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Example Problems 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
  • Example I: Make a Plot of the Fraction of CO Molecules in Various Rotational Levels 0:10
  • Example II: Calculate the Ratio of the Translational Partition Function for Cl₂ and Br₂ at Equal Volume & Temperature 8:05
  • Example III: Vibrational Degree of Freedom & Vibrational Molar Heat Capacity 11:59
  • Example IV: Calculate the Characteristic Vibrational & Rotational temperatures for Each DOF 45:03

Transcription: Example Problems II

Hello, welcome back to, welcome back to Physical Chemistry.0000

Today, we are going to continue our example problems in statistical thermodynamics.0004

Let us jump right on in.0008

The first question says make a plot of the fraction of carbon monoxide molecules 0012

in various rotational levels versus J at 298 K and 1000 K.0018

We are going to have 2 graphs.0025

We are going to use one on top of the other.0026

One is going to be in 298, one is going to be a 1000.0029

We are going to graph a fraction of the molecules vs. That J value.0033

Let us go ahead and let us do it.0044

A fraction of the molecules in a given rotational state J is equal to the fraction in that level 0049

divided by the partition function which is the sum of all of the rotational levels.0058

It is going to be G sub J E ⁻E of J/ KT divided by the rotational partition function itself.0065

G sub J is 2J + 1, this is the energy of the Jth rotational level is E ^- the characteristic rotational temperature × J × J + 1 divided by the KT.0085

And the partition function itself is we had that simplification that we can use,0111

whenever the rotational characteristic temperature is a lot less in the actual temperature at which we happen to be working.0118

Now, for carbon monoxide, our characteristic rotational temperature is equal to 2.77.0129

2.77 is a lot less than 298.0138

We are justify it in using this simplified version of the rotational partition function in the denominator.0144

If it were not the case, then we have to use the actual summation, the full formula for the partition function.0151

Therefore, when we do this we get the fraction in J =2J + 1.0160

In other words, you are just simplifying this equation.0169

I can live it like that, I can enter myself to have plot a formula like that but 2J + 1, I will go ahead and do it this way.0171

Θ R/ T × E ⁻θ × J × J + 1/ T.0179

The θ R includes the K, my apologies.0198

This is the equation that we are going to graph and when we put in our software for our 2 values of T.0208

For one value, we are going to put T = 298.0214

For the second plot, we are going to put T = 1000 and what we will get is the following.0218

This is the plot that we get.0225

This is the plot for T = 298 K.0227

And this is for T = 1000 K.0233

X axis, these are all that J values.0240

We see that it is at 298 K, the J values are becoming more populated 1, 2, 3, 4, 5.0242

It looks like it is a maximum of about J7.0249

The J = 7 at 298 is the one that actually most populated.0253

At 1000 K, that looks like the most populated or somewhere around 13 to 15.0259

Somewhere around, maybe 13, something like that.0265

It goes down as the J goes up.0268

More populated, least populated.0272

Notice, it is exactly what we hope to notice.0276

Clearly, at higher temperatures, higher values of J become more populated.0280

In other words, higher rotational states become occupied.0296

Become occupied or I will just say populated.0305

Let us say at certain temperature is this.0310

As we raise the temperature, notice more of them become populated.0311

Here, this temperature drops off 15, 20, 25 are not too many.0318

As we raise the temperature, now the 15, 20, 25, 30, 35, are all becoming significantly populated.0325

And of course, it is exactly what you expect.0332

As more and more become populated, the ones in the lower levels are moving to the upper levels.0335

These graphs actually go down here and bulge out here.0341

Again there is only a certain number of molecules beginning to be gathered in 5 levels0346

or you can spread them out for 20 levels.0351

They go down over here, they go more over here.0354

As we said before, if the θ, the characteristic rotational temperature is not significantly less than T, 0356

I would say somewhere in the range of 1 to 5%.0378

Generally, 5 - 6% is about where I cut it off and use the whole partition function instead of the simplified version.0382

