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

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

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### Statistical Thermodynamics: The Various Partition Functions 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
- Lesson Overview 0:19
- Monatomic Ideal Gases 6:40
- Monatomic Ideal Gases Overview
- Finding the Parition Function of Translation
- Finding the Parition Function of Electronics
- Example: Na
- Example: F
- Energy Difference between the Ground State & the 1st Excited State
- The Various Partition Functions for Monatomic Ideal Gases
- Finding P
- Going Back to U = (3/2) RT

### Physical Chemistry Online Course

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

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

*In the last lesson, we introduce statistical thermodynamics and we gave you a broad overview of what our intent was.*0004

*Today, we are going to continue discussing this statistical thermodynamics*0011

*and we are going to introduce the various partition functions.*0015

*Let us jump right on in.*0019

*Let me see, should I work in blue or black?*0021

*I think today I’m going to work in black.*0025

*We said Q was equal to 1/, let me write it as Q ⁺nth/ N!.*0028

*Where Q, the molecular partition function is equal to the sum / the I states of the G sub I of E ⁺E sub I/ K × T.*0046

*Let me make this T a little bit better here, K × T.*0081

*In this particular case, I is the quantum level.*0086

*G sub I is the degeneracy of that level.*0095

*E sub I that is the energy of the I level.*0112

*In this lesson, we see the expressions for the Q, the molecular partition function for the Q of various systems.*0126

*In particular, we seek the Q of translation which is going be symbolized as QT.*0155

*We seek the Q of vibration, the translational partition function.*0182

*The vibrational partition function, this is going be Q sub V.*0187

*We seek the rotational partition function which will symbolized with Q sub R.*0193

*And we seek the electronic partition function Q electronic which will usually symbolize with Q sub E.*0200

*Occasionally, we will use Q vib, thing like that.*0209

*In this particular lesson, we limit ourselves to mono atomic and diatomic gases.*0214

*Each particle of an N particle system has energy E.*0244

*The energy of a particle is distributed among the different ways that the energy can be stored.*0268

*The energy is equal to the particle has translational energy, the energy of its motion.*0275

*It has a vibrational energy, in the case of a diatomic molecule, how much it is vibrating?*0284

*It has a rotational energy, if it is a diatomic molecule.*0290

*If it is a polyatomic it is rotating.*0293

*And it has electronic energy, how much energy is just by virtue of the electrons in that system.*0296

*Now, since E is the sum of 4 energies, the translational, vibrational, rotational, and electronic,*0303

*the molecular partition function is the product.*0327

*That is how it works.*0335

*It is the product of individual partition functions.*0336

*The partition function for translation.*0353

*The partition function for vibration.*0354

*The partition function for rotation.*0356

*And the partition function for energy.*0357

*The energy is the sum of 4 energies.*0359

*The molecular partition function, the total molecular partition function of a molecule*0362

*is the product of the individual partition functions for each component of the energy.*0366

*In other words, Q is equal to Q of translation × Q of vibration × Q of rotation × Q electronic.*0385

*We would begin with monoatomic ideal gases.*0400

*We begin with monoatomic ideal gases.*0404

*Energy = energy of translation + energy of electronic.*0424

*The reason is an atom does not rotate.*0432

*The energy of an atom consists of only two components.*0437

*It has the translational energy, its energy of motion and the electronic energy.*0440

*And atom does not rotate and an atom does not vibrate.*0445

*Does not rotate nor vibrate.*0455

*The energy of the atom is made of translation and electronic energies.*0467

*When we deal with diatomic molecule, it is going have all 4.*0494

*We are going to find the Q of translation first.*0499

*We find QT first.*0504

*The atom can move in 3 directions, X, Y, and Z.*0516

*X direction, Y direction, and Z direction.*0533

*Each has a Q of its own each.*0538

*Each has a Q, Q sub X, Q sub Y, Q sub Z.*0544

*Therefore, the Q of translation is equal to Q of X, Q of Y × Q of Z, the product.*0551

