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Example Problems I

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: Calculate the Fraction of Potassium Atoms in the First Excited Electronic State 0:10
  • Example II: Show That Each Translational Degree of Freedom Contributes R/2 to the Molar Heat Capacity 14:46
  • Example III: Calculate the Dissociation Energy 21:23
  • Example IV: Calculate the Vibrational Contribution to the Molar heat Capacity of Oxygen Gas at 500 K 25:46
  • Example V: Upper & Lower Quantum State 32:55
  • Example VI: Calculate the Relative Populations of the J=2 and J=1 Rotational States of the CO Molecule at 25°C 42:21

Transcription: Example Problems I

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

Today, we are going to start doing our example problems for statistical thermodynamics.0004

Let us jump right on in.0009

Using the data table below, calculate the fraction of the potassium atoms0013

in the first excited electronic state at 298 K, 1500 K, and 2500 K.0018

Let us take a look at this data table.0032

We have our ground state here, I will go ahead and do this in red.0034

We have our ground state and again, the ground state is the 0 energy.0043

The first excited state is actually this one right here.0049

Notice, this actually has 2, there is a doublet P 1/2 and a doublet P 3/2.0052

The first excited state is actually this one.0059

This is our second excited state, maybe this is our third excited state.0062

We want to know what the fraction of potassium atoms is in the first excited state at these different temperatures.0069

Let us go ahead and work this out.0082

The fraction is the same as any other fractions.0084

It is a part over the whole.0088

The fraction is going to be the degeneracy of that particular level × E ⁻energy of that level.0090

I’m going to go ahead and call this level 2, I’m going to call this level 3, and I’m going to call this level 4.0101

We are interested in the first excited state which is going to be level 2.0112

The degeneracy E raised to that divided by KT all over the electronic partition function.0119

We will use the first 4 terms of Q sub E, the electronic partition function in the denominator.0132

Now, if you are wondering how I came up with 4, it is just a random pick.0148

I will just pick the first 4 terms.0152

For electronic energy states, actually you can just go ahead and go with the first 2,0156

it is not a problem but I figure what the hell.0159

In the expression for the molecular partition function, it is the sum.0162

You can take 3, 4, 5, whatever you want.0167

Electronic state, I just decide on 4.0170

For rotational state, it is going to be a lot more than that.0174

For vibrational state, 2, 3, 4, that is usually enough.0177

It is a random choice on my part.0181

You are probably using math software, you can use 50 terms if you want.0184

It is going to give you an answer just as quickly.0188

I decided to just go ahead and use my calculator for this one.0192

This is the definition.0194

This is how you are going to do it.0196

The energy in that state divided by the total partition function, that gives us the fraction.0198

Q sub E that is equal to G1 + G2 E ⁻E2/ KT.0204

It is just the sum, the sum of the different + the degeneracy of the 3rd level × its exponential E ⁻E3 ⁺KT + the degeneracy of the 4th level E ^- E4/ KT.0220

That is what the molecular partition function is.0239

I have decided to just take 1 term, 2 terms, 3 terms, 4 terms.0242

I have my energies E2, E3, and E4, that is the energy of the 2nd state,0247

that is the energy of the 3rd state, that is the energy of the 4th state.0253

I have K, that is Boltzmann constant.0258

I have the different temperatures so I have all the numbers that I need.0260

The rest is just arithmetic, really.0263

Let us talk about some degeneracies here.0267

The degeneracies are important.0269

This degeneracy, this is doublet S ½.0271

The degeneracy is the 2J + 1.0274

G sub I is equal to 2J + 1.0280

2 × ½ + 1 gives me degeneracy of 2 for this one.0287

Here, I have a degeneracy of 2.0292

Here, 2 × 3/2 + 1 that is going to give me a degeneracy of 4.0295

And here, this is a doublet S ½, this is going to give me a degeneracy of 2.0302

When I put all of these values into E and everything, here is what I get.0310

I get Q sub E is equal to 2 + 2 × E⁻¹²⁹⁸⁵ and these are in inverse cm, over KT.0316

Let me go ahead and put the values in.0333

The Boltzmann constant, the 1.381 × 10⁻³²³ J/ K, because we are using inverse cm,0336

