New User Special Price Expires in

Let's log you in.

Sign in with Facebook


Don't have a StudySoup account? Create one here!


Create a StudySoup account

Be part of our community, it's free to join!

Sign up with Facebook


Create your account
By creating an account you agree to StudySoup's terms and conditions and privacy policy

Already have a StudySoup account? Login here

Physical Chemistry

by: Mr. Clementine Gottlieb

Physical Chemistry CHEM 3520

Mr. Clementine Gottlieb
GPA 3.68


Almost Ready


These notes were just uploaded, and will be ready to view shortly.

Purchase these notes here, or revisit this page.

Either way, we'll remind you when they're ready :)

Preview These Notes for FREE

Get a free preview of these Notes, just enter your email below.

Unlock Preview
Unlock Preview

Preview these materials now for free

Why put in your email? Get access to more of this material and other relevant free materials for your school

View Preview

About this Document

Class Notes
25 ?




Popular in Course

Popular in Chemistry

This 29 page Class Notes was uploaded by Mr. Clementine Gottlieb on Wednesday October 21, 2015. The Class Notes belongs to CHEM 3520 at Tennessee Tech University taught by Staff in Fall. Since its upload, it has received 14 views. For similar materials see /class/225696/chem-3520-tennessee-tech-university in Chemistry at Tennessee Tech University.


Reviews for Physical Chemistry


Report this Material


What is Karma?


Karma is the currency of StudySoup.

