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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 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

Transcription: Entropy & Probability II

Hello and welcome back to 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

We will see you next time, bye.2105