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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

Transcription: Statistical Thermodynamics: The Various Partition Functions I

Hello and welcome back to, 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 and2016

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

We will see you next time, bye.2842