For more information, please see full course syllabus of Physical Chemistry

For more information, please see full course syllabus of Physical Chemistry

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### Entropy & Probability II

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Volume Distribution
- Distributing 2 Balls in 3 Spaces
- Distributing 2 Balls in 4 Spaces
- Distributing 3 Balls in 10 Spaces
- Number of Ways to Distribute P Particles over N Spaces
- When N is Much Larger than the Number of Particles P
- Energy Distribution
- Volume Distribution
- Entropy, Total Entropy, & Total Omega Equations

- Intro 0:00
- Volume Distribution 0:08
- Distributing 2 Balls in 3 Spaces
- Distributing 2 Balls in 4 Spaces
- Distributing 3 Balls in 10 Spaces
- Number of Ways to Distribute P Particles over N Spaces
- When N is Much Larger than the Number of Particles P
- Energy Distribution
- Volume Distribution
- Entropy, Total Entropy, & Total Omega Equations 27:34
- Entropy, Total Entropy, & Total Omega Equations

### Physical Chemistry Online Course

### Transcription: Entropy & Probability II

*Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.*0000

*Today, we will go to continue our discussion of entropy and probability.*0004

*Let us dive right on in.*0008

*I think I will start in blue here.*0011

*I think not, I will go to black.*0017

*In the previous lesson we discussed the energy distribution.*0025

*Now we discuss the other distribution that contributes to entropy which is volume because*0049

*we said that there are two independent ways of changing the entropy of the system.*0082

*You can change the energy and you can change the volume.*0085

*Let us do some examples and we will generalize.*0092

*Let us say we have 3 spaces and we have to 2 balls, how many ways can I distribute those 2 balls in the 3 available spaces, that is the question.*0101

*In how many ways can I distribute 2 balls among the 3 spaces?*0112

*The number of balls represents the number of particles of the system and the spaces represent the volume of our system.*0130

*3 spaces that is the question, let us actually do this by brute force.*0137

*We have 3 spaces available, it turns out I can put the 1 ball here and 1 ball there.*0141

*I can put 1 ball here and 1 ball there or I can put 1 ball here and 1 ball there, 3 ways.*0150

*Given 3 volume elements and 2 particles there are 3 ways of distributing those 2 particles among those 3 volume elements.*0161

*I would slow down with my writing.*0179

*There is a symbol for this 3 2 and we say 3 choose 2.*0181

*You may have seen it and you may remember it from these is the binomial coefficients.*0190

*The ones that show up when you do binomial expansion, remember the Pascal’s trinomial 111 211 331 14641 that is what these things are.*0195

*The 14641151015 that is the binomial coefficients but they show up in all other ways.*0204

*In this particular case, they show up as the number of ways of distributing certain number of particles in a certain number of empty spaces so 3 choose 2.*0211

*3 choose 2 = 3 is the 3 ways of doing it.*0220

*Let us go to the next level, in how many ways can we distribute, now I would increase the volume but same number of particles,*0226

*can we distribute 2 balls in 4 spaces?*0242

*The same number of particles I just increase the volume, I have increase the number of available places to put those particles.*0250

*Let us do it 1234 1234 1234.*0257

*I did know how many this is 123456, I did know that before hand.*0270

*When you are doing this in the brute force method you have to write out 4 spaces, 4 spaces, 4 spaces,*0275

*to see how many combinations you can come up with.*0280

*As it turns out, I can put 1 here , 1 here, 1 here, 1 here, 1 here, 1 here, that is 3.*0283

*I can start up here and go here and here.*0289

*I can go here and here or go here and here 123456.*0292

*If I try to get the combination it is going to be some variation of this because the order does not matter.*0299

*123456 there are 6 ways.*0305

*We just went from 3 ways to 6 ways just by adding one more volume element.*0310

*This is going to be 4 spaces, 2 balls, 4 choose 2, its symbolize this way 4 choose 2.*0313

*Let us increase the volume and fill in one more particles.*0327

*What about 3 balls among 10 spaces?*0331

*That is going to be 10 choose 3.*0344

*I’m going to do it and I’m going to give you the general expression for it.*0349

*It is going to be 10! / 3! × 7! = 120 ways.*0352

*That is a lot of ways.*0361

*In general, this is what we are after.*0367

*In general, if we have 10 spaces we have to be careful.*0370

*I was looking for this lesson right before I started, after recording this lesson and*0381

*I realize that I use this variable n in this lesson to represent the number of spaces not the number of particles.*0387

