Wrk and eat Deinitins FL- Surrundings: Everything utside system + q -q + System: he part S the rld e are bserving. Wrk, : transer energy as a result unbalanced rces - eat, q: transer energy resulting rm a temperature dierence eat Sign cnventin: Psitive heat is input t the system Negative heat is utput rm system Wrk Sign cnventin: Psitive rk dne n the system Negative rk dne by the system
Expansin and cmpressin FL- Figure 9. EXPANSION M h M OMPRESSION M Remve pins Remve pins h M Questins. What is system? Surrundings?. des initial and inal pressure crrespnd t P ext? 3. What is sign rk r bth cases? 4. Exactly h much rk has been dne? Mgh Mg A Ah P Sign Δ?? rce*distance pressure*vlume (rce/area)*(area*height) extδ
EX-FL What i P ext is nt cnstant? FL-3 I P ext is nt cnstant during the expansin (r cmpressin), the rk is the integral ver the path rm i t and e need t kn h P ext varies ith : i P ext d General expressin. P ext is a unctin.
Wrk is the area under P ext vs... FL-4 nsider an isthermal cmpressin at cnstant pressure, P ext P P ext P < P ext P P P ext Δ P ext Δ he curve is r an ideal gas, at cnstant : P nr he rk is equal t the shaded area: nte h it depends n P ext. Figure 9.
Reversible isthermal cmpressin FL-5 Wrk depends n the path taken rm t. Fr cmpressin, the abslute minimum rk is dne alng the reversible path. Reversible path: At every ininitesimal step P ext is made ininitesimally larger than P. At every step, P ext is equal t the equilibrium gas pressure,. nr P gas rev rev P gas nr d d Figure 9.3 rev nr ln Is rk psitive r negative? mpressin: ln < 0 +
Reversible isthermal expressin EX-FL FL-6 Wrk depends n the path taken rm t. Fr expansin, the abslute maximum rk is dne n surrundings alng the reversible path. nr ln rev (Ideal gas) Is rk psitive r negative? Expansin: ln > 0 - Figure 9.3
State Functins vs Path Functins FL-7 As e ve seen, the rk depends n the path taken beteen initial and inal state. Wrk and heat are path unctins. State unctins dn t depend n the path taken but nly upn the state the system. Energy, U r, are state unctins. Why is this imprtant??? (See Math hapter r mre detail) he dierentials path unctins are inexact and can t be integrated nrmally! he dierentials state unctins are exact and can be integrated nrmally! State unctins du U U ΔU Path unctins δ δ (nt Δ r ) q q (nt Δq r q q )
he First La FL-8 he First La hermdynamics: Energy is nserved. du q δ + δ Dierential Frm ΔU q + Integral Frm Even thugh δq and δ are path unctins (inexact dierentials), their sum is a state unctin (exact dierential).
Let s explre state vs. path mre deeply! FL-9 3 reversible paths t the same place P,, P,, Path A: Reversible isthermal expansin Path B+: Reversible adiabatic expansin lled by heating at cnstant vlume Path D+E: Reversible cnstant-pressure expansin lled by cling at cnstant vlume All three paths are reversible, but ill they all invlve the same rk? ΔU?
Path A: Reversible Isthermal Expansin FL-0 P,, P,, A he energy an ideal gas depends nly n the temperature Recall U (3/)R du A 0 du δ q + δ δ rev, A δqrev, A Since the prcess is reversible δ rev, A Pgasd nr d rev, A nr ln q rev, A nr ln
Path B: Reversible Adiabatic Expansin FL- P,, P 3,, Adiabatic: N energy transerred as heat. S q 0. du δ Recall rm BZ/PFIG slides: We can get rm du ( ) Since ideal gas U depends nly n : Put it all tgether rev, B ΔU B U ( ) ( B du d du ( ) d ) d Frm t r
Path : eat at nstant FL- P 3,, P,, nstant vlume NO P Wrk!! Δ U qrev, + rev, qrev, We need t ind q rev, and ΔU q rev, ΔU ( ) d Frm t Path B + Path ΔU B + ΔU rev, B + ( ) d ( ) d rev, ( ) d 0
Paths D and E FL-3 Path D: nstant pressure expansin P,, P,, 3 ΔU q D 3 ( ) d P ( ) rev, D rev, D ΔU D rev, D Path E: ling at cnstant 3 D ( ) d + P ( ) E P,, 3 P,, 0 ΔU E q 3 rev, E ( ) d ΔU E rev, E rev, E 3 ( ) d
Path D + E FL-4 D E q rev rev q + q P ( ), D+ E rev, D rev, E + P ( ), D+ E rev, D rev, E Δ U q + D + E 0
Summary q,, ΔU FL-5 P ( ) rev, D+ E q rev, D+ E P ( ) Δ U rev +E, D 0 rev, A nr ln rev, A nr ln q ΔU A 0 rev, B+ ( ) d qrev, B+ ( ) d Δ U rev, B+ 0 ΔU, state unctin, is same r all paths but q rev and rev, path unctins, dier based n path.
