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ЊS №о № ƒ №0€„М™ƒПРџ №`А`€№У ™ №fŸ *ЁbјЊ  №р № ƒ №0€ŒЬ™ƒПРџ №`Аа€№У  ™ №hŸ *ЁbњЊ  №р № ƒ №0€\б™ƒПРџ №` а€№У ™ №hŸ *ЁbиЊ  №H № ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВ КDefault Design№.ё€ 0 Ў№І №Р№>№( № №Р№Ь №Р ƒ №0€xЦƒПРџ №69№У  Ц №TŸ *Ё љЊ №ь №Р ƒ №0€аш!ƒПРџ №l Ђ9№У Ц №tŸ *Ё&b  јЊ  №d №Р c №$‡Пџ ?№ж4o №У Ц№ №Р ƒ №0€x№!ƒПРџ №Ÿ ЊљЁ№У Ц №ˆŸ 8USћQdkY‘ЭkHr‡e,g7h_ ,{ŒNЇ~ ,{ NЇ~ ,{лVЇ~ ,{”NЇ~Ђ Њ №р №Р “ №6€lч!‡ƒПРџ №=6v№У  Ц №bŸ *Ё њЊ  №т №Р “ №6€|#Ц‡ƒПРџ №=l Ђv№У Ц №dŸ *Ё иЊ  №H №Р ƒ №0ƒ“™Cg”ПЮ—Пџ ?№ џџџ€€€Лру33™™™™Ьˆ8Š0К___PPT10‹ы.ePЪљŸМЩЊ :№2Р№Ф№Ъ№( № №Ф№Ь №Ф ƒ №0€ЄDЦƒПРџ №69№У  Ц №TŸ *Ё љЊ №ь №Ф ƒ №0€ЦƒПРџ №l Ђ9№У Ц №tŸ *Ё&b  јЊ  №р №Ф “ №6€l™!‡ƒПРџ №=6v№У  Ц №bŸ *Ё њЊ  №т №Ф “ №6€SЦ‡ƒПРџ №=l Ђv№У Ц №dŸ *Ё иЊ  №H №Ф ƒ №0ƒ“™Cg”ПЮ—Пџ ?№ џџџ€€€Лру33™™™™Ьˆ8Š0К___PPT10‹ы.ePЪ№aУЧюcя€ 0 № 0№№Ѓ№( № №№Ѓ № “ №6€јHљПƒПРџ№€А~ №Žˆ†Š4К___PPT10‹Б ŠBК___PPT9‹$Ќ №ЇŸЈлThermodynamic Equations   (a)Heat, Work and the 1st Law   PV=nRT equation of state for ideal (perfect) gas   work done against external pressure   work done by ideal gas in isothermal reversible expansion/compression Ё‚Ф ffbb bjbb bb f*bb bb $bb bb bfџ3ўbb `Њ&=  œ №\В № S № AС ??№ш : № С №\В № S № AС ??№ `к № С №H № ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВюDя€ 0 є№ь@№№„№( № №№X № “ №6€L†љПƒПРџ№€  r №Žˆ†Š4К___PPT10‹Б ŠBК___PPT9‹$Ќ №\ŸЈК  First Law of Thermodynamics   definition of enthalpy   change in enthalpy at constant pressure   molar heat capacity at constant pressure   molar heat capacity at constant volume   ЁfЙ bb bb bb bb bb (bb bb (bb bb &bb bb b`Њ Й №\В № S № AС ??№pP €H№ С №\В № S № AС ??№€p hX№ С №\В № S № AС ??№PР 6š№ С №\В № S № AС ??№0 РVЮ № С №\В № S № AС  ??№@ &о № С  №H № ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВюзя€ 0 ‡№P№ №№( № № №\В №  S № A С  ??№` шB № С  №\В №  S № A С  ??№ @H№ С  №Ђ № ƒ №0€ љПƒПРџ№Аа №ЇŸЈ'Molar internal increment from T1 and T2ЁP( 2dlчџdlчџЊ ( №Ђ № ƒ №0€РІљПƒПРџ№€€  №ІŸЈ&Molar enthalpy increment from T1 to T2ЁP' 2dlчџdlчџЊ ' №њ № Г №B€И­љ…‡ПƒПРџ№А рна №ˆŸЈ<Q: How to calculate enthalpy change for ideal gas expansion?Ё= 2=dџ3ўЊ = №H № ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВю8я€ 0 ш№р`№(№x№( № №(№” №( “ №6€ОљПƒПРџ№‡€№Žˆ†Š4К___PPT10‹Б ŠBК___PPT9‹$Ќ №˜ŸЈŽ(b) Thermochemistry   Standard enthalpy of reaction from standard enthalpies of formation      Kirchhoff equation     For H2+1/2O2=H2O ЁІ{ bb bb Dbb bb bb bb bb bb bb dџ3ўlџ3ўчџdџ3ўlџ3ўчџdџ3ўlџ3ўчџdџ3ў`Њ@  C  $ №\В №( S № A С ??