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### Statistical Thermodynamics: The Various Partition Functions II

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• Intro 0:00
• Diatomic Gases 0:16
• Diatomic Gases
• Zero-Energy Mark for Rotation
• Zero-Energy Mark for Vibration
• Zero-Energy Mark for Electronic
• Vibration Partition Function
• When Temperature is Very Low
• When Temperature is Very High
• Vibrational Component
• Fraction of Molecules in the r Vibration State
• Example: Fraction of Molecules in the r Vib. State
• Rotation Partition Function
• Heteronuclear & Homonuclear Diatomics
• Energy & Heat Capacity
• Fraction of Molecules in the J Rotational Level
• Example: Fraction of Molecules in the J Rotational Level
• Finding the Most Populated Level
• Putting It All Together 46:06
• Putting It All Together
• Energy of Translation
• Energy of Rotation
• Energy of Vibration
• Electronic Energy

### Transcription: Statistical Thermodynamics: The Various Partition Functions II

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

Today, we are going to continue our discussion of the various partition functions and0004

we are going to talk about diatomic molecules.0008

In the last lesson, we talked about monoatomic ideal gases.0010

Let us go ahead and get started.0017

I will go ahead and stick with black, it is not a problem.0022

For a diatomic gas, the total energy is equal to the energy of translation +0025

the energy of vibration + the energy of rotation + the electronic energy.0038

Therefore, the total partition function is equal to partition function of translation ×0044

the partition function of vibration × the partition function of rotation × the partition function for the electronic energy.0051

The partition function of the system is the individual partition function over that.0062

This Q is this Q, which is composed of these 4Q.0068

Which = 1/ N! Of 2T QV QR QE ⁺nth.0075

We defined the QT, the QV, the QR, and the QE.0092

The translational partition function for a diatomic molecule is actually the same as for a monoatomic.0096

QT, the Q of translation that is just equal to 2 π.0104

The only difference is now we are just going to add the masses M1 + M2.0109

The individual masses of the atoms that make up the diatomic molecule.0114

In other words, the total mass × KT/ H²³/2 × V.0118

That takes care of the translational partition function.0134

We must now choose, now we are going to deal with the rotation, vibration, and electronic.0149

We must now choose our 0 energy marks for these 3 energies because we are going to measure everything from that.0156

We need to pick a 0.0168

We must now choose our 0 energy marks for rotation, vibration, and electronic.0170

The rotation part is easy.0193

The rotation, we will just take the J = 0 quantum state as our 0 energy.0195

That is nice and easy.0201

For the vibration, let me go ahead and draw a little picture here.0203

We have our potential energy curve, looked something like that.0211

For the vibration, this was our ground state energy.0217

The R = 0, it does not equal 0.0222

We can either choose this state, we can choose as the 0 mark the R = 0, or we can choose this.0226

We can choose the minimum of the potential energy curve as the 0 energy mark.0240

We can choose the min of the potential energy curve.0250

Those are two choices.0263

We can pick that as our 0 energy or we can pick that as our 0 energy, that we are going to measure everything relative to.0264

We are going to choose the minimum.0274

We are going to choose that as our 0 energy.0276

For vibration, we choose the min of the potential energy curve.0283

The R = 0 state which we did not pick for our 0 energy, the R = 0 vibrational state has energy R = 0 = H ν × 0 + ½ = ½ H ν.0310

That is the energy of the 0 ground state vibration of energy, not the 0 point energy.0337

That takes care of the vibration, let us talk about the electronic.0346

Let me go back to black here.0351

For electronic, I’m going to draw 2 pictures.0357

Actually 1 picture, 2 potential energy curves.0363

I got 1 potential energy curve here and let us say I got another one.0369

Here is how we can do it.0379

We are going to take our 0 point energy right here.0382

We are going to take the fully dissociated diatomic molecule.0389

When the atoms in infinite distance apart from each other at rest.0392

That is what we are going to take this our 0 point energy.0396

This up here, the first excited state, that is going to be energy 2.0400

Now, we have to energies that we can take.0407

From the 0 down to here, we call that the dissociation energy from the 0 point vibration.0415

The R = 0 vibrational state.0430

We also have this one, we also have this energy.0433

From the 0 point energy, all the way to the energy minimum.0437

This we call the DV.0441

There are 2, right there.0443

We take this as our 0 energy mark for electronic.0446

For the electronic partition function, the fully dissociated molecule is going to be our 0 energy mark.0455