*Be very careful because it confuses me a little bit.*0395

*I know in the previous lesson I was using N as the number of particles.*0397

*In this lesson N is the number spaces.*0401

*N space it represents the volume, N spaces and P particles.*0405

*P is going to be used for the number of particles and N is going to be used for the number of spaces.*0412

*N is volume P is particles.*0416

*In general, if we have N spaces and P particles the number of ways of distributing P particles/ N spaces is N choose P = N!/ P! × N - P!.*0420

*That is why had 10!/ 3! 7 because 10 -3 is 7.*0466

*That is why I came up with this, this is the general expression.*0471

*The number of spaces which in this case is N, it represents the volume available.*0476

*Clearly, as I increase the volume a little bit, here you are going to find increase in the number of particles a little bit.*0487

*It jumped up from 1 to 6 to 120.*0502

*It is a just huge jump.*0506

*In general, the space available is going to be much larger than the number of particles.*0510

*The space available, let me just use variables since I have introduced the variables.*0525

*N which is the space available is much larger than the number of particles.*0534

*You can have something like 10/ 3 but you might have something like 10,000,000,000/50,000.*0551

*10 billion is so much larger than 50,000 so you would have 10 billion spaces available and you have 50,000 particles.*0564

*How many different ways are there of actually distributing those 50,000 particles over those 10 billion spaces that are available for you?*0567

*N is much larger than the number of particles but you know this already.*0574

*In general, if I have 1 mol of gas and I put it into a 1 L flask, these individual spaces available if*0578

*I actually divided the total volume by the volume of the individual atom, the sheer number of spaces is going to be so huge.*0584

*The volume is always going to be much larger than the number of particles available for the volume.*0594

*In general, N is much larger than the number of particles P.*0600

*In this case there is an approximation to this expression which actually works quite well and it is perfectly good*0608

*because we are going to be dealing with a huge numbers of particles on the order of 10, 20, 30, 40.*0615

*in these cases, there is an approximation we can use.*0620

*The approximation is this N choose P = approximately 10 ⁺P ÷ P!.*0631

*You are welcome to use this or this.*0652

*This works a little bit better because there is an approximation.*0655

*Let us confirm this by doing just a couple of quick examples just so you see this approximation is valid.*0661

*Let us say we have 500 spaces so let N = 500 and let number of particles just = 5.*0669

*Clearly 500 is a lot bigger than 5.*0678

*If we do 500 choose 5 and if we do the exact version we end up with 2.55 × 10¹¹.*0681

*In other words, if I have 500 spaces and I have 5 different particles, how many different ways can I distribute those 5 particles in those 500 spaces?*0692

*There are 2.55 × 10¹¹ ways, massive.*0701

*Let us do the approximation.*0707

*The approximation is N is 500⁵ ÷ 5!.*0710

*When I do that I get 2.6 × 10¹¹ this is the approximate.*0717

*Clearly, for something on the order of something like this it is still 2.55 2.6 × 10¹¹.*0726

*They are virtually the same, it really does not matter.*0732

*For all practical purposes, we will more than likely use that expression.*0735

*We will use N ⁺P/ P!.*0739

*Therefore, O for the volume distribution = N choose P = N!/ P! × N- P! which is approximately = N ⁺P ÷ P!.*0744

*There you go this is our general expression for O for a volume distribution just like we have a general expression for the energy distribution.*0766

*Let us go back to entropy, S = Boltzmann constant × log of O.*0778

*Let us let N sub 1 be a certain number of spaces and P particles be one set of circumstances.*0790

*I had N1 and I have P, O is going to be N sub 1 choose P.*0809

*If we increase N sub 1 to N sub 2, in other words if we increase the number of spaces available, in other words if we increase the volume,*0819

*this is analogous to increasing the volume available for these P particles, increasing the volume from V1 to V2.*0833

*In other words, creating more space to distribute the same number of particles P.*0854

*Let us go ahead and do some calculations.*0878

*Our entropy 1 for our first circumstance is going to be KB × the nat log of N sub 1 ⁺P/ P!.*0881

*We are going to go ahead and use the approximation, that = KB × the nat log of N sub 1 ⁺P - the nat log of P!.*0892

*Let us go ahead and calculate the entropy for circumstance 2 where we increased the space available from N1 to N2.*0914

*The entropy for 2 is going to be = KB × the nat log this time is going N sub 2 ⁺P/ P! = K sub P × the nat log of N sub 2 ⁺P – ln of P!.*0922

*So far so good.*0948

*Let us do a δ, δ S is final – initial.*0950

*δ S the change in entropy is going to be the entropy at 2 - the entropy at 1 = I’m going to go ahead and do the distribution.*0954

