WEBVTT chemistry/physical-chemistry/hovasapian
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Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.
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Today, we will go to continue our discussion of entropy and probability.
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Let us dive right on in.
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I think I will start in blue here.
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I think not, I will go to black.
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In the previous lesson we discussed the energy distribution.
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Now we discuss the other distribution that contributes to entropy which is volume because
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we said that there are two independent ways of changing the entropy of the system.
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You can change the energy and you can change the volume.
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Let us do some examples and we will generalize.
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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.
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In how many ways can I distribute 2 balls among the 3 spaces?
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The number of balls represents the number of particles of the system and the spaces represent the volume of our system.
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3 spaces that is the question, let us actually do this by brute force.
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We have 3 spaces available, it turns out I can put the 1 ball here and 1 ball there.
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I can put 1 ball here and 1 ball there or I can put 1 ball here and 1 ball there, 3 ways.
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Given 3 volume elements and 2 particles there are 3 ways of distributing those 2 particles among those 3 volume elements.
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I would slow down with my writing.
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There is a symbol for this 3 2 and we say 3 choose 2.
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You may have seen it and you may remember it from these is the binomial coefficients.
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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.
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The 14641151015 that is the binomial coefficients but they show up in all other ways.
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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.
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3 choose 2 = 3 is the 3 ways of doing it.
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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,
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can we distribute 2 balls in 4 spaces?
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The same number of particles I just increase the volume, I have increase the number of available places to put those particles.
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Let us do it 1234 1234 1234.
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I did know how many this is 123456, I did know that before hand.
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When you are doing this in the brute force method you have to write out 4 spaces, 4 spaces, 4 spaces,
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to see how many combinations you can come up with.
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As it turns out, I can put 1 here , 1 here, 1 here, 1 here, 1 here, 1 here, that is 3.
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I can start up here and go here and here.
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I can go here and here or go here and here 123456.
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If I try to get the combination it is going to be some variation of this because the order does not matter.
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123456 there are 6 ways.
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We just went from 3 ways to 6 ways just by adding one more volume element.
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This is going to be 4 spaces, 2 balls, 4 choose 2, its symbolize this way 4 choose 2.
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Let us increase the volume and fill in one more particles.
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What about 3 balls among 10 spaces?
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That is going to be 10 choose 3.
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I’m going to do it and I’m going to give you the general expression for it.
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It is going to be 10! / 3! × 7! = 120 ways.
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That is a lot of ways.
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In general, this is what we are after.
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In general, if we have 10 spaces we have to be careful.
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I was looking for this lesson right before I started, after recording this lesson and
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I realize that I use this variable n in this lesson to represent the number of spaces not the number of particles.
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Be very careful because it confuses me a little bit.
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I know in the previous lesson I was using N as the number of particles.
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In this lesson N is the number spaces.
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N space it represents the volume, N spaces and P particles.
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P is going to be used for the number of particles and N is going to be used for the number of spaces.
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N is volume P is particles.
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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!.
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That is why had 10!/ 3! 7 because 10 -3 is 7.
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That is why I came up with this, this is the general expression.
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The number of spaces which in this case is N, it represents the volume available.
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Clearly, as I increase the volume a little bit, here you are going to find increase in the number of particles a little bit.
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It jumped up from 1 to 6 to 120.
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It is a just huge jump.
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In general, the space available is going to be much larger than the number of particles.
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The space available, let me just use variables since I have introduced the variables.
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N which is the space available is much larger than the number of particles.
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You can have something like 10/ 3 but you might have something like 10,000,000,000/50,000.
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10 billion is so much larger than 50,000 so you would have 10 billion spaces available and you have 50,000 particles.
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How many different ways are there of actually distributing those 50,000 particles over those 10 billion spaces that are available for you?
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N is much larger than the number of particles but you know this already.
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In general, if I have 1 mol of gas and I put it into a 1 L flask, these individual spaces available if
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I actually divided the total volume by the volume of the individual atom, the sheer number of spaces is going to be so huge.
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The volume is always going to be much larger than the number of particles available for the volume.
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In general, N is much larger than the number of particles P.
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In this case there is an approximation to this expression which actually works quite well and it is perfectly good
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because we are going to be dealing with a huge numbers of particles on the order of 10, 20, 30, 40.
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in these cases, there is an approximation we can use.
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The approximation is this N choose P = approximately 10 ⁺P ÷ P!.
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You are welcome to use this or this.
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This works a little bit better because there is an approximation.
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Let us confirm this by doing just a couple of quick examples just so you see this approximation is valid.
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Let us say we have 500 spaces so let N = 500 and let number of particles just = 5.
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Clearly 500 is a lot bigger than 5.
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If we do 500 choose 5 and if we do the exact version we end up with 2.55 × 10¹¹.