Our energy 1 is actually going to be –DE.0463

These values have been tabulated.0467

This is the dissociation energy.0470

It is dissociation energy of the molecule from the potential minimum.0472

-D0 = - DE + ½ H ν, this energy is just this energy + that little bit.0480

Therefore, the electronic partition function by definition is equal to the sum is equal0493

to the degeneracy of that level × E ^- E1/ KT E + the energy degeneracy of the second level × E ⁻E2/ KT, and so on.0501

But we can stick with two terms.0514

Again E2, the electronic energy is much higher than E1 that we can essentially ignore the second term.0517

We can usually ignore the second term because this energy here is much higher than this,0550

that most atoms of the ordinary temperatures are not going to be in the state.0555

Our electronic function partition function is equal to the degeneracy of that particular level E ⁺D0/ KT.0563

That is the electronic partition function of the diatomic molecule.0574

Let us flip that page and let us see if I can move to the next page here.0582

For Q vibration, let us find the Q of vibration.0588

The diatomic of molecule vibrating harmonic oscillator.0599

The energy of a harmonic oscillator in the vibrational quantum state R is equal to H ν × R + ½.0604

R = 0, 1, 2, and so on.0620

QV is equal to the definition of the partition function.0627

The sum of G sub I E ^- E sub I/ KT.0632

The vibrational levels are none degenerate.0640

G sub I is always equal to 1.0642

The vibrational levels are non degenerate for G sub I=1 always.0652

Therefore, Q sub V = the sum R goes from 0 to infinity of E ⁻H ν R + ½ divided by KT.0666

We go through a bunch of math, I will just go ahead and write it off, which is absolutely not altogether that important.0684

And we end up with QV is equal to E ^- H ν/ 2 KT divided by 1 - E ⁻H ν/ KT.0692

And now this is our vibrational partition function.0712

Now we define something called the characteristic vibrational temperature.0716

We define the characteristic vibrational temperature, it signified θ sub V is equal to H ν/ K.0722

For that particular ν is the fundamental vibration frequency of that molecule.0752

Therefore, expressed in terms of this thing here I can put the H ν and K together.0757

I end up with the vibrational partition function is equal to E ⁻θ/ 2T divided by 1 – E ^- θ/ T.0763

There you have your vibrational partition function for a diatomic molecule.0779

Let us see if I want to actually do, that is fine.0786

I will go ahead and do this part.0801

If the temperature, if the 10th is very high or very low, this last expression can be simplified.0804

I will go ahead and write the simplified expression but the simplified expressions are not the ones that you are going to use.0825

It is only for very high or very low temperatures.0830

I’m just listing them here for the sake of completion.0832

Where it can be used when you use the vibrational partition function is the expression that we just wrote.0835

That is the one that you want to use, not be simplified versions.0839

This expression can be simplified.0844

Let us see, case 1 temperature is very low.0846

If the temperature is very low then the θ of V divided by one of the temperature is going to be a lot greater than 1.0854

That implies that E ⁻θ sub V/ 2T is going to be a lot less than 1.0866

The second term in the denominator can be ignored.0878

What we end up with is E ⁻θ/ 2T, this is the original expression E ⁻θ/ T.0894

We can ignore this term and we end up with QV is equal to E ^- θ/ 2T, if the temperature is very low.0905

In other words, if θ V over whatever temperature it happens to be.0916

10 K is a lot greater than 1.0920

The other one is, if the T is very high.0925

If T is very high and the θ of V/ T is a lot less than 1 which implies that we can write E ^- θ V/ T as - θ/ T,0932

that implies that the denominator is -1 -θ of V/ T which = θ V/ T.0957

Q of V = E ⁻θ V/ 2T/ θ V/ T which = T × E ⁻θ V/ 2T/ θ V.0976

This gives us the high temperature partition function.0997

In general again, try to avoid this unless you are talking about really very higher or very low temperatures.1000

In general, use QV is equal to E ^- θ V/ 2T / 1 –E ⁻θ V/ T.1015

Note, you do not have to wait until the final Q to find U = KT² D LN Q DT which is equal to N KT² D Q DT.1039

