1 / 76

PPT 107

PPT 107. PH Y SICAL CHEMI S T R Y Sem e s t er 2. Academic session 201 2 /201 3 Dr Hayder Kh . Q. Ali. CHAP T ER 1 THERMO D YNAM I CS. CHA P TER 1 THERMO D YNAMICS. C O N TEN T :. 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8. P h y si c al Che m i s try T he r mod y nam i cs

mirit
Download Presentation

PPT 107

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. PPT107 PHYSICALCHEMISTRY Semester2 Academic session 2012/2013 Dr Hayder Kh. Q. Ali

  2. CHAPTER 1 THERMODYNAMICS

  3. CHAPTER 1 THERMODYNAMICS CONTENT: 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 PhysicalChemistry Thermodynamics Temperature TheMole Ideal Gases DifferentialCalculus EquationsofState IntegralCalculus

  4. 1.1 PhysicalChemistry

  5. • What is Physical Chemistry? – Physicalchemistryisthestudyoftheunderlyingphysicalprinciples thatgovernthepropertiesandbehaviorofchemicalsystems. • WhatisChemical Systems? – Achemicalsystemcan macroscopicviewpoint. be studied from either a microscopic or a Themicroscopicviewpointisbasedonthe conceptofmolecules. Chemical Systems Themacroscopicviewpointstudieslarge-scale propertiesofmatterwithoutexplicituseofthe moleculeconcept. The firsthalf of this bookusesmainlya macroscopicviewpoint;the second halfusesmainly a microscopic viewpoint.

  6. • "microscopic"impliesdetail at the atomic or subatomiclevels whichcannot be (even witha microscope!). seendirectly • The macroscopic worldis the one we can know by directobservationsofphysicalproperties suchas mass,volume,etc.

  7. 4 Branchesof Physical Chemistry Thermodynamics Quantum Chemistry Statistical Mechanics Kinetics Thermodynamicsisa macroscopic sciencethatstudies the interrelationshipsof the variousequilibrium properties of a systemandthe changesinequilibrium propertiesinprocesses. Moleculesandthe electrons andnuclei thatcompose them do notobeyclassical mechanics.Instead, theirmotionsare governed bythelaws of quantummechanics. Applicationof quantum mechanics toatomic structure,molecular bonding, andspectroscopygives us quantumchemistry. Themolecularand macroscopiclevelsare related toeachother by thebranchof science calledstatistical mechanics. Statistical mechanics givesinsight intowhythelaws of thermodynamicshold andallowscalculation of macroscopic thermodynamic propertiesfrom molecularproperties. Kinetics is thestudyof rateprocesses such as diffusion,and electrochemicalcell. portionsof quantumchemistry, mechanics. chemicalreactions, the flowofchargeinan Kinetics uses relevant thermodynamics, andstatistical

  8. 1.2 Thermodynamics

  9. THERMODYNAMICSYSTEM Animportantconceptinthermodynamicsisthethermodynamicsystem.Athermodynamic systemisonethatinteracts andexchanges energywiththeareaaroundit(transformation of energy). A system could be as simple as a block of metal or as complex as a compartment fire. Outside the system are its surroundings. The system and its surroundingscomprisetheuniverse. Systems: Aregionoftheuniversethatwe directourattentionto. Surroundings: Everythingoutsideasystemiscalledsurroundings. Boundary: Theboundaryorwallseparatesasystemfromits surroundings. UNIVERSE

  10. Forexample,tostudythevapor pressureofwaterasa functionof temperature,wemightputa sealedcontainerofwater(with any airevacuated)ina constant- temperaturebathandconnecta manometer to thecontainerto measurethepressure.Here, the systemconsistsof theliquidwater andthewatervaporinthe container,andthesurroundings aretheconstant-temperaturebath andthemercuryinthe manometer. A key propertyin thermodynamicsis temperature, and thermodynamicsis sometimesdefined asthestudy of therelationof temperatureto the macroscopicpropertiesof matter. For example we might consider a burning fuel packageasthesystemandthecompartmentasthe surroundings.Onalargerscalewemightconsider thebuildingcontainingthefireasthesystemandthe exteriorenvironmentasthesurroundings.