You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/21/15
CHEM 3520 Spring 2004 The Properties of Gases Ideal Gases 7 the case of any gas in nitely suf ciently dilute so one can ignore the interactions between molecules Sum 7 v lsmm mum 7 Ideal perfect gas equation of state PVnRT or P17RT quot V 1 mm 7 The ideal gas constant R 83145 JmolK R 0083145 LbarmolK R 0082058 L atmmol K Pressma a Pressure 9 0 7 Perfect gas law is obtained by combining quot191 7 Boyle s Law PV ct 7 GayLussac Law V ctx t 27315 or V cth 7 Charles Law P cth c gt a g u E 3 g 3 g u g lncreaslng g 3 temperature T 9 a Increasmg temperamm T E rrauarau un E lvapolanun E Lrapolaimn 0 a u u 0 Volume v 0 IN 7273 Temperature we 0 Temperature WK 7 Ifa gas obeys the ideal gas equation it is called an ideal gas or the gas is said to behave ideally 7 Combined gas equation T1 T 2 7 Dalton s law ofpartial pressures each gas in a mixture ofideal gases acts as ifthe gases were not present anT n Ptotal ZPJ Where P V total ythotal quotJ CHEM 3520 Spring 2004 Real Gases 7 To determine if a gas behaves ideally one could plot denoted Z and called compressibility factor versus P Z 1 for ideal gas Z 1 for real gases at low pressures Z 1 for real gases at high pressure Z gt 1 repulsion forces 17ml gt ideal Z lt 1 attraction forces 17ml lt ideal 2m x 7 The ideal gas equation does not account for the intermolecular interactions m 7 As the temperature increases the real gas I 5 behave more like the ideal gas M 0 200 011 00 800 INN PI Iul 7 Introduce another very popular equation of state to account for intermolecular interactions This is the van der Waals equation a P 7 21V 7b RT Where a and b are called van der Waals constants b Lmol re ects the size ofthe molecules a L2 barmol2 re ects how strongly the molecules interact Pressure a L2 barmolz b Lmol H2 024646 0026665 CH4 23026 0043067 N2 13661 0038577 C6H6 18876 011974 7 Rewrite the equation as P 77i2 I7 a 7 The compressibility factor Z 7 7 RT V 7b RT V CHEM 3520 Spnng 2004 eMolarvolurne rs obtatned by solvrng the eubre equatron W 7b 72 1771 0 P P P e Other equataon ofstates rnore accurate1ncludng a1so only two pararneters dhcherong equatron e PengrRobmson equataon RT p 4 7 V78 V7B 78 5 to at mal Lquot elnvestagate sorne enpenrnenta1 plots of Pversus V at eonstant Tea11ed rsotherrns eThe ease of co2 e Cnncal pornt e Cntreal ternperature Tc rs the ternperature above whreh agas eannot be Tc 304 14 K 30 99 quotc hque edregardless ofthe pressure Z 0 094 Lrno1 eThe conespondmg pressure and molarvolume at the entrea1 pornt are the entrea1 pressure PE and the errtaeal volume Z 372 CHEM 3520 Sprlng 2004 Wth eaeh other The olasheol euwe eonneeung the enols of honzontal hnes ls ealleolthe coexlstence eurve o the lelt there ls only llquld present on the nght there ls only gas on the honzontal hne there are both gas andllquld e The hne beeornes a polnt an ln eeuon polnt at 7C There the rnenlseus apor dlsappears and there ls no dlstlnetlon between ul39face tenslon dlsappears and the gas both have the sarne olenslty ealleol enueal olenslty between llquld andlts v gas andllquld phases eEquataons ofstate that ean be wntten as euble equataons m 7 van olerWaals Redllcherong ean olesenbe both the gaseous and the llquldreglons ofa substanee eThese equatlons show spunous loops at T lt7c eThe DA hne ls olrawn so that the DCE area and EBA area are equal ealleol Maxwellrequalrarea eonstruetlon eThe Gas 7A eornpresslon othe gas eThe AeEeD hne llquld and gas are ln equlllbnum e The Dequulol ehange ofvapor of h quld wlth lnereaslng pressure eThe ArB segrnent rnetastable reglon correspondlng to superheated vapor eThe C7D segrnent rnetastable reglon correspondlng to supercooledllquld eThe BeEec segrnent T gt 0 unstable reglon not observedfor equlllbnum systems eAeEeD hne has 3 solutaons hlgher one polnt A comespondmg to gas lower one pomtD correspondlng to hqulol and the mlddle one pomt E spurlous CHEM 3520 Spring 2004 7 Densities molL of CO 2 experimental vs calculated at 142 69 K and 3 50 atm Experimental van der Walls RedlichKwong PengRobinson p 22491 1414 2013 2361 pg 5241 4615 5147 5564 7 At critical point in ection point 2 0 and 0 T W T 7 Use these conditions to nd the critical constants in terms of van der Waals constants a and b 7 van der Waals equation 73 b 172 2711 0 P P P 7 AtT To there is only one solution of the equation 7 V 3 0 173 4170172 3I7CZI7 I7c3 0 31703bP a2 8quot 27 27bR c 7 Similar but mathematically more complicated one obtains for Redlich Kwong equation 23 13 23 70 384733 Pc 0029894A R T0 034504 1 353 BR 7 Law of Corresponding States PCVL RT 0 7 The values of are