WEBVTT chemistry/physical-chemistry/hovasapian
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Hello and welcome to www.educator.com, welcome back to Physical Chemistry.
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Today, we are going to start doing our example problems for statistical thermodynamics.
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Let us jump right on in.
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Using the data table below, calculate the fraction of the potassium atoms
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in the first excited electronic state at 298 K, 1500 K, and 2500 K.
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Let us take a look at this data table.
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We have our ground state here, I will go ahead and do this in red.
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We have our ground state and again, the ground state is the 0 energy.
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The first excited state is actually this one right here.
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Notice, this actually has 2, there is a doublet P 1/2 and a doublet P 3/2.
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The first excited state is actually this one.
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This is our second excited state, maybe this is our third excited state.
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We want to know what the fraction of potassium atoms is in the first excited state at these different temperatures.
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Let us go ahead and work this out.
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The fraction is the same as any other fractions.
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It is a part over the whole.
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The fraction is going to be the degeneracy of that particular level × E ⁻energy of that level.
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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.
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We are interested in the first excited state which is going to be level 2.
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The degeneracy E raised to that divided by KT all over the electronic partition function.
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We will use the first 4 terms of Q sub E, the electronic partition function in the denominator.
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Now, if you are wondering how I came up with 4, it is just a random pick.
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I will just pick the first 4 terms.
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For electronic energy states, actually you can just go ahead and go with the first 2,
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it is not a problem but I figure what the hell.
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In the expression for the molecular partition function, it is the sum.
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You can take 3, 4, 5, whatever you want.
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Electronic state, I just decide on 4.
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For rotational state, it is going to be a lot more than that.
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For vibrational state, 2, 3, 4, that is usually enough.
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It is a random choice on my part.
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You are probably using math software, you can use 50 terms if you want.
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It is going to give you an answer just as quickly.
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I decided to just go ahead and use my calculator for this one.
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This is the definition.
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This is how you are going to do it.
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The energy in that state divided by the total partition function, that gives us the fraction.
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Q sub E that is equal to G1 + G2 E ⁻E2/ KT.
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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.
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That is what the molecular partition function is.
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I have decided to just take 1 term, 2 terms, 3 terms, 4 terms.
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I have my energies E2, E3, and E4, that is the energy of the 2nd state,
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that is the energy of the 3rd state, that is the energy of the 4th state.
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I have K, that is Boltzmann constant.
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I have the different temperatures so I have all the numbers that I need.
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The rest is just arithmetic, really.
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Let us talk about some degeneracies here.
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The degeneracies are important.
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This degeneracy, this is doublet S ½.
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The degeneracy is the 2J + 1.
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G sub I is equal to 2J + 1.
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2 × ½ + 1 gives me degeneracy of 2 for this one.
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Here, I have a degeneracy of 2.
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Here, 2 × 3/2 + 1 that is going to give me a degeneracy of 4.
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And here, this is a doublet S ½, this is going to give me a degeneracy of 2.
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When I put all of these values into E and everything, here is what I get.
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I get Q sub E is equal to 2 + 2 × E⁻¹²⁹⁸⁵ and these are in inverse cm, over KT.
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Let me go ahead and put the values in.
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The Boltzmann constant, the 1.381 × 10⁻³²³ J/ K, because we are using inverse cm,
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Boltzmann constant K is equal to 0.6950 inverse cm/ K.
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We have to watch our units very carefully.
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0.6950 × T + 4 × E⁻¹³⁰⁴³/ 0.6950 × T + QE⁻²¹⁰²⁷ divided by 0.6950 T.
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The numerator of the fraction is going to equal 2 × E⁻¹²⁹⁸⁵,
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because it is the first excited state that we are interested in, over 0.6950 T.
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Therefore, the fraction at 298 K is equal to the numerator at 298.
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I put this value, this 298 here divided by Q sub E at 298.
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In other words, I put 298 in here, here, and here, and I calculate this.
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I do the sum and I do the division.
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What I end up with is 1.18 × 10⁻²⁷ divided by 2.
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This is going to equal 5.9 × 10⁻²⁸.
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You notice a partition function is around 2.
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The partition function is a numerical measure of the number of quantum states that are accessible to a molecule,
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to an atom at that temperature.
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In this case, it is really only 2.
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However, it is a degenerate so it is actually this 2.
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Most basically, what this says is 5.9 × 10⁻²⁸.
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5.9 × 10⁻²⁶ % of the atoms are in the first excited state.
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That is what this says.
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That is reasonable higher.
