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by: Durward DuBuque


Marketplace > University of Pittsburgh > Chemical Engineering > CHE2101 > FUNDAMENTALSOFTHERMODYNAMICS
Durward DuBuque
GPA 3.94


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This 18 page Class Notes was uploaded by Durward DuBuque on Monday October 26, 2015. The Class Notes belongs to CHE2101 at University of Pittsburgh taught by Staff in Fall. Since its upload, it has received 47 views. For similar materials see /class/229429/che2101-university-of-pittsburgh in Chemical Engineering at University of Pittsburgh.




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Date Created: 10/26/15
Lecture 1 1 Lecture 1 Objectives Students will be able to U FPJEOE 1 Describe the terms classical thermodynamics quantum mechanics statistical mechanics De ne the terms intensive and extensive variables ldentify different notational conventions Derive the Gibbs phase rule De ne the four laws of thermodynamics Overview In this class we will be concerned with three mechanics namely classical thermodynamics contin uum mechanics quantum mechanics and statistical mechanics We can brie y describe these as follows 1 to DJ Classical thermodynamics This is the observational science dealing with heat and work It was developed based on empir ical observations without assumptions about the make up of matter It describes macroscopic quantities such as heat work internal energy enthalpy entropy Gibbs free energy etc It does not contain any information about the state or even existence of molecules Classical thermodynamics tacitly assumes that the world is made up of a continuum Quantum mechanics Quantum mechanics lies on the other end of the spectrum from classical thermodynamics It deals with nanoscopic properties ie length scales on the order of 10 9 m Quantum mechanics gives rise to concepts such as the particle wave duality which states that all energy and all matter behaves both like a wave and like a particle It tells us that energy and other quantities are not continuous but discrete The governing equation is the Schrodinger equation The state of any system is described by the wave function which is the solution of the Schrodinger equation Quantum chemistry typically deals with solving the Schrodinger equation for single molecules giving information on the electronic structure how atoms are bonded together how electrons are shared to make up chemical bonds and the geometrical structure Recent experiments have now been able to image individual molecules identifying the same sorts of information through atomic force microscopy AFM as one can compute from quantum chemistry See Figure 1 as an example the bonds in the pentacene molecule In principle one could solve the Schrodinger equation for a macroscopic system such as the state of a polymer mixture inside an extrusion reactor but in practice these systems are much too difficult to solve Therefore quantum mechanics is limited to isolated molecules or perfect crystals usually at absolute zero temperature Hence quantum mechanics does not tell us anything about the thermodynamics of a macroscopic system Statistical mechanics Many of the methods of statistical mechanics were developed by arguably the greatest Ameri can scientist J Willard Gibbs 1839 1903 Gibbs even coined the term statistical mechanics The basic idea is that one can take the properties energy levels probabilities of individual Mm 1 molmuln mm qmm mam ma waxe um m an animate W m chum ri mummy uuuumu ummwmw mun mu m I vianulk m ansumumm mu m c And my mmm N Am n 92 m p m Drmmr zn p may 09 parametersam as mm m 21 pm 711nm v mam an my nmsmmmmmmm nzzi nmam mnuaMuxytmammm n mnnmmlzwm mwmwMurman I m A 1h mulrwk is pmmm auxll Lyayumm mmquot HeirMM lt0 mm a mum m mth m alvudn 3951 ad mule AF39MmsEDfpamAmem mm mm Glosst quotSlum 225 mm mm Hawa magma mms 5 the badge bmm quantum mms my Ma a markW5 mmm m5 2 Basics of Thermodynamics bet mswe ma manswe yam bet a mp x5 dm aawe helm l ammw As wmddyunbeamad Iwassumswgveyuntkuigdd7l sammwheso dwas mumps maoissxwwymwmddptobzblvb 15mm m mmm m m mg 5 quot2 q t We an t t v pm ampla mp nme 5 memimbMom 5 at am Cum adding mama mm blmkdns m m mpaaznre 43 c m n dag m um as numh my the Wm WM m mewx Dice We Mm 1 22 a w q02 lt PM 20 m qDS 7 q B q10 crime poims 123 1 H 16 12 000 005 010 015 020 025 030 P mu 2 Compll39ed hqwdrllqwd mum Phasedugxunsim maa mpoxym pm and Pm anasimmwnDimen mufthePEG39Dptmansuenumq Memmmhmd I am m awsmumz Lecture 1 4 03 0 Name some other intensive and extensive variables Note that any extensive property can be made into an intensive property by dividing by another extensive property Example V a VN Notation Inconsistent notation has plagued thermodynamics perhaps since the beginning The key in reading any text is to nd out what the notation means