independent of the gas P I7 7van der Waals iL 27bR 2 0375 RTc R 27132 8a 8 P I7 7RedlichKwong c c 03333 RTc P075 7 PengRoblnson 03074 RTc P VC 026 030 RTc 7 Experimentally Spring 2004 CHEM 3520 5 The properties of all gases are the same if we compare them under the same conditions relative to their critical point Law of Corresponding States 7 van der Waals equation in terms of reduced parameters 27 RT 2 RT 7 Substituting a and b 64Pc 8P 7 Divide by Pc and 7 use PJ7c RTC and rearrange 5373 Z RT RT AL P V2 7 3 PJZ m 3r L T R To ll VI 7 Introduce reduced quantities PR 1 7R c gt PR VR 7lj TR VR2 3 3 7 This equation does not have any parameters that are gas dependent it is a universal equation for all gases 7 Different gases at the same values ofreduced parameters same PR7RTR behave the same are and are said to be at corresponding states This is an example of the law of corresponding states All gases have the same properties if they are compared at corresponding conditions same values of PR7RTR 7Example CO2 at P 143atm 719 Lmol T 456 K and N2 at P 658 atm 7 l 8 Lmol T 189 K are equivalent because they have the same reducedparameters PR 0196 7R 20 TR 15 7 Expression of compressibility factor in terms of reduced parameters 5 Z VR 7 9 universal equation VR 7 VRTR 7 Z can be expressed as a function of any two reduced parameters VR andTR PR andTR PR and VR H 7 Book has an example of the law of corresponding we r states Z vs PR at various TR should look same quota for any gas M 7 Everything is relative to critical point M CHEM 3520 Spnng 2004 eThe vlnal equauoh of state elt ls the rhostfuholarhehtal equatroh ofstate a ap l u ruralul are ealleol vlnal expanslons P7 1 l 2 z 2 1 3 3 RT arm V 3mm where the eoemerehts are ealleol vlnal eoemerehts 32V ls the seeohol vlnal eoemereht 33V ls the thrd vlnal eoemereht eAhothervrhal equatroh of state ls wrrtteh by expandmg wrth respeetto P 2 l32pmp33pmpz RT eThe vlnal eoemerehts m the two equauohs are relateol 3W0 Rmzpm re eet the rst devrauoh from rolealrty as the pressure thereases eThe 32pm ear be oleterrruheol from the slope Zversus P rThe gym ls hegatrve at low temperatures theh beeorhes L posmve wth a shallow maxrrhurh eThe 3mm 0 atBoyle temperature the repulslve and x m attraeuve parts othe rhterrholeeularrhteraetrohs eaueel eaeh other 2 gas appears to behave ldeally E 1 rhteraeuohs e Conslderrto be the dlstance between the eehters oftwo rholeeules eAssurhe that all posslble ohehtauohs average out 7 Conslder ho to be the potehtral ehergy rhteraetroh energy oftwo rholeeules at dlstance t 2 3mm 227 e 37 2112 u U0 0 2 32V 73 0 ldeal behavlor ho rhteraetroh Spnng 2004 CHZEM 3 520 r Lemmerones potential 7 An amacuve component 512 r A repulsrve componmt 7 7 512 Comlnmng me two no 7 4r l2 5 l2 c 450 r Rearrangmg no 4s 3 r 3 where 2 6 0 r r 6 40 3 quot a Paramztcrs Ens a measure ofhow strongly me molecules amact a 15 me depth omne well 7 Is a measure omne srze omne molecules a where no so r Subsumung Lemeroms potential m me expressron for 32 B T 2rNTex 3 Ef gy 71 2dr 2quot 7 A0 P kET r r Reorgamze use Tquot M x L s a r Drvrde by 2ra3NA 3 7 Introduce reduced second vmal coefficient depends only on T 8230 82VT 2zra3NA3 82quot 7quot 73J39exprixquot2 r x 571x2 lt T o i solved nurnencally exlsts m tables rPlol of 83 vs Tquot for dclrerenl gases grves only one curve uner ammple omne law ofcorrespondcng states tally and from merr values 7 The 32 are determined expenmen me Lemmerones parametns are deterrnmzd CHEM 3520 Spring 2004 7 Hardsphere potential 7 The attractive part of the potential is missing 00 rlt039 ur 0 rgt039 3 3 BZV T 27m NA 4 times the volume of NA spheres 7 This potential predicts BZV to be independent of T but experimentally the molecules are not really hard 7 Squarewell potential ur 8 0 rlt039 altrltla rgt ta 3 2 BZVT 1 13 AXEW 4 7 Give good agreement with experiment for BZV vs T 7 Combination of hardsphere and LennardJones potentials 00 r lt 039 0 Cs VgtO39 r 27ra3N 27rN 6 32V T A Ag 3 3kBT039 7 Interpretation of BZV T 7 neglecting the higher order terms of the virial expansion 1 2 1 w RT RT 3 BZVT V l7ideal P 7 Van der Waals constants in terms of molecular parameters 7 Rewrite van der Waals equation as RT 1 a RT 1 b2 a 1 T l V 1 bV V2 V 1 2234 b L RT RTV b2 m CHEM 3520 Spring 2004 7 Comparing with virial equation of state a B T b 2V RT 7 Comparing with the expression of BZV obtained for a combination of hard sphere and LennardJones potentials 27ra3N 27rN 6 32V T A 3 3kBTo 3 2 27z39N 3 a 1 c6 3039 a is proportional to 06 is a measure of the interaction between molecules 3 272039 N 3 b 2 TA b is four times the volume of N A spheres 