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I have raised the temperature higher, the fraction at 1500 K.
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We raised the temperature a little bit, now we are going to calculate the numerator at 1500,
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and we are going to put 1500 there.
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The numerator at 1500 divided by Q sub E, or we put in 1500
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and it is going to be 3.90 × 10⁻⁶/ approximately 2, just slightly above 2.
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What we end up with here is 1.95 × 10⁻⁶.
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That is a lot different.
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Now, the percentage is a lot higher, a lot higher than 10⁻²⁸.
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By raising the temperature from 300 K to 1500 K, we actually made a greater fraction
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of these atoms are actually in the first excited state.
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If we do fraction at 2500, we are going to do numerator at 2500 divided by Q sub E at 2500.
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And what we end up with is 1.14 × 10⁻³ by approximately 2.
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It is going to be approximately 5.7 × 10⁻⁴.
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We see as the temperature is rising, the fraction of atoms in the first excited state is going up.
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More of the atoms are jumping from the ground state.
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They are spending more time in the first excited state.
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Not a lot, it is still small but it is still pretty reasonable.
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Just by going up to 2500 K, we have pushed them up into and spend more time in the first excited state.
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Not too high.
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What is happening here, as the temperature rises more atoms are moving from the ground state to the first excited state.
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In other words, the population distributions changing.
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We said that the partition function is a measure of the states that are thermally accessible to a particle at a given T.
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At these temperatures, level 1 which is 2 fold degenerate, is still the most populated.
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Especially, for the 1500 and 2500, it is for the fraction of 298, the denominator, the partition function was 2.
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2 is the degeneracy of the first level or the ground state.
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Therefore, that basically says that at 298, pretty much the first excited state is not all that accessible.
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However, but the slightly larger, the slightly bigger than 2 partition function for the 1500 and 2500 degrees,
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I have put it approximately equal to 2, they are slightly bigger than 2.
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The slightly bigger than 2 partition function shows that at least one other level,
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the first excited level, this one other level is reasonably accessible.
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And that is how a partition function works.
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As the partition function rises, it is giving you a measure of the states that are starting to become accessible.
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It is going to be discreet.
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It is not going to be 2 to 3 to 4 to 5, 2 to 2.1, 2 to 2.06, something like that.
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We are starting to go up like that.
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That is what is happening.
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Show that each translational degree of freedom contributes all over 2 to the molar heat capacity.
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Let us see what we have got here.
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Let me ahead and work this in blue.
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Each translational degree of freedom, let us pick one translational degree of freedom.
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The partition function to the X direction is equal to A over H × 2 π M KT¹/2.
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The system partition function is equal to the molecular partition function raised to the nth/ N!.
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Let us go ahead and take the natlog of Q.
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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.
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D LN Q, let me go ahead and take D LN Q, when I take D of LN Q DT at constant volume,
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it is going to end up equaling M D LN of Q of X DT at constant volume.
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Therefore, U is going to be which is KT² D LN Q DT.
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V is going to equal M KT² D LN Q of X DT under constant V.
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Now, the constant volume heat capacity is equal to DU DT K of Q of X.
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CV is DU DT, I'm going to take the derivative of this expression and this expression with respect to T.
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I need to find LN of Q of X.
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Let us go ahead and do this one.
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LN of Q of X is equal to LN of A + ½ LN of 2 π M KT - LN of H.
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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.
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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.
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N × K, I will divide those number × the Boltzmann constant is equal to R.
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It = RT/ 2.
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CV, we said was the derivative of this with respect to T.
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Remember, we said we are going to find expression for the energy and we are going to take the derivative of that.
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DU DT constant V is equal to R/ 2.
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Each translational degree of freedom, in other words each energy of linear motion,
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each translational degree of freedom contributes R/ 2, to the molar heat capacity.
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In other words, if I have some gas, some monoatomic gas, neon gas or something, it has 3 translational degrees of freedom.
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The gas can move in the X direction, the Y direction, the Z direction.
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Its heat capacity is going to be 3/2 R.
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Half of that R, half of that 3/2 comes from the energy of motion in the X direction.
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Half of that R comes from the motion in the Y direction, half of that R comes from motion in the Z direction.
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You add them up and you get the 3/2 R.
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That is where it comes from.
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For something that is moving, its total more heat capacity R/2 comes from linear motion in one direction.
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All other two comes from linear motion in another direction.
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It might have electronic energy, that is going to contribute something else to the overall heat capacity.
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There might be vibrational energy, that is going to contribute something else to the heat capacity.