  11. Energytransferisstudiedinthreetypes of systems: Opensystems Open systems can exchangeboth matterand energywithan outsidesystem.Theyare portionsoflarger systems and inintimatecontactwiththelargersystem.Your bodyis an opensystem. Closedsystems Closedsystemsexchangeenergybut not matterwithan outsidesystem.Though theyare typically portionsoflargersystems,theyare notin complete contact. The Earth is essentiallyaclosedsystem;it obtains lots ofenergyfromtheSunbuttheexchangeofmatterwith the outsideisalmostzero. Isolatedsystems Isolatedsystemscan exchangeneitherenergynormatterwith anoutsidesystem.Whiletheymay be portionsoflargersystems,theydo notcommunicatewith the outside in anyway.The physicaluniverse is an isolatedsystem;aclosedthermos bottleis essentiallyan isolatedsystem(thoughitsinsulationis notperfect). Heatcanbetransferredbetweenopensystemsandbetweenclosedsystems,butnotbetween isolatedsystems.

  12. Example

  13. Example Forexample,infigureabove,thesystem ofliquidwaterpluswatervaporin the no not sealed container is closed (since but matter can enter or leave) isolated (since it can be warmed or cooledbythesurroundingbath and by can the be compressed or expanded mercury). Athermodynamicsystemiseitheropenorclosedandiseitherisolatedornon-isolated. Mostcommonly,weshalldealwithclosedsystems

  14. WALLS Asystemmaybeseparatedfromitssurroundingsbyvariouskinds ofwalls. 1. Awallcan be either rigidornonrigid(movable). In Fig. 1.2, the system is separatedfromthebathby thecontainerwalls 2. A wall maybepermeableorimpermeable. Impermeablemeansthatit allowsnomattertopassthroughit. 3. A wall maybeadiabatic ornonadiabatic. Anadiabaticwallisonethatdoesnotconductheatatall, whereasanonadiabaticwalldoesconductheat.

  15. EQUILIBRIUM Anisolatedsystemisinequilibriumwhen its macroscopic propertiesremainconstantwith time. • • Anonisolatedsystemisinequilibriumwhenthe followingtwoconditionshold: – Thesystem’smacroscopicpropertiesremainconstantwith time; removalofthesystemfromcontactwithitssurroundings causesnochangeinthepropertiesofthesystem. – • If condition(a)holdsbut (b)does nothold, the system is in a steadystate.

  16. Types ofEquilibrium: 1. Mechanicalequilibrium • No unbalancedforcesactonor withinthesystem;hencethe systemundergoesno acceleration,andthereis no turbulencewithinthesystem. 2. Materialequilibrium • No netchemicalreactionsare occurringinthe system,noris there anynet transfer of matter from onepart of thesystemto anotherorbetweenthesystemandits surroundings;the concentrationsof thechemicalspeciesinthevarious parts of the systemareconstant intime. 3. Thermalequilibriumbetweena systemand itssurroundings • Theremust be nochangein thepropertiesofthesystemorsurroundings whenthey are separatedby athermally conductingwall. Likewise,we caninsertathermally conductingwallbetweentwopartsofa systemto test whethertheparts are inthermalequilibriumwitheachother. Forthermodynamic equilibrium,allthreekinds of equilibriummust be present.