*This thing - this thing, I'm going to get this first one is KB × the nat log of N2 ⁺P! - distribute my KB × the nat log of P! - this thing.*0966

*It is going to be - KB nat log of N1 ⁺P! + KB × the nat log of P! – N +, those two cancel.*0990

*I'm going to end up with δ S = I’m going to go ahead and bring, this is nat log functions so I can go ahead and bring this down here.*1009

*I get P × K sub B × the nat log of N2 - P × K sub P the nat log of N1.*1022

*δ S = P × KB nat log of N sub 2 ÷ N sub 1.*1038

*Just using the properties of logarithms.*1047

*N sub 2/ N sub 1 we said that the number of spaces available, this is the number spaces available.*1053

*Let us say I started with 500 spaces available and I increased it to 1000 spaces available.*1061

*This is going to be 1000/ 500 which is going to equal to.*1065

*This is going to be analogous to volume 2/ volume 1.*1069

*I have 500 ml and I changed it to 1000 ml it is going to be 1500, you are still going to get 2.*1075

*I'm just changing volume.*1083

*When I change a space, change volume, I’m going to rewrite this as just that.*1085

*The number actually ends up being the same, N2/ N1 was nothing more than V2/ V1.*1090

*Let us go ahead and write it this way.*1099

*Now that the number of particles equal the other number, let P =1 mol and the other number N sub A.*1104

*We get δ S = N sub A × K sub B × the nat log of V2/ V1.*1118

*I put V2/ V1 for that and I put N sub A for that.*1139

*Let us go to the next page and we rewrite this and do this in red.*1148

*We have δ S = N sub A.*1153

*The Avogadro’s number × Boltzmann constant × nat log of the final volume/ the initial volume.*1166

*The Avogadro’s number × Boltzmann constant = R the gas constant/ Avogadro’s number.*1176

*When I multiply I get Avogadro’s number × Boltzmann constant actually = the gas constant.*1189

*Therefore, the change in entropy = the gas constant × that nat log of V2/ V1.*1198

*This is for 1 mol of Avogadro’s number.*1207

*We are talking about 1 mol here.*1209

*This expression right here should look really familiar to you, if not, let me remind you.*1215

*For an ideal gas δ S = CP × the ln of T2/ T1 - N × R × the nat log of V2 / V1.*1221

*The change in entropy of an ideal gas under conditions of temperature and volume this was the expression for it, we derived it already.*1241

*Let us say we held the temperature constant we would not have to worry about this term.*1250

*For 1 mol N =1 and we end up with R ln V2/ V1.*1257

*Statistically, we ended up with R ln V2/ V1 the change in entropy.*1262

*If I have a certain number of particles, 1 mol of particles in this case and if I make entropy of volume 1,*1268

*if I increase the volume to volume 2, the δ S is going to be this expression right here.*1275

*We did this statistically with probability and accounting.*1282

*It is the same expression that we achieved when we did it empirically and analytically.*1285

*These two expressions are exactly alike.*1291

*These expressions are the same.*1296

*We derived a relation for the entropy increase with respect to volume, the probabilistic methods used to expressions are the same.*1307

*We derive the relation for the entropy increase with respect to volume via*1352

*statistical probabilistic methods that is identical to the one we derived empirically, analytically.*1356

*That confirms what we already knew about entropy.*1382

*We derive this expression for entropy for an ideal gas.*1387

*The entropy increase if you go from volume 1 to volume 2 was given by this part of the expression right there.*1393

*The number of moles × R × the log of the ratio of volume 2 to the volume 1.*1400

*Using a statistical methods and the definition of which was S = KB ln O, using that we ended up deriving the same relation.*1406

*They should corroborate that this is a good thing, this is right, this is correct.*1418

*As we increase V the volume for a given number of particles N chose P which is the O increases.*1425

*When O increases, entropy increases but we knew this already.*1460

*This should be a + not -, I think I have the CV.*1470

*I'm sorry this is not a P this is a V.*1490

*The temperature and pressure is the one that actually different, increases as the entropy increases.*1495

*Let us put it together.*1508

*For energy distribution we have the following we had O = N!/ N sub 1! N sub 2! N sub 3! And so on, subject to the following constraints.*1510

*Subject to the sum of N sub I = N the number of particles and the sum of the number of particles in each energy bin,*1546

*E sub I when I add those together is the total energy of the system for the volume distribution.*1554

*For the volume distribution we have the following for O, O = N choose P which is approximately = N ⁺P/ P!.*1563

*Here N is not the number of particles but N is the number of spaces available,*1586

*P is the number of particles that you have to distribute among those N spaces.*1590