1 - 5% of T, then this Q of R = T/ θ of R cannot be used, which is not a problem0391

because we have software, we have calculators.0407

For example, in the case of hydrogen, it is like that.0417

I think the θ of rotation hydrogen is really very high.0420

In this case, we use the full partition function for rotation.0424

In other words, we use the sum.0441

We use Q of R = the sum from J equal 0 to J = infinity of 2J + 1 E ⁻θ RJ × J + 1/ T.0443

We just use this.0462

In the case of rotation, it depends on the temperature.0464

Most normal temperatures up to about 11500 K, you might want to use maybe 7500 terms.0468

Just to be on the safe side.0476

You are using software, it does not really matter.0477

That is it, nice and simple.0481

Calculate the ratio of the translational partition functions for CO₂ and Br₂ at equal volume and temperature.0487

Once again, for ratio of the translational partition functions.0495

The partition function for Br₂ or CO₂ divided by the partition function for CO₂ at equal volume and temperature.0498

Let us write out what the translational partition function is.0506

We already know what that is.0508

The translational partition function is 2 π × the mass of the molecule × Boltzmann constant × 0510

the temperature/ planks constant² all raised to the 3/2 × the volume of the container.0517

The Q of translation for CO₂/ the Q of translation for Br₂ that =, put them in.0529

2 π the mass of CO₂ × KT/ H²³/2 divided by 2 π × the mass of Br₂ not CO₂, that is on top.0541

The mass of Br₂ × K × T/ Plancks constant²³/2.0559

Everything cancels.0566

You are left with the ratio is just the mass CO₂ divided by the mass of Br₂.0569

Therefore, the ratio is equal to 5.89 × 10⁻²⁶ kg divided by 1.33 × 10⁻²⁵ kg.0578

And again, I got these numbers by just taking the molar mass and dividing by Avogadro’s number.0597

I actually get the mass of one molecule and I converted into kg.0600

I did not mean to, the kg cancel.0606

I can leave these grounds if I want to, it is not a problem.0608

That = 0.4429 and I hope that I get my arithmetic correctly, 0.4429.0611

I like to see it in terms of a number larger than 1.0618

I’m going to go ahead and take the reciprocal of this.0625

1/ 0.4429, in other words the ratio of the Br to the CO.0628

Br₂ and CO₂, I just flipped the ratio.0638

That = 2.26, that just makes more sense to me personally.0641

What this says is, at the same volume and temperature, Br₂ has about 2 ¼ ×, in another words 2.26 × 0645

as many translational states accessible to the molecules.0677

That is all this is, accessible to the molecules.0690

If CO₂ had 100 translational states available to it, at this volume and temperature, 0698

what this says is the bromine has 226 states available to it.0705

That is all, nice and simple.0709

That takes care of that one.0717

Now, calculate the contribution of each vibrational degree of freedom of H2O to the vibrational molar heat capacity at 500 K.0719

When I do the lessons for statistical thermodynamics, I talked about monoatomic gases and0732

I have talked about diatomic gases, I did not talk about polyatomic molecules.0736

I just decided to go ahead and use these example problems to talk about polyatomic molecules.0745

I will go ahead and do that.0750

I’m going to use these examples, in particular this example, to discuss the partition functions of polyatomic molecules.0753

Let me go to blue here.0794

A polyatomic molecule with N atoms, by N we just mean the number of atoms.0800

3, 4, 5, 6, 7, whatever, in the case of H₂O N =3.0813

A polyatomic molecule with N atoms has 3N degrees of freedom.0817

A degree of freedom is just a fancy way of saying a way to move.0827

In the case of water, a water molecule has 3 atoms.0843

3 × 3 is 9.0849

It has 9 different ways that this molecule can move.0850

It has translational movement that accounts for 3 of them.0858

In other words, it can move along the X axis, it can move along the Y axis, and it can move along in the Z axis.0861

That takes care of 3° of freedom, that 3.0866

It has rotational movement.0872

In other words, it has rotational degrees of freedom.0876

In the case of a linear polyatomic molecule, there are 2 rotational degrees of freedom.0879