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There might be rotational energy, that is going to contribute something else to the heat capacity.
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All 4 of those added together give you the total heat capacity for the gas, that is what is happening.
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Let us take a look at example 3.
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For the molecule NO, the dissociation energy from the 0 point, the R = 0 vibrational state is 627 kj/ mol.
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The θ of vibration, the characteristic temperature of vibration is 2719 K.
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Use this data to calculate the actual dissociation energy from the minimum of the potential energy curve.
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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.
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Θ V is equal to H ν/ K.
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That implies that H ν is equal to K θ V.
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D sub E is equal to D sub E0 + K θ V/ 2.
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Watch your units very carefully.
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D₀ kj/ mol 2.
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K θ V/ 2 is in J.
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K is J/ K, θ V is in K, that gives you Joules.
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These have to match.
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I'm going to go ahead and multiply this K θ V/ 2 by N.
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N, which is Avogadro’s number, N K θ V/2, that gives me units in J/ mol.
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We will take N K θ V/ 2 divided by 1000.
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Now, it is in kj/ mol.
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I have to make the units match and I have to do what is necessary to make the units match.
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DE is equal to D0 + K θ V/ 2 becomes DE = D₀ + NK θ V/ 2 × 1000.
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Therefore, DE = 627 + 6.02 × 10²³ × 1.381 × 10⁻²³ × 2719 all divided by 2000.
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DE = 627 + 11.3, DE = 638 kj/ mol.
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Using the vibrational partition function, calculate the vibrational contribution
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to the molar heat capacity of oxygen gas of 500 K.
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Using the vibration partition function, calculate the vibrational contribution to the molar heat capacity.
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Oxygen gas has a vibrational energy.
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Of the total heat capacity, how much does the vibrational energy contribute, that is what we are asking, at 500 K.
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For oxygen, the θ V, the characteristic vibrational temperature is 2256 K.
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We have that the Q of vibration is equal to E ^- θ V/ 2T divided by 1 - E ^-θ V/ T.
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That is our definition for the vibrational partition function.
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The energy U is equal to N KT² D LN Q of V DT.
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And of course, the heat capacity is just DU DT.
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We are going to find an expression for energy and I would differentiate with respect to T.
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LN Q of V, we need to find this.
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Here is QV, let us take ELN of QV is going to equal -θ V/ 2T - the natlog of 1 – E ^-θ V/ T.
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Therefore, the D LN Q sub V DT = the D DT of this expression that I just got.
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It is going to be θ/ 2T² + E ^-θ V/ T/ 1 – E ^-θ V/ T × θ/ T².
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U is just equal to N KT² × this.
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We have N KT² × θ V/ 2T² + E ^-θ V/ T.
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It is all just a bunch of algebra.
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Ultimately, that is all signs really comes down to in the end, of θ/ T².
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Therefore, U is equal to N × K × θ V/ 2 + θ V × E ^-θ V/ T/ 1 - E ^-θ V/ T.
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There you go, that is U.
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We have found expression for U, now I need to just differentiate that with respect to T.
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It is going to look more comfortable here.
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CV is equal to DU DT constant V.
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It is equal to R ×, I wonder if I should go through all this.
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Just put this in your math software.
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That is fine, I will just do it here.
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We have 1 –E ^-θ V/ T × θ × E ^-θ V/ T × θ V/ T² + 0 -θ V E ^-θ V/ T
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× θ/ T² E ^-θ V/ T/ 1 – E ^-θ V/ T².
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This is just calculus, it is all it is.
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I will skip one step here.
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When you put all of this together, you are going to end up with CV is equal to,
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I will write it all out, that is fine.
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R θ V²/ T² × E ^-θ V/ T - E⁻² θ/ T + E⁻² θ V/ T/ 1 –E ^-θ V/ T².
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Therefore, our constant volume heat capacity is going to be R θ V²/ T² × E ^-θ V/ T/ 1 –E ^-θ V/ T².
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I just worked it all, I will go ahead and put all my values in here.
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The constant volume heat capacity is going to equal 8.314 × 2256².
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At 500 K, it is 500² and this is going to be E⁻²²⁷⁶/ 500 divided by 1 – E⁻²²⁷⁶/ 500².
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And when I do all that, I end up with CV = 1.90 J/ mol K.
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There we go, that is the answer.
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At 500 K, oxygen gas, the vibrational contribution to the total molar heat capacity is 1.90.
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Let us take a couple of deep breaths here, gather our thoughts.