  17. THERMODYNAMICPROPERTIES -usedto characterizeasystem inequilibrium extensive Isone whosevalueisequaltothe sumof itsvalues forthepartsof thesystem. Thus,ifwedividea system intoparts,themassofthe systemisthesumofthemasses oftheparts; mass isanextensive property.Soisvolume. intensive Is one depend system, remains whose value does not on the size of the provided the system Density ofmacroscopic. and pressure are examples of intensive properties. We can take a drop of water or a swimmingpoolfullofwater,and bothsystems willhavethesamedensity. Phase Aphaseisaregionofspace(athermodynamicsystem),throughoutwhichall physicalpropertiesofamaterialareessentiallyuniform.Examplesofphysical propertiesincludedensity,indexofrefraction,andchemicalcomposition

  18. • Extensive Parameters: – Parameterswhichvaluesfor thecompositesystemare thesum of thevaluesfor eachof thesubsystems.These parametersarenon-localin thesensethattheyreferto theentiresystem. Examplesare: Volume,internalenergy,mass, length. – • IntensiveParameters: –Theseparametersareidenticalfor eachsubsysteminto whichwemight subdivideour system. –Examplesare: Pressure,temperature,anddensity.

  19. • Homogenoussystem: –A systemis homogenouswhenit has compositionthroughout. –e.g.mixture ofgasesor true solution somechemical ofsolidin liquid. • Heterogenoussystem: –Twoor more differentphaseswhicharehomogenous butseparatedby aboundary. –e.g.Icein water.

  20. 1.3 Temperature

  21. • To determine whetheror notthermal equilibrium existsbetweensystems. • By definition,twosystemsinthermalequilibrium witheachotherhavethesametemperature;two systemsnotin thermalequilibriumhavedifferent temperatures. • Symbolized byθ (theta).

  22. The ZerothLaw Twosystems thatareeachfoundto bein thermalequilibriumwitha thirdsystem willbefoundto beinthermal equilibriumwitheachother. Itisso called becauseonly afterthefirst,second,andthird lawsof thermodynamicshadbeenformulated wasitrealized thatthezerothlawis needed for thedevelopmentof thermodynamics. Moreover,a statementof thezerothlaw logicallyprecedestheotherthree.The zerothlawallowsustoasserttheexistenceof temperatureasa state function.

  23. 1.4 The Mole

  24. Relative Atomic Mass,Ar The ratioof theaveragemassofanatomof anelement tothemassofsomechosenstandard. TheRelativeAtomicMassofachemicalelementgivesus anideaof how heavyitfeels(theforce it makes when gravitypullsonit). Therelativemassesofatomsaremeasuredusingan instrumentcalled amassspectrometer. Lookattheperiodictable,thenumberatthebottomof thesymbolistheRelativeAtomicMass(Ar): • • • •

  25. Relative Molecular Mass,Mr • Mostatoms existin molecules. • ToworkouttheRelativeMolecularMass,simplyadd up theRelativeAtomicMassesof eachatominthemolecule: • Arelativemolecularmasscanbecalculatedeasilyby adding together the relative atomic masses of the constituent atoms. For example, ethanol, CH3CH2OH, has a Mr of 46 (Try it!).

  26. Grammolecularmass Molecularmassexpressedingrams is numericallyequalto gram molecular mass substance. Molecularmassof O2= 32Gram • ofthe •

  27. Calculation of MolecularMass Molecularmassis equalto sum of the atomic • masses ofall atoms thesubstance. present in one molecule of • Example: –H2OMassof H atom= 18g –NaCl = 58.44g ThestatementthatthemolecularweightofH2Ois18.015meansthatawatermolecule hasonthe averageamassthat is18.015/12timesthe massofa12Catom.

  28. Why unitless?Findout! Remember that relativeatomic mass/relativemolecularmassisa ratio andhasnounitswhilegrammolecular massandgram atomicmassare expressedin grams.

  29. Mole Concept and Avogadro’s Number • It is convenient to consider the number of atoms needed to make 12g of carbon and for this number to be given a name - one mole of carbon atoms. • Avogadro's number and the mole are very important to the understanding of atomic structure. • The Mole is like a dozen. You can have a dozen guitars, a dozen roosters, or a dozen rocks. If you have 12 of anything then you would have what we call a dozen. The concept of the mole is just like the concept of a dozen. • You can have a mole of anything. The number associated with a mole is Avogadro's number. Avogadro's number is 602,000,000,000,000,000,000,000 (6.02 x 1023).