In other words, if this is a polyatomic molecule, let us say 3 atoms that is linear like CO₂, carbon oxygen, 0883

it has 2 rotational degrees of freedom .0895

It can rotate this way or it can rotate this way.0898

This degree of freedom that rotates along the axis itself does not count.0900

It has 2 rotational degrees of freedom, that is linear.0906

If it is non linear as in the case of water, it has 3 rotational degrees of freedom.0910

It has 3 translational degrees of freedom, 3 rotational degrees of freedom that leave vibrational degrees of freedom.0917

How can it actually vibrate?0924

In the case of linear, we said it has a total of 3N degrees of freedom.0928

3N -3 – 2, we have a total of 3N -5 ways that this molecule can vibrate.0934

In the case of nonlinear, it has 3N - 6 ways that this molecule can vibrate.0943

The polyatomic molecule has 3N ways that it can move.0956

3 of them is taken care by translation.0960

Either 2 or 3 is taken care of by rotation.0961

The rest is vibrational motion, that I sit.0964

Each of these degrees of freedom are called,0970

Now we are talking about vibrational.0978

It is getting a little crazy here.0983

Each of these vibrational degrees of freedom is 3N -5 and 3N -6, 0992

vibrational degrees of freedom act as an independent harmonic oscillator.1000

Instead of one vibrational degree of freedom for a diatomic like we had, 1026

one vibrational degree of freedom for diatomic which gave us 1 vibrational partition function.1042

Each vibrational degree of freedom gives us a particular vibrational partition function.1056

Each degree of freedom contributes its own partition function.1064

That is why.1074

In the case of a diatomic molecule, something like that, we have 2 atoms, N = 2.1075

3 × 2 is 6.1083

6 – 3, the translational degrees of freedom that leaves us with 3° of freedom .1088

A diatomic molecule is automatically linear.1095

Therefore, it has 2 rotational degrees of freedom.1099

3 -2 that leaves us 1 vibrational degree of freedom.1103

That vibrational degree of freedom is something that we just see.1107

That is it, back and forth.1110

This is one oscillator, it can only oscillate back and forth.1111

Water molecules is not about that way.1114

The water molecule has 3 atoms.1117

3 × 3 is 9, if I take away 3 of those degrees of freedom as translation, that leaves you with 6.1120

It is nonlinear so it has 3 rotational degrees of freedom.1127

That take away another 3that leaves me with 3 vibrational degrees of freedom.1130

Now notice, the water molecule has 2 bonds.1134

The bonds do not account for the degrees of vibrational freedom.1141

It is not vibration here, vibration here.1143

It is us something else.1146

It actually has 3 different ways that it can vibrate.1148

I do not want to go into that.1153

That actually falls under the purview of spectroscopy but just know that these are the numbers.1156

In the case of water, we have 3 vibrational degrees of freedom.1161

Because we have 3 vibrational degrees of freedom, 3 ways that it can vibrate independently.1165

We have 3 partition functions.1171

We have 3 vibrational partition functions for water.1173

Each one contributes, each vibrational mode has its own fundamental vibration frequency.1176

Therefore, it has its own characteristic vibrational temperature.1183

It has its own partition function.1187

That is what is happening here.1190

As we said, the energy of vibration is H ν × R + ½, for each degree of freedom.1199

The energy of the vibration total is equal to the sum of H ν sub Y × R sub I + ½.1214

And as I goes from 1 to either 3N - 5 to 3N -6 depending on how many degrees of freedom you have.1245

The total energy of vibration is the energy of one oscillator, the energy of the other oscillator,1253

the energy of the other oscillator.1258

In the case of water, depending on how ever many degrees of freedom you have,1260

you just add up the energies of the individual degrees of freedom.1263

Because the energy is the sum of the energies of the different degrees of freedom, 1269

a partition function is the product of the partition functions for each degree of vibrational freedom.1274

The total vibrational partition function is the product of the individual partition functions for each degree of freedom.1285