Boltzmann constant K is equal to 0.6950 inverse cm/ K.0344

We have to watch our units very carefully.0355

0.6950 × T + 4 × E⁻¹³⁰⁴³/ 0.6950 × T + QE⁻²¹⁰²⁷ divided by 0.6950 T.0360

The numerator of the fraction is going to equal 2 × E⁻¹²⁹⁸⁵,0391

because it is the first excited state that we are interested in, over 0.6950 T.0404

Therefore, the fraction at 298 K is equal to the numerator at 298.0412

I put this value, this 298 here divided by Q sub E at 298.0427

In other words, I put 298 in here, here, and here, and I calculate this.0436

I do the sum and I do the division.0441

What I end up with is 1.18 × 10⁻²⁷ divided by 2.0444

This is going to equal 5.9 × 10⁻²⁸.0456

You notice a partition function is around 2.0464

The partition function is a numerical measure of the number of quantum states that are accessible to a molecule,0468

to an atom at that temperature.0477

In this case, it is really only 2.0480

However, it is a degenerate so it is actually this 2.0483

Most basically, what this says is 5.9 × 10⁻²⁸.0488

5.9 × 10⁻²⁶ % of the atoms are in the first excited state.0495

That is what this says.0503

That is reasonable higher.0506

I have raised the temperature higher, the fraction at 1500 K.0510

We raised the temperature a little bit, now we are going to calculate the numerator at 1500,0514

and we are going to put 1500 there.0519

The numerator at 1500 divided by Q sub E, or we put in 15000521

and it is going to be 3.90 × 10⁻⁶/ approximately 2, just slightly above 2.0530

What we end up with here is 1.95 × 10⁻⁶.0540

That is a lot different.0545

Now, the percentage is a lot higher, a lot higher than 10⁻²⁸.0548

By raising the temperature from 300 K to 1500 K, we actually made a greater fraction0553

of these atoms are actually in the first excited state.0560

If we do fraction at 2500, we are going to do numerator at 2500 divided by Q sub E at 2500.0566

And what we end up with is 1.14 × 10⁻³ by approximately 2.0584

It is going to be approximately 5.7 × 10⁻⁴.0602

We see as the temperature is rising, the fraction of atoms in the first excited state is going up.0608

More of the atoms are jumping from the ground state.0618

They are spending more time in the first excited state.0621

Not a lot, it is still small but it is still pretty reasonable.0625

Just by going up to 2500 K, we have pushed them up into and spend more time in the first excited state.0630

Not too high.0638

What is happening here, as the temperature rises more atoms are moving from the ground state to the first excited state.0641

In other words, the population distributions changing.0689

We said that the partition function is a measure of the states that are thermally accessible to a particle at a given T.0705

At these temperatures, level 1 which is 2 fold degenerate, is still the most populated.0755

Especially, for the 1500 and 2500, it is for the fraction of 298, the denominator, the partition function was 2.0791

2 is the degeneracy of the first level or the ground state.0800

Therefore, that basically says that at 298, pretty much the first excited state is not all that accessible.0803

However, but the slightly larger, the slightly bigger than 2 partition function for the 1500 and 2500 degrees,0811

I have put it approximately equal to 2, they are slightly bigger than 2.0829

The slightly bigger than 2 partition function shows that at least one other level,0833

the first excited level, this one other level is reasonably accessible.0848

And that is how a partition function works.0857

As the partition function rises, it is giving you a measure of the states that are starting to become accessible.0860

It is going to be discreet.0870

It is not going to be 2 to 3 to 4 to 5, 2 to 2.1, 2 to 2.06, something like that.0871

We are starting to go up like that.0877

That is what is happening.0881

Show that each translational degree of freedom contributes all over 2 to the molar heat capacity.0888

Let us see what we have got here.0894

Let me ahead and work this in blue.0898

Each translational degree of freedom, let us pick one translational degree of freedom.0903

The partition function to the X direction is equal to A over H × 2 π M KT¹/2.0907

The system partition function is equal to the molecular partition function raised to the nth/ N!.0924

Let us go ahead and take the natlog of Q.0934

What we end up getting is, we got 4, we get N LN of Q of X – N LN N + N, when we actually work this out.0941

