fδ W system = 0 then dfsystem 0 dfsystem = d( U TS) = du TdS 0 Therefore F reaches a mnmum at equlbrum A system that s sothermal wth a bath, and that can only exchange heat but O WORK wll try to mnmze ts free energy F. Any sothermal process that ncreases the nternal never occur spontaneously. Irreversble sothermal process happens spontaneously untl df = 0 (F reaches a mnmum) Usng the Enthalpy: Joule Thompson Effect In laboratory processes we are mostly usng a constant pressure and constant temperature, so that the Gbbs potental s the best choce. However there s one mportant stuaton P f,v f,t f P,V,T In whch the process occurs at constant Enthalpy. As strange as ths may sound, t s the 64

65 basc prncple behnd ndustral coolng. Consder the last fgure above, where a gas s forced trough a porous plug or obstructon n the ppe. The cylnder and the pstons are A volume V gas s forces though at an ntal pressure and temperature P, T. The cylnder and the pstons are nsulatng. Upon passng through the porous membrane, real gases usually cool down. So Tf < T, however the temperature change depends on the ntal pressure. To understand why ths happens, we start by calculatng the nternal energy before and after the gas goes though: U f = U + pv pfvf work done on the system left sde work done by the system rght sde from here we get by rearrangng U f + pfvf = U + pv H f H Therefore ths process s Iso-Enthalpc. It occurs at constant enthalpy. We now calculate the temperature change by frst wrtng the temperature change as a dervatve: dt dt = dp dp H At ths stage, we would not be able to go any further, because we do not know what to do wth the dervatve at constant Enthalpy. What we do next s qute standard, although t mght appear qute snazzy at ths stage. The dervatve manpulatons wll be dscusses shortly. ow lets fnsh ths dervaton: We frst wrte the dervatve as: H p dt T, dt = dp = dp snce both and H are constant. dp H H, T p, ext, we recognze the dervatves. The one n the numerator can be wrtten n terms of the fundamental relaton for the enthalpy: dh = TdS + VdP + μd from whch we see that H ds dp d = T + V + μ ± p p p p T, T, T, T, because s constant the last term vanshes. The second term equals V : The dervatve n the denomnator can also be connected to some known quantty: H = Cp T p, Wth these replacements we get: 65

66 ds T dt p + V T, ds dt = dp = dp. How can we replace by somethng dp C H, p p T, recognzable? We do ths by wrtng the Gbbs potental whch s a state functon, and therefore ts second dervatves satsfy the relaton ds dv = p T T, P, dv We fnally note that = αv the sobarc expanson coeffcent. T P, ds T V p + T, αvt + V V Then dt = dp = dp = ( αt ) dp Cp Cp C p Snce n our experment P on the sde we are pushng the pston s larger that the pressure on toe other sde, 0 αt > 0 so that dt < 0 ths mples that lower than Tnverson dp <. For ths experment, the gas wll cool down only f ( ) T nverson > If the gas starts at a temperature equal or α t cools down. Else t gets hotter. Inverson curve for Joule-Thompson coeffcent. Scanned from H. Callen. Thermodynamcs and an ntroducton to thermostatstcs nd ed. The last fgure shows some senthalps. The maxma of the senthalps s connected by the nverson curve. 66

67 Dervatves and ther relatons: () Implct functons A functon ψ ( x, yz, ) s held constant. Then the varatons of x, y and z are not ndependent: ψ ( xyz,, ) = constant Then there s a functon zxywhch (, ) s mplct by the statement dψ ( x, y, z) = 0 ψ ψ ψ dψ ( x, y, z) = dx + dy + dz x yz, y z xz, xy, () Settng dz = 0 and dvdng by dx ψ ψ y ψ = constant 0 = + Ths approprately says x yz, y x xz, ψ, z z = constant () solvng ψ y x yz, = x ψ, z ψ y A VERY USEFUL RELATIO () x, z Another two formulas can be obtaned analogously: y settng dy = 0 and dvdng by dx ψ ψ z 0 = + x yz, z xy, x ψ, y ψ z x yz, = x ψ ψ, y z x, y (4) y settng dx = 0 and dvdng by dy ψ ψ z 0 = + y z xz, x, y z ψ, x ψ y z x, z = y ψ ψ, x z x, y (5) Returnng to ψ ψ ψ 0 = dx + dy + dz x yz, y z xz, xy, Settng dz = 0 and dvdng by dy (nstead of dx ), we get 67

68 ψ x ψ 0 = + x y y yz, ψ, z x, z (6) solvng, ψ y x x, z = y ψ ψ, z x yz, but before (Settng dz = 0 and dvdng by dx ) we had that ψ ψ y 0 = + x yz, y x ψ, z xz, ψ ψ x = replacng ths n (7) x y y ψ yz, x, z, z ψ x y xz, = = y ψ x x ψ, z y y y x, z ψ, z ψ, z (7) (8) x = y y ψ, z x ψ, z VERY USEFUL (9) Usng (), and (9) we get ψ x y x, z = y ψ ψ, z x yz, (0) () and (5) we get ψ y z x, y = () z ψ, x ψ y x, z Combnng (0), () and (4), we get 68

69 ψ ψ ψ x y z y z x y z, z ψ, x x ψ ψ ψ ψ, y ψ x y z x y z y z x ψ, z ψ, x ψ, y xz, xy, yz, = = yz, xz, xy, = Some tmes useful relaton ow consder that ψ ( x, yz, ) has ts varables x, yzdependng, on a varable u x = xu ( ), y= yu ( ), z= zu ( ) then ψ dx ψ dy ψ dz dψ = du+ du+ du x yz, du y du z xz, xy, du f ψ s now held constant then there s a relaton between x, yz, and also between dx dy dz,, and du du du ψ dx ψ dy ψ dz 0 = + + x yz, du ψ y du z xz, ψ xy, du ψ f we further assume that z s constant, ndependent of u (.e. x= x( u), y = y( u), z = constant ) ψ dx ψ dy 0 = + x du y du rearrangng: yz, ψ, z xz, ψ, z dy ψ du ψ, z x yz, dy = = dx ψ dx du y ψ, z xz, ψ, z where we have used () dy dx ψ, z dy du = dx du ψ, z ψ, z VERY USEFUL RELATIO 69

