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Equation.30,Microsoft Equation 3.0B@ Equation.30,Microsoft Equation 3.0CA Equation.30,Microsoft Equation 3.0WEquation Equation.30,Microsoft Equation 3.0/ 00DTimes New Roman{u xO 0xD[SOes New Roman{u xO 0x DSymbolew Roman{u xO 0x0DArialew Roman{u xO 0x@ .  @n?" dd@  @@`` 3U   %''$#%(*)2$(, Ed $$$$$$2$8X` i7Vp$$2$IƱlКUk2$%\rOU$$$$$$$$$$$2$E?67AOŨ2}Z2$1G%>$!z{7Mm 2$: x'Y* 2$^lN^2n82$d|$뉤o)22$>,EPT ^12$e`L'NXG 2$2aHἐpq2$0j?WqrRD2$#˱:S t"Ҭ2$ yֵ X$.2$,{ҁ3JΜ2$4ɪlV_e"y2$ "2$ߒF M:C$2$DBRx?J& 0AA3@uʚ;2Nʚ;g4FdFd~u O 0$ppp@ <4!d!dv 0{u <4ddddv 0{u  ___PPT10D[SOes New RomanO 0DTimes New RomanO 0 pp?  O  =0 1 2 34579BCDEFGHX$   0` ` ̙33` 333MMM` ff3333f` f` f` 3>?" dd@,|?" dd@   " @ ` n?" dd@   @@``PR    @ ` ` p>> VN (    6pRu  P u  X Click to edit Master title style!!   0Uu   u  RClick to edit Master text styles Second level Third level Fourth level Fifth level!    S   0x\u  `` u  f*b    0du  `  u  h*b    0iu  `  u  h*b  H  0޽h ? ̙33 Default Design. 0 >(    0tG  P   u  T*    0hx     u  t*&b    d  c $ ?  u   0 x   0 u  8USQdkYkHre,g7h_ ,{N~ ,{ N~ ,{V~ ,{N~    6x  _P  u  b*     6x  _  u  d*   H  0޽h ? 3380___PPT10.fP0F :2(    0Dx  P   u  T*    0pKx     u  t*&b      6xݔ  _P  u  b*     6t  _  u  d*   H  0޽h ? 3380___PPT10.fPUL 0 F(    689 @~v0___PPT106___PPT9 "fEntropy and Probability Background  Classical approach to thermodynamics. What entropy  means What is happening in terms of the atoms and molecules when we change the state of the system DS>0, there is always an increase in  randomness or  disorder . d fb b `b b c b'bb b b b b b c bbb b b b b c b\bb b b b cb@b    &/   H  0޽h ? ̙33D 0 8(  8 8 6q `4___PPT10 B___PPT9$  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 h bb b bb bb bb /bjb bb bb b bb bb bb `X +  5  -  \ 8 S A  0??PP P 0\ 8 S A  1?? z^  1H 8 0޽h ? ̙33 0 <5(  <5 < 6   4___PPT10 B___PPT9$ 9 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. N bb b,bb bb bb bb bb Bbjbjbb bb `$   \ < S A 2??0v  2\ < S A 3??p 0  3H < 0޽h ? ̙33} 0 -%@(  @ @ 6L P5 4___PPT10 B___PPT9$ G 1. Mixing gases e.g. mix 1 mole of O2 with 3 moles of N2 $F bb b bb bb bb bjbjb bb bb `$ E  \ @ S A 4?? B 4\ @ S A 5?? 0.  5H @ 0޽h ? ̙33 0  D((  D D 6(  4___PPT10 B___PPT9$ d . ==Enthalpy of Fusion e.g. melt 10 moles of Pb at its melting point 600K, =4812 J.mole-1 c bb bb bbb bb bb @bjb bb `> 1  -  \ D S A 6?? "2 6\ D S A 7?? f 7\ D S A 8??p  8\ D S A 9?? P 9 D 0l 00&4___PPT10 B___PPT9$ Melting solid B  2 bb `  H D 0޽h ? ̙33 0 }PH(  H H 6 `4___PPT10 B___PPT9$ [1. Oxidise a metal to its oxide e.g. oxidise 2 moles of Al(s) to Al2O3(s) at 298K $8 b bff bb #bjbj bb bb b` \ \ H S A :??   :\ H S A  ;??@ `  ; H B  @~v0___PPT106___PPT9 &=50.99-2(28.33)-3/2(205.0)=-313.2JK-1.6'#bjb ' H H 0޽h ? ̙33  0   @LH (  L  L 6H  P4___PPT10 B___PPT9$ 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. H? >`b `b bV`b bk`b b`b bC`b bv`b b`b `f    i    H L 0޽h ? ̙33 0 JB0T(  T T 6< ~v0___PPT106___PPT9 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 Wf (larger number of different arrangements). The less disordered state has a lower probability Wi (smaller number of different arrangements). Example: two colour balls mixingNk =`df `b  `4h_h, b !d&   H T 0޽h ? ̙33 0 p\R(  \ \ 60 The more disordered state has a higher probability Wf (larger number of different arrangements). The less disordered state has a lower probability Wi (smaller number of different arrangements). { `b abj`b `b `` `  4h_h, X       H \ 0޽h ? ̙333 0 ``s(  ` ` 60w @ ,4___PPT10 B___PPT9$ -These are related by the Boltzman Equation X+ *`b `b `2      \ ` S A \??  \H ` 0޽h ? ̙33 0 pdN(  d d 6 H 4___PPT10 B___PPT9$ 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. X `b `b `  H d 0޽h ? ̙33 0 @8h(  h4 h 6DG pr <___PPT10R___PPT94, `The equation proposed by Boltzman is as follows: Where Wfinal = Number of ways of achieving the final state (final probability) Winitial = number of ways of achieving the initial state (initial probability). Wfinal>Winitial, then DS>0 then process will occur spontaneously in an isolated system. v1 0`b `b `b `hB hG  hh@` &    \ h S A" =??  =H h 0޽h ? ̙33 0 XPl(  lL l 6u  ` 4___PPT10 B___PPT9$ PExample: 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 = ,( '`b `b R`b `b =`h`h-`h`h'`h`h`h`h9`h `h`h`h`b `b %`h`b `@       \ l S A# >??p " >H l 0޽h ? ̙33 0 D<p(  px p 6u @~v0___PPT106___PPT9 For N molecules, chance of all N molecules being in V1=, which is the probability of initial state W initial. tn 5`h-` ` n  \ p S A$ ???  ? p 6pu  `@~v0___PPT106___PPT9 Probability of final state=1 since all of the molecules must be somewhere inside the volume (V1+V2) i.e. W final =1 d ^`h`h`b ` ` t  H p 0޽h ? ̙33 0 h`t(  t t 6Du P<~v0___PPT106___PPT9 NApply Boltzman Equation to obtain DS x% "``  2     \ t S A% @??  @L t 6G    ~v0___PPT106___PPT9 `for 1 moles i.e. N=6.0231023. Thus we get the same result as before. ) `` ' G  H t 0޽h ? ̙33 0 xF(  x x 6u 4___PPT10 B___PPT9$ >Conclusion: There are two ways to calculate entropy changes X< ;`b `b `>   \ x S A& A??   A\ x S A' B??  B\ x S A( C??p P&  CH x 0޽h ? ̙33  0   @ (    <u F>\___PPT10<4___PPT9         What 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? 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