In other words, Q of V is equal to this P which is like the big Greek symbol S for sum, Π means product.1323

I = 1 either 2, 3N -5 or 3N - 6 of E ^- θ of V sub I/ TT/ 1 - E ⁻θ V sub I/ KT.1337

Do not worry, this will make sense when we actually do a problem.1358

I will capitalize this DOF.1361

Each degree of freedom, each motor movement, each way of moving has its own fundamental vibration frequency ν.1371

I will put a tilde over it because it is a ν sub I.1397

Therefore, each vibrational degree of freedom has its own characteristic vibrational temperature.1402

It has its own θ of V sub I which is equal to H ν sub I/ K.1421

In this case, I left it without a tilde.1430

Vibrational frequency just depends on,1433

That is fine, you should not cause too much of confusion.1437

When you do not see a tilde, it is in Hertz.1439

When you see a tilde, it is in inverse cm.1443

Just take the one in Hz divide by the speed of light and you get the one in inverse cm.1446

Energy is equal to NKT² D LN Q of V DT constant V.1455

It = N × K × the sum.1469

I = 1 to either 3N - 6 or 3N - 5 of, I'm just adding the ones for each degree of freedom.1476

Θ of V sub I/ 2, I already have seen these formulas, + θ of V sub I E ⁻θ V sub I/ T/ 1 – E ⁻θ V sub I/ T.1492

CV, an expression for the constant volume heat capacity is equal to N × K × the sum 1, 2, 3, N – 5/ 3N -6.1520

This time it is going to be θ V sub I/ T² × E ⁻θ V sub I/ T/ 1 – E ⁻θ V sub I/ T².1538

Let us deal with the water.1563

We might do this in black.1565

Returning to water, H₂O, we have 3 atoms so N = 3.1570

That implies that we have 9° of freedom, that was nonlinear.1581

3N - 6 = 9 -6 = 3 vibrational degrees of freedom.1595

In other words, the water molecules vibrate in the 3 different ways.1606

There are 3 individual harmonic oscillators for this.1609

There are θ of vibration 1, θ of vibration for the other degree of freedom,1619

and θ of vibration for the third degree of freedom.1625

These are tabulated for us, all we have to do is look them up.1630

The θ of vibration for the first degree of freedom is 2290 K.1636

The θ of vibration 2 = 5160 K.1641

Let me make my 5 a little bit clear here, 5160. 1649

My θ of vibration for the third degree of freedom is equal to 5360.1652

I just put these values in.1661

CV is equal to R.1665

For each degree of freedom, I just use the same expression as I had back when1670

I was working with the diatomic molecule that adds them up.1674

That is what this is.1678

It just says add them up.1680

R θ V/ T² × E ⁻θ V/ T/ 1 – E ⁻θ V/ T².1684

I just put these values in.1702

CV, the constant volume heat capacity for the 2290 value = 8.314 × 2290/ 5001705

because this is happening at K, × E⁻²²⁹⁰/ 500/ 1 – E⁻²²⁹⁰/ 500².1721

I end up with 1.826 J/ mol K.1737

That is the contribution of one of the degrees of freedom.1745

We will take the second degree of freedom, this one.1750

The contribution of the heat capacity for the 5160 =, again same thing, 0.314.1756

It is the same formula, we are just doing them one at a time and this just says add them up.1763

8.314 × 5160/ 500² × E⁻⁵¹⁶⁰/ 500/ 1 -E⁻⁵¹⁶⁰/ 500².1770

I get 0.0292 J/ mol K.1791

Then, I have my third contribution which is CV of 5360, that is equal to 8.314.1799

Let us slow down a little so I can write it more clearly.1812

Take a deep breath.1816

8.314 × 5360/ 500² × E⁻⁵³⁶⁰/ 500/ 1 –E⁻⁵³⁶⁰/ 500².1818

And I end up with 0.0211 J/ mol K.1839

I just add them up if I need to, not a big deal.1854

I found the contribution.1857

Water has 3 vibrational degrees of freedom.1861

Each level of vibrational degrees of freedom contributes this to the overall heat capacity.1864