70 Response Functons: We now note some propertes of the second dervatves of the thermodynamc potentals, exemplfyng ths wth the Gbbs potental. Common manpulatons n thermodynamcs We wll llustrate the use of the relatons between dervatves. dg = SdT + Vdp + μd at constant composton d = 0, the second dervatves are G G = S = V T p, p T, there are second dervatves: G S G V G V = = = T T p, p, p p T p T T, T, p, p, For reversble processes at constant pressure, we had δ q = nc dt and rev TdS = nc dt so that S T p, G T p p, nc = T v p nc = T thus, p Ths s the change of Volume wth pressure at constant Temperature, but t depends on V. Then we defne: V κt = It s V p T, called the sothermal compressblty (kappa- T) and so G = κtv The p T, mnus sgn s because the volume decreases when p ncreases. Ths s the change of Volume wth Temperature at constant pressure. It s also proportonal to V, so we defne V α p = the V T p, sobarc expanson coeffcent. Then, G = Vα p T p p, All these parameters measure how the system responds to a change n some ntensve varable such as pressure, r temperature. ecause of ths, they are called response functons It turns out that C p, α p and κt convenent measurable quanttes wth a clear physcal meanng. Thus t s possble to elmnate the potentals from the thermodynamc 70

71 expressons such as the fundamental relatons, and wrte equatons n terms of these famlar second dervatves (the response functons). efore we shall dscuss another mportant set of relatons between the second dervatves. Maxell Relatons y vrtue of beng state functons, the thermodynamc potentals have dentcal crossed second dervatves. We have made use of ths property already. ow lets see what can one get from t. We have the followng fundamental relatons: du = TdS pdv + μd () df = SdT pdv + μd () dh = TdS + Vdp + μd () dg = SdT + Vdp + μd (4) If we take (), and calculate the second dervatves at constant composton, we have U U T = T and = S V S V V, S, S, U U p = p and = - V S, S V S V, V, Whle unfamlar to us, the crossed second dervatves must be equal: U U = S V V S V, S, whch leads us to p T - = S V V, S, Ths s an example of what s called Maxwell Relatons whch turn out to be extremely useful n thermodynamc calculatons. Another equalty can be obtaned from U U μ = μ and = SV, V V S, S, U U p = p and = - V V S, V, p V, p μ p from whch we get = - another Maxwell relaton. V S, V, p Fnally you can get U U μ = μ and = SV, S S V, V, 7

72 U U T = T and = S S, S S, p Sp, μ T from whch we get = yet another Maxwell relaton. S V, S, p ote that we have wrtten the fundamental relaton lmted to a system that does not nclude magnetc feld and magnetzaton work, or electrc work and so on. There wll be another set of Maxwell relatons n systems descrbed by such fundamental relatons. Exercse: Dervate the 9 maxwell relatons from the fundamental equatons: df = SdT pdv + μd (I) dh = TdS + Vdp + μd (II) dg = SdT + Vdp + μd (III) Mnemonc rule: Snce we would not be able to quckly transform expressons fluently wthout rememberng the by now- many equatons and relatons, t s tme for some sort of ad. orn came up wth a very useful graphcal way for generatng the Maxwell relatons. V F T U G S H P The dagram shown above s an example. Ths partcular example apples for processes at constant, and therefore does not appear n t. The order s mportant for the rule to work, so we wll need to remember that.(*) The square has the thermodynamc potentals n the center of each sde. The natural varables for these potentals are on the corners closest to each potental. 7

73 Wth ths dagram you can easly recall the expresson of the fundamental relaton for each potental and ts natural varables: du = TdS pdv + μd () df = SdT pdv + μd () dh = TdS + Vdp + μd () dg = SdT + Vdp + μd (4) For each potental, the dervatves wth respect to the varables that are next to t n the square allow us to buld (-4). The sgn of the dervatve s gven by the drecton of the arrow: If the arrow ponts away from a natural varable gves a postve (+) dervatve. An arrow pontng towards the natural varable gves a negatve dervatve. To get the Maxwell relatons from the dagram, all you have to do s to look at two consecutve corners and n ths example; V T V T = = S p, p S, arrow ponts arrow ponts away of S away of p S P S P V T thus, we read the Mawxell relaton = S p p, S, To get another relaton we frst rotate the dagram: V F T U G S H P ow we focus on the lower corners: 7

74 S S = p V, arrow ponts away of p (+) V V = - T p, arrow ponts towards T (-) p T p T Therefore we read the Maxwell relaton: S V = - p T V, p, y successvely rotatng the square by 90 degrees and by readng out the lower consecutve corners as dervatves, we have a way to generate the Maxwell relatons. Exercse: y rotatng the square agan and by readng the bottom corners obtan the followng two relatons: p S = T V, V T, T p = - V S S, V, The four Maxwell relatons that can be obtaned from ths dagram are by far the most commonly used. Thus the student s urged to become famlar wth them. Of course ths dagram can also be adapted to yeld other Maxwell relatons. For nstructons to do so, please see H. Callen book, or W. Grener s ook. (*)ote: Here s a mnemonc to remember the mnemonc other: Very Fancy Tools Usually Grant Suffcently Hgh Productvty 74