Adiabatic Expansin OOL FL-6 Adiabatic s q 0 δ and du ( ) d du δ (nte that i either q 0 r 0 then the remaining dierential becmes exact) δ d Fr an ideal gas and reversible expansin: Putting them tgether nr d Pd nr ( ) d d d R I e kn h depends n, e can take bth integrals. Fr ideal mnatmic gas, 3R / B d d
Adiabatic vs. Isthermal Expansin FL-7 Adiabatic expansin (ideal mnatmic gas): 3/ P nr P P 3/ 5/3 P 5/3 P mpare ith Byle s La r isthermal prcesses: P P
hermchemistry: Δ and ΔU FL-8 Recall: Δ q p, ΔU q, U + P Δ ΔU + PΔ + ΔP ΔU + PΔ Melt ice at 0 and atm given q p 6.0 kj/ml Δ q p 6.0 kj/ml Given the mlar vlumes, slid: liquid: s l 0.096 L/ml 0.080 L/ml ΔU alculate L bar 00 J L atm 0.3 J
hermchemistry: Δ and ΔU FL-9 nsider vaprizatin (biling) ater at 00 and atm. Given: q p 40.7 kj/ml, 0.080 L/ml, l g 30.6 L/ml Δ q p 40.7 kj/ml alculate ΔU ΔU Δ PΔ ΔU q the heat t vercme the intermlecular rces hlding ater tgether.
q p hermchemistry (cn t) We ill cus n since mst chemistry is dne at cnstant P nsider the absrptin and evlutin energy (heat) assciated ith chemical reactins. Δ r < 0 exthermic (ex ut) Δ Release energy as heat. eat is ne the prducts. r reactants prducts enthalpy prd Figure 9.8 Recall dems rm last eek?! react prducts reactants q p Δ r FL-0 > 0 endthermic (end in) Absrb energy as heat. eat must be supplied t drive the reactin.
able 9. Subscripts We, in class and text, adpt the lling cnventins: FL- Reactin aprizatin, evapratin Sublimatin Melting, usin ransitin beteen phases Mixing luids Adsrptin mbustin Frmatin General reactin Subscript vap sub us trs mix ad c r
Δ is additive: ess s La FL- Δ is a state unctin, and this means it is an additive prperty I e kn () and (), () () ( s) O ( g ) + O + O ( g ) ( g ) O O ( g ) ( g ) Δ ( ) 0.5 r ( ) 83. 0 Δ r kj kj We can add them t ind (3) (3) + O O ( s) ( g ) ( g ) EX-FL 3&4
Setup r EX-FL 3 and 4 FL-3 EX-FL3 Given: P( s) + 3l( g ) Pl3( l) Δ 640 r kj Find: P( s) + 5l( g ) Pl5( s) Pl + l Pl 3( l) ( g ) 5( s) Δ 887 r Δ r kj EX-FL4 Given: Find: Fe + 3 O Fe O ( s) ( g ) 3( s) 3Fe( s) + O( g ) Fe3O4( s) 4FeO3( s) + Fe( s) 3Fe3O4( s) Δ 06 r Δ 36 r Δ r kj kj
Δ r he standard reactin enthaply FL-4 is extensive: it depends n the number mles reactants acilitate the tabulatin reactin enthalpies IUPA has prpsed use the standard reactin enthaply (intensive) Δ r Deined as: ne mle reagent and all reactants and prducts in their standard states (r a gas this is ne bar at the temperature interest) Fr example: + O O ( s) ( g ) ( g ) (ne mle is cmbusted) Δ r 393.5 (Intensive) kj ml - Use t get: ( s) + O( g ) O( g )
Standard mlar enthalpy rmatin FL-5 he enthalpy rmatin ne mle rm the cnstituent elements is the standard mlar enthalpy rmatin (intensive) Δ this means all reactants and prducts in their standard states ( g ) + O ( g ) O ( l) Δ 85.8 kj ml - at 98.5 K Standard states at bar and 98.5 K One mle O (l) is 85.8 kj dnhill in enthalpy rm the cnstituent elements
Δ r elements a big zer FL-6 tabulate values r Δ the values Δ r each pure element in its stable rm at ne bar and the temperature interest is set t zer. At 5 : (g) O (g) l (g) Br (g) I (g) (diamnd) Δ 0 kj ml - Δ 0 kj ml - Δ 0 kj ml - Δ 30.907 kj ml - Δ 6.438 kj ml - Δ.897 kj ml - See able 9. r mre Why nt zer?
Using Δ t get Δ r FL-7 aa + bb yy + zz aδ [A] Δ [B] yδ [Y ] b zδ [Z] # mles per mle Δ r Δ ( prducts) Δ ( reacants) Δ r ( ) ( yδ [ Y ] + zδ [ Z] aδ [ A] + bδ [ B] )
Which ne d yu expect t be larger, r P? Why? EX-FL5 eat apacity FL-8 eat capacity is a path unctin. Fr example, the amunt energy required t raise the temperature a substance ne degree is dierent i dne at cnstant r cnstant P At cnstant, the energy added as heat is q (ΔU q ) U ΔU Δ q Δ At cnstant P, the energy added as heat is q P (Δ q P )
p and emperature Dependence Δ FL-9 We can calculate the dierence in enthalpy at t dierent temperatures rm the heat capacity at thse temperatures: P P Integrate ( ) ( ) P ( ) d Prblem: Phase transitins!!! At phase transitins, P. hat is, there is n change in temp as yu add heat. hus, the enthalpy assciated ith the phase transitin must be added in ( ) (0) us s ( ) d + Δ + P us 0 us l P ( ) d
Benzene FL-30 Benzene: us 78.7 K, vap 353. K Figs 9.5 and 9.6 slid liquid gas liquid gas slid Fr > vap,
Δ r at dierent FL-3 I yu kn Δ r at ne temp,, and ant t kn it at anther,, e can use the heat capacities: Reactants Δ r ( ) Want t kn Prducts Δ r ( ) Knn: Path
Summary FL-3 Energy is cnserved. du δ q + δ U and are state unctins. q,, P and are path unctins. here are many ays t discuss P rk. Understanding the heat cnsumed r evlved can prvide perul insight int chemical reactins. We can use the act that U and are state unctins t tabulate and calculate thermchemical values. NEX: Energy is nt enugh t predict the directin a spntaneus prcess (reactin) the rld preers disrder