*The energies of translation of the energies of a particle in the box.*0565

*Translation, motion.*0569

*The energies of translation are the energies of the particle in a box.*0577

*Energy sub X is equal to H² N²/ 8MA².*0600

*N is the quantum number, 1, 2, 3, and so on.*0611

*A is for the particle in the box, the length of the box.*0616

*Q of X, the partition function is equal to, always begin with the definition of the partition function.*0633

*The sum of the J sub I E ⁻E sub I/ KT where J sub I is the degeneracy of the level.*0641

*I’m not going to go through the math here.*0667

*I will just right math in brackets, our result is the following.*0674

*Our Q of X is going to equal A/ H × 2 π MKT ^½.*0686

*Q sub Y is going to equal B/ H.*0700

*B is the length of the box in the Y direction.*0703

*2 π M KT ^½ and the partition function in the Z direction is going to equal C / H.*0707

*C is the length of the box in the third dimension.*0720

*In other words, the box is A × B × C, the volume.*0723

*2 π M KT ^½, M is the mass of the atom.*0727

*K is the Boltzmann constant.*0733

*T is the temperature in Kelvin.*0735

*We said that Q of translation is equal to QX × QY × QZ.*0738

*Therefore, the Q of translation is equal to 2 π,*0747

*Let us go ahead with the M first.*0753

*2 π MKT, we tend to keep tied together with K/ H²³/2 × A × B × C.*0756

*Or Q of translation is equal to 2 π M KT/ H²³/2 × V, the volume.*0773

*There you go.*0790

*This is an expression for the translational partition function.*0793

*We found an expression for the translational partition function.*0800

*Let us find the electronic function.*0807

*We find Q electronic next, Q sub E.*0812

*Once again, the definition Q is equal to the sum/ I of J sub I E ⁺E sub I/ KT.*0827

*This is what is important, the E sub I.*0838

*We have an expression for the electronic energy.*0839

*And again, G sub I is the degeneracy.*0845

*E is the energy of the I electronic state.*0847

*We set E sub 1 to 0.*0852

*We need a 0 so we can choose our 0.*0859

*We set E sub 1 to 0.*0862

*If you remember those tables that we saw back when we are doing atomic spectroscopy,*0868

*that first energy level, the ground state, that was set at 0.*0872

*Everything is measured from the ground state.*0876

*We set the ground state to equal 0.*0879

*It just makes that makes it easier for us.*0881

*We measure energy or we measure everything relative to the ground state.*0884

*Relative to the ground electronic state.*0898

*Q of E = G1 because of this is 0, this is 1 + V2 E ⁻E2/ KT + G3 E ⁻G3/ KT, and so on.*0913

*Note that in this case, Q is a function of T, it is not a function of V.*0939

*The way the translational function was.*0954

*T not the Q of T was a function of both T and V.*0957

*For most atoms at room temperature, I have to say normal temperatures.*0966

*Normal meaning not too high of room temperature, maybe 100 or 200 K.*0982

*The 200 above room temperature.*0987

*For most atoms at normal temperatures, the first two terms, in other words this one and this one.*0990

*The first two terms of Q sub E are enough.*1008

*When you are doing your problems, you do not have to calculate.*1014

*You have to do this entire sum.*1017

*By the way, you have math software so you can certainly calculate this + that, very easy.*1018

*If for any reason you need the third term, the fourth term, the 5th term, you can just add them.*1024

*That is all this is, this is just the sum are enough.*1029

*In !, the first term alone is enough because the second term is very small.*1034

*As an example, let us take a look at sodium.*1061

*Let me do this one in red.*1064

*As an example, let us look at sodium gas.*1071

*In those spectroscopic tables that we looked at, we have something like this.*1085

*Sodium, you have 2P6 3S1, the ground state is a doublet S1/2.*1088

*It has a degeneracy of 2, its energy is 0.*1097

*The first excited state is 2P6 3P1.*1101

*It is a doublet P ½, its degeneracy is 2, and its energy is 16,956.*1107

*This is an inverse centimeter, by the way.*1116

*Therefore, Q of E is equal to G1 + G2 × E ⁻E2 divided by KT + any of the terms that I want to take.*1127