7 From molecular point of view the van der Waals equation is based on intermolecular potential that is hardspheres potential at small distances and a weak attraction potential at larger distances Intermolecular Interactions 7 are the source of the deviations from the ideal gas behavior 7 The form of the r712 repulsion term is not well established 7 Intermolecular forces responsible for the attractive term r76 term in LennardJones potential 7 Interaction between the dipole moments 1 and 2 of polar molecules dipole dipole interaction 7 depends on orientation 7 overall average interaction potential 151510 212 i 4723980 2 3kBT r6 7 Interaction between a dipole moment and an induced dipole dipoleinduced dipole interaction CHEM 3520 Spring 2004 The Second Law of Thermodynamics The second law of thermodynamics shows that isolated systems not in equilibrium evolve such to increase their disorder It introduces another state function called entropy that gives a quantitative measure of the disorder of the system 7 What is the force that determines if a process will proceed spontaneously or not 7 A criterion for a reaction or process to proceed spontaneously is to be exothermic or to evolve energy 7 This idea is coming from the classical idea of rolling down hill The variational principle of quantum mechanics also said that a system seek its lowest energy state 7 There is another condition or force that drives spontaneous processes The following processes obey first law of thermodynamics but cannot be explained by it An analysis in a microscopic or molecular point of view shows that in these processes there is an increase in disorder or randomness of the system 7 Expansion against vacuum two bulbs one with gas and one empty separated by a stop clock By opening the stop clock the gas occupies both bulbs although AU or AH is essentially zero 7 Mixing of gases two bulbs with different gases By opening the stop clock the gases mix and both occupy both bulbs although AU or AH is essentially zero 7 Melting of ice at temperature higher than 0 C AquHO 60 kJmol endothermic process but still spontaneous 7 Reaction of BaOH2 with NH 4NO3 BaOH2 s 2NH4NO3 s gt BaNO32 s 2H20l 2NH3 g reaction is endothermic and the system can cool down to 20 C 3 The systems evolve spontaneously in a direction that lowers their energy and that increases their disorder Entropy a quantitative form of disorder 7 It is a state function 7 Consider the heat transfer associated with a reversible process in which both T and Vof an ideal gas are changed 5qu dU 5me CV TdT PdV CVTdT nRT VdV CHEM 3520 Spring 2004 62 f 62 f the cross derivative are not equal I Eqrev 1s an 1nexact d1fferent1al axay an quotRT depends on T CV T does not depend on Vbut V Eqrev CVTdT dV T T V Egrev T is an exact differential so it is the derivative of a state function 7 De ne this function as entropy and denote it by S dS agrev T 7 The entropy S is a function of T and V 7 At lower Tthe increase in S is higher than higher T for the same amount of heat 7 For a cyclic process AS 0 or dS 0 where means cyclic process 3 fgquev 0 7 The formulas are general applicable to both ideal gas and other systems 7 Examples of entropy calculations for various reversible transformations of ideal gas P 7 Isothermal transformation path A 2 V2 nRT V ASA j Eqm A ji 1dV mm 2 1 T V T1 V V1 7 Adiabatic transformation path B 2 EquVB ASE l T 0 7 Isochoric transformation path C and path E 2 T1 T2 T2 C T C T ASczj39gqrewCzquot V dTJ39 V dej39 1 T T T T T 2 1 2 T1 C T ASE J396qrevE J V dT T T 1 T3 7 Isobaric transformation path D 2 T3 C T V ASD j 5qrew j V dTann 2 1 T T T V1 1 CHEM 3520 Spring 2004 172 1 171 1 7Fora P Vie Tb gas A A Rlnbb A BC 0Rln 7 Entropy change for processes in an isolated system 7 By adding energy as heat to a system then its entropy increases because its thermal disorder increases didn t show explicitly that the entropy is related to the disorder of the system 7 The entropy of an isolated system increases as a result of a spontaneous process 7 Consider an isolated system and two compartments of this system separated by a divider that allows heat ow Also no work can be done UA UB constant S SA SB dUA Egrev wrev TAdSA dUB Egrev wrev TBdSB But wrev 0 because VA ct amp VB ct dU dU T T 13 Mgg UAUBct3dUA dUB TB TA 3dSdSA dSB If TB gtTA 3dUB lt0 3 dSgt0 If TB ltTA3dUB gt0 3 dSgt0 If T B T A 3 dS 0 7 For an in nitesimal spontaneous change in an isolated system energy remains constant so driving force is the increase in entropy But because system is isolated the increase in entropy is created within the system itself so the entropy is not necessarily conserved