75 Usually we get expressons wth dervatves that do not have a drect physcal nterpretaton, or a smple measurable quantty assocated wth them. It would be convenent to be able to transform any formula contanng dervatves nto an expresson that contans only response functons, and laboratory controllable varables. We now dscuss a recpe for dong transformatons n thermodynamc expressons: The followng rules must be appled n the order they are gven, to transform an expresson. Rule (I) If a dervatve n an expresson contans any thermodynamc potental, brng them one by one to the numerator, and replace ts dfferental wth the correspondng defnton from: du = TdS pdv + μd () df = SdT pdv + μd () dh = TdS + Vdp + μd () dg = SdT + Vdp + μd (4) p Example: let us say we have the dervatve n an expresson. Ths dervatve U G, does not mmedately mean anythng measurable n the lab. Partcularly, ths dervatve mples that we vary U at constant G. We do not n prncple fnd ths as a smple task. We can transform that dervatve nto somethng more appealng: ow we brng U to the numerator: p = U G, U p G, - - p U S V = = T p U G, p p p G, G, G, then, we use the relaton between dervatves frst on G S p S, V = T p p G p S p, G, G, S p G, 75

76 V and then on n order to remove the constant G condton from these p G, dervatves: G G S p p S, V, T p = p G G G, S p, V p, We can repeat the use of the rule (I) on G to elmnate t (t s ok to have S but the dea s to remove the potentals U, F, H and G), T T S V S V p + p + S S, V, = T p p T p T p G, + V S + S p, S p, V p, V p, = 0 = 0 constant p constant p whch fnaly does not contan any of the potentals. However there are stll many dervatves n t. Often the dervatves can also be replaced. Rule (II) If the dervatve contans the chemcal potental μ, brng t to the numerator, and S V elmnate t by the Gbbs-Duhem relaton: dμ = dt + dp= SmdT + Vmdp Example: μ S T V p = + V V V Rule (III) S, S, S, If the dervatve contans the entropy, brng t to the numerator and elmnate t usng one of the most common Maxwell relatons: V T = S p, p S, S V = - p T V, p, p S = T V, V T, T p = - V S S, V, 76

77 If ths does not elmnate the entropy, then put a dt under ds, and then use the dy dx du ψ, z relaton: = wth u = T. Then the numerator wll contan one of the dy dy ψ, z du specfc heat capactes In our example we have: ψ, z C p or C v. ds ds dv dt dp dp dt = = = dp ds Cp C dt T T S, We need to brng S to the numerator frst and fnally we get: T, T, P, p, p dt T dv T = = dp C dt C S, p P, = αv p αv Rule (IV) rng the Volume to the numerator. The remanng dervatve wll be expressble n terms of α and κ T Example: dv dp dt T, VκT VκT κt = = = = dp dv dv Vα α V, dt dt Rule (V) p, p, The orgnal gven dervatve has now been expressed n terms of the four quanttes: C v,α and κ T. ow elmnate Cv and wrte t n terms of C p,α and κ T, usng the V α relaton (to be shown very soon). Cv = Cp T κt We do not need rule V strctly speakng, but t s customary to use t to elmnatec. v C p, 77

78 Relaton between C andc : p v We start from the total dfferental of U as a functon of the varables VT,, du du du du = dv + dt + d dv T, dt V, d T, V at constant we have: du du du = dv + dt = πt dv + CvdT dv T, dt V, π T The coeffcent πt measures the change n U wth V at constant temperature. For one mole of gas, H U U ( PV) U Cp Cv = = + T p, T V, T p, T p, T V, () U H = U + PV Ths s T p, ote that we do not have the dervatve of U wth respect to T at constant VOLUME. from () we get after dvng trough by dt U V U π = + T T T then we get, T p, p, V, αv U = πα T V + Cv T p, now we replace ths n Cp Cv (): H U U ( PV) U Cp Cv = = + T p, T V, T p, T p, T V, p v T T T T p, αv () () Ths dfference s ( PV ) πα = T V + Cv Cv + T p, Snce p s constant, V C C = παv + p = παv + pαv = αv( π + p) Ths s a general equaton, vald for any substance. 78

79 ote that; U U () T p, T V, = απtv αv change n V wth T. πt change U wth V. Ths product converts the change n V wth T nto a change n nternal energy. ( PV ) V () = p = α pv The volume change wth T T p, T p, Δ V = αv pδ V = W the work aganst pressure p (a constant pressure). The varaton of U wth V: du du du du = ds + dv + d ds dv d V, S, S, V at constant we have; du du ds du ds = + = T p dv T, ds V, dv T, dv S, dv T, Impose constant T p ow we use the Maxwell relaton S p = V T T, V, V T U p T p, TαV = T p= p= p V T, T V, V κtv p T, and then U Tα πt = = p for one mole of gas we havecp Cv = αv( πt + p), and thus, V κ T, Tα VTα Cp Cv = αv( p+ p) = κ κ as we wanted to show C T T p T VTα Cv = A very mportant relaton to be remembered. κ T 79

80 Exercse: H Reduce the dervatve p T, We apply rules (I-V) T T p p H H, H, T H = = = p T p T T, H, H T p, H p, p, The dervatve T p H, s famlar to us from the dscusson of Joule-Thompson effect T p H, V = ( αt ) C p Usng ths, H p T, V = ( αt ) Cp = ( αt) V C p The orgnal dervatve s the sothermal Joule-Thompson Coeffcent. The fnal formula provdes an alternatve way of measurng the Joule-Thompson effect. (see P.W. Atkns, physcal Chemstry). Heater Isolaton P,T P f,t f Gas Gas Isolaton The heater mantans the two temperatures equal at all tmes. Knowng the volume of gas that passed, and the pressure change we can equate the amount of electrc work necessary to keep T = Tf as n: H ΔH Welectrc V( αt) = = p Δp Δp T, whch allows us to get α. 80