*G1 is equal to 2.*1139

*This first term is equal to 2, that is the degeneracy.*1150

*This term, when I put it in, this is equal to 2 × E⁻¹⁶⁹⁵⁶.*1159

*These energies have already been tabulated.*1170

*All I have to do is put the energies in.*1171

*The energy of the first excited level.*1173

*The energy 2 in this case is the thing that I just wrote 16,956 divided by 0.6950.*1175

*The reason I use 0.6950, this is in inverse cm.*1185

*Boltzmann constant in inverse cm is 0.6950 in J/ K.*1189

*It is in inverse cm/ K.*1194

*In J/ K, it is 1.381 × 10⁻²³.*1197

*Watch your units, it is going to be the biggest problem with physical chemistry.*1202

*Let us just pick some temperature.*1207

*Let us just pick 500 K, like that.*1210

*This number is equal to 1.3 × 10⁻²¹.*1212

*Notice that the you have got 2 and you have got 1.3 × 10⁻²¹..*1220

*Now, the fraction of the atoms in the first excited state which is the doublet P1/2 is, the fraction is the part / the whole.*1226

*The part is this G2 E ⁻E2/ KT/ Q sub E.*1253

*This is equal to 1.3 × 10⁻²¹/ 2 + 1.3 × 10⁻²¹.*1266

*When I do this math, I get the fraction is equal to 6.5 × 10⁻²².*1279

*That means of all the atoms in our system, only 6.5 × 10⁻²², that fraction 6.5 × 10⁻²⁸%*1290

*of those atoms is actually in that second electronic state, or the first excited electronic state.*1301

*Virtually, no atoms are in the first excited state.*1309

*In other words, the doubled P1/2 excited state, all of the atoms are in the ground electronic state*1329

*which is this is that you normally find for most atoms and molecules electronically.*1337

*Almost all of them, all the atoms are going to be in the ground state at normal temperatures.*1342

*This is 500 K, this is not even room temperature.*1346

*At room temperature, it is even less.*1349

*500 K is reasonably high.*1351

*In the case of NA, in the case of sodium, the first term alone is enough.*1357

*What I have done here by calculating this fraction, because it is a fraction of the second level or*1373

*the first excited level is small, it means I can just completely ignore the second term.*1378

*I do not even need the second term.*1383

*I can just stick with G1 for my partition function, that is what I'm doing.*1385

*Let us take a look at halogen.*1392

*Let us look at Fluorine.*1400

*For fluorine of chlorine, I’m not exactly sure of which values I have.*1410

*I have F but here I have CL.*1419

*Anyway, it is a halogen, it does not really matter.*1421

*Let us go ahead and take a look of the, we have 2S2 2P5.*1423

*That ground state is a doublet P3/ 2 and has a degeneracy of 4.*1430

*Remember the degeneracy is twice this + 1 and its energy is 0.*1436

*That is the ground state of a halogen, of all the halogens.*1441

*There is another state here, it is doublet P1/2.*1446

*It has the degeneracy of 2 and it is at 404 inverse cm.*1453

*That is actually the first excited state.*1458

*The second is 2S2 2P4 3S1.*1462

*This is a quadruplet P5/2 state and its degeneracy is 6 and it is at 800 and 2406.*1471

*We have the ground state, the first excited state, the second excited state.*1485

*Let us see what is populated and what is not.*1490

*We want to find the fraction of which of the state I'm dealing with.*1493

*The fraction, in this particular case of the second level, the first excited state.*1500

*We have 2 E ⁻E2/ KT all divided by G1 + G2, the electronic partition function.*1510

*E ⁻E sub 2 KT +, and so on.*1524

*When I actually put these numbers in, I can use 2 terms, I can use 3 terms, it does not really matter.*1531

*I will do it up here, it is going to equal 2 × E.*1538

*I think I got some of my values wrong, that is okay, it does not really matter.*1555

*I will just go ahead and go with the numbers that I got here.*1559

*I will use the numbers that I have.*1563

*Here is what is going on, -404 divided by KT/ G1 which was 4 + 2 × E⁻⁴⁰⁴ divided by KT + any other term if I want to add it.*1569