D LN Q, let me go ahead and take D LN Q, when I take D of LN Q DT at constant volume,0958

it is going to end up equaling M D LN of Q of X DT at constant volume.0976

Therefore, U is going to be which is KT² D LN Q DT.0985

V is going to equal M KT² D LN Q of X DT under constant V.0996

Now, the constant volume heat capacity is equal to DU DT K of Q of X.1011

CV is DU DT, I'm going to take the derivative of this expression and this expression with respect to T.1024

I need to find LN of Q of X.1040

Let us go ahead and do this one.1044

LN of Q of X is equal to LN of A + ½ LN of 2 π M KT - LN of H.1048

Therefore, the D LN Q of X DT is equal to 0 + ½ × 1/ 2 π M KT × 2 π MK -0, which = 1/ 2T.1066

The energy is equal to N KT² × what I just got D LN Q of X DT, which is equal to N KT² × 1/ 2T is equal to N KT over 2.1103

N × K, I will divide those number × the Boltzmann constant is equal to R.1124

It = RT/ 2.1133

CV, we said was the derivative of this with respect to T.1144

Remember, we said we are going to find expression for the energy and we are going to take the derivative of that.1149

DU DT constant V is equal to R/ 2.1155

Each translational degree of freedom, in other words each energy of linear motion,1163

each translational degree of freedom contributes R/ 2, to the molar heat capacity.1182

In other words, if I have some gas, some monoatomic gas, neon gas or something, it has 3 translational degrees of freedom.1198

The gas can move in the X direction, the Y direction, the Z direction.1212

Its heat capacity is going to be 3/2 R.1216

Half of that R, half of that 3/2 comes from the energy of motion in the X direction.1222

Half of that R comes from the motion in the Y direction, half of that R comes from motion in the Z direction.1230

You add them up and you get the 3/2 R.1235

That is where it comes from.1238

For something that is moving, its total more heat capacity R/2 comes from linear motion in one direction.1241

All other two comes from linear motion in another direction.1251

It might have electronic energy, that is going to contribute something else to the overall heat capacity.1257

There might be vibrational energy, that is going to contribute something else to the heat capacity.1262

There might be rotational energy, that is going to contribute something else to the heat capacity.1267

All 4 of those added together give you the total heat capacity for the gas, that is what is happening.1272

Let us take a look at example 3.1282

For the molecule NO, the dissociation energy from the 0 point, the R = 0 vibrational state is 627 kj/ mol.1285

The θ of vibration, the characteristic temperature of vibration is 2719 K.1297

Use this data to calculate the actual dissociation energy from the minimum of the potential energy curve.1302

Our definition was - DE = – D0 -1/2 H ν, where ν is the fundamental vibration frequency, or DE is equal to D0 + ½ H ν.1310

Θ V is equal to H ν/ K.1346

That implies that H ν is equal to K θ V.1355

D sub E is equal to D sub E0 + K θ V/ 2.1369

Watch your units very carefully.1379

D sub 0 kj/ mol 2.1389

K θ V/ 2 is in J.1403

K is J/ K, θ V is in K, that gives you Joules.1412

These have to match.1417

I'm going to go ahead and multiply this K θ V/ 2 by N.1420

N, which is Avogadro’s number, N K θ V/2, that gives me units in J/ mol.1427

We will take N K θ V/ 2 divided by 1000.1444

Now, it is in kj/ mol.1455

I have to make the units match and I have to do what is necessary to make the units match.1461

DE is equal to D0 + K θ V/ 2 becomes DE = D sub 0 + NK θ V/ 2 × 1000.1471

Therefore, DE = 627 + 6.02 × 10²³ × 1.381 × 10⁻²³ × 2719 all divided by 2000.1495

DE = 627 + 11.3, DE = 638 kj/ mol.1526

Using the vibrational partition function, calculate the vibrational contribution1548

to the molar heat capacity of oxygen gas of 500 K.1553

Using the vibration partition function, calculate the vibrational contribution to the molar heat capacity.1558

Oxygen gas has a vibrational energy.1564

Of the total heat capacity, how much does the vibrational energy contribute, that is what we are asking, at 500 K.1569

For oxygen, the θ V, the characteristic vibrational temperature is 2256 K.1577

We have that the Q of vibration is equal to E ^- θ V/ 2T divided by 1 - E ⁻θ V/ T.1585

