hermodynamcs and Gases Last tme Knetc heory o Gases or smple (monatomc) gases Atomc nature o matter Demonstrate deal gas law Atomc knetc energy nternal energy Mean ree path and velocty dstrbutons From ormula or E nt, can get specc heats Specc Heats o Smplest Gases Constant olume Constant Pressure oday Specc Heats or more complex gases Adabatc Expanson Entropy slde 1
revew Internal Energy rom Atomc Nature pnr understood rom atomc nature o matter pnk s equvalent orm Both are generally applcable (up to small van der Waals correctons) or all gases p knetc energy o atoms Internal energy o the gas s a sum o all the energy orms (ncludng knetc energy) o the molecules smplest s monatomc gas (one atom n the molecule, rotatonally symmetrc) -> energy all translatonal real world: coecent, 3/, only apples to noble gases ANY gas p Nk nr K atom atom 3 3 k E N K Nk nt monatomc gas E nt nr 3 p Nk nr E atom C R Ent NEatom N k E nt nc k C R slde
revew Specc Heat at Constant olume (sochorc) 1st Law o herm. de dq -dw nt monatomc gas E nt 3 nr No change n volume mples no work done: dw 0 Heat ntroduced proportonal to temperature change when no work Q n C D Snce dw0, then the heat added must equal the change n nternal energy DE nt Q n C D And we predct: Monatomc (bllard ball) gases have C 3R/ E any nt gas nc slde 3
revew Specc Heat at Constant Pressure (sobarc process) For process shown (n xed) Q n C P D Here, as expanson occurs, work s done and nternal energy ncreases Q D E + W P nc D + p D nc D nt ( ) C C + R + nrd Q nc + R D slde 4
How about molecules that are not monatomc? We ound that sphere-lke molecules (sngle nert atoms) had C 3R/ what about the rest? Came rom consderatons o knetc energy (E nt nc ) dervng gas law Our sngle atoms had only one knd o avalable energy Knetc energy rom translaton n 3 dmensons More generally molecules can have other degrees o reedom rotatons vbratons,,, Maxwell: molecule E ( k ) nt 1 #deg o reedom slde 5
Molecular Specc Heats molecule E ( k ) nt 1 #deg o reedom derent ndependent ways that molecule can contan and exchange energy Need enough thermal energy to excte the modes to be seen (Quantum Mechancs) g - H gas C /R nt nt gas 1 E n( R ) C E P C any nc 1 R R( + ) 1 slde 6
Emprcal check Real Gas at low pressures C v C p g C p / C v ranslatonal Only 3 3R/ 5R/ + Rotatonal datomc + Rotatonal polyatomc 5R/ 7R/ 1 3R 4R 5 3 1.67 7 5 1.40 4 3 1.33 C any gas # actve d.. 1 R C R( + 1) P Cp g 1 + C slde 7
Asde: How about Solds? Generally expect C v R/ Whle gases are unbound, solds have atoms bound to adjacent atoms no net translatons n 3D space but there does exst bratonal knetc energy -- 3 deg o reedom potental energy (o bndng) -- 3 deg o reedom Recall table 19-3: many solds tend to have molar heat capacty C ~ 5 J/mol-K Navely expect 6 and C v 3R ~3(8.31 J/mol-K ) hs value approached n most solds as temperatures rse slde 8
Adabatc Expanson o Gas Adabatc no heat enters or leaves system (gas) Q0 But temperature changes as gas does work p nr pd + dp nrd nd pd + dp pd + dp R ( C C ) P de de nt nt nc d nd dq dw dw pd pd C pd + dp pd c c c P slde 9
Adabatc Expanson - algebra pd + dp pd c c c P 1 1 dp pd + c c c c c P P c cp + pd 0 c P pd c ( c P c ) c P c dp dp d C + g 0 g p C P dp Derental equaton states that ractonal change n pressure plus g tmes ractonal change n volume must equal zero n any adabatc expanson slde 10
Adabatc Expanson - calculus dp d + g 0 g p lnp + g ln const. g ln( p ) const. g g p g p 1 1 Cp 1 + C ( + 1) R 1 + R g R C C C P p g nr Ł ł const g -1 g -1 1 1 co nst Conclude that adabatc change (no heat enter or leave) requres new relatons among parameters rato (g ) spec. heats mportant Ideal gas law also stll true g slde 11
revew Gases: Isothermal Expanson p Unts nr Isotherms p nr const R 8.31 J/(mol-K) kn A Work (constant temperature) done obtaned rom ntegral n p- (see sample prob 0-1) nr ln Ø Œ ø œ W pd d nr º ß slde 1
Reversblty All these (sothermal, sobarc, sochorc, adabatc) are reversble processes each pont on P, dagram s a possble state state can be changed n any drecton along a curve Contrast ree expanson o gas not reversble p p but state cannot return to state slde 13
Sample Problem 0-9a Recall sample problem 0-, sothermal expanson o 1 mole O at 310K & p.0 atm rom 1 lters to 19 lters gave P 1.6 atm & WQ1180 J. Now adabatc expanson o same sample. Fnal pressure? How much work done? p p p 1 p Ł 1 ł g 1 0. Ł 19 ł 105. atm Q W W 0 140. E 1 Ł ł 1 nc g -1 1 310 Ł 19 ł 58K D -5.K 0 nt 040. (1)(0.85)( 5.0) 1085 Joules slde 14