before you read There are basically four types of variables a Extensive eg V U N or V Tester amp Modell Walas Vt Van Ness amp Abbott b Molar eg VN UN sometimes denoted V Sandler v Denbigh V Van Ness amp Abbott Walas note that this is the same as other author s extensive and V c Speci c eg VM where M is the total mass sometimes denoted V d Partial molar eg W Denbigh Vi Van Ness amp Abbott Walas Sandler Pure component and partial molar notation occur when we are talking about mixtures Partial molar properties are de ned as V ZNW 139 Example If you mix 1 liter of 0014 and 1 liter CGHG you actually get more than 2 liters of the mixture The partial molar properties tell you how each component behaves in the mixture We will use V for extensive V for molar V for speci c and V for partial molar Gibbs Phase Rule We observe that there are many variables for a given system eg pressure density internal energy viscosity thermal conductivity composition volume etc How many of these prop erties do we have to specify to x the system ie put it in a de nite state of being For example you are designing a system to pump some uid up a hill You want to make sure that the uid is in a liquid state not a two phase or a vapor state because that would make it much more dif cult to pump How many variables do you have to specify Consider a uid of C independent components and 73 coexisting phases How many variables do we have It is suf cient to specify a quantity for each component and the temperature and pressure in each phase ie T PD 71f Note that specifying 71f n is the same as specifying m1N where N is the total number of moles in that phase Now one does not need to specify the total number of moles or the extent of the system in order to specify the state of the system Therefore specifying T P zf properties is suf cient This gives us C 1 variables for each phase 0 m071 mtenswe Num variables C 1 How many constraints do we have for equilibrium Thermal equilibrium requires that T is the same in each phase Mechanical equilibrium requires that P is the same in each phase Chemical equilibrium means that the chemical potential of each component must have the Lecture 1 5 same value in each phase T T Tpthermal equilibrium P P P73 mechanical equilibrium M MiaMi3 Inc 7 MC 7 7 Inc chemical equilibrium 73 7 1C 2 num constraints Then the number of intensive degrees of freedom left to specify is fC17377371C2 f C 7 73 2 What does it mean a If you specify f intensive properties then you can calculate all others if you are clever enough b Total differentials for changes in state pure components 8P 8P PP VT dP dV dT 8V T 8T 7 Example Suppose you want to separate a mixture of ortho 7 meta 7 and pam 7 xylene by vapor liquid equilibrium Can 1 specify the temperature and pressure of the system and guarantee that the system is in the two phase region 3 The Four Laws of Thermodynamics The Laws of thermodynamics are not laws7 but principles derived from observation that have never known to have been violated There are theoretical underpinnings but no rigorous proof of these laws All four laws give rise to distinct concepts Additionally7 each of the rst three laws of thermodynamics give rise to a certain thermodynamic function What are these functions 1 The Zeroth law of thermodynamics lf system A is in thermal equilibrium with system B7 and system B is in thermal equilibrium with system C7 then system A is in thermal equilibrium with system C lmportance This gives rise to the idea of temperature A7 B7 C are at the same temperature Hence the zeroth law gives rise to the concept of to The First law of thermodynamics The rst law deals with work and heat Recall from physics that work is the integral of some force over some distance WFd S Lecture 1 6 where S is some path This means that work is dependent on the path taken For example what is the work required to carry a book from the rst oor to the 12th oor of BEH That depends on how you get there In contrast what is the increase in potential energy is independent of the path and so is an exact differential 6W is inexact but dU internal energy is exact An exact differential of two variables is one which can be written F FXY 8F 8F dF 7 1X 7 dY lt8X gt Y 83 gt X The next concept is heat Consider a container with rigid walls Rigid walls means no work can be done on or by the system The system initially has ice and a liquid water Time goes by and the system now has only liquid water What happened The internal energy of the system stuff inside the container changed The change in internal energy must have been due solely to heat interactions dU 662 Now consider the same system but with adiabatic walls Result No change in internal energy therefore only heat and work can change internal energy of a closed system Now consider a gas inside a cylinder with adiabatic walls and tted