81 Revew of equlbrum condtons: Assume a system + thermal reservor, and both of them nsde an nsulatng enclosure.. When an nternal constran s removed n the system, we seek for the mathematcal condton to predct the equlbrum state. Mnmum energy prncple. In equlbrum the total energy of the {system + thermal reservor} s a mnmum. du ( + U res ) = 0 and d ( U + U res ) = d U > 0 (mnmum) ut the entre system s solated. Then ds ( + S res ) = 0 Ths statement requres a closer look. Thermal Reservors Our system q U res Reservor The reservor sze s much larger than the system sze. Then res res du res = Tres ds res pres dvres + μ d res but d = 0 (no matter flows n and out of the reservor. ThenV = constant and therefore du = T ds or res res res U res Tres = S res V, ow suppose that we add heat δ q to the reservor. Tres wll change by Δ Tres when δ q= T ds s added: res res 8

82 T res Δ Tres = ds S res V, res T the dervatve of wth respect to S s, T res U res U res = so that Δ Tres = S res S res V, S V, res V, ds ut U res and S res both extensve varables, whch means that they scale wth T Ths s why the reservor acts as such, the second dervatve T S res res res Δ res = V, res (*) Then T lm Δ = lm = lm = 0 res res res T res res S res V, res The concluson s, smply that f the reservor s suffcently large, ts temperature wll not change ( Δ = 0 ) when a lttle bt of entropy s added to t by vrtue of a heat transfer. T res (*) ote To understand ths you need to recall that the dervatve s the rato of the change n the functon dvded by the change n the varable, and that the second dervatve s the change n the dervatve wth the change n the varable: y Δy unts of ( y) slope = f '( x ) has Δx unts of ( x) and f ''( x) wll have unts of unts of ( y) unts of ( x ) (thnk of the velocty and the acceleraton for example). x A vew of the equlbrum from the stand pont of the dfferent thermodynamc potentals Consder a system that s made up of several nternal sub-systems. Ths composte system s n contact wth a thermal reservor. 8

83 If the nternal wall separatng the system nto subsystems s movable, and mpermeable we wll have the followng boundary condtons: d = d = 0 for all j () () j j () () dv ( + V ) = 0 whereas f the nternal wall was rgd and permeable to component k we would had: d () () ( k k ) 0 d + = component k can pass through the wall = d = 0 for all j dfferent from k () () j j () () dv = dv = 0 These equatons together wth du ( + U res ) = 0 determne the equlbrum condton. ut the term du n du ( + U res ) = 0nvolves a sum of the type: () () () () T ds + T ds whch arses from heat fluxes wthn the system (the nternal parts of t). Also, () () () () p dv p dv arsng from the volume changes and μ d system. μ d due to changes n the number of partcles or moles of k wthn the () () () () k k k k ut certanly, the heat that the system exchanges wth the reservor has to satsfy that: δ q + δ q = 0 so that sys res () () () () () () () () T ds T ds Tres ds res T ds T ds Tres ds sys + + = + = 0 () () () () () () = T ds + T ds Tres d( S + S ) = 0 then () () () () () () ( T ds Tres ds ) + ( T ds Tres ds ) = 0 () T = Tres () T = Tres It follows that at equlbrum each of the parts of the system s at the same temperature T. res If we rewrte the equaton du ( + U res ) = 0 as du ( + U res ) = du+ TresdSres = 0 (system enclosed by a reservor) () We dropped the sub ndex sys for the quanttes related to the system so from now ds = ds sys ecause we have ds ( + S res ) = 0 whch mples ds = ds res, replacng ths n () 8

84 du ( + U res ) = du ( TresS) = du ( TS) = df= 0 at equlbrum And we also have the condton of mnmum energy, whch states that t must also be true that d ( U + U res ) > 0 thus, snce du ( + U res ) = du ( TS) at equlbrum, t must also be true that d ( U TresS) > 0. The concluson s that at equlbrum df = 0 and d F > 0 So that at equlbrum the Helmholtz free energy s a mnmum. The energy of the system plus the reservor s of course a mnmum at equlbrum, but df = 0, df> 0 s sayng the same thng by referrng to the F potental of the system ALOE. In df = d( U TS) the term d( TS) represents the change n energy of the reservor snce Tres = T and ds = ds res. ow consder a pressure reservor and a system wth walls that are non-restrctve wth respect to volume changes. An nternal constrant wthn the system s removed, and the system evolves to equlbrum. d( U + U res ) = du presdvres = du + presdv = 0 snce dvres = dv In ths case for an equlbrum we must have pres = p. Ths means that d( U + U res ) = 0 = du + pdv = d( U + pv ) = 0 and ths wll lead us to dh = 0 and d H > 0 (mnmum enthalpy) For a system n equlbrum wth a pressure reservor, the enthalpy s a mnmum. The most mportant pont of vew s based on the Gbbs potental, for whch we consder a system n smultaneous contact wth a pressure reservor and a heat reservor. ow we have du ( + U ) = du+ T ds p dv = 0 (so far mnmum nternal energy) res res res res ds = ds res and dvres f we replace the condtons Tres = T and pres = p for equlbrum, so we can wrte du ( + U ) = du ( TS+ pv) = dg= 0and res = dv we have after acceptng that d ( U + U res ) > 0mples that dg> The value of the Gbbs potental for a system n equlbrum wth a pressure and heat reservors s a mnmum. So that when an nternal constrant s removed n the system, the unconstraned varable wll seek for a value that mnmzes the Gbbs potental. Systems of many components: We can use the Gbbs potental to descrbe systems that contan more than one component. y defnton, G = U TS + PV Recall the Euler relaton for one component, we had: U = TS PV + μ In ths case, combnng G wth Euler relaton, we have G = TS+ PV + TS PV + μ = μ so that G = μ The molar Gbbs potental equals the chemcal potental 0 84