All nvolve expanson o gas at atm, 310K, and 1lters to a volume 19lters n derent ways Free Expanson Contrast Gas Expansons (sample 0- & 0-9) arable Isothermal Adabatc Free P 1.0 atm.0 atm.0 atm 1 1 L 1 L 1 L 1 310 K 310 K 310 K P 1.6 atm 1.05 atm 1.6 atm 19 L 19 L 19 L 310 K 58 K 310 K W 1180 J 1085 J 0 J reversble rreversble Check you get the answers or these Bg derence or rreversble: ENROPY slde 15
S S S Q Entropy Denton or dq xed precse quanttatve denton measurng heat low energy at specc temperature (SQ/, unts J/K) Sgn conventon: heat added (DQ>0), then entropy ncreases (DS>0) smplest underlyng concept underlyng thermodynamcs: Closed system total entropy change zero or postve Reversble process: an entropy ncrease s compensated by an entropy decrease n another part o system entropy recoverable Irreversble: entropy ncreases and cannot be recovered can calculate entropy change or rreversble process by connectng ntal & nal states wth (seres o) reversble processes slde 16
Second Law o hermodynamcs Smplest (and smplstc) expresson: Heat energy always lows rom hotter bodes to cooler bodes ntutvely obvous ake a more complete statement (though more abstract) wth concept o entropy In any closed system, the entropy always ncreases or rreversble processes and s constant or reversble processes. Entropy never decreases n closed systems, and typcally ncreases n real processes! Heat low rom hot to cold s a consequence. Found to be provde expermentally correct predctons n wde array o crcumstances Has a physcal nterpretaton n terms o the system s order Provdes an understandng o where our sense o the arrow o tme arses. slde 17
reversble sothermal expanson Reversblty: sothermal expanson o gas pd D S Q W nr d D S + nr gas expanson Entropy ncrease o the gas durng expanson occurs as a consequence o heat (Q) nto gas at temp An equal and opposte decrease n the entropy o the reservor occurs at the same tme I the gas s sothermally compressed, the gas entropy decreases and the reservor entropy ncreases D S -nr gas compresson ln ln slde 18
Isochorc temperature change o gas Add heat wth dong work by ncreasng reservor temp emperature change o gas n contact wth temp reservor s also reversble D D nt D S nc Q E d D S + nc ln D S + nc ln -DS slde 19
Entropy (more) Entropy s a physcal measurable property o a system (lke,, p, ) that s calculable n terms o the others state uncton How does one calculate entropy change when the net entropy ncreases? (n an rreversble process lke ree expanson o a gas) Answer: Calculate entropy change or rreversble process by connectng ntal & nal states wth (magned seres o) reversble processes. Examples Smple heat low between two solds. Free expanson. slde 0
Entropy Change n Irreversble Process DQ DQ D - > 350 650 t 0 t 0 S t 0 0 Note that entropy wll ncrease so long as heat lows rom hot body to cold body. he reverse (cold to hot) s not possble rom nd Law. Smple heat low rom hot body to cold s rreversble Calculate entropy change n rreversble process: connect ntal & nal states wth seres o pretend reversble processes slde 1
Entropy change: solds Net entropy change o any process o a closed system s ether zero (reversble) or greater than zero (rreversble) Example sample problem 1-: equal mass (m) solds at derent temps equlbrum when both at nal temperature what s entropy chg? dq d D S mc mc ln Ø ø D Srev D SL +D SR mc Œln + ln L œ º R ß D S rev D S DS rev mc ln rrev L R slde
60 0 C 0 0 C D S D S +DS L Ø mc Œln + Sample Problem 1- R ln 40 0 C ø DS Equal masses: œ 1 º L R ß [ L + R ] D S m 1.5 kg c 386 J/kg-K rev DS rev mc ln DS rrev L R 0 40 C (.kg)( 15 386J/kgK)ln Ł DS 37. J/ K ( 93) ( 333) Same net entropy ncrease or the two masses Lost n rreversble process Stored n reservor or reversble process 313 ł slde 3
rreversble n a closed system Entropy ncrease o ree expanson o gas reversble sothermal equvalent Q W nr D S ln Derved prevously or sothermal expanson o the gas Snce,n the two processes (reversble and rreversble), the gases have same ntal states and same nal states the entropy change o the gas s the same or both processes Note that ree expanson process s rreversble Note that net entropy o closed system n sothermal expanson s zero DS nr ln slde 4
nd Law o hermodynamcs: gases Net entropy change o any process o a closed system s ether zero (reversble) or greater than zero (rreversble) dscusson secton 1- Entropy s a state uncton; depends only on parameters o the system: calculate general change or gas dq de + dw dq nc d + pd rev D S S -S D S DS nt dq rrev D S 0 entropy change or deal gas dq d d D S nc + nr D + S nc ln nr ln slde 5
hermodynamcs and Gases oday Specc Heats o Smplest Gases Constant olume (sochorc) Constant Pressure (sobarc) Specc Heats more generally Adabatc Expanson Reversble and Irreversble Processes Entropy More on Entropy Next Uses n engnes, heat pumps, rergerators Orgn o Entropy slde 6