№@Ов№ С №\В №( S № AС ??№€  b № С №\В №( S № AС (??№А аJ № С (№xВ №( Ѓ №<A +?ƒПРџ?№а ˜ ш№ С +№H №( ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВю”я€ 0 D№<p№,№д№( № №,№ш №, “ №6€XољПƒПРџ№эР€ы №~ˆvŠ0К___PPT10‹БŠ6К___PPT9‹Ќ №ќŸЈl  Љ 2nd Law and Entropy Changes     2nd law   where   entropy change for isothermal adsorption of heat  ЁTk bb bjbb bb bb bjbb bb bb 2bb bb Њ k №\В №, S № AС ??№№ˆ № С №\В №, S № AС ??№€ €: p № С №фЂ №, ƒ №0€ ёљПƒПРџ№№ pі №„ŸЈ8Q: How to calculate the entropy of phase transformation?Ё9 29dЊ 9 №H №, ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВюpя€ 0  №€№”№А№( № №”№А №” “ №6€ ПƒПРџ№  `њ №Žˆ†Š4К___PPT10‹Б ŠBК___PPT9‹$Ќ №ДŸЈf Entropy change on heating from T1 to T2 at constant pressure.   if Cp is constant from T1 to T2. Ёg 2!bjтџbjтџbb bb bjчџbjтџbjтџbb `Њ$ e  №\В №” S № A)С "??№€РН b№ С "№\В №” S № AС #??№ Аc № С #№H №” ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВю#я€ 0 г№Ы€№0№c№( № №0№п №0 “ №6€ј ПƒПРџ№№Р№ №~ˆvŠ0К___PPT10‹БŠ6К___PPT9‹Ќ №ѓŸЈO Molar entropy change for isothermal expansion/compression of an ideal gas   Ё\K Jbb bb bЊ$ N  №\В №0 S № AС ??№РА`Б№ С №\В №0 S № A С r??№` Pх № С r№| №0 “ №6€€иљПƒПРџ№P0ж]№~ˆvŠ0К___PPT10‹БŠ6К___PPT9‹Ќ №ŸЈ0e.g. 10 moles ideal gas expanded from 1 to 10 m3Ё01/fnЊ 1 №H №0 ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВюЂя€ 0 R№J№€№т№( № №€№т №€ “ №6€м" ПƒПРџ№NАрЅ №Žˆ†Š4К___PPT10‹Б ŠBК___PPT9‹$Ќ №цŸЈЪEntropy change for random mixing of nA moles of A and nB moles of B   Entropy change for formation of one mole of an binary ideal gas mixture, or an ideal liquid solution or an ideal solid solution. ЁЌЫ 2%bjтџbjтџ bb bb ƒbb `ЊL#    ’  №\В №€ S № AС p??№А 0С№ С p№\В №€ S № AС q??№` `№Ќ№ С q№H №€ ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВюЈя€ 0 X№P №№ш№( № №№А № “ №6€X7 ПƒПРџ№Ар R №Žˆ†Š4К___PPT10‹Б ŠBК___PPT9‹$Ќ №ДŸ ЌCalculate the change in entropy when a block of iron (heat capacity=0.48J.g-1.K-1) of 500g mass at 473K is brought into contact 1000 g of water at 293K after thermal equilibrium is reached. Tf= 302.7K D№S=29.7JK-1ЁОзKbjbjlbbjчџ b тъjЊ&Р   №H № ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВюя€ 0 Ж№Ўр№№F№( № №№ № “ №6€Иs ПƒПРџ№рР@№~ˆvŠ0К___PPT10‹БŠ6К___PPT9‹Ќ №"Ÿ fEntropy and Probability   Background       З       Classical approach to thermodynamics.     