85 For many components, ths can be generalzed to U = TS PV + μ+ μ + μ μrr = TS PV + μ now the Gbbs potental wll be the sum of the contrbutons for all the components: G = μ+ μ + μ μr dvdng trough by = G r = μ + μ + μ μr defnng the ratos χ = the mole fracton of component we see that the partal molar Gbbs potental of component I s the chemcal potental for that component We fnd that the equlbrum s determned by = dg = SdT + VdP + μ d 0 n partcular μ d = 0 at constant T and p μ We wll return to ths when we descrbe chemcal reactons. For now, we wll return to the energy descrpton of a system to analyze one last mportant aspect. We would lke to know how many state varables are necessary to determne the state of the system. We start wth an solated system, contanng C dfferent partcle types of speces (components). The system also contans a number φ of phases movable walls =... = =4 =φ q = = 85

86 In addton to the walls between all the phases beng free to move, components and heat are also allowed to pass from one phase to another. The system s enclosed n an nsulatng contaner, so no heat or component molecules enter or leave the system. The frst law of thermodynamcs apples to each phase. For reversble changes of state we have: c () () () () () () () μ j j j= du = T ds p dv + d () the ndex represents the phase, and the ndex j represents a component. ecause we can () wrte an equaton lke () for each of the φ phases we have for each phase that U s a () () () () () () state functon of (the extensve) varables U = U U, V,... c varables c varables So we have c+ varables n total for ths phase (). In a system wth magnetc felds and magnetzaton, there wll be more varables. Snce we have φ phases, the total number of varables are ( c + ) φ. When the system reaches thermodynamc equlbrum we have the followng addtonal condtons: T = T =... = T () () ( φ ) p = p =... = p () () ( φ ) μ = μ =... = μ () () ( φ ) μ = μ =... = μ () () ( φ ) μ = μ =... = μ () () ( φ ) c c c thermal equlbrum and mechancal equlbrum: ( φ -) equatons chemcal equlbrum c( φ -) equatons ecause we have a total of ( c + ) φ varables and ( φ -) + c( φ -) equatons we can elmnate ( c + )( φ -) varables. ( c + ) φ -( c + )( φ -) =( c + ) Ths leaves us wth ( c + ) ndependent varables. Ths number does not depend on the number of phases present n the system. When the system s n equlbrum the szes of each phase stays constant. The volumes of these phases s therefore self-determned, and t s not a varable that we are free to adjust. So we subtract the a number of varables equal to the number of phases to obtan the true number of free varables (degrees of freedom): f = c+ φ 86

87 Let us consder some examples. gas lqud here we have two phases and one component. f = c+ φ = + = If we fx T then P s determned. If we fx P then T s determned gas here we have one phase and one component. f = c+ φ = + = Here we are free to set p and T and the system wll be at equlbrum. We have degrees of freedom. 87

88 Entropy and stablty We wll now dscuss the mplcaton of the curvature of the entropy as a functon of ts varables. Consder the fgure: S Ths system would be unstable because there s a regon where > 0. U We should also consder smultaneous varatons of U and V. What s the general condton of system stablty? We begn wth the expresson of the change n the entropy for a system when U and V change: S( U +Δ U, V +Δ V, ) + S( U ΔU, V ΔV, ) S( U, V, ) () ow we expand the two terms on the left as a Taylor seres to second order dfferentals: Remnder: For the frst term we wll have somethng lke df df f ( x + dx, y + dy, z) = f ( x, y, z) + dx + dy + dx dy yz, xz, d f d f d f d f ( dx) ( dy) dxdy dxdy dx dy dxdy dydx yz, xz, yz, xz, And also we have for the second term: 88

89 df df f( x dxy, dyz, ) = f( xyz,, ) dx dy+ dx dy yz, xz, d f d f d f d f ( dx) ( dy) dxdy dxdy dx dy dxdy dydx yz, xz, yz, xz, because of the sgn of dx and dy. ote that: I)- When we expand SU ( 0 +Δ UV, 0 +ΔV, ) and SU ( 0 ΔUV, 0 Δ V, ), and term by term as we have n equaton (), the frst order dfferental terms wll cancel exactly. II)- ecause S s a state functon the crossed second dervatves are dentcal: ds ds = wth these two thngs n mnd, we wrte the expanson for (): dudv dvdu V, U, ds ds ds S( U0, V, ) + ( ) ( ) ( du + dv dudv S U + 0, V, ) du dv dvdu V, U, U, but S 0 always, then: () ds ds ds ( ) ( ) 0 du + dv dudv + du dv dvdu V, U, U, ds ote that ths condton predcts that f dv 0 0 and du V, ds du 0 0 n agreement wth what we found n a prevous class. dv U, ut ths s sayng that only one varable s changng. In general both may change so ΔU 0 and ΔV 0. Then the condton for stable equlbrum must account for these cases. Equaton () can be rewrtten n a matrx form: ds ds du dvdu du ( du, dv ) = ds ds dv () dvdu dv ds ds ds ds ( du, dv ) du + dv + du + dv 0 du dvdu dvdu dv ds ds ds ( du ) + dvdu + ( dv ) 0 du dvdu dv for a real matrx that s symmetrc, t s sad that the matrx s negatve defnte f all ts egenvalues are negatve numbers. 89

90 Alternatvely, the expectaton value of the matrx V AV = va, jvj < 0 for all, j, ( v, v,... v ), v 0 M j Ths last defnton s what we have n equaton (), for any ( du, dv ) vector wth non zero du, dv. Also because the determnant equals the product of the two egenvalues we have: ds ds ds 0 > (4) Ths s the crteron for local stablty. du dv dvdu The system s unstable n the regon CDEF. Then we replace the underlyng fundamental relaton by the tangent lne HF. CDE fals the LOCAL stablty condton (4). CDEF fals the GLOAL stablty condton (). When a system s unstable t spontaneously separates nto two stable sub-systems (phases). In ths case these subsystems are phase and phase F. Ths consttutes the process of PHASE transton. Thus the overall fundamental relaton s really the famly of tangents that at all ponts are above the curve (the underlyng fundamental relaton). In analogy wth the entropy formulaton we have the local condtons for U, H, F and G but all these functons are convex (the entropy s concave) U T 0 = S S U U U and for the cooperatve varaton of 0 U p V S V S = 0 V V 90

