WEBVTT chemistry/physical-chemistry/hovasapian
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Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.
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Today, we are going to continue our example problems for the first law and for energy and things like that.
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Let us just jump right on in.
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Our first example of this lesson is the following.
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We have 1 mol of Van Der Waals gas at 298 K and it expands isothermally and reversibly from 25 dm³ to 65 dm³.
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The Van Der Waals constant is 0.366.
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This is Pascal m⁶/ mol².
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These are just the units, do not worry about those and B = 0.0429 dm³/ mol.
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For the gas, the DU DV this should be sub T, my apologies, at constant temperature.
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The rate of change of energy with respect to volume at temperature = an²/ V².
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We would like you to find Q, W, Δ U, and δ H for the transformation.
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There is a lot of information in this particular problem.
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Things that are important, we have the temperature isothermal and reversible.
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Remember, reversible means that the pressure external is actually equal the pressure of the system.
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The change in volume we have the constants and we have this thing so this is a Van Der Waals gas, it is not an ideal gas.
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For an ideal gas, this DU DV = 0 that term drops out and that expression for the total differential which I will write in just a minute.
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Let us go ahead and see what we can do.
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Now the expression I was talking about,
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Let me do this in blue.
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The expression I’m talking about is the following.
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Our basic mathematical expression, what is handful equations that we definitely want to know and use in order to derive everything else.
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DU = CV DT + DU DV at constant temperature × DV.
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This is the general expression for the change in energy of a system.
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For the differential change in energy of the system.
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We would want the full change which we just integrate this expression.
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Let us see what we can do to simplify this for us and let the problem itself put constraints on this.
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This is isothermal.
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Isothermal means that the temperature is constant.
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That implies that as DT term = 0 because there is no change in temperature, that means this one drops out.
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What we are left with is the following.
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We are left with DU = DU/ DV sub T DV that is going to give us the differential change in energy.
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Therefore, when we integrate this we end up with a total change in energy for the system =
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the integral from volume 1 to volume 2 of this DU DV T DV.
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Fortunately, they give us the DU DV sub T.
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It