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

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

## Discussion

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Statistical Thermodynamics: The Various Partition Functions 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
- Diatomic Gases
- 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

- 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

### Physical Chemistry Online Course

### 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 and*0004

*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 equal*0493

*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/ KT*1356

*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

*It is about 0.003%.*1549

*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 values*1630

*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 R*2933

*- 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 V*3074

*× 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

## Start Learning Now

Our free lessons will get you started (Adobe Flash

Sign up for Educator.com^{®}required).Get immediate access to our entire library.

## Membership Overview

Unlimited access to our entire library of courses.Learn at your own pace... anytime, anywhere!