91 Smlar condtons apply to all the thermodynamc potentals. Recall that when we dscussed the Legendre transforms Potental dψ dp d Ψ P = whch mples that = = (*) (by propertes of the param dx dx dx dx Varable d Ψ dervatves). dψ( P) On the other hand, we have x = dervate ths wth respect to p to get dp dx d Ψ( P) = = = dp dp dp d Ψ whch can be combned wth (*) to get d Ψ( P) = dp d Ψ dx dx dx OTE THE MIUS SIG From ths relatonshp we get F F 0 (concave) T V V, H H 0 (convex) S p p, G G 0 (concave) T p p, T, S, T, (convex) df = SdT pdv + μd (concave) dh = TdS + Vdp + μd (concave) dg = SdT + Vdp + μd Convex n extensve varables, concave n ntensve varables. Physcal consequence of local stablty: F p 0 = = = = V V T, T, V VκT VκT p T, From the nternal energy, whch mples that κ T > 0 S d T 0 = = = = U du T T U U V, V, T ncvt T V, mplyng CV 0 Snce C C 0 then C 0 as we know p V p Cp CV = TV α 0 κ T 9

92 Propertes of the Gbbs Potental G = U + PV TS = H TS dg = du + pdv + Vdp TdS SdT replacng du we get dg = SdT + Vdp + μd. G = S 0 always negatve! T p, The larger the entropy of a pure substance the more G changes wth T. G = V 0 Always postve. The p T, larger the molar volume of a substance, the more G changes wth T. G T, p = μ we can not conclude anythng from ths. Varaton of G wth Temperature H G G = U + PV TS = H TS S = T but G G H G H G G H S = = = = () T p, T T T T p, T T we can wrte ths n a compact way by usng the dervatve G T G T G G G G = G + = = T T T p, T T T p, T T T p, T p, p, now we combne ths wth () to get G G G G H T T H T = = = T p, T T T T T p, p, So we obtan the Gbbs-Helmholtz equaton. Ths equaton s useful because we do not need to know the entropy of a system to predct varatons of G wth T. We can use H whch can be measured n a calormetrc experment. Snce H and G are both state functons we can see that for a process that nvolves a change n G Δ G = G G then f 9

93 G f G T H f T H = and = T T T T p, p, ecause these are lnear dfferental equatons we can wrte ( Gf G) ΔG T ( H f H ) T ΔH = or more brefly = T T T T p, p, Pressure dependence of the Gbbs potental At constant composton, dg = Vdp SdT f we know G at one pressure, what would be ts value another pressure when T s constant? Gf dg = Vdp SdT dg = Vdp = G G G pf f we know the equaton of state theng G Vdp p f f p f = + p for lquds and solds (also called condensed phases) V m changes lttle wth pressure, so we can take t out of the ntegral, (ths would not be true f the pressure change s too bg, such as n some geophyscal processes). For condensed phases we may have, p f G = G + V dp= G + V ( p p ) = G + V Δp mf m m m m f m m p furthermore, VmΔp Gm unless we are lookng at extremely hgh pressures. Thus, we can say that G s approxmately constant wth pressure for condensed phases. Example: Calculate ΔG for pressure changes n deal gases pf pf pf nrt G = G + Vdp= G + dp= G + RT mf m m m p p p p so that p Δ G = nrtln note that Δ G > 0 f p f > p p f dp p The Chemcal potental 9

94 For a system of one component the functon GT (, p, ) combned wth Euler relaton gves: G = U TS + pv = TS pv + μ TS + pv = μ Thus, for a system that has only one component, the chemcal potental s smply molar G Gbbs potental: μ = G Also note that μ = Thus we already know how μ depends on T and P by the pt, prevous secton. For an deal gas, at constant temperature we wll have p G (, ) (, ) ln f mf pf T = Gm p T + RT p We can defne a standard state at p = p =.0 bar for each temperature so that p Gm ( p, T) = μ( p, T) = μ + RTln where μ = μ ( p, T) p Then at constant temperature, for an deal gas μ s a logarthmc functon of the pressure: For real gases the equaton μ( pt, ) = μ RTln p + s not vald but we can replace p by a p functon of p to be able to use the same logarthmc dependence. Ths s done by defnng a functon called fugacty f ( p ) such as at very low pressures f ( p 0) p and for real gases then we can use 94

95 f ( p) μ( pt, ) = μ + RTln f ( p ) but the dffculty s to know an equaton of state n terms of fugacty. In general there are emprcal relatons that provde the fugacty for a gas as a functon of pressure usng some expermental parameters. We are not gong to dscuss ths further. For an n depth dscusson, see P.W. Atkns n Physcal Chemstry. Systems of Varable composton: We should extend the expresson dg = Vdp SdT to a system wth several components: G G G G dg = dt + dp + dn + dn +... T p n n pn, Tn, T, p, n j T, p, n j where the terms refer to dervatves wth respect to the number of moles of one component at all other numbers of moles kept constant. dg = Vdp SdT + μdn+ μdn +... or more compactly: dg = Vdp SdT + μdn wth the sum runnng over all the components of the mxture. Equlbrum condton n terms of chemcal potental Suppose that we have a system n whch the chemcal potental of a component s A dfferent n two dfferent regons of t, say μ and μ. Then keepng the temperature and pressure and all the other numbers of moles constant we transfer dn moles of component from regon A to regon. A A Then the change n the potental G s dg = μ ( dn ) and dg = μ dn (because dn moles are added to regon A, and + dn moles are added to regon. The total change n the Gbbs potental s then A A A dg = dg + dg = μ ( dn ) + μ dn = ( μ μ ) dn A We now observe that f μ < μ then dg < 0 and ths transfer of matter (component ) decreases the Gbbs potental. We have seen that ths functon s a mnmum at equlbrum. Thus, f G decreases by transferrng I from regon A to regon, G decreases, whch says that we are not yet at equlbrum, but the system s evolvng towards equlbrum. Ths wll naturally occur, and the net transfer wll stop once we have reached the equlbrum. Thus at constant pressure and temperature, a substance wll flow spontaneously from the regon of hgh chemcal potental to the regon of low chemcal potental. Ths flow wll contnue untl the chemcal potental s unform through the system. Furthermore, ths condton must be satsfed for all the components of a system to be at thermodynamc equlbrum. 95