*I probably use different variables, forgive me.*1595

*Let me write it here when N is the number of spaces.*1598

*In other words, the volume and P is the number of particles.*1604

*Here the constraint is the volume itself, the number of spaces that is the constraint.*1612

*We have two independent ways of actually increasing the entropy of the system.*1623

*We can change the energy and we can change the volume.*1628

*In each case we have this expression S = KB ln O.*1633

*In any given system, there are 2 distributions.*1642

*There is the energy distribution and there is the volume distribution.*1646

*Each of these contributes to that total entropy of the system.*1650

*S = KB × the nat log of O.*1659

*The question is why the log of O?*1668

*Why not just O?*1671

*Why did we just say S = KB × O?*1675

*here is why, we just said that in any given system there are 2 distributions,*1680

*the particle distribute themselves/ the energy and the particles distribute themselves/ the volume.*1687

*Both of those contribute to the total entropy.*1693

*The total entropy of the system = the entropy contribution from the energy and the entropy contribution of the volume.*1697

*Entropy is an extensive property, extensive means additive.*1707

*In other words, if I have this much entropy from this contribution then this much entropy from this contribution,*1710

*the total should be I just add them together.*1716

*we want this to be true, we want the total entropy = the entropy of one contribution + the entropy of the other contribution, this just makes sense.*1719

*Let us go ahead and do this.*1728

*The total entropy = Boltzmann constant × the nat log of the O, the total O.*1733

*The total O = the O of the energy contribution × the O of the volume contribution.*1742

*Remember what we did earlier, we said if I have a certain number of ways of distributing 3 particles of distributing 2 particles*1752

*I just multiply those 2 together to give me the total number of ways of distributing particles.*1758

*It is the same thing.*1763

*The number of ways of distributing particles / the energy and a number of ways of distributing the particles / the volume,*1765

*if I multiply those 2 ways that gives me the total number of ways of distributing the particles / the volume and the energy.*1771

*I multiply them together that is how we get it.*1779

*If the total entropy = KB × ln OT and a total OT which accounts for both contributions is the energy ω and the volume O I get the following.*1783

*I will go to the next page.*1797

*I get ST, I get the total entropy = Boltzmann constant × the nat log of O of energy × O of volume.*1800

*The log of something × something is the log of something + the log of something.*1818

*Therefore, ST = KB × the nat log of OE + KB × the nat log of OV ST, this is the entropy of the energy contribution.*1823

*This is the energy of entropy of the volume contribution.*1845

*This is supposed to be a checkmark.*1853

*The total entropy does actually end up equaling the entropy contribution from the energy + the entropy contribution of volume.*1856

*The log of function allows us to do that, that is why it is a logarithm and it is not just KB × O.*1865

*We needed the total entropy to equal the entropy, the sum of the individual entropy.*1875

*If we get it the other way, it would not work out.*1880

*We needed V ST = the sum that is very important.*1888

*The natural logarithm of the function allows this and natural logarithm of function allows for this.*1904

*If Boltzmann would have written the following.*1915

*If we had written that S = KB × O without the log function, here is what would happen.*1918

*The total would be KB × O total which is going to be KB × OE × OV.*1927

*We will just set that aside for a second.*1945

*There is not going to be an easy way to separate these two.*1951

*The contribution of the energy is going to be KB × OE.*1955

*The entropy contribution of the volume is going to be KB OV.*1964

*This is based on something that is not true.*1972

*If we had defined it this way here is what happened, this would be one thing.*1974

*When we add these together + KB OV.*1981

*Notice this FT here gives us this right here.*2002

*When we do them individually, when we separate them out the some of them gives us this right here.*2009

*They are both ST the KB × OE × OV is not the same as KB OE + KB OV.*2015

*If we use this definition of entropy without the logarithm function I do not get an extensive property.*2023

*Some of the individual entropy is not equal the total entropy but if I use the natural logarithm property, if I define it as we did,*2033

*Boltzmann defined it as he did as KB × the log of O function allows for the possibility that*2041

*the total entropy actually equals the sum of the energy contribution + the volume contribution.*2050

*Later on, if I happen to mix one gas with another gas, the entropy of the total system now*2058

*the mixture should be because entropy is an extensive property.*2065

*It should be the entropy of the one gas A + the entropy of the other gas B separately.*2069

*Using the log to define this entropy allows for this property is equality to be maintained.*2076

*I hope that makes sense.*2084

*In this case, ST does not = S of E + S of V, but when we use the log it actually does, that is why we use the log function to define entropy the way that we do.*2085

*I hope that that make sense.*2101

*Thank you so much for joining us here at www.eudcator.com.*2103

*We will see you next time, bye.*2105

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