That is our definition for the vibrational partition function.1599

The energy U is equal to N KT² D LN Q of V DT.1607

And of course, the heat capacity is just DU DT.1620

We are going to find an expression for energy and I would differentiate with respect to T.1627

LN Q of V, we need to find this.1632

Here is QV, let us take ELN of QV is going to equal -θ V/ 2T - the natlog of 1 – E ⁻θ V/ T.1638

Therefore, the D LN Q sub V DT = the D DT of this expression that I just got.1659

It is going to be θ/ 2T² + E ⁻θ V/ T/ 1 – E ⁻θ V/ T × θ/ T².1672

U is just equal to N KT² × this.1693

We have N KT² × θ V/ 2T² + E ⁻θ V/ T.1700

It is all just a bunch of algebra.1712

Ultimately, that is all signs really comes down to in the end, of θ/ T².1714

Therefore, U is equal to N × K × θ V/ 2 + θ V × E ⁻θ V/ T/ 1 - E ⁻θ V/ T.1722

There you go, that is U.1742

We have found expression for U, now I need to just differentiate that with respect to T.1745

It is going to look more comfortable here.1757

CV is equal to DU DT constant V.1760

It is equal to R ×, I wonder if I should go through all this.1767

Just put this in your math software.1777

That is fine, I will just do it here.1780

We have 1 –E ⁻θ V/ T × θ × E ⁻θ V/ T × θ V/ T² + 0 -θ V E ⁻θ V/ T1783

× θ/ T² E ⁻θ V/ T/ 1 – E ⁻θ V/ T².1815

This is just calculus, it is all it is.1830

I will skip one step here.1836

When you put all of this together, you are going to end up with CV is equal to,1839

I will write it all out, that is fine.1847

R θ V²/ T² × E ⁻θ V/ T - E⁻² θ/ T + E⁻² θ V/ T/ 1 –E ⁻θ V/ T².1855

Therefore, our constant volume heat capacity is going to be R θ V²/ T² × E ⁻θ V/ T/ 1 –E ⁻θ V/ T².1887

I just worked it all, I will go ahead and put all my values in here.1910

The constant volume heat capacity is going to equal 8.314 × 2256².1914

At 500 K, it is 500² and this is going to be E⁻²²⁷⁶/ 500 divided by 1 – E⁻²²⁷⁶/ 500².1927

And when I do all that, I end up with CV = 1.90 J/ mol K.1944

There we go, that is the answer.1957

At 500 K, oxygen gas, the vibrational contribution to the total molar heat capacity is 1.90.1960

Let us take a couple of deep breaths here, gather our thoughts.1975

All kinds of symbols all over the place, my head is spinning.1982

The relative population of 2 quantum states that is the ratio of one state to another is given by this expression right here.1986

The number in one state divided by the number in the other state.1997

It is equal to the probability of finding the fraction in the I state over the fraction in the J state.2002

This is the definition of the relative population of 2 quantum states.2012

The ratio of the population of one state to another.2016

When we say relative population of 2 quantum states, we are talking about the ratio.2019

For 2 level system, both none degenerate.2024

With energy of the first level = 100 inverse cm and the energy of the second level = 400 inverse cm,2026

at what temperature would you find a population of the upper state in 1/4 not in the lower state?2035

They are saying that the population of the upper state is 1/4 that of a lower state.2044

This is going to be 1/4/ 1 which is just ¼.2054

N/2 / N1 is ¼, that is equal to P2/ P1.2059

Let us see now , P sub I/ P sub J, the fraction in the I state/ the fraction2074

in the other state is equal to E ⁻E sub I/ KT divided by Q, the partition function/ P sub J.2090

That is E ⁻E sub J/ KT/ Q.2115

The Q go away and what we are left with is E ⁻E sub I/ KT/ E ⁻J/ KT.2121

Therefore, the relative population, we said that N2 and N1 is 1/4 to 1, that is what this part says.2138