96 The chemcal potental of a mxture of deal gases For a pure deal gas the chemcal potental s gven by μ( p, T) = μ ( T) + RTln p Where p s a pure number; p /bar. At a gven temperature, the pressure s a measure of the chemcal potental of the deal gas. Lets assume that we have a contaner dvded nto two parts by a membrane that allows gas to pass but prevents gas j to pas through. We start wth gas on one sde of the membrane(a), and gas j on the other (). The chemcal potental μ measured by the pressure of gas s larger on one sde of the membrane than on the other. The gas wll A flow through the membrane and become mxed wth gas j untl μ = μ. The chemcal potental of s then A A μ = μ ( T) + RTln p on sde (A). (pure ) Snce at equlbrum ths value should be unform throughout (A-), t must be concluded that μ = μ ( T) + RTln p on sde () mxture + j y usng the partal pressures p = χp where χ s the mole fracton of component gas and p s ts partal pressure. Wth ths we wrte a more general expresson; μ = μ ( T) + RTln p = μ ( T) + RTln p+ RTlnχ The frst two terms on the rght sde are the chemcal potental for pure gas at pressure p μ = μ( pt, ) + RTlnχ snce χ < RT ln χ < 0 showng that G wll always decrease when deal gases mx, and we know ths from everyday experence. Ths s also true for lquds. A pure deal lqud n contact another deal lqud wll spontaneously form a mxture. In ths case we can defne (by generalzng the result from the deal gas mxture) the partal chemcal potental for each component n the deal lqud mxture as: φ μ = μ ( T, p) + RTlnχ (***) Entropy of Mxng In formng a mxture of several components we wll have ntally (snce all the components are PURE) φ φ φ φ Gntal = G+ G + G = nμ + nμ + nμ = nμ after they mx we wll have Gntal = G+ G + G = nμ + nμ + nμ = nμ The change n the Gbbs potental due to the mxng s 96

97 G G = n ( μ μ ) + n ( μ μ ) + n ( μ μ ) = n ( μ μ ) φ φ φ φ fnal ntal φ Replacng μ μ from the equaton (***) n ths we get Δ G = G G = RT( n lnχ + n lnχ + n lnχ = RT n lnχ mx fnal ntal usng the expresson n = χn Δ G = RT χ nln χ = nrt χ ln χ < 0 mx The sum s always negatve, snce each term s negatve. If there are only two components n the mxture then, χ = χ, χ = χ and mx [ ln ( ) ln( )] Δ G = nrt χ χ + χ χ Ths functon has a mnmum at χ = /. The mxture wth the smallest Gbbs potental s that wth 50% of each component. For a ternary mxture, the lowest G s for a mxture wth / mole fractons. Δ G = G G wth respect to T gves us the entropy of mxng. Dfferencatng mx fnal ntal ΔGmx G fnal Gntal = = ( Sfnal Sntal ) = ΔSmx T pn, T T pn, pn, dfferencatng both sdes of Δ G = nrt χ ln χ wth respect to T we have mx ΔGmx = nr χln χ so that Δ Smx = nr χ ln χ > 0. Ths shows that deal T pn, gases (lquds) ALWAYS spontaneously mx. (they are always mscble!!) Problem: Calculate the entropy of mxng for two gases. What s the mxture that has the maxmum entropy?. Plot the entropy of mxng s a functon of the composton. χ = χ, χ = χ Phase transtons of pure substances: We saw that for a system n equlbrum entre system. If there are several phases n the system, then phases where s present. μ for each component has to be the same n the μ wll be the same n all the 97

98 In the zones of the dagram we have μg μ = l μ Sm( g), = s Sml () T = S T p T p p μg μ = l μg Vm( g), = Vml () = Vms ( ) p p T p T T Snce the molar entropes are n the order Sm( g) > Sm( l) > Sm( s) the varaton of the chemcal potental wll show negatve slopes as n the fgure: m( s) The phase wth the lowest chemcal potental s the stable one. 98

99 Heat of transton For a sample of water n the form of ce, at pressure p0 we ncrease the temperature by supplyng heat at constant rate: The temperature frst rses, but then t stays constant. At ths stage we notce that we have a mxture of lqud water and sold water. We contnue to supply heat, and then we fnd that all the sold melted. After ths the temperature contnue to rse. Latent heat of meltng To melt one mole of ce, we need to supply a fxed amount of heat, l sl whch depends on the entropes of the lqud and the sold. lsl = Tm( Sm() l Sm() s ) = TΔ Sm (the sub-ndex m stands for molar) (&%&) Ths s a general property of the so-called frst order phase transtons. G H TS = = μ = Hm TSm At equlbrum μs = μl. Ths mples that Hm( s) TSm( s) = Hm( l) TSm( l) rearrangng ths Hm() l Hm( s) =Δ H( s l) = TmeltSm() s + TmeltSm() l Δ H( s l) = TmeltΔ S( s l) combnng ths wth (&%&) we see that the molar heat of meltng (transton) s therefore lsl = Δ H( s l) 99