  30. A mole of marbles would spread over the surface of the earth, and produce a layer about 50 miles thick. A mole of sand, spread over the United States, would produce a layer 3 inches deep. A mole of dollars could not be spent at the rate of a billion dollars a day over a trillion years. This shows you just how big a mole is. • Probably the only thing you will ever have a mole of is atoms or molecules. One mole of magnesium atoms (6.02 X 1023) magnesium atoms weigh 24.3 grams. 6.02 X 1023carbon atoms weigh a total of 12.0 grams. 6.02 X1023molecules of CO2 gas only weigh a total of 44.0 grams. • The actual number of atoms that is needed to give the relative atomic mass expressed in grams is called Avogadro's number. 1023 Avogadro'snumber=6.02x 1023 1MoleCatom=6.02x Catoms=12g 1023 1MoleMgatom=6.02x Mgatoms=24.3g

  31. Example How manyatoms are there in • 24g carbon? 24gof carbon= 24/12= 2 moles 1023 1 mole ofatoms = 6.02x Therefore 2molesof carboncontains: = 1.204x 1024atoms 1023 2 x 6.02x atoms

  32. TryThis! How many atoms andmolesof siliconarein sample of siliconthat has a massof5.23g? • a • • Answers = = 0.186mol Silicon;and 1023 1.12x atoms

  33. Molar Mass,M Themole is just anumber; it canbe usedforatoms, molecules, ions, electrons, or anything elsewewish torefer to. Becauseweknowtheformulaof wateris H2O, forexample, thenwe cansay one moleof watermolecules containsone moleof oxygen atoms andtwomoles of hydrogenatoms. Onemole of hydrogenatomshas a mass of 1.008 g and 1molof oxygenatomshas a mass of 16.00 g,so 1molof waterhasa mass of (2 x 1.008g)+16.00g =18.02g. The molarmass of wateris 18.02 g/mol. • • • 1molofoxygenatomshasamassof16.00gMolarmassofO=16g/mol MolarmassofH2O=2molofH+1 molofO =(2x1.008g/molofH)+(16g/molofO) M = mass= m =18.02 g/mol mole n

  34. 1.5 IdealGases

  35. Ideal Gas Law Anidealgasisdefinedas one inwhich all collisionsbetween atoms ormoleculesare perfectlyelastic and inwhich there areno intermolecularattractiveforces.One can visualizeit asa collectionofperfectlyhardsphereswhich collidebutwhich otherwisedo not interact with each other.In such agas, all the internal energyisin the formofkineticenergyandanychangein internal energyisaccompaniedbya changeintemperature. Anideal gas canbe characterized bythree statevariables: absolute pressure(P),volume(V),andabsolutetemperature(T). Therelationship between them maybe deduced fromkinetic theoryandiscalled the Where: n=numberofmoles R=universalgasconstant=8.3145J/molK N=numberofmolecules k=Boltzmannconstant=1.38066x10-23J/K=8.617385x10-5eV/K k=R/NA NA=Avogadro'snumber=6.0221x1023

  36. The Ideal Gas Law PV = nRT P= Pressure(inkPa) V = Volume (in L) T=Temperature(in K) n = moles R =8.3145kPa • L mol • K R is constant.If we are given three of P, V, n, or T,we can solve for the unknown value. R = 82.06 cm3.atm mol. K or

  37. FortheVolume-Pressurerelationship: Boyle’sLaw • n1= n2and T1=T2therefore the n's and T's cancelin the above expressionresulting followingsimplification: in the • P1V1=P2V2 or PV =constant (mathematical expression of Boyle'sLaw)

  38. Forthe Volume-Temperaturerelationship: Charles'sLaw • n1= n2and P1 = P2 thereforethe n's and cancelin the originalexpression resulting followingsimplification: the P's in the • V1T2= V2T1 or V / T=constant (mathematical expression of Charles'sLaw)