it actually increases until the system reaches equilibrium dS 0 at equilibrium 7 Not only dS 0 at equilibrium but also dS 0 for a reversible process that remains essentially in equilibrium dS gt 0 spontaneous process 7 In an 1solated system dS 0 revers1ble process 7 Entropy change for processes in a closed system dS dSprod dSexch dSprod dSexch is due to exchange of energy as heat with the surroundings dSprod is created by irreversible process CHEM 3520 Spring 2004 7 But the in nitesimally change in heat 5g can be 7 for a reversible process Eqmv 7 for an irreversible process qm 7 For reversible processes dSProd 0 and g Eqrev 3 dS quiev 7 For irreversible processes dSprod gt 0 and Eq Eqirr 3 dS gt 5 7 Combining the two expressions dS Z T or AS 2J T Inequallty of Claus1us where equal for reversible and greater for irreversible process 7 This is one way of expressing the Second Law of Thermodynamics 7 Formal Statement of the Second Law of Thermodynamics 7 There is a thermodynamic state function of a system called the entropy S such that for any change in the thermodynamic state of the system dS Z where the equality sign applies if the change is carried out reversibly and the inequality sign applies if the change is carried out irreversibly at any stage 7 Example of entropy increase during a spontaneous irreversible process 7 Assume two transformations between state 1 and state 2 one done irreversibly from 1 to 2 and one done reversibly from 2 to l 2 641 1 641 IA 0 because the system is isolated J rev S1 S2 lt 0 T T 1 2 3 AS S2 S1 gt 0 S increases going irreversibly from state 1 to state 2 7 Considering Universe as an isolated system and all naturally occurring processes as irreversible the second law states that the entropy of Universe constantly increasing 7 Clausius summarizes rst two laws of thermodynamics The energy of the Universe is constant the entropy is tending to a maximum CHEM 3520 Spring 2004 7 Molecular interpretation of entropy 7 The entropy is related to the disorder of a system but how to measure the disorder 7 Consider an ensemble of A isolated systems each with energy E volume V and number ofparticles Nwhere E EN V 7 Systems have same energy but they may be in different quantum states because of degeneracy Consider the degeneracy associated with energyE to be QE and label the degenerate levels as j l2QE 7 Consider that the A systems are distinguishable and let 611 be the number of systems in the ensemble that are in state j the number of ways of having all systems in state 1 a2 in state 2 is given by I I A39 A39 where ZajA Wa1502 a1a2a3 Haj J f can be taken as a quantitative measure of ensemble disorder 7AllA systems in one state 611 A ell 0 3 W 1 7 All aj are equal highest disorder 3 highest value for W 7 Boltzmann proposed the relationship between S and W as S kB an 7 the most famous equation of statistical thermodynamics 7 correlates a thermodynamics quantity S with a statistical quantity W 7 When W l 3 S 0 3 completely ordered system 7 Completely disordered system 3 W maximum 7 The logarithm dependence has advantage that for a system with 2 parts X and Y Stotal SX SY because WXY WxVVY 7 The relationship between S and degeneracy Q 7 Consider Q degenerate quantum states occurring in an ensemble with equal probability An ensemble of isolated systems should contain equal number of systems in each degenerate quantum state Sensemble AS nQS Ssystem S kB In Q system system CHEM 3520 Spring 2004 7 Example A system of N distinguishable spins or dipoles that can be oriented in one of two possible directions with equal probability 7 Each spin has degeneracy of 2 7 Degeneracy of N spins 2N 3 Entropy S NkB ln2 7 Example Ideal gas QE CN f E VN Determine the entropy change during an isothermal expansion of 1 mole of ideal gas Q AS kB 11192 7kB 1an kB 111 2 Q1 N E VN V AS k 1nM 22N R1n2 CN f E1 V1 V1 7 Example Change in S per mole for mixing of 2 gases AmiR y11ny1 y2 lnyz Amix gt 0 because ln yl lt 0 7 Calculations of entropy for various processes 7 To determine the entropy changes assume a reversible process for which use equality 7 Example Isothermal expansion of ideal gas against vacuum V1 gt V2 gqgev Eqrev dU gwrev gwrev PdV dV 2 Asj 1 V2 3ASann7 V2gtV1ASgt0 1 7 A similar result is obtained for real gas 7 This result is same for both reversible and irreversible processes What is the difference Look at ASsm 7 Reversible process 2 qrev V2 ASsm T ann V1 2 AStotal 0 7 Irreversible process V2 ASsm 0 3 AStOtal quotR1117 gt 0 1 9m 0 because AU 0 5W 0 CHEM 3520 Spring 2004 7 Example Mixing of two ideal gases 7 Each gas act independently of the other so each gas