100 Slope of the coexstence curves ote that S ( A) S ( A') m m m A A' S ( ) S ( ') but μ = μ and ' μ = μ because on the coexstence curve we have equlbrum between the phases on each sde. p the slope s T Coex. Curve For a change n pressure keepng the system n equlbrum Δp Δ T = Tf T = p T m Coex. Curve because of the condtons μ = μ () and A A' μ = μ () and subtractng () from () ' A ' A' μ μ = μ μ () ow we can use the Gbbs-Duhem relatng the ntensve varables: SdT Vdp + dμ = 0 SmdT Vmdp + dμ = 0 because A are related by dp and dt, ther chemcal potentals dffer by A dμ = μ μ μ = μ μ = + for the phase on the left sde of the coexstence curve A d SmdT Vmdp ' A' dμ μ μ S' mdt V ' mdp = = + for the phase on the rght sde of the CC because of () SmdT + Vmdp= S' mdt + V ' mdp or ( S ' m Sm) dt = ( V ' m Vm) dp then, p ( S' m Sm) = T ( V ' V ) CC m m More smply wrtten p T CC ΔS = ΔV m m 00

101 ote that Δ Sm and Δ Vm are dscontnuous at the coexstence curve. We can combne ths result wth the latent heat of transton ltrans = Δ H( trans) = TΔ Sm( trans) p ΔH = T T ΔV mtrans ( ) CC ( trans) m( trans) Clapeyron Equaton for a sold-lqud equlbrum ΔVm should be accounted for and ths can be done from the densty of the lqud and the sold. In the case of a sold-gas or lqud-gas transton we can make an approxmaton: Δ Vm = Vm( g) Vm( l) Vm( g) p ΔHm( vap) ΔHm( vap) pδhm( vap) = ( ) = T RT CC TbolVm ( vap) gas bol T RTbol bol pvap p pδhmvap ( ) ths s the Clausus-Clapeyron equaton. T RT lg bol From ths equaton we can rearrange and ntegrate dp ΔH mvap ( ) dt = p R T ntegratng pf Tf pf dp ΔHm( vap) dt ΔH m( vap) ΔH m( vap) ln = p = p = = R T R T p T f T R T T f from here we can obtan the coexstence curve: pt ( ) = pt ( ) e f ΔHmvap ( ) R T T f In the case of the sold-lqud transton p ΔH m( fus) = T T ΔV pf p sl ( fus) m( fus) Tf ΔHm( fus) ΔH m( fus) T f ( fus) Δ H m( fus) T ( fus) + Tf ( fus) T ( fus) dp = dt = ln = ln = T T ( fus) ΔVm( fus) ΔV m( fus) T ( fus) CC V m( fus) T Δ ( fus) s very steep ΔH T T ΔH T T ln + ΔV m( fus) T ( fus) ΔVm ( fus) T ( fus) m( fus) f ( fus) ( fus) m( fus) f ( fus) ( fus) (ln(+x)~x f x s small) Then 0

102 p f ΔH T T ΔH ΔT p = = ΔV T ΔV T m( fus) f ( fus) ( fus) m( fus) fus m( fus) ( fus) m( fus) ( fus) f we know the pressure change we can calculate the correspondng metng pont change Δ T. fus The absolute entropy of a gas could then be approxmated accountng for the entropy changes durng the transtons and usng the heat capactes: Tfus Tbol Tf Cp( sold) ΔH m( fus) C p( lqud) ΔH m( vap) C p( gas) ST ( ) = S0 + dt+ + dt dt T T + + T T T 0 ( fus) Tfus ( bol) Tbol There s a dffculty n measurng the heat capacty at very low temperatures. However, t has been establshed that near absolute zero temperatures, the heat capacty of most substances vares wth the thrd power of the temperature. Ths s known as Debye law: Cp( sold) Cv( sold) = at Usng ths, we can estmate the absolute entropy of a materal at any temperature. We must know of all the phase transtons that the materal undergoes durng the temperature varaton from T=0K to the fnal temperature. Ths last statement may be dffcult to ensure. 0

103 PART II STATISTICAL THERMODYAMOCS Knetc theory of the deal gas We wll now calculate the pressure and the nternal energy of an deal gas. *) There s one partcle n the box **) v s parallel to the box length and perpendcular to A ***) after bouncng of the wall the velocty changes from v to v After a perod of tme the partcle returns to the left hand wall. The tme nterval between collsons wth A on the left s l t = () v n a box wth many partcles the pressure wll be the average of the forces by an enormous number of collsons on the area A. The absolute value of the force actng on the partcle on each collson s mdv d( mv) dp F = ma= = = dt dt dt () so that the change n momentum per unt tme s the force F on the partcle. The force on the wall would be F. Snce the momentum before the collson s mv and after the collson s mv the net change s Δ p = mv mv = mv The total change n momentum per unt tme s the change n momentum per collson multpled by the number of collsons per unt tme. Δ p v mv = ( mv) = = F Δt l l We now get the pressure due to ths partcle by dvdng the force on the wall -F by the area A: F mv mv p = = = () A la V 0

104 where the volume of the box replaced the product of the area A tmes the box length. If many partcles are present n the box travelng parallel to the length of the box, we wll have addtve effects of the forces wth re resultng pressure mv ( + v + v +...) p = (4) V ecause there wll be many molecules t s the average of the squared velocty that measures the pressure ( v + v + v +...) v over all = ( v over all partcles + v + v +...) = v partcles n the box n the box and we use ths n the pressure: m v p = (5) V whch of course s only the pressure of an unphyscal undmensonal gas. To make thngs more realstc we stretch the model one more tme by ncludng the effect of collsons that are not perpendcular to the area A: The x component of the velocty remans unchanged n the collson. Only the normal component of the velocty contrbutes to the force on the wall. Thus we restate (5): m v p = V Wth the rather good assumpton that the gas s sotropc (t behaves dentcally n the x, y, and z drectons) we rewrte the total velocty as v = v + v + v so that the average velocty s x y z v = vx + vy + vz = vx + vy + vz = vx + vx + vx v = v v = v = x x and then the pressure becomes: v sotropc gas 04