That is it, that is all.1870

If I want to know what the total contribution of the vibrational mode is, I just add these up, these 3 values.1872

However, I want to do it.1880

Let me go back to black.1882

Notice, θ E = H ν/ K.1889

Where ν is the fundamental vibration frequency.1901

The higher the fundamental vibration frequency, the higher θ V is.1922

The higher θ V corresponds to a stiffer oscillator.1949

It is hard to get it going.1966

Therefore, higher temperatures are needed before these modes, 1972

before these degrees of freedom contribute significantly to the overall CV.1996

Notice these last 2, .02, .03, the first one, the big one.2014

The first one, the θ V was very small.2019

It means that the particular mode of oscillation, it is not very stiff.2022

It oscillates rather freely.2025

Because that is the case, under reasonably low energy conditions than 500 K,2027

most of the heat capacity is going to come from that.2033

In other words, mostly energy goes towards that mode of oscillation.2038

These last two, because they are stiff, if I wanted to actually activate them, 2043

if I wanted to make them contribute more to the overall heat capacity, I actually have to raise the temperature.2049

Once I raise the temperature then these stiffer oscillators, they start to engage.2057

That is all that is happening here.2063

We just talked about the vibration of a polyatomic molecules.2069

Now, let us talk about the rotational partition function of polyatomic molecules.2075

If we had a linear polyatomic, there is actually no difference.2104

It is actually exactly the same as it is for a diatomic molecule.2109

In the case of something like CO₂ which is a linear triatomic molecule, the partition function is going to be the same.2112

A linear polyatomic same as the diatomic.2121

In other words, the Q sub R is equal to the sum as J goes from 0 to infinity 2J + 1 × E ⁻θ R × J × J + 1/ T.2131

We said that θ R is equal to this characteristic of temperature of rotation.2153

Where θ sub R is equal to H ̅²/ 2 π × K.2163

It is also equal to HB/ K.2174

It is equal to HCV ~/ K.2178

All of those depending on what you are given, rotational constant, inverse cm, and Hz, or just the rotational inertia depending.2182

I is the rotational inertia of the molecule.2190

As before, when the θ of R is a lot less than the temperature which we are operating, which is for most molecules,2197

we can use Q of R = T over σ of θ of R, where σ is the symmetry number of the molecule.2203

Again, these have been tabulated.2225

Once again, if we have a linear symmetric molecule, the σ is going to equal 2.2226

If we have a linear asymmetric molecule, the σ is going to equal 1.2250

Some examples, CO₂ is a linear symmetric molecule.2259

Its symmetry number is 2.2269

Something like HCN hydrogen cyanide or maybe N2O, these are linear asymmetric.2273

Their symmetry numbers are going to be 1.2280

Nonlinear polyatomic like water.2287

Once again, each rotational degree of freedom has its own partition function.2305

Each rotational degree of freedom has its own rotational inertia.2315

It has its own moment of inertia.2326

Therefore, each rotational degree of freedom has its own characteristic rotational temperature.2332

Θ R sub I = H ̅²/ 2 I sub I K = HB sub I/ K = HC B~ sub I/ K.2350

Each rotational degree of freedom has its own moment of inertia, rotational moment of inertia, 2373

its own rotational inertia.2377

Therefore, each degree of freedom has its own characteristic θ of rotation.2379

It has its own characteristic rotational temperature.2384

They are independent rotated, that is it.2386

It is just rotating, this axis rotating along another axis.2388

If the θ of rotation A is equal to the θ of rotation of B axis is equal to the θ of rotation of the C axis,2391

we call that a spherical top.2407

It all depends on what these θ are.2414

If none of them, if all of them are equal, we call that type of molecule a spherical top.2417

In that particular case, the rotational partition function is actually 2421

the product of the rotational partition functions of each degree of freedom.2429

Just like before.2433

Our final answer, our final result as part of the ½ over the symmetry number × temperature/ θ sub RA ^½ of the temperature/ θ sub RB ^½, these are all the same.2434

