For more information, please see full course syllabus of Physical Chemistry

For more information, please see full course syllabus of Physical Chemistry

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

- 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

### Physical Chemistry Online Course

### Transcription: Example Problems I

*Hello and welcome to www.educator.com, 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 atoms*0013

*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 1500*0521

*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 fraction*0553

*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 contribution*1548

*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/ T*1783

*× 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 fraction*2074

*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 is*2157

*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 degeneracy*2404

*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 populated*2512

*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 states*2784

*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 www.educator.com.*2905

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

*Take care, bye.*2911

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