expands from Vnitial t0 V nal V V V V ASA nARlnM ASE nBRlnM V VB ASASAASB 7 Transform volume to number of moles and this to mole fraction A S 3 JAIHJ A yB lnyB 7 More general N AmiXS RZyj lnyj AmiXS gt0 j1 7 Example Two equal pieces of metal at different temperatures 7Let Tc cold Th hot T T CVTh TCVT TcT d3 6gquot although process is irreversible Eqrev dU CVdT because no work done C dT T AS IV CV ln 2 assuming CV ct between T1 and T2 T T1 T1 T T T T ASh CV1n ASC CV 1mg 2Th 2T T T 2 ASAShASc CV1n h gt0 4Tth because Th Tc2 gt 4TthcgtTh Tc2 gt 0 7 Example Entropy changed during isobar heating 7 Given CP fT T2 C T AS 63 T T1 CHEM 3520 Spring 2004 Heat steam engines and the second law of thermodynamics 7 The second law of thermodynamics was determined by Hm source investigation of the ef ciency of the steam engines Tn 7 The steam engine withdraws energy as heat from a high temperature thermal reservoir uses some energy to do work and discharges rest of energy as heat to a lower temperature thermal reservoir 7 The amount of work is in practice even smaller due to 7 the fact that reversible processes are idealized 7 the iction and other mechanical factors 7 The steam engine is based on a cyclic process AUengine W 4th qrevc 0 Cold Slnk 4th 4m c AS v 0 engine h To 5 7 w qmvyh qmvyc 7 wis positive Adlabat 7 The ef ciency of a steam engine i lsothen n 7 The maximum ef ciency is obtained in a heat engine following an ideal circle called Carnot circle consisting on two isothermal and two Pressure 9 adiabatic transformations W 7 qrevh qrevc 7 Tc 7 Th 7Tc 77mm 7 717 4mm 4mm T h T h Volume v 7 Example heat engine working between 273 K and 573 K 5737 273 17m T 534 7 The ef ciency is higher as Tc is lower and T11 is higher 7 If T c T h 17 0 ltgt no net work can be obtained in isothermal cyclic process 7 Kelvin statement of second law 7 A closed system operating in an isothermal cyclic manner cannot convert heat into work without some accompanying change in the surroundings CHEM 3520 Spring 2004 Thermodynamic Functions and Thermodynamic Equations The Helmholtz energy 7 Criterion for spontaneous processes for systems at constant volume and temperature dU5q w w7 extalV EqSTdS 7 At constant volume SW 0 3 dU S TdS 3 dU TdS S 0 7 For isolated systems dU 0 61 2 0 7 Both T and Vare constant dU TS S 0 7 De ne a new thermodynamic state function called Helmholtz energy A U TS dA S 0 constant T and V where dA 0 at equilibrium 7 If a system is held at constant T and V the Helmholtz energy decreases until all the possible spontaneous processes have occurred at which time the system will be in equilibrium andA will be a minimum 7 Isothermal change from one state to another AA AU TAS S 0 constant T and V 7 If AA gt 0 the process cannot take place spontaneously at constant T and Vand work should be done on the system to make the change process AA represents the compromise between the tendency of a system to decrease its energy and increase its entropy 7 If AU lt 0 and AS gt 0 both contribute to AA lt 0 AU more important at low T AS more important at high T 7 Example mixing of two gases at constant T and V AU 0 and A y1Rlny1 y2Rlny2 3 AA RTy11ny1 y21ny2lt 0 3 spontaneous process 7 Physical interpretation of the Helmholtz energy A AA AU TAS lt 0 for an irreversible process but consider the reversible path connecting the two states TAS qrev 3 AA AU qrev wrev isothermal reversible work CHEM 3520 Spring 2004 AA lt 0 wrev maximum work that can be obtained from the system work done by system in a reversible process 7 less if irreversible due to friction AA gt 0 wrev the work needed to be done on the system to produce the change reversibly 7 irreversibly the work required is even bigger The Gibbs energy 7 Criterion for spontaneous processes for systems at constant pressure and temperature dU S T dS PdV dU TdS PdV S 0 dU TS PV S 0 at constant Tand P 7 De ne a new thermodynamic state function called Helmholtz energy GU TSPVH TSAPV AG S 0 constant T and P where dG 0 at equilibrium 7 If a system is held at constant T and P the Gibbs energy decreases until all the possible spontaneous processes have occurred at which time the system will be in equilibrium and G will be a minimum 7 The Gibbs energy G is analog of the Helmholtz energyA for processes that take place at constant T and P 7 Isothermal change from one state to another AG AH TAS S 0 constant T and P 7 If AG gt 0 the process cannot take place spontaneously at constant T and P and work should be done on the system to make the change process AG represents the compromise between the tendency of a system to decrease its energy and increase its entropy 7 If AH