with a massless frictionless piston Now compress the gas perform work on the system and look at the energy change dU6W Consider performing work by heating a cylinder of gas tted with a piston You put heat into the system and get work out The amount of heat and work depend on factors such as friction etc The change in energy is an exact differential dU6W6Q dU is exact but 6W and 662 are inexact differentials Therefore the inexactness must cancel Note also that you can turn work into heat with this device If you compress the gas the gas will give up heat to the surroundings The rst law states that the energy in these processes is conserved and that heat can be converted into work and work into heat You can t get more work out than heat you put in and you can t get more heat out than work you put in Here are other statements of the rst law 0 The energy of a closed system can only be changed through heat and work interactions 0 Heat and work are equal 0 Due to Clausius Die Energie der Welt ist konstant The energy in the universe is constant The rst law of thermodynamics give rise to the concept of Lecture 1 7 3 The Second law of thermodynamics Consider two blocks of copper block 1 at T1 and block 2 at T2 where T1 gt T2 We put the blocks in thermal contact The nal temperature of the system is Tf where T1 gt Tf gt T2 What about the inverse event We have two blocks of copper at Tf we bring them together and their nal temperatures are T1 gt T2 The rst law says this is ne because the total energy of the system is constant Why doesn t this happen Statements of the second law include o Spontaneous processes always increase the entropy of the universe 15 Z 0 where S is the entropy the measure of disorder in the universe 0 Heat and work are equal but work is more equal than heat with apologies to George Orwell 0 Due to Clausius Die Entropie der Welt strebt einem Maximum zuThe entropy in the universe tends to a maximum Consider a container with rigid walls Only thermal interactions change the energy If the system initially contains two phases ice and water then after some passage of time the ice will melt as heat is transfered into the system The ice is more ordered so that the nal state has a higher entropy than the initial state From this we see that the entropy is related to the heat interactions 6Qrev TdS dU Only for a reversible process Now consider a massless frictionless piston adiabatic 6Wrev iPdV dU Putting these together we have dU6Q6WTdSiPdV 6Qrev 6Wrev iPdV Note that from the Gibbs phase rule we know that we only need two intensive variables to specify the state of a pure substance Hence Lecture 1 F Comparing we get 8U 7 T 85 V 8U 7 7p 8V 5 Signi cance These equations can be used to nd changes of state temperature pressure energy The second law gives rise to the concept of Third law of thermodynamics The Third law has to do with an absolute value of the last function entropy In simple terms the Third law states that the absolute entropy is zero for any perfect crystalline substance at a temperature of absolute zero This is only true when the crystalline phase does not have any energetic degeneracies The third law also de nes the concept of the zero of temperature Hence the Kelvin scale comes about because of the third law of thermodynamics Without it all temperature scales would be arbitrary Lecture 3 1 Lecture 3 Objectives Students will be able to U FPJEOE H Describe the rst law in terms of heat and work interactions Describe the second law in terms of adiabatic and reversible processes Identify the difference between internal and total energy ldentify specialized thermodynamic equations and tell when they can be applied Describe ideal gas thermodynamic properties Heat work and the rst law We observe that for a closed system the only way to change the energy of a system is by heat and work interactions with the system We write the differential rst law as dE6w6Q where E is the total energy of the system Note that dE is an exact differential whereas 6w and 662 are inexact The integrated form is AEwQ For an adiabatic process AE wad and Q 0 so wad is path independent but you could also take a different path to get the same AE so wad u Q Likewise if the work interactions are zero then there must be some path independent heat interaction dE 6Qrev where Qrev is the reversible heat Now suppose that the whole system is translating and accelerating m w wstationary mgAy 0 7 U W 2 APE AXE So AE 7 APE 7 AKE wsmonary Q state function De ne the internal energy AU AEi APEi AKE so that U is also a state function then AU wstationary l Q Lecture 3 2 2 The second law Why does a basket ball bounce lfl drop a basket ball from a height of 1 m will it return to the same height when it bounces Why not Analyze the basket ball bounce in terms of the rst law AUAPEAKEwQ We are converting potential