З      What entropy  means     З      What is happening in terms of the atoms and molecules when we change the state of the system   З   D№S>0, there is always an increase in  randomness or  disorder .   З    Ёd fb b `b b c b'bb b b b b b c bbb b b b b c b\bb b b b cbт@bт т т у ттЊ&/   Іџ№H № ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВюDя€ 0 є№ь№№8№„№( № №8№„ №8 “ №6€T— ПƒПРџ№ `ї№Žˆ†Š4К___PPT10‹Б ŠBК___PPT9‹$Ќ №ˆŸЈ–  1.     Expansion of a gas at constant T     e.g. 10 moles ideal gas expanded from 1 to 10 m3     2.     Compression of gas at constant T     Ёh• bb b bb bb bb /bjb bb bb b bb bb bb `ЊX +  5  -  Іџ№\В №8 S № A С 0??№PP P№ С 0№\В №8 S № A С 1??№Р аz^ № С 1№H №8 ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВюѕя€ 0 Ѕ№№<№5№( № №<№5 №< “ №6€˜З ПƒПРџ№ €и №Žˆ†Š4К___PPT10‹Б ŠBК___PPT9‹$Ќ №9ŸЈ•  1.     Heat up a solid, liquid or gas at constant P   If Cp=const     e.g. Heat 2 moles of Al (~54g) from 298K to 373K, Cp(Al,s)=24.36JK-1mol-1.   ЁN” bb b,bb bb bb bb bb Bbjbjbb bb `Њ$ “  Іџ№\В №< S № AС 2??№0v а№ С 2№\В №< S № AС 3??№p 0˜ № С 3№H №< ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВю}я€ 0 -№%№@№Н№( № №@№Н №@ “ №6€xж ПƒПРџ№P€5 №Žˆ†Š4К___PPT10‹Б ŠBК___PPT9‹$Ќ №СŸЈG  1.     Mixing gases     e.g. mix 1 mole of O2 with 3 moles of N2     Ё$F bb b bb bb bb bjтџbjтџb bb bb `Њ$ E  Іџ№\В №@ S № AС 4??№№ Bі№ С 4№\В №@ S № AС 5??№ 0І. № С 5№H №@ ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВюшя€ 0 ˜№ №D№(№( № №D№ж №D “ №6€$ПƒПРџ№Ра №Žˆ†Š4К___PPT10‹Б ŠBК___PPT9‹$Ќ №кŸЈd  .   ==Enthalpy of Fusion   e.g. melt 10 moles of Pb at its melting point 600K, =4812 J.mole-1   Ёc bb bb bbb bb bb @bjb bb `Њ> 1  -  Іџ№\В №D S № AС 6??№А№ "2№ С 6№\В №D S № AС 7??№а f№ С 7№\В №D S № AС 8??№p€ЦК № С 8№\В №D S № AС 9??№№ PЈŽ№ С 9№‚Ђ №D ƒ №0€ј ПƒПРџ№Р00&№Žˆ†Š4К___PPT10‹Б ŠBК___PPT9‹$Ќ №ŒŸЈMelting solid ЁB  2 bb `Њ  №H №D ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВюея€ 0 …№}P№H№№( № №H№‰ №H “ №6€дПƒПРџ№`€Й№Žˆ†Š4К___PPT10‹Б ŠBК___PPT9‹$Ќ №ŸЈ[1.     Oxidise a metal to its oxide   e.g. oxidise 2 moles of Al(s) to Al2O3(s) at 298K    Ё$8 b bff bb #bjтџbjтџ bb bb b`Њ \ №\В №H S № AС :??№ АК ј № С :№\В №H S № A С ;??№@ `о № С ;№„ №H Г №B€to …‡ПƒПРџ№€ @г№~ˆvŠ0К___PPT10‹БŠ6К___PPT9‹Ќ №ŒŸЈ&=50.99-2(28.33)-3/2(205.0)=-313.2JK-1.Ё6'#bjbЊ ' №H №H ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВю я€ 0 И №А @№L№H №( № №L№  №L “ №6€ЄEПƒПРџ№ P€К№Žˆ†Š4К___PPT10‹Б ŠBК___PPT9‹$Ќ №Ÿ 4Now examine these in terms of changes in degree of  disorder .   