  39. For the Pressure-TemperatureRelationship: Gay-Lussac'sLaw • n1= n2and V1 = V2thereforethe n's and the cancelin the above originalexpression: V's • P1T2= P2T1 or P / T=constant (mathematical expression of Gay Lussac'sLaw)

  40. For the Volume-Molerelationship: Avagadro's Law • P1 = P2 and T1 = T2thereforethe P's and T's cancelin the above originalexpression: • V1n2=V2n1 or V / n = constant (mathematical expression ofAvagadro'sLaw)

  41. Boyle’sLaw At constanttemperature, the volumeofa givenquantityofgasis inversely proportionalto itspressure: V 1/P Soatconstanttemperature,ifthevolumeofagasisdoubled,itspressureishalved. OR Atconstanttemperatureforagivenquantityofgas,theproductofitsvolumeandits pressureisaconstant: PV= constant,PV=k • • Atconstanttemperatureforagivenquantityofgas:PiVi=PfVf wherePiistheinitial(original)pressure,Viisitsinitial(original)volume,Pfisitsfinal pressure,Vfisitsfinalvolume Pi andPfmustbe inthe sameunits ofmeasurement(eg,bothin atmospheres),Vi andVfmustbeinthesameunitsofmeasurement(eg,bothinlitres). AllgasesapproximateBoyle'sLawathightemperaturesandlowpressures.A hypotheticalgaswhichobeysBoyle'sLawatalltemperaturesandpressuresiscalled anIdealGas.ARealGasisonewhichapproachesBoyle'sLawbehaviourasthe temperatureisraisedor the pressurelowered. • •

  42. Boyle’sLaw P1V1=P2V2

  43. CharlesLaw Atconstantpressure,thevolumeofa given quantityofgasisdirectlyproportional tothe absolute temperature: V T (inKelvin) So atconstantpressure,if thetemperature(K) is doubled,thevolumeofgasis alsodoubled. OR Atconstantpressurefora givenquantityofgas,theratioofitsvolumeandtheabsolute temperatureis a constant: V/T= constant,V/T = k Atconstantpressurefora givenquantityofgas : Vi/Ti=Vf/Tf whereViis theinitial(original)volume,Tiis itsinitial(original)temperature(inKelvin),Vfis its final volme,Tfisitsfinaltempeature(inKelvin) ViandVf mustbeinthesame unitsofmeasurement(eg,bothinlitres),TiandTfmustbein KelvinNOTcelsius. temperatureinkelvin= temperatureincelsius+ 273 (approximately) AllgasesapproximateCharles'Lawathightemperaturesandlowpressures.Ahypotheticalgas whichobeysCharles'Law atalltemperaturesandpressures iscalledanIdealGas.A RealGasis onewhichapproaches Charles'Lawasthetemperatureis raisedorthepressurelowered. As a RealGasis cooledatconstantpressurefroma pointwellaboveitscondensationpoint,its volumebeginsto increaselinearly.As thetemperatureapproaches thegasescondensation point,thelinebeginsto curve(usuallydownward)sothereisa markeddeviationfromIdeal Gas behaviourcloseto thecondensationpoint.Once thegascondensesto a liquiditisno longera gasandso doesnotobey Charles'Law atall. Absolutezero(0K,-273oCapproximately)isthetemperatureatwhichthevolumeofa gas wouldbecome zeroif itdidnotcondenseandif itbehavedideallydownto that temperature. • • • • •

  44. CharlesLaw V1/V2=T1/T2 P1V1/T1=P2V2/T2

  45. Pressure PRESSURE P(Pressure)=F(Force) and Volume Units VOLUME 1L=1dm3=1000cm 3 A(Area) ATMOSPHERE 1 atm=760torr=1.01325x105Pa InSI: 1Pa(Pascal)=1N/m 2 Chemistsuse: 2 1torr=133.322Paor 2 =133.322N/mor =133.322kg/ms 5 1bar=10Pa=0.986923atm=750torr

  46. Example1.1: Density of an IdealGas Page16 • Findthedensity of F2gas at 20.0°Cand 188torr.

More Related