lt 0 and AS gt 0 both contribute to AG lt 0 AU more important at low T AS more important at high T 7 Example NH3g HClg gt NH4Cls at 298 K AIH 71762kJ 3 ArG 9l21kJ 3 spontaneous process AIS 70285kJK 7 Example H20l gt H20g CHEM 3520 Spring 2004 AvapG AvapH TA S vap A H 4065kJmol vap 3Avap G Oldmol at 373 K and 1 atm AvapS 10891dKmol At 363 K AvapG 110kJmol 1 g not spontaneous At 383 K AvapG l081dmol 1 g spontaneous 7 Physical interpretation of the Gibbs energy G G U TS PV 3 dGdU TdS SdTPdVVdP But dU TdS w 3 dG 7SdT VdP wrev PdV rev But wrev PdV EWHOHPV work other than PVwork electrical work 3 dG 7SdT VdP wnoan 3 dG wnonPV reversible constant T and P 3 AG Wman reversible constant T and P AG lt 0 Wman maximum work exclusive of PVwork that can be done by system AG gt 0 Wman the work exclusive of PVthat can be done on the system 7 Example H2 1 02 g gt H20l AG 237 l kJmol AG is the maximum nonPVwork that can be given by system 7 Example H20l gt 02 g H2 g AG 237l kJmol AG is the minimum nonPVwork that can be put for the reaction to occur Maxwell relations and dependences of state functions 7 Not all thermodynamic functions can be measured directly so some quantities are expressed in terms of others that can be experimentally determined 7 The temperature and volume dependence of the Helmholtz energy AU TS 3 dAdU TdS SdT For a reversible process dU TdS PdV 3 dA PdV SdT 6A 6A But AAVT 3 dA dV dT 6V T 6T V 3 l nuoi spnng 2004 CHEM3520 43 EJ 4 8V T 37 VT AA7J39PdV at constant 7 K V dV erRTln VI 7 TAS V V K rIdeal gas AA RT also because AU 0 at constant Tforideal gas 7The Volume dependence ofthe entropy 82A 7 82A aVaT 7 876V 15ng But i E density at constant temperature a Maxwell relation 87 t n t t we V AS jdV at constant 7 37 Vl T choose V as a Volume when gas behave ideally STVr S d 1 ij Em 24545 Jmolquot Kquot for ethane at 400 K V Ideal gas E AS nRId V nlean 2 lsothennal process 37 V V V V l P77b RT gas hi Rln 1 V Tb TIhe Volume dependence ofthe energy TFohldeal gas Udepmds only on Tand not on V What about real gases7 J 4 r r r AUTS 7 8V 8V 8V E and 1 1 8V T 8V T 37 T CHEM 3520 Spnng 2004 epwf av T aT V emtegrate between avolume where there rs an xdeal behamor andthe eurrerrt volume V aP UTvrU d 77 7P v ateorrstantn w 3quotquot V rExamples Idealgas ePT ePP0 av T v 51775 RT gas Redhcherong gas p r t j 3A m rebate m 37 T 2T127 ray p g av 7 a7 P ar V aT P we we Idl CrC T R eagas P V VP 1 2 specny E a7 av T K 1 v where 527 L rstheeoemerentomermalexpansror v aT P g the rsothermal eompressrbrlrty v aP T rThe temperature and volume dependenee ofthe Gbbs energy G Ur 79 Pv m we SdTerSVdPPdV but 411 mu PdV m exam WP 8G ButGGTP dG d7 j dp aT P aP T CHEM 3520 Spnng 2004 E V and E S T 8T P e e thbs energy oleereases wth lncreaslng Tandlncreases wth lnereasln P eThe pressure olepenolenee ofthe entropy But aPaT BTBP 2 e E 2 a Maxwell relatron 3P 7 8739 P 7 Useful relataonshrp beeause allow deterrnrnatron ofthe ehange tn Swth P at eonstantternperature AS2r dP attonstantD 5 8T P 24 7 Choose P as apressure where gas behave ldeally p m sages I j dp tn pm 57 p F m id 246 45 J rnolquot Kquot for ethane at400 K o no mu mquot P Ideal gas 2 AS 21Pan lsotherrnal proeess 8T P P 7 Same as the olepenolenee wth respeet to V eThe pressure olepenolenee ofthe enthalpy eForroleal gas Holepenols only on Tand not on P What aboutreal gases7 Genem f r 2 yand 7 8P 7 31quot p 8P 8H av 2 i y 7 me 3P T 37 P T p a n 2 AH ver jdP at eonstant 7 8T e F to not no F mo 0 Mutual l on rum CHEM 3520 Spring 2004 Natural independent variables for various thermodynamic functions 7 U H S A G depends upon natural sets of variables that allow simple equations to connect these quantities 7 Energy Consider S and Vto be independent variables of U UUSV3dU a U dS6 U dV 6S V 6V S 7 Compare with dU TdS PdV 3 6 U Tand6 U P 6SV BVS 3 Natural variables of U are S and V 7 Criterion for spontaneous processes AU lt 0 dU lt 0 for systems at constant S and V Consider T and Vto be independent variables of U 3 dU 46 13 7P dVCVdT 6T V 7 Enthalpy Consider S and P to be independent variables of H HHSP3dHai dS6 dP 6S P 6P S 7 Compare with dH TdS VdP 3 32 Tand 31 V 6S 6PS 3 Natural variables of H are S and P 7 Criterion for spontaneous processes AH lt 0 dH lt 0 for systems at constant S and P Consider T and P to be independent variables of H 3dH V T a V dPCPdT 6T P l P 7 Entropy Rewrlte the above equatlon as dS dU dV 6S 1 6S P 3 and 6U V T 6V U T 3 Natural variables of S are U and V CHEM 3520 Spring 2004 7 Criterion for spontaneous processes AS gt 0 dS gt 0 for systems at constant U and V Rewrite the above equation as dS dH gdP 6S 1 6S V 3 and 6H P T 