You do not have to wait until that.1087

You can do anytime.1089

Any time you find any partition function vibration, rotation, electronic, translational,1090

you can go ahead and use that Q in here to find that contribution to the energy.1095

It is that simple.1101

You can do at any time to find U C V, etc. for any component.1109

For example, for the vibrational component.1130

The energy of vibration is equal to the average energy of the vibration is equal to N KT² D LN Q of vibration/ T.1149

CV for vibration is nothing more than a derivative of what we just get above, the derivative of U of vibration DT.1173

If I want to know how much the vibrations of a molecule contribute to the overall energy heat capacity whatever,1192

let us just pick that one, overall heat capacity, I simply find the heat capacity of the vibrational component.1229

Let us find the fraction of molecules in the rth vibrational states.1253

We are always concerned with the fraction that is in a particular state.1258

We found the vibrational partition function, now let us find the fraction.1262

Fraction of molecules in the R quantum state, in the rth vibrational state.1271

Let me go back to black here.1285

A fraction is equal to the term for that energy divided by the total partition function E ^- H ν R + ½ divided KT/ Q of vibration.1290

It = E ^- H ν R + ½ divided by KT/ E ⁻H ν/ 2 KT/ 1 - E ^- H ν/ 2 KT.1313

Once again, we go through a bunch of math and we get that the fraction is equal to 1 - E ^- H ν/ KT × E ^- H ν R/ KT1356

or in terms of this thing called the characteristic vibrational temperature, we have 1 - E ⁻θ V/ T × E ⁻R θ V/ T.1380

This gives me the fraction of molecules in the rth vibrational state.1398

If I want to k now which fraction is in the 3rd vibrational state, I put 3 in there and I work this out for the molecule.1402

For example, let me go to red.1410

At 298 K, what fraction of carbon monoxide molecules are in the R = 0 and R = 1 states?1423

Θ V for carbon monoxide is equal to 3103 K, that is tabulated.1457

The fraction in the R = 0 state = 1.1467

I will just put R in there.1475

1 - E⁻³¹⁰³ divided by 298 × E⁰ × 3103 divided by 298 and this is going to be approximately 1.1479

Let us see.1505

When I do the same thing, when I do the fraction of the R = 1 state,1508

that is going to equal 1 - E⁻³¹⁰³ divided by 298 × E⁻¹ × 3103 divided by 298.1512

I get the 3.0 × 10⁻⁵.1534

Once again, the fraction that is in the first excited electronic state is only this.1544

Virtually, all of the molecules at 298 are going to be in the first, are going to be in the ground electronic state.1553

Let us find the Q of rotation.1570

The energy of rotation E sub J is equal to H ̅²/ 2I × J × J + 1.1581

Let us make this J a little bit better.1593

J × J + 1, or J takes on the values of 0, 1, 2, and so on.1596

These are the energies of a diatomic molecule.1601

These are the rotational energies.1604

G sub J, t he degeneracy is equal to 2J + 1.1606

Well, the Q of rotation = the sum/ I of the G sub I × E ⁻E sub I / KT.1612

Q sub R, when I put this into there, this expression into there, I get the sum J takes on the values1630

0 to infinity G sub J × E ⁻E sub J/ KT is equal to the some as J goes from 0 to infinity 2J + 1 × E ⁻H ̅² J × J + 1/ 2I KT.1641

We are now going to define something called a characteristic rotational temperature, θ sub R.1682

That is equal to H ̅²/ 2 I K, Boltzmann constant K.1692

Therefore, we get Q sub R is equal to the sum J = 0 to infinity 2J + 1 × E ^- θ R × J × J + 1 divided by T.1702

This is the rotational partition function.1724

There is no close form expression for the summation.1732

I simply have to take however many terms I think I need, 2, 4, 6, 8, 10, 20, 30, 40.1735

It does not really matter.1740

Let us stick with blue, it is a happy color.1745

There is no closed form expression for the summation.1760

However, because θ R is usually a lot less than the temperature at which we happen to be dealing,1778

θ R is a lot less than T.1791

In most cases, we can make simplifying approximations to give Q sub R = T/ θ sub R which is equal to 2I KT/ H ̅².1793

Most of the time we can go ahead and use this.1834

If θ sub R is a lot less than the particular temperature that we happen to be dealing with.1836

Now when θ sub R is not much less than T, we simply use the expression we have.1842