and kinetic energy into internal energy of the gas molecules in the ball and the material of the ball along with heat and work The rst law does not preclude the ball bouncing to the same height and bouncing forever However we observe that this never happens Likewise you can drop a weight into water and convert the potential energy of the weight into internal energy of the water however we never observe the converse although the rst law says that it is possible Hence the need for the second law of thermodynamics There is a directionality to spontaneous processes this directionality gives rise to another state function which we call entropy We de ne entropy as 7 6Qrev 15 T the differential entropy is the reversible heat divided by the absolute temperature Note that reversible heat requires that there be no temperature gradients just as reversible work required no force imbalances The second law can be written as d5 dssys dsm 2 0 The Clausius statement of the second law Heat cannot be converted into an equal amount of work through a cyclical process Proof For a cyclical process AU 0 and AS 0 Add heat to the system reversibly then extract work from the system reversibly and adiabatically AU 0 Qrev wad H wad Qrev OK by 1st law Now check second law AS ASrev ASad 0 but ASad 0 and ASrev QrevT 7 0 so the second law is violated 3 Generalized equations a Closed system no translations dU6Q6w Lecture 3 In Open systems A 2 dE 6Q6WZKHmgzmUfgt5mml A U2 7 Z ltHout l 92mm l 021 6mout out where 2 dEdUmgzm2U c Steady state7 d lav 0 A 1 267774 31 91 6Q 6w i d Compressors amp turbines pressure change lt gt work AH Q ws where wS is shaft work e Throttling devices pressure drop AH Q f Nozzles AKE lt gt AP A 1 AH 7w 7 a g Heat exchangers AH Q 4 Ideal Gas State Properties T U 8 w 0V V Both 0p and CV are only functions of temperature Lecture 3 5 Open system example Filling a storage tank A storage tank is pressurized by rapid lling from a much larger storage reservoir The tank is well insulated The tank volume is 2 m3 and the gas in the tank is initially at 1 bar and 280 K The air in the reservoir is at 100 bar and 300 K What is the temperature of the gas in the tank when the pressure reaches 5 bar Approach This is an open system with negligible change in kinetic and potential energy There is one inlet stream and no outlet streams Heat and work interactions are zero dE dU Hintsnm with Him a constant and 671m dn dU ndU udn HUM We can rearrange this to du dn Hin U 77quot We can integrate this between states 1 and 2 U a 7 2 lnnl2 a ln M ln U1 1 Hm U2 771 7 ln Him 7 Rearranging and assuming ideal gas behavior of the air Him 7 i 7 712 7 P2V2 RTl 7 P2T1 Hin7T72 711 T RTZ P1V1 T P1T239 Let us assume that the heat capacities are constant The enthalpy andinternal energy can then be written as referenced to some temperature To H H0 CpT 7 To and U U0 CVT 7 To We can now write Hin l 7 Ho aPCEn 7 To 7 30 VT1 7 TM Ho 7 70 pTin 7 VTl 7 GP 7 av To But H07 U0 PV RTO and 6 70V R for an ideal gas Therefore HoiljoiRTo O and Him 7 1 pTin 7 VTl Therefore 3me 7 3le P2T1 CPTm 7 VTZ T PITZ39 Lecture 3 5 Solving for T27 GPTin vT1gtP1T2 GPED 7 C39VT2 132T1 T2 P1 me C39VT1 PZTICWV PZTIGPED T2 PZPl T111115 m where a C39pC39V Using C39V 52R and a 14 we get T2 3818 K The number of moles at this point is n 327 315 moles Lecture 3 1 Lecture 4 Objectives 1 Understand the need for auxiliary functions 2 Be able to derive the differential functions from the Gibbsian equations 3 Explain the chemical potential in physical terms 1 Begin by reviewing the rst and second law fundamental equation for a closed pure system dU TdS 7 PdV dU 662 6 W Recall that dU is an exact differential Note that U is a homogeneous function of rst order That is U US V is a complete function completely described by S and V and that UmS mV mUS V to Auxiliary energy functions We have some kind of feel for the internal energy For a closed pure component system with no chemical reactions the functions V U and S form a complete basis that is we can describe all of thermodynamics by manipulating these three functions Unfortunately nature will not let us get away with just using U S and V The problem is that we can t easily measure S so it is difficult to develop an equation for U US V Suppose that we measure U as a function of say T and V ls that good enough U UT V Lecture 3 2 Does this function carry the same amount of information as U US V Suppose that we need to know the value of S for some reason eg we want to know if dS gt 0 Can we nd S from our function U UT V Solve for 2 W MW e 375 Faiv am 5 i or ST WdT c We can t nd S because we only know how S changes at constant volume The unknown constant of integration in the last equation is actually a function of V Therefore we don t know how S depends on volume In other words U UT V contains less information