*When I run this at different temperatures, if I do the fraction of 298,*1595

*in this particular case I decided to just use two terms.*1601

*The first excited state/ the two terms of the electronic partition function.*1606

*I get 0.066 with a fraction at 1000 K, it jumps all the way to 0. 218.*1612

*These numbers, in this particular case, when I said earlier, I think I end up with slightly different numbers.*1627

*I use 581 instead of 404.*1633

*I think 581 came from the use of chlorine, instead of fluorine.*1637

*It is 0.066.218 might be slightly off, as far as actual calculations are concerned.*1641

*But the G is the relevant number, this is what we are looking at.*1646

*The fraction of 1500, if I’m raising the temperature, the fraction 0.253.*1653

*21% of the molecules are in the first excited state.*1662

*Out of the 150 K, 25% of the molecules are in the first excited state.*1664

*At 298, its 6%, 0.06 are actually in the first excited state.*1669

*That is pretty significant.*1675

*What this means is even at room temperature, the first excited state is reasonably populated.*1679

*0.6% is the 6%, it is reasonably populated.*1701

*Therefore, when calculating Q, when using Q sub E, when using the electronic partition function,*1710

*it is best to include two terms.*1730

*In this particular case, I do a fraction thing, I find that the fraction is 0.066.*1738

*That is reasonably high based on the energy, the energy of the first excited state.*1744

*Because it is 0.066 that is reasonably populated.*1748

*Because it is reasonably populated, I cannot ignore it in the expression of the actual molecular partition function.*1751

*When I go on to use the electronic partition function for this particular case, in the case of a halogen,*1757

*I'm going include at least two terms.*1762

*In general, at room temperature, just look at the energy difference between the ground state and the first excited state.*1768

*If this Δ E is in the 10² range, 500, 600, 700, 800, it is in the hundreds, it is in the 10² range,*1808

*then use 2 terms for G sub E.*1832

*If the range is in the thousands or above, 10³ or above, you are safe ignoring the second term and on.*1843

*You are safe using only one term.*1857

*Again, using math software, if you want to be accurate, just use 3 terms.*1863

*That is fine, if you want to, not a big deal.*1867

*At higher temperatures, 1000, 1500, 2000, things like that, you have to check like we just did.*1874

*At higher temperatures, check the fraction to see if the excited state is reasonably populated at that temperature.*1886

*At that T, then decide on the number of terms that you want.*1923

*Totally your choice.*1938

*Let us go back to black.*1941

* For monoatomic gases, we have the translational partition function = 2 π M KT/ H²³/2 × V.*1947

*And we have a Q electronic partition function = G1 + G2 × E ⁻E2/ KT.*1972

*Again, you just decided if you will use 2 terms in general.*1983

*Our total partition function is equal to Q of translation × Q of electronic is equal to 2 π M KT/ H²³/2 × V × G1 + G2 E ⁻E2/ KT.*1987

*In the last lesson, we said that we are mostly to be concerned with the energy and*2016

*the constant volume heat capacity, in the case of pressure.*2032

*In the last lesson, we said we will concern ourselves with U and CV.*2035

*U = KT² D LN Q DT V and Q = Q ⁺N/ N!.*2065

*LN Q = N LN q – N LN N + N.*2086

*Therefore, D of LN Q DT under constant volume is equal to the D of DT holding V constant of this Q - N +*2101

*N equal to N × D LN q DT under constant V.*2130

*-0 + 0, U is equal to KT² D LN Q DT which is equal to N KT² D ln q DT constant V.*2142

*Energy is this expression, in terms of Q.*2175

*We just found Q, let us go ahead and put that in.*2178

*Let us find D LN Q DT.*2187

*K was equal to 2 π M KT/ H²³/2 × V × G1 + G2 E ⁺E2/ KT.*2195

*I need DQ DT, this is Q.*2225

*I’m going to take LN Q first and I'm going to differentiate with respect to T.*2226