I’m just going to end up multiplying them to the second.2454

T/ θ sub RC ^½ K.2457

Since, θ R = θ RB = θ RC, for a spherical top this simplifies 2 π ^½/ σ × T/ θ sub R, whatever it happens to be³/2.2463

That is the first case.2485

If two of them are equal, in other words if θ sub RA = θ sub RB, but neither one is equal to the 3rd θ sub R of C, 2486

we call this asymmetric top.2506

The molecule is called asymmetric top.2510

We have a spherical top, when all the θ of R are equal.2513

It is a fully symmetric molecule.2518

And we have asymmetric top, when two of them are equal but not to the other.2520

In this particular case, the Q of R is the same thing.2528

It is π/ ½ × σ, I’m going to call this equation 1.2533

I’m going to call this equation 1.2540

This is the basic equation.2541

It is π¹/2/ σ, the symmetry number × the temperature/ the θ RA × the temperature of RB.2544

Each on is to the ½ power.2551

This is the basic equation.2552

Since, these are all equal, it simplifies to this.2553

In this case, it simplifies to, I will just say equation 1 which ends up simplifying to,2556

I can multiply two of them together, I will just use A.2567

It ends up being that.2577

Our final case is if none of them are equal.2584

If θ sub RA is different than θ sub R of B, it is different than θ sub R of C, in other words,2588

if the rotational inertias are different.2600

The X axis, the Y axis, the Z axis, then what you have is just equation 1.2605

Q sub R = equation 1.2611

I will write that again.2615

Π ^½/ σ × T/ θ of RA ^½ T/ θ sub RB ^½ T/ θ sub RC ^½.2618

Now, the energy is equal to NKT² D LN Q of R DT.2638

I work all that out and I get 3/2 RT.2652

When I take DDT of this, I get 3/2 RT.2658

When I take the heat capacity, I get 3/2 R.2664

Each rotational degree of freedom contributes R/2 to the overall rotational heat capacity.2671

There you go.2696

Let us do a problem.2698

The following spectroscopic data was determined for sulfur dioxide SO₂, fundamental frequencies,2704

the final vibration frequency there are 3 of them.2711

SO₂ is a nonlinear molecule.2715

It is a nonlinear molecule, it is polyatomic.2723

There are 3 atoms in it, that is 9.2727

We have 3 translational degrees of freedom, those we get rid of.2731

We have 3 rotational degrees of freedom and there is 1, 2, 3.2734

And we have 3 vibration al degrees of freedom 1, 2, 3.2739

The fundamental frequencies are ν 1, ν 2, ν3, notice the tildes.2744

These are in inverse cm.2748

These are actually rotational constants not the rotational inertias, I apologize.2751

Let us go ahead and do that out.2762

These are rotational constants, that just come from the rotational inertias via that formula.2763

The rotational constants for one rotational degree to 2.02, the other is 0.344, .293.2769

Calculate the characteristic vibrational and rotational temperatures for each degree of freedom.2777

It is going to be 6 that we are going to do, then calculate the heat capacity at 750 K.2781

Let us see what we can do.2792

The vibrations, let us talk about vibrations first.2795

Let me do this in blue.2799

Let us talk about the vibrations first.2801

We will deal with vibrations first.2805

The θ of vibration is equal to H ν/ K is equal to HC ν~/ K.2810

Therefore, θ of vibration for the first one, I just put the values in.2823

I just put them in. 6.626 × 10⁻³⁴ × 2.998 × 10⁻¹⁰ cm/ s.2830

I will write it out, not a problem.2839

6.626 × 10⁻³⁴ J/ s × 2.998 × 10⁻¹⁰.2842

This is cm/ s because we had here 1154 inverse cm.2853

1154.03, sorry about that.2863

It will make a middle bit of difference.2866

0.03 inverse cm, because we want that to cancel with that.2868

The second cancels with a second here, it is a J/ s divided by K which is 1.381.2873