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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?
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There are 2.55 × 10¹¹ ways, massive.
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Let us do the approximation.
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The approximation is N is 500⁵ ÷ 5!.
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When I do that I get 2.6 × 10¹¹ this is the approximate.
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Clearly, for something on the order of something like this it is still 2.55 2.6 × 10¹¹.
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They are virtually the same, it really does not matter.
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For all practical purposes, we will more than likely use that expression.
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We will use N ⁺P/ P!.
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Therefore, O for the volume distribution = N choose P = N!/ P! × N- P! which is approximately = N ⁺P ÷ P!.
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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.
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Let us go back to entropy, S = Boltzmann constant × log of O.
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Let us let N₁ be a certain number of spaces and P particles be one set of circumstances.
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I had N1 and I have P, O is going to be N₁ choose P.
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If we increase N₁ to N₂, in other words if we increase the number of spaces available, in other words if we increase the volume,
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this is analogous to increasing the volume available for these P particles, increasing the volume from V1 to V2.
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In other words, creating more space to distribute the same number of particles P.
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Let us go ahead and do some calculations.
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Our entropy 1 for our first circumstance is going to be KB × the nat log of N₁ ⁺P/ P!.
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We are going to go ahead and use the approximation, that = KB × the nat log of N₁ ⁺P - the nat log of P!.
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Let us go ahead and calculate the entropy for circumstance 2 where we increased the space available from N1 to N2.
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The entropy for 2 is going to be = KB × the nat log this time is going N₂ ⁺P/ P! = K sub P × the nat log of N₂ ⁺P – ln of P!.
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So far so good.
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Let us do a δ, δ S is final – initial.
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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.
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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.
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It is going to be - KB nat log of N1 ⁺P! + KB × the nat log of P! – N +, those two cancel.
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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.
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I get P × K sub B × the nat log of N2 - P × K sub P the nat log of N1.
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δ S = P × KB nat log of N₂ ÷ N₁.
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Just using the properties of logarithms.
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N₂/ N₁ we said that the number of spaces available, this is the number spaces available.
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Let us say I started with 500 spaces available and I increased it to 1000 spaces available.
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This is going to be 1000/ 500 which is going to equal to.
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This is going to be analogous to volume 2/ volume 1.
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I have 500 ml and I changed it to 1000 ml it is going to be 1500, you are still going to get 2.
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I'm just changing volume.
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When I change a space, change volume, I’m going to rewrite this as just that.
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The number actually ends up being the same, N2/ N1 was nothing more than V2/ V1.
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Let us go ahead and write it this way.
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Now that the number of particles equal the other number, let P =1 mol and the other number N sub A.
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We get δ S = N sub A × K sub B × the nat log of V2/ V1.
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I put V2/ V1 for that and I put N sub A for that.
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Let us go to the next page and we rewrite this and do this in red.
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We have δ S = N sub A.
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The Avogadro’s number × Boltzmann constant × nat log of the final volume/ the initial volume.
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The Avogadro’s number × Boltzmann constant = R the gas constant/ Avogadro’s number.
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When I multiply I get Avogadro’s number × Boltzmann constant actually = the gas constant.
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Therefore, the change in entropy = the gas constant × that nat log of V2/ V1.
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This is for 1 mol of Avogadro’s number.
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We are talking about 1 mol here.
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This expression right here should look really familiar to you, if not, let me remind you.
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For an ideal gas δ S = CP × the ln of T2/ T1 - N × R × the nat log of V2 / V1.
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The change in entropy of an ideal gas under conditions of temperature and volume this was the expression for it, we derived it already.
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Let us say we held the temperature constant we would not have to worry about this term.
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For 1 mol N =1 and we end up with R ln V2/ V1.
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Statistically, we ended up with R ln V2/ V1 the change in entropy.
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If I have a certain number of particles, 1 mol of particles in this case and if I make entropy of volume 1,
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if I increase the volume to volume 2, the δ S is going to be this expression right here.
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We did this statistically with probability and accounting.
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It is the same expression that we achieved when we did it empirically and analytically.
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These two expressions are exactly alike.
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These expressions are the same.
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We derived a relation for the entropy increase with respect to volume, the probabilistic methods used to expressions are the same.
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We derive the relation for the entropy increase with respect to volume via
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statistical probabilistic methods that is identical to the one we derived empirically, analytically.
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That confirms what we already knew about entropy.
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We derive this expression for entropy for an ideal gas.
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The entropy increase if you go from volume 1 to volume 2 was given by this part of the expression right there.
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The number of moles × R × the log of the ratio of volume 2 to the volume 1.
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Using a statistical methods and the definition of which was S = KB ln O, using that we ended up deriving the same relation.
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They should corroborate that this is a good thing, this is right, this is correct.