1.     Gas molecules occupy large volume. Each molecule has more space. Increase in disorder. 2.     Gas molecules occupy less volume. Each molecule as less space. Decrease in disorder, ie. Increase in order. 3.     For crystals, atoms vibrate about fixed lattice points. When the crystal is heated up, amplitude of vibrations has increased, ie. More disordered. 4.     When gases have formed a mixture, there is an increase in disorder. 5.     When a crystal is melted, ordered array of atoms on crystal lattice become free movement of atoms within liquid phase. 6.     When gas reacts with solid to form another solid, the net result is removal of gas phase. Therefore, it leads to decrease in disorder, ie. Increase in order. ЁH?м >`b `b bV`b bk`b b’`b bC`b bv`b b`b `Њfњ  ™  i    Іџ№H №L ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВюšя€ 0 J№B0№T№к№( № №T№Ђ №T “ №6€М4ПƒПРџ№€А Ў№~ˆvŠ0К___PPT10‹БŠ6К___PPT9‹Ќ №ЖŸ pWhat do we means by  disorder and how do we quantify this? By  probability   A  disordered state is one where there are a large number of possible equally probable arrangements, ie a high probability state.   The more disordered state has a higher probability W№f (larger number of different arrangements). The less disordered state has a lower probability W№i (smaller number of different arrangements). Example: two colour balls mixingЁќNk =`df `b ‚рт `4рhтџ_рhтџ,рт b !dЊ&Е   №H №T ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВюъя€ 0 š№’№œ№*№( № №œ№ђ №œ “ №6€”dПƒПРџ№pАо№Žˆ†Š4К___PPT10‹Б ŠBК___PPT9‹$Ќ №іŸ жWhat do we mean by  probability ? Let s consider throwing a dice. What are the chances (ie the probability) of throwing a six?   For two dice, what are chances of throwing 2 sixes?   For three dice -3 sixes? The chance is 1 in 216. ЁЪ"Ъ 2!df ]`b `b 4`b `b 1`b `Њ2X  ‘  №H №œ ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВюя€ 0 Ю№Ц`№X№^№( № №X№& №X “ №6€Ь{ПƒПРџ№аpв №~ˆvŠ0К___PPT10‹БŠ6К___PPT9‹Ќ №:Ÿ ЄTherefore, the probability of several events occurring simultaneously is the product of the probabilities of the individual events   1/6Д№1/6Д№1/6=1/216   n dice-probability of throwing n sixes =1/6­n   Thus a  disordered state is one where there are a large number of possible equally probable arrangements, ie a high probability state. Ё@ƒа ‚`b `b `р`р `т рт +рштџшт рт ˆрт рЊ24    №H №X ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВюя€ 0 Т№Кp№\№R№( № №\№ №\ “ №6€ ЙПƒПРџ№яр0з№~ˆvŠ0К___PPT10‹БŠ6К___PPT9‹Ќ №.