The population of the upper state to the 1/4 of the lower state, 1/4 to 1.2144

P2/ P1 is equal to E ^- energy of the second level 400/ KT/ E⁻¹⁰⁰/ KT is2157

equal to E¹/ KT × - 400 – 100.2178

This is basic math here, we get E⁻³⁰⁰/ KT.2196

1/4 is equal to E⁻³⁰⁰/ KT.2214

Let us take the log of both sides.2222

We get LN of 1 – LN of 4 = – 300/ KT.2225

The energies are in wave numbers, inverse cm.2241

K is equal to 0.6950 inverse cm/ K.2246

T, when I solve for T, T is equal to -300 inverse cm divided by LN 4.06950 inverse cm/ K.2256

I get T equal to 311.4 K.2290

At 311.4 K, for this particular 2 level system, 2 energy system, we have 100 inverse cm, 400 inverse cm.2302

At 311.4 K, I find that the 2nd level is actually ¼ as populated as the 1st level.2312

Let us switch colors here, just for the hell of it.2326

More generally, when levels are degenerate, we just have to include the degeneracy.2333

Nothing more than that numbers R, the degenerate, the equations are as follows.2344

N sub I/ N sub J = G sub I E ⁻E sub I/ KT/ the degeneracy.2359

Let us make it a small j E ⁻J/ KT.2373

It is going to end up equaling G of I or G of J E¹/ KT × -E sub I + J.2381

There you go.2402

In these cases, even when the upper level is higher in energy because of the degeneracy2404

or because of the greater degeneracy of higher levels.2445

I will leave the degeneracy ahead.2450

Because of degeneracy it can and will often,2453

In these cases, even when the upper level is higher in energy because of the degeneracies,2469

the upper level tends to actually be more populated than the lower level.2482

In general, we know that the lower levels are more populated at given temperature.2505

It is harder to get it up to the upper levels.2509

However, because the degeneracies of upper levels are so high sometimes, they are populated2512

that it ends up inverting this population.2519

You will end up with higher energy levels but they actually have more population than the lower energy levels.2522

It is actually the mostly be the case like this for rotation, not so much for vibration and electronic.2529

Let us go to example 6.2537

Calculate the relative population for the J = 2 and J = 1 rotational states of the carbon monoxide molecule at 25°C.2544

The temperature here, the 25° C, the temperature = 298 K.2555

The N J = 2/ N J = 1, that = the G of 2 E ⁻E of 2/ KT/ G of 1 E ⁻E/ 1/ KT.2563

These are rotation levels, rotational states.2589

The degeneracy of level 2 = 2 × 2 + 1 = 5.2593

The degeneracy of level 1 is 2 × 1 + 1 is equal to 3.2600

N of 2/ N of 1 is equal to 5/3 E⁻¹/ KT - E2 + E1.2610

We need to find what the energies are.2630

The energy sub J is equal to the rotational constant × J × J + 1.2633

The energy of 2 is equal to that × 2 × 2 + 1 = 6B~.2645

And energy 1 = B~ 1 × 1 + 1 = 2B~.2658

-6B~ + 2B~ = -4B~.2670

B~ for carbon monoxide, just go ahead and look it up.2682

I did not supply it here, just go ahead and look it up in your book.2687

It is 1.931 inverse cm.2690

N2/ N1 = 5/3 E⁻⁴ × 1.931 divided by KT divided by 0.6950 × 298.2698

And when I do that, I get 1.606.2725

Let me go ahead and do it over here.2735

1.606 is N of 2 over N of 1.2738

This means that the J = 2 rotational state is 1.606 × as populated as the J = 1 rotational state.2749

As we mentioned in the previous problem, where the degeneracies are involved, the higher energy states2784

because they have the higher degeneracies, the higher energy states will often be more populated than the lower energy states.2823

This mostly applies to rotational energies because rotational energies tend to be very close.2849

It will give you a very small scale.2866

You might have an electronic state that has a much higher degeneracy than the lower electronic state.2868

But because of the difference in energy between electronic states is huge, it is not going to make a difference.2876

In the case of rotation, it makes a difference.2881

This mostly applies to rotational levels, as we see here.2883

The rotational level of 2 has a degeneracy of 5.2890

The rotational level J = 1 has a degeneracy of 3.2893

The 2 is higher in energy, however because the degeneracy, it is more populated.2897

It has more things in that particular state.2901

Thank you so much for joining us here at

We will see you next time for more examples on statistical thermodynamics.2907

Take care, bye.2911