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All kinds of symbols all over the place, my head is spinning.
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The relative population of 2 quantum states that is the ratio of one state to another is given by this expression right here.
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The number in one state divided by the number in the other state.
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It is equal to the probability of finding the fraction in the I state over the fraction in the J state.
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This is the definition of the relative population of 2 quantum states.
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The ratio of the population of one state to another.
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When we say relative population of 2 quantum states, we are talking about the ratio.
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For 2 level system, both none degenerate.
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With energy of the first level = 100 inverse cm and the energy of the second level = 400 inverse cm,
00:33:55.200 --> 00:34:04.200
at what temperature would you find a population of the upper state in 1/4 not in the lower state?
00:34:04.200 --> 00:34:14.200
They are saying that the population of the upper state is 1/4 that of a lower state.
00:34:14.200 --> 00:34:19.300
This is going to be 1/4/ 1 which is just ¼.
00:34:19.300 --> 00:34:34.200
N/2 / N1 is ¼, that is equal to P2/ P1.
00:34:34.200 --> 00:34:49.700
Let us see now , P sub I/ P sub J, the fraction in the I state/ the fraction
00:34:49.700 --> 00:35:15.100
in the other state is equal to E ⁻E sub I/ KT divided by Q, the partition function/ P sub J.
00:35:15.100 --> 00:35:21.300
That is E ⁻E sub J/ KT/ Q.
00:35:21.300 --> 00:35:37.800
The Q go away and what we are left with is E ⁻E sub I/ KT/ E ⁻J/ KT.
00:35:37.800 --> 00:35:44.600
Therefore, the relative population, we said that N2 and N1 is 1/4 to 1, that is what this part says.
00:35:44.600 --> 00:35:56.800
The population of the upper state to the 1/4 of the lower state, 1/4 to 1.
00:35:56.800 --> 00:36:18.600
P2/ P1 is equal to E ^- energy of the second level 400/ KT/ E⁻¹⁰⁰/ KT is
00:36:18.600 --> 00:36:35.700
equal to E¹/ KT × - 400 – 100.
00:36:35.700 --> 00:36:54.600
This is basic math here, we get E⁻³⁰⁰/ KT.
00:36:54.600 --> 00:37:02.400
1/4 is equal to E⁻³⁰⁰/ KT.
00:37:02.400 --> 00:37:04.900
Let us take the log of both sides.
00:37:04.900 --> 00:37:21.100
We get LN of 1 – LN of 4 = – 300/ KT.
00:37:21.100 --> 00:37:26.500
The energies are in wave numbers, inverse cm.
00:37:26.500 --> 00:37:36.000
K is equal to 0.6950 inverse cm/ K.
00:37:36.000 --> 00:38:09.800
T, when I solve for T, T is equal to -300 inverse cm divided by LN 4.06950 inverse cm/ K.
00:38:09.800 --> 00:38:21.700
I get T equal to 311.4 K.
00:38:21.700 --> 00:38:32.200
At 311.4 K, for this particular 2 level system, 2 energy system, we have 100 inverse cm, 400 inverse cm.
00:38:32.200 --> 00:38:46.000
At 311.4 K, I find that the 2nd level is actually ¼ as populated as the 1st level.
00:38:46.000 --> 00:38:53.000
Let us switch colors here, just for the hell of it.
00:38:53.000 --> 00:39:04.400
More generally, when levels are degenerate, we just have to include the degeneracy.
00:39:04.400 --> 00:39:18.700
Nothing more than that numbers R, the degenerate, the equations are as follows.
00:39:18.700 --> 00:39:33.000
N sub I/ N sub J = G sub I E ⁻E sub I/ KT/ the degeneracy.
00:39:33.000 --> 00:39:41.100
Let us make it a small j E ⁻J/ KT.
00:39:41.100 --> 00:40:02.400
It is going to end up equaling G of I or G of J E¹/ KT × -E sub I + J.
00:40:02.400 --> 00:40:04.600
There you go.
00:40:04.600 --> 00:40:44.900
In these cases, even when the upper level is higher in energy because of the degeneracy
00:40:44.900 --> 00:40:50.000
or because of the greater degeneracy of higher levels.
00:40:50.000 --> 00:40:53.300
I will leave the degeneracy ahead.
00:40:53.300 --> 00:41:09.200
Because of degeneracy it can and will often,
00:41:09.200 --> 00:41:21.900
In these cases, even when the upper level is higher in energy because of the degeneracies,
00:41:21.900 --> 00:41:44.900
the upper level tends to actually be more populated than the lower level.