Ÿ ј  З   Thus entropy changes can be regarded as a measure of the increase in the  degree of mixing such that for   Less disordered stateЎ№more disordered state   D№S=Sfinal -Sinitial >0   The more disordered state has a higher probability W№f (larger number of different arrangements). The less disordered state has a lower probability W№i (smaller number of different arrangements). ЁЌ{ `b abj`b `b `р`рт р`штџрштџрт рт 4рhтџ_рhтџ,рт рЊX Ђ    Ш  Іџ№H №\ ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВю3я€ 0 у№л`№`№s№( № №`№з №` “ №6€hЇПƒПРџ№@А ,№Žˆ†Š4К___PPT10‹Б ŠBК___PPT9‹$Ќ №лŸЈ-These are related by the Boltzman Equation   ЁX+ *`b `b `Њ2      №\В №` S № AС \??№р№Ц № С \№H №` ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВюя€ 0 О№Жp№d№N№( № №d№ №d “ №6€dжПƒПРџ№Hц №Žˆ†Š4К___PPT10‹Б ŠBК___PPT9‹$Ќ №Ÿ †When we use the term  disorder , we are referring to a situation which can be realised in many different ways whereas  ordered states refer to those which can only be obtained in a few ways.   ЁXС Р`b `b `ЊУ  №H №d ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВюя€ 0 @№8€№h№а№( № №h№4 №h “ №6€рПƒПРџ№p№Рr №ІˆžŠ<К___PPT10‹БŠRК___PPT9‹4Ќ, № Ÿ `The equation proposed by Boltzman is as follows:     Where W№final = Number of ways of achieving the final state (final probability) W№initial = number of ways of achieving the initial state (initial probability).   W№final>W№initial, then D№S>0 then process will occur spontaneously in an isolated system. Ёv1 0`b `b `b `рhтџBрт рhтџGрт рт рhтџрhтџр@`т рЊ&    №\В №h S № A"С =??№№€ № С =№H №h ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВюЈя€ 0 X№PА№l№ш№( № №l№L №l “ №6€!ПƒПРџ№ №`ё №Žˆ†Š4К___PPT10‹Б ŠBК___PPT9‹$Ќ №PŸЈРExample: expansion of gas into a vaccum   We consider the situation after opening the valve and before the gas has expanded.   What is the probability of finding a particular molecule in V1 as opposed to finding it in V2. Probability must be related to volumes of V1 and V2, ie. Chance of finding a molecule in V1 is V1/(V1+V2). Now add a second molecule, chance of finding this in V1 is also V1/(V1+V2).   Probability of finding them both in V1 = Ё,(™ '`b `b R`b `b =`hтџ`hтџ-`hтџ`hтџ'`hтџ`hтџ`hтџ`hтџ9`hтџ `hтџ`hтџ`hтџ`b `b %`hтџ`b `Њ@    ш  Ќ №\В №l S № A#С >??№p ј"№ С >№H №l ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВю”я€ 0 D№<Р№p№д№( № №p№x №p “ №6€T-!