than U US V and we no longer have complete knowledge of the system For this reason people have introduced at least three auxiliary energy functions They are enthalpy H E U PV Helmholtz free energy A E U E TS and Gibbs free energy G E U PV E TS The set of equations for U H A and G are called the Gibbsian equations Here is a mnemonic to help you remember how these four energy functions are related U H TS T l 7T5 A iV G Here is a mnemonic device to remember the mnemonic device U Have Great Articulation Go clockwise around then do PV on top EPV on bottom TS on left ETS on right For example HUPV Lecture 3 3 03 Work out the derivative form of A given the derivative form of U and the relationship between A and U given above A U 7 TS dA dU 7 TdS 7 SdT TdS 7 PdV 7 TdS 7 SdT 7PdV 7 SdT Likewise dH TdS VdP dG 7SdT VdP Mixtures and open systems A new function the chemical potential Chemical potential and the related functions activity activity coefficient fugacity and fu gacity coefficient are among the more difficult functions to understand I will attempt to rationalize not derive the chemical potential in physical terms a Consider a mixture of H2 02 and Pt in a closed rigid container with adiabatic walls What happens Does the temperature change If we consider the pressure low enough and the temperature high enough then the gases will be ideal Does the internal energy change Are there work andor heat interactions How does this conform to dU 662 6W We need another function to describe the system A T V Consider an open system and write down 16 7SdT VdP Now add small amount of one of the components but do so at constant temperature and pressure Does the Gibbs free energy change Yes because G is extensive How is this accounted for in 1G 7SdT VdP We deduce that for a mixture of C components or an open system U US7V7n17n277n0 dU is still an exact differential so 8U 8U C 8U dU lt7 15 lt7 W lt7 dn as V i 8V S i 87 SgtVgtnj i Note that the subscript 71hr means that all mole fractions except i are held constant eg if C 3 i 2 then j 1 3 The temperature and pressure de nitions are still valid ie ltgtV i T and ltgtSm 7P Let us now de ne the Chemical potential as E M 87 534nm Lecture 3 4 Then our fundamental equation becomes 0 dU TdS 7 PdV E Main i1 The equation can be written in terms of intensive variables 0 dU dnU Tdn 7 Pam Z pidxm i1 where z This can be re written using the product rule as C C 71 def Td Pdfi Zu m dn U7 T Pv7 2pm 0 i1 i1 Since dn and n are independent and arbitrary each grouping of variables must independently vanish Hence the integrated form of the internal energy equation is C IT 7PlZlulii i1 or C UT57PVZulm i1 The chemical potential plays an analogous role to temperature and pressure A temperature gradient gives rise to the ow of heat ie a temperature difference is like a potential for heat transfer Likewise a pressure gradient is a potential for mechanical work no pressure differences no PV work A gradient or difference in chemical potential is the potential to perform chemical work either by chemical reaction or by mutual diffusion In other words if all the components of a mixture have the same chemical potential no chemical reactions will take place and no mutual diffusion will occur because there is no driving force AT potential for heat ow AP potential for mechanical work AIM potential for chemical work An electrical analogue is the voltage A voltage difference 120 V relative to ground is the potential to do electrical work turn a fan light a room but voltage and chemical potential are intensive so it is the amount of electricity number of amps and the amount of chemical number of moles reacted that tells you how much work is done Lecture 3 5 4 What are the integrated and differential Gibbsian equations for multi component open sys tems Use the mnemonic device to derive them H a A dH dG 1A 5 Returning to the problem of H2 02 and Pt in a closed rigid container with adiabatic walls Start with the integrated combined rst amp second law Assume that the reaction goes to completion and that we have a stoichiometric mixture of H2 and 02 C 0 AU T252 7 T151 7 VP2 7 P1 2 mm 7 2mm 0 3331 556 Where the last equality comes from the rst law For an ideal gas we can write 1 RTlnyi 7 T RTiny Also VP2 7 P1 R nH2QT2 7 mTl where m nH2 7102 So 0 7R nH20T2 7 niT1 T252 7 T151 nHZo HH20T2 7 T2 H20T27 P2 RTz 1111 77m HH2T1 7 T1 H2T1P1 RT1 ln szl 77102 HO2T1 7 T1 02T1 P1 RT1 ln 3402 Now we note that 2 nH20 H20T2 P2 and 1 7m SH2T1P1 7 R ln szl 7102 O2T1 P1 7 R ln 3402 Therefore R anoTz 7 mTi HH20T2 7 HH2 7 Ho2 Now if T1 298 K P1 1 bar then we can write HH20T2 7 HH2 7 Ho2 ano T2 AHZH2O T CPH20dT 1 Using AHH2O 72418x105 Jmol and BEHZO 291631449 x10 2T7202 x10 6T2 in Jmol nHZO 1 mol and n 15 mol we can solve for T2 Matlab gives T2 62036 K


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