00:41:44.900 --> 00:41:49.500
In general, we know that the lower levels are more populated at given temperature.
00:41:49.500 --> 00:41:52.500
It is harder to get it up to the upper levels.
00:41:52.500 --> 00:41:59.500
However, because the degeneracies of upper levels are so high sometimes, they are populated
00:41:59.500 --> 00:42:02.200
that it ends up inverting this population.
00:42:02.200 --> 00:42:09.200
You will end up with higher energy levels but they actually have more population than the lower energy levels.
00:42:09.200 --> 00:42:17.100
It is actually the mostly be the case like this for rotation, not so much for vibration and electronic.
00:42:17.100 --> 00:42:23.800
Let us go to example 6.
00:42:23.800 --> 00:42:35.200
Calculate the relative population for the J = 2 and J = 1 rotational states of the carbon monoxide molecule at 25°C.
00:42:35.200 --> 00:42:43.300
The temperature here, the 25° C, the temperature = 298 K.
00:42:43.300 --> 00:43:09.300
The N J = 2/ N J = 1, that = the G of 2 E ⁻E of 2/ KT/ G of 1 E ⁻E/ 1/ KT.
00:43:09.300 --> 00:43:13.300
These are rotation levels, rotational states.
00:43:13.300 --> 00:43:20.400
The degeneracy of level 2 = 2 × 2 + 1 = 5.
00:43:20.400 --> 00:43:30.100
The degeneracy of level 1 is 2 × 1 + 1 is equal to 3.
00:43:30.100 --> 00:43:50.400
N of 2/ N of 1 is equal to 5/3 E⁻¹/ KT - E2 + E1.
00:43:50.400 --> 00:43:53.100
We need to find what the energies are.
00:43:53.100 --> 00:44:05.500
The energy sub J is equal to the rotational constant × J × J + 1.
00:44:05.500 --> 00:44:17.700
The energy of 2 is equal to that × 2 × 2 + 1 = 6B~.
00:44:17.700 --> 00:44:30.400
And energy 1 = B~ 1 × 1 + 1 = 2B~.
00:44:30.400 --> 00:44:42.600
-6B~ + 2B~ = -4B~.
00:44:42.600 --> 00:44:47.200
B~ for carbon monoxide, just go ahead and look it up.
00:44:47.200 --> 00:44:50.200
I did not supply it here, just go ahead and look it up in your book.
00:44:50.200 --> 00:44:58.300
It is 1.931 inverse cm.
00:44:58.300 --> 00:45:25.100
N2/ N1 = 5/3 E⁻⁴ × 1.931 divided by KT divided by 0.6950 × 298.
00:45:25.100 --> 00:45:35.400
And when I do that, I get 1.606.
00:45:35.400 --> 00:45:38.100
Let me go ahead and do it over here.
00:45:38.100 --> 00:45:49.200
1.606 is N of 2 over N of 1.
00:45:49.200 --> 00:46:24.100
This means that the J = 2 rotational state is 1.606 × as populated as the J = 1 rotational state.
00:46:24.100 --> 00:47:03.500
As we mentioned in the previous problem, where the degeneracies are involved, the higher energy states
00:47:03.500 --> 00:47:29.100
because they have the higher degeneracies, the higher energy states will often be more populated than the lower energy states.
00:47:29.100 --> 00:47:46.200
This mostly applies to rotational energies because rotational energies tend to be very close.
00:47:46.200 --> 00:47:48.600
It will give you a very small scale.
00:47:48.600 --> 00:47:55.900
You might have an electronic state that has a much higher degeneracy than the lower electronic state.
00:47:55.900 --> 00:48:01.000
But because of the difference in energy between electronic states is huge, it is not going to make a difference.
00:48:01.000 --> 00:48:03.400
In the case of rotation, it makes a difference.
00:48:03.400 --> 00:48:10.500
This mostly applies to rotational levels, as we see here.
00:48:10.500 --> 00:48:13.200
The rotational level of 2 has a degeneracy of 5.
00:48:13.200 --> 00:48:16.700
The rotational level J = 1 has a degeneracy of 3.
00:48:16.700 --> 00:48:21.600
The 2 is higher in energy, however because the degeneracy, it is more populated.
00:48:21.600 --> 00:48:25.400
It has more things in that particular state.
00:48:25.400 --> 00:48:27.100
Thank you so much for joining us here at www.educator.com.
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We will see you next time for more examples on statistical thermodynamics.
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Take care, bye.