ПƒПРџ№№Р@м№~ˆvŠ0К___PPT10‹БŠ6К___PPT9‹Ќ №ŒŸ мFor N molecules, chance of all N molecules being in V1=, which is the probability of initial state W№ initial. Ёtn 5`hтџ-`р `т рЊn  №\В №p S № A$С ???№рр ’ № С ?№И №p “ №6€Ш6!ПƒПРџ№а `@Ђ№~ˆvŠ0К___PPT10‹БŠ6К___PPT9‹Ќ №ЬŸ шProbability of final state=1 since all of the molecules must be somewhere inside the volume (V1+V2) i.e. W№ final =1 ЁЈd ^`hтџ`hтџ`b `р `т рЊt  №H №p ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВюИя€ 0 h№`а№t№ј№( № №t№ №t “ №6€\I!ПƒПРџ№P а<№~ˆvŠ0К___PPT10‹БŠ6К___PPT9‹Ќ №Ÿ NApply Boltzman Equation to obtain D№S   Ёx% "`р`т рт рЊ2     №\В №t S № A%С @??№Ріт № С @№L №t “ №6€T!ПƒПРџ№ Р  №~ˆvŠ0К___PPT10‹БŠ6К___PPT9‹Ќ №`Ÿ Žfor 1 moles i.e. N=6.023Д№1023.  Thus we get the same result as before. Ё–) `р`шрт 'рт рЊG  №H №t ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВюя€ 0 Ж№Ўр№x№F№( № №x№т №x “ №6€b!ПƒПРџ№аМ№Žˆ†Š4К___PPT10‹Б ŠBК___PPT9‹$Ќ №цŸЈ>Conclusion: There are two ways to calculate entropy changes   ЁX< ;`b `b `Њ>  І Арў№\В №x S № A&С A??№ њЎ № С A№\В №x S № A'С B??№АЈ Ў № С B№\В №x S № A(С C??№p P& Т № С C№H №x ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВю• я€ 0  №§ @№ №• №( № № №Г №  Ѓ №<€T љ‡ПƒПРџ№‡ќ№Fˆ>Š\К___PPT10‹<Б4ŠвК___PPT9‹ДЌЌ         №љŸЈЛWhat is the reversible process? [5 marks] Give an example of the reversible process. [5 marks] Explain why a gas expansion to vacuum is not reversible process? [5 marks] For the reaction Ё№*‘1 џ0 џ b bb+b b b Ab bbbb b ЊЛ  І @№jВ №  S № A!С †??"ё?№hwЏ^ № С †№0 №  Ѓ №<€(љ‡ПƒПРџ№2 wСь №ˆўŠXК___PPT10‹8Б0Š–К___PPT9‹xЌp №ЖŸ вCalculate the heat of reaction at 1000 K if the heat capacities in J.mol-1K-1 are given by Cp= 25.7 for CO(g), Cp=24.8 for CO2(g) Cp=24.6 for O2(g) Enthalpies of formation for CO(g)  105.6 kJ.mol-1, for CO2(g) -376 kJ mol-1 at 298 K.Ё€[0 џHbjbj bb"bjтџb b  bjтџbb/bjbjтџbj bbЊ@h  D  6 №H №  ƒ №0ƒ“ŽŸ‹”оНhПџ ?№ џџџ€€€Ь™33ЬЬЬџВВВˆ8Š0К___PPT10‹ы.QшХ@ЁтqGxœэWЭoUŸнMмxm“ФDU=lV* !{зЂ‰Б­T‰+Ej+ŽЌJPEўxv^хѕ:йMьєРЅRХ‘# qр BBBЈ мИPўЧ^Ъ štfї­ыD8EыљНљи7ofg~ЯўщсфЯŸ~qљ8EyPрш8 Ё™$†GВрŽёё=WєЧˆx‡Ѓb‰љ‚ўџД6~\а mœw`џ4ќ%НЃ§ž'<”=љЏО1hлYљш‡јЏпK ЎFLYƒАsљ$di0žaŸ›‚Рџ"ЦoAЯQ…;чіGџ„ЮАўЩО!жŠ№ЛŒйoрIžС?œ7~Eœ—шHМ7š  ўП„c G(ЧQ1a;оpЯ9Iў5~‚^yqh<АеЏ§гНџ2~­№кŽэи W+lяV\nЗЕйЄщ•вRЉ/ѓЪ+`’Гp8џхі0•TѓIŒяяž&› $С С4@ ~J&v‚ЂzщКтiЉSТrАђQ€V’Lni•”І…,вЗ‹Ъ]˜Œуъ•а8\˜At€ЋФM<ˆјG6QXЉИ[ћЩOdђ§‡ §Ояc 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