6P H T 3 Natural variables of S are H and P 7 Criterion for spontaneous processes AS gt 0 dS gt 0 for systems at constant H and P 7 Helmholtz energy dA SdT PdV 3 Sand P 6T V 6V T 3 Natural variables of A are T and V 7 Criterion for spontaneous processes AA lt 0 61A lt 0 for systems at constant T and V 7 Gibbs energy dG SdT VdP 3 E Sand E V 6TP 6PT 3 Natural variables of G are T and P 7 Criterion for spontaneous processes AG lt 0 dG lt 0 for systems at constant T and P 7 All the equation can be deduced quite easily dU TdS PdV Add dPV PdV VdP to both sides dU PV TdS PdVPdVVdP 3 dH TdS VdP Subtract dTS TdS SdT from both sides dH TSTdSVdP TdS SdT 3 dG 7SdT VdP Subtract dPV PdV VdP from both sides dG PV 7SdT VdP PdV VdP 3 511 7SdT PdV CHEM 3520 Spring 2004 Thermodynamic Energy Natural Variable Differential Expression Maxwell Relation U SandV dUTdS7PdV 6 T 6 13 6V S as V H SandP dHTdSVdP 6 T 6 V 6P S as P A T and V dA 7SdT 7PdV 6 13 6V 6T G TandP dG7SdTVdP g 6 V 6P 6T 7 Application the correction to the entropy due to nonideality behavior at 1 bar 7 The standard state for a gas is hypothetical ideal gas at one bar and if the gas does not behave ideally at 1 bar some corrections should be made Real gas Correction for Hypothetical ideal gas gt at 1 bar non39ideahty at 1 bar Use real gas Use ideal gas equation of state equation of state Real Ideal gas at very low pressure Pid lbar lbar 39 6V 6V SP1dS1bar J dP J dP So1barSP1d J39 56113 lbar 6T P Pid 6T P Pid P lbar 3 SO at 1 bar at 1 bar J Bid P dL M P P J where S O at 1 bar is the standard molar entropy of the gas hypothetical ideal gas and at 1 bar is the molar entropy from heat capacity data and heat of transition B T 7 Example l 2V P equation of state RT 2 6 V 5dBZV 6T P P dT dB 3 S atlbar Sat lbar 6172 X l bar P1d is neglected 7 For N2 the correction is 002 Jmol 1 K 1 CHEM 3520 Spring 2004 Temperature and pressure dependence of the Gibbs energy 7 The Gibbs 7Helmholtz equation 7 give the temperature dependence of G GH7TS 39 7S T T BGT H 1 6H 6S atconstantP 6T P T2 T 6T P 6TP Wig 0 T 6T P 6T P BGT H 3 2 6T P T2 The Gibbs 7Helmholtz equation BAG T g 6T P T 2 7 The temperature dependence of G can be determined directly 7 The case ofa gas that has only one solid phase Tfus Tvap T HT H0 ICISTdTAquH ICPTdTAvapH IC TdT 0 Tfus Tvap Tfus s Tvap 1 A H T Cg T sT j CPTdT AfusH j CPTdT Vap j P dT 0 T T Tm T T Twp T 2 Tm 170 17T 170 T T T7I70 G T is a continuous function of T AH because Ttrs AHSS 3 AHSG 0 E S so the slope is smaller for solid state 6T P than for liquid state than for gas state and is discontinuous in going from one phase to another T 7 The H T H 0 S T and G T H 0 are tabulated quantities CHEM 3520 Spring 2004 7 The pressure dependence of G P2 V 3 AG GP2T GPIT IVdP at constant T 6P T Pl Similarly lgl 17 WP Jr PZdP P P 7The ideal gas case AG RTJ RTln 2 AH 0 and AS Rln 2 Pl P P1 Pl 7 Usually P1 1 bar 3 C7TP G TRTlnP1 bar where GO T is the standard molar Gibbs energy the Gibbs energy of one mole of ideal gas at 1 bar and depends only on temperature 3 C7TP GO T increases logarithmically with P 7 For liquids and solids Vis almost constant 2 ACT 7P2T CP1T Ilzl7dP 17P2 7P1 P1 7 The nonideal gas case 1BZPTPB3PTP2 P P 6113 P P 2 j 516 RT j RTBZPT deRTB3PT deP Pid Pid P Pid Pid P RTB TP2 RTBZPTPLN Pld 2 Pid Po RTB TP2 PRTBZPPW Po 7TPCTPidRT1n But TPid G TRT1n 2 7TPG TRT1n 7 It is more convenient to rewrite this equation by de ning a new thermodynamic function called fugacity and denoted f P T 2 6TPG Tern where fPT P o exp32PTP B3PTP2


Buy Material

Are you sure you want to buy this material for

25 Karma

Buy Material

BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.


You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

Why people love StudySoup

Steve Martinelli UC Los Angeles

"There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

Jennifer McGill UCSF Med School

"Selling my MCAT study guides and notes has been a great source of side revenue while I'm in school. Some months I'm making over $500! Plus, it makes me happy knowing that I'm helping future med students with their MCAT."

Bentley McCaw University of Florida

"I was shooting for a perfect 4.0 GPA this semester. Having StudySoup as a study aid was critical to helping me achieve my goal...and I nailed it!"

Parker Thompson 500 Startups

"It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

Become an Elite Notetaker and start selling your notes online!

Refund Policy


All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email


StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here:

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.