*LN Q = 3/2 LN 2 π M KT - 3/2 H² + LN V + LN of G1 + G2 E ⁻E2/ KT.*2231

*It looks complicated, it is actually not.*2259

*The derivative with respect to T all of these is 0, holding V constant.*2261

*When I take the derivative of this with respect, I have written it out here*2269

*but I decided actually not to go ahead and write it.*2273

*That is fine, I have written it out, let us go ahead and do it.*2278

*D LN Q DT of V = 3/2 × 1/2 π M KT × 2 π MK -0 + 0 + 1/*2281

*G1 + G2 E ⁻E2 / KT × 0 + G2 B ⁻E2/ KT/ KT².*2309

*When I put all that together, I get Q.*2332

*Therefore, U is equal to KT² D LN Q DT of V, that is equal to N KT² × 3/ 2T + N KT² T2 not G².*2345

*T2 E ⁻E2/ KT divided by G1 + G2 × E ⁻E2/ KT × KT².*2371

*This is Q sub E.*2397

*We end up with a final expression of U is equal to 3/2 N KT + N × G2 × E ⁻E2/ KT/ Q sub E.*2403

*This is the average translational energy.*2429

*In other words, I found an expression for the energy.*2438

*This is it, the energy is made up of a translation component + an electronic component.*2440

*The average translational energy is this.*2448

*This part is the average electronic energy above and beyond the ground state energy which is 0.*2451

*The average electronic energy in excess of the ground state.*2458

*The electronic contribution of the energy to the total energy is very small at normal temperatures.*2476

*The electronic energy contribution is very small at ordinary temperatures.*2490

*We skip this term.*2520

*We can take U equal to 3/2 N KT.*2522

*We have N is equal to 6.02 × 10²³.*2534

*N × K is equal to R, the gas constant.*2547

*The average energy is equal to 3/2 RT.*2554

*There you go, now the constant volume heat capacity we said is equal to the time,*2561

*the temperature derivative of this holding volume constant is equal to 3/2 R.*2567

*Remember, we said we will find an expression for the energy and we will just take the derivative of it directly.*2573

*Let us see what have here.*2581

*We found energy of the monoatomic gas and I found the heat capacity of the monoatomic gas.*2589

*It might be nice to find P, the pressure.*2597

*In this case, let us see what it might look like.*2608

*P is equal to, we have an expression for that.*2613

*KT D LN Q DV at constant T equal to N × KT.*2616

*Everything is the same except you are replacing LN Q.*2629

*You are replacing the Q with q D LN Q DT.*2634

*That got a little carried away there, anticipating everything.*2639

*Recall that LN of Q was equal to 3/2 × the natlog of 2 π M KT -3/2 H² + the natlog of V + the natlog of G1 + G2 is Q sub E.*2646

*Let us go ahead and write that.*2683

*Therefore, the derivative of LN Q with respect to V holding T constant is actually 1/ V.*2685

*P is equal to KT N KT is equal to N KT/ V.*2699

*NK is R, P = RT/ V.*2715

*PV = RT.*2727

*Again, we are talking about one mol of gas.*2728

*This is the ideal gas law.*2729

*When we calculated for pressure, we ended up recovering the ideal gas law.*2732

*We found a partition function for a monoatomic gas.*2738

*We found an expression for the pressure.*2742

*We put that partition function in and when we solve for what the pressure is in a monoatomic gas,*2747

*we end up with P = RQ/ V.*2751

*We ended up deriving the ideal gas law from a partition function.*2753

*As opposed to finding the ideal gas law empirically which is how it was originally done.*2757

*This is amazing.*2763

*Let us look at one final thing here.*2768

*Let me go ahead and let us go back to the energy = 3/2 RT.*2774

*This is the translational energy.*2800

*This 3/2 RT = ½ RT + ½ RT + ½ RT.*2804

*This is the energy in the X direction.*2819

*This is the energy in the Y direction.*2821

*This is the energy in the Z direction.*2824

*Each directional degree of freedom contributes ½ RT to the energy.*2827

*It contributes ½ R to the specific heat capacity, that is the translational heat capacity.*2832

*Thank you so much for joining us here at www.educator.com.*2840

*We will see you next time, bye.*2842

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