In this particular case, we leave it as Joules.2882

1.381 × 10⁻²³ J/ K, this is J/ s.2885

The seconds cancel seconds, Joules cancel Joules.2893

We are left with 1660 K.2896

I do the same for θ V2.2902

When we put the 2 up here, I get 752 K.2907

When I do θ V of 3, I get 1960 K.2911

Those are the 3 characteristic vibrational temperatures.2916

Let us go ahead and do rotation.2921

Our rotation, our θ of R = HCB~/ K.2928

Therefore, θ R of A is equal to 6.626 × 10⁻³⁴ J/ s × 2.998 × 10¹⁰ cm/ s ×,2941

in this particular case the rotation constant for A was 2.02 inverse cm/ 1.381 × 10 ⁻²³ J/ K.2967

Cm and cm, J and J.2986

You are left with 2.292 K.2990

I will draw this to make it look a little bit better.2998

2.92 K.3000

I do the same thing for θ R of B.3003

I’m left with 0.495 K and θ R of C is equal to 0.422 K.3007

We want the total molar heat capacity which means the entire partition function.3024

The Q of SO₂, the whole molecule is equal to the partition function of translational, 3033

the partition function for rotation × the partition function for vibration × the partition function for the electronic.3041

We have the Q of SO₂ = the partition function for translation, it is the same.3053

2 π MKT/ H²³/2 V.3065

Polyatomic, diatomic, linear, does not matter.3078

The translational partition function is the same.3081

The rotation partition function π ^½/ σ, this is going to end up being T³/ θ R of A θ R of B θ R of C ^ ½.3085

The vibrational partition function, their product, the 3 vibrational degrees of freedom. to be 1 - 3 of E ⁻θ V sub I/ T/ 1 – E ⁻θ E sub I/ T².3121

No, this is the partition function not the heat capacity.3154

There is no squared here, it is very easy to lose your way with statistical thermodynamics.3158

× the electronic G1 E ⁺D sub E/ KT.3166

The energy of SO₂ is equal to N KT² D LN Q DT.3180

The LN Q is the LN of this whole thing.3193

When we take the heat or we take the derivative of the U, the CV is equal the DU DT.3198

I’m not going to go through all the math.3209

What we end up getting is the following.3212

We end up getting CV equation is 3/2 R + 3/2 R + the sum of 1 to 3 of θ of V sub I/ T² × E ⁻θ V sub I/ T/ 1 – E ⁻θ V sub I/ T².3217

Therefore, here is what it ends up being.3256

Constant volume heat capacity is going to be 3/2 R + 3/2 R +, I’m just going to pull the R out of this one.3260

It is going to be R ×, 3272

I forgot my R, that is fine, I’m good.3278

It is going to be R ×, θ V, the first one is 1660/ 750 K² × E⁻¹⁶⁶⁰/ 750/ 1 – E⁻¹⁶⁶⁰/ 750² + the next one, 3285

the next θ sub V was 750/ 750² × E⁻⁷⁵⁰/ 750/ 1 – E⁻⁷⁵⁰/ 750² + 1960/ 750² 3315

× E⁻¹⁹⁵⁰/ 760/ 1 –E⁻¹⁹⁵⁰/ 760².3348

The CV/ R = 3 + 0.6014 that is the first one, + 0.5820 + 0.5403.3363

The CV/ R = 4.724 J/ mol K.3388

When I multiply by R, I get a final heat capacity of 39.273 J/ mol K.3396

By using partition functions, I was able to calculate the heat capacity of SO₂ gas at 750 K.3408

39.273, the actual value is very close to this number.3420

It is absolutely extraordinary.3425

We did not have to use kilometric methods.3427

We do not have to use anything except just some good nice, theoretical, math.3430

Thank you so much for joining us here at www.educator.com3437

and thank you so much for joining us for this absolutely amazing course in Physical Chemistry.3439

It is been my pleasure teaching you.3446

Thank you, take good care, and I wish you the best of luck, bye.3448