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As we increase V the volume for a given number of particles N chose P which is the O increases.
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When O increases, entropy increases but we knew this already.
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This should be a + not -, I think I have the CV.
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I'm sorry this is not a P this is a V.
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The temperature and pressure is the one that actually different, increases as the entropy increases.
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Let us put it together.
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For energy distribution we have the following we had O = N!/ N₁! N₂! N₃! And so on, subject to the following constraints.
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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,
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E sub I when I add those together is the total energy of the system for the volume distribution.
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For the volume distribution we have the following for O, O = N choose P which is approximately = N ⁺P/ P!.
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Here N is not the number of particles but N is the number of spaces available,
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P is the number of particles that you have to distribute among those N spaces.
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I probably use different variables, forgive me.
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Let me write it here when N is the number of spaces.
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In other words, the volume and P is the number of particles.
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Here the constraint is the volume itself, the number of spaces that is the constraint.
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We have two independent ways of actually increasing the entropy of the system.
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We can change the energy and we can change the volume.
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In each case we have this expression S = KB ln O.
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In any given system, there are 2 distributions.
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There is the energy distribution and there is the volume distribution.
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Each of these contributes to that total entropy of the system.
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S = KB × the nat log of O.
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The question is why the log of O?
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Why not just O?
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Why did we just say S = KB × O?
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here is why, we just said that in any given system there are 2 distributions,
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the particle distribute themselves/ the energy and the particles distribute themselves/ the volume.
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Both of those contribute to the total entropy.
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The total entropy of the system = the entropy contribution from the energy and the entropy contribution of the volume.
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Entropy is an extensive property, extensive means additive.
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In other words, if I have this much entropy from this contribution then this much entropy from this contribution,
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the total should be I just add them together.
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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.
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Let us go ahead and do this.
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The total entropy = Boltzmann constant × the nat log of the O, the total O.
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The total O = the O of the energy contribution × the O of the volume contribution.
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Remember what we did earlier, we said if I have a certain number of ways of distributing 3 particles of distributing 2 particles
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I just multiply those 2 together to give me the total number of ways of distributing particles.
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It is the same thing.
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The number of ways of distributing particles / the energy and a number of ways of distributing the particles / the volume,
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if I multiply those 2 ways that gives me the total number of ways of distributing the particles / the volume and the energy.
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I multiply them together that is how we get it.
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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.
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I will go to the next page.
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I get ST, I get the total entropy = Boltzmann constant × the nat log of O of energy × O of volume.
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The log of something × something is the log of something + the log of something.
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Therefore, ST = KB × the nat log of OE + KB × the nat log of OV ST, this is the entropy of the energy contribution.
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This is the energy of entropy of the volume contribution.
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This is supposed to be a checkmark.
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The total entropy does actually end up equaling the entropy contribution from the energy + the entropy contribution of volume.
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The log of function allows us to do that, that is why it is a logarithm and it is not just KB × O.
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We needed the total entropy to equal the entropy, the sum of the individual entropy.
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If we get it the other way, it would not work out.
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We needed V ST = the sum that is very important.
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The natural logarithm of the function allows this and natural logarithm of function allows for this.
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If Boltzmann would have written the following.
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If we had written that S = KB × O without the log function, here is what would happen.
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The total would be KB × O total which is going to be KB × OE × OV.
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We will just set that aside for a second.
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There is not going to be an easy way to separate these two.
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The contribution of the energy is going to be KB × OE.
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The entropy contribution of the volume is going to be KB OV.
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This is based on something that is not true.
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If we had defined it this way here is what happened, this would be one thing.
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When we add these together + KB OV.
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Notice this FT here gives us this right here.
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When we do them individually, when we separate them out the some of them gives us this right here.
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They are both ST the KB × OE × OV is not the same as KB OE + KB OV.
00:33:42.900 --> 00:33:53.100
If we use this definition of entropy without the logarithm function I do not get an extensive property.
00:33:53.100 --> 00:34:01.500
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,
00:34:01.500 --> 00:34:10.500
Boltzmann defined it as he did as KB × the log of O function allows for the possibility that
00:34:10.500 --> 00:34:17.700
the total entropy actually equals the sum of the energy contribution + the volume contribution.
00:34:17.700 --> 00:34:25.200
Later on, if I happen to mix one gas with another gas, the entropy of the total system now
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the mixture should be because entropy is an extensive property.
00:34:28.800 --> 00:34:36.300
It should be the entropy of the one gas A + the entropy of the other gas B separately.
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Using the log to define this entropy allows for this property is equality to be maintained.
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I hope that makes sense.
00:34:45.600 --> 00:35:00.800
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.
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I hope that that make sense.
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Thank you so much for joining us here at www.eudcator.com.
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We will see you next time, bye.