Q sub R = the sum J = 0 to infinity 2J + 1 × E ^- θ R × J × J + 1/ TT and take as many terms as necessary.1865

Let us see.1908

For example, at 298 K for carbon monoxide, the θ of R = 2.77.1912

This is definitely a lot less than 298.1933

For H2, R θ R is equal to 85.3.1940

This is not a lot less than 298.1947

For this case with carbon dioxide, we can use this approximation.1951

In the case of hydrogen, we can use this approximation.1954

We have to use the actual partition function itself.1957

Here, we must use the sum.1963

It is not a problem.1970

It is an easy thing, not a big deal.1971

You have math software to do this for us.1974

This expression QR = T/ θ R is valid for hetero nuclear diatomic molecules, No, Co, Hcr, Hbr,1995

For homonuclear or homonuclear diatomics, CL2, BL2, N2.2025

QR is actually equal to T/ σ θ R.2041

There is a σ! or that we have to put in.2045

The reason we have to put in is called a symmetry number.2049

The reason it is there is because a homonuclear has extra symmetry but a heteronuclear does not.2052

In general expression, for all diatomic molecules is Q sub R = T/ σ θ sub R = 2I KT/ σ H ̅².2084

Σ is the cemetery number of the molecule that has been tabulated for various molecules.2122

Symmetry number of the molecule, it is really easy.2130

For a homonuclear, σ is equal to 2.2138

For heteronuclear, σ is just equal to 1.2149

And again, some of the numbers have been tabulated for various molecules.2157

Let us see what we have got here.2163

Let us go back to blue.2165

Our U of rotation or energy of rotation is equal to N KT² D LN Q of rotation DT constant V.2168

Using Q of R = T/ θ sub R, we get LN of Q of R = LN of T - LN of θ sub R.2187

DDT constant V of LN Q sub R = the derivative of this which is 1/ T – 0, we get 1/ T.2210

N KT² × 1/ T gives you equal N KT or RT.2230

The heat capacity is just equal to the derivative of this.2248

At constant volume, the derivative of this is just R.2255

The rotational contribution of energy is RT.2260

The rotational contribution, the heat capacity is R.2262

Now, a diatomic molecule has 2 rotational degrees of freedom.2267

In other words, if this is a diatomic molecule, it can rotate this way or it can rotate this way.2274

This rotation does not count, only this and this.2278

A diatomic has 2 rotational degrees of freedom.2292

A degree of freedom is just a fancy word for motion.2301

It has 2 ways it can move.2305

When we say something has 3° of vibrational freedom that means we are saying it has 3 ways that it can actually vibrate.2305

That is it, that is all degree of freedom means.2313

If it has 3 translational degrees of freedom, it has 3 ways it can actually move in space.2315

There you go.2321

A diatomic molecule has 2 rotational degrees of freedom.2324

U is equal to RT, it is equal to ½ RT + ½ RT.2327

Each degree of freedom contributes RT to the energy.2334

CV = R, each degree of freedom contributes ½ R to the heat capacity.2338

Let me see, it is a little long here, we are almost there.2349

Fraction of molecules in the various rotational states.2357

Fraction of molecules in the Jth rotational level.2370

Again, we choose level because we have the degeneracy.2378

There are going to be many states in that particular energy level.2382

That is equal to this.2386

A fraction in the Jth energy level, the equation is going to be 2J + 1 × E ⁻θ R ×,2387

The fraction in the Jth level is equal to 2J + 1 × E ^- θ R × J × J + 1/ T divided by T/ θ R.2408

The particular energy provided by the partition function.2428

For example, at 298 K.2433

Let me see, the fraction of N2 molecules in the first 12 rotational levels is,2461

in this particular case we have θ of rotation of N2 is equal to 2.88.2492

In this particular case, the T/ θ R molecule of rotational level.2499

Do not forget the σ.2504

Let me go ahead and we have these values.2507

We have J and we have F of J.2509

I’m just going to go ahead and do the,2516

You know what, I do not think I have enough room here.2518

I’m just going to do the even ones, the even values of J.2520

At 0, 0.0097, 2nd 0.0456, 4, 6, 8, 10, 12.2525

The 4th is 0.0717, 0.08, 6th 0.0837, 0.0819, 0.0701, 12th 0.0535.2542

Notice how it goes up, that it peaks out and then it starts coming down again.2567

That means that somewhere around the 6th and 7th rotational level that is the one that is most heavily populated.2573

The fraction of 0, the fraction of 2nd level, the fraction of the 4th level, fraction of the 6th level.2581

You also have 12, 1, 3, 5, 7, 9.2585

I did not have enough room here on this page and I want to do it on one page.2588

I just put the even numbers.2591

At normal temperatures, let me go back to black.2595

At normal temperatures, most molecules are in the excited rotational states.2602

That is what these numbers prove.2623

Under normal temperatures, for vibrational states they are in the ground state.2628

For electronic states, they are in the ground state.2633

For rotational states, they are in excited rotational levels.2635

Most molecules that are in the ground state, they are in excited states rotating very fast.2638

Most molecules are in excited rotational states.2645

Now, if we treat F of J as a continuous function, it is not a continuous function.2647

J is discrete 0, 1, 2, 3, 4, 5 but we treat it as a continuous function.2661

We can take the derivative, we can take D of F of J with respect to J and we consider it equal to 0 to find the maximum.2672

What J value is the most highly populated, = 0.2688

If we know what function, we can take that and find the most populated level.2694

It ends up being the J max is equal to T/ 2 θ R ^½ - ½.2713

For our example above which was nitrogen, the J max of N2 is equal to 298/ 2 × 2.288 ^½ - ½, we got 7.2727

The most populated rotational level is J = 7 at 298 K.2755

Let us go ahead and put all this together now.2763

Let us see what we have got.2767

Let us put it all together.2769

I will do blue.2784

Q is equal to Q of translation, Q of vibration, Q of rotation, Q of electronic.2792

Therefore, Q is equal to 2 π M KT.2802

M is the mass of the molecule/ H².2808

3/2 V × T/ σ θ R ×, in this particular case, V is the approximation if θ R is a lot less than T.2816

If not then we have to use the summation, this is the rotational partition function.2835

I’m sorry, let me switch orders here.2842

Let me make sure.2844

Rotation vibration, θ × vibrational partition function which is E ⁻θ of V/ 2T/ 1 - E ⁻θ V/ T × G of 1 E ⁺D of E/ KT.2846

Q = Q ⁺nth/ N! Is this whole thing.2876

This is crazy.2888

Let me see, I think I do not have to go through all of the math here.2892

You know what, that is fine.2901

Q is equal to KT² D LN Q DT = N KT² D LN Q DT LN Q.2907

This whole thing is equal to 3/2 LN of 2 π M KT – 3/2 LN H² + LN V + LN T - LN σ θ R2933

- θ of V/ 2T - LN of 1 - E ^- θ V/ T + LN of G1 + DE/ KT.2967

LN Q is all of that.2987

D of LN Q DT, I’m going to differentiate all of this with respect to T.2992

Not all of these are constants not a problem.2995

D LN Q DT holding V constant is going to equal 3/ 2T - 0 + 0 + 1/ T - 0 + θ V/ 2T² +2998

θ V/ T × E ⁻θ V/ T/ 1 – E ⁻θ V/ T + 0 - DE/ KT².3027

I think I’m going to skip a couple of steps because there is a lot of math here.3062

There is a lot of algebra, it is all algebra.3065

What is it that I wrote, if you to simplify everything, multiply a few things out and collect terms,3068

what you will end up with is the following.3072

You end up with 3/2 RT + RT + R × θ of V/ 2 + R × θ V3074

× E ^- θ/ T/ 1 – E ⁻θ V/ T – N DE.3093

Let us call this term 1, term 2, term 3, term 4, and term 5.3107

Term 1 is the energy of translation.3113

We have RT/ 2 for each degree of freedom, translation of X, translation of Y, translation of Z.3125

They add up to 3/2 R.3137

Term 2, that is the energy of rotation.3142

We have RT/ 2 for each degree of freedom.3153

A diatomic molecule has 2° of rotational freedom.3160

Number 3 is the energy of vibration.3165

This one is the 0 point energy.3176

This is also energy of vibration at beyond the 0 point.3187

And this is the electronic energy, the first term.3217

This is relative to 0 mark, we chose which was a fully dissociated molecule.3225

There you have it, those are the partition functions for monoatomic and diatomic molecules.3241

Thank you so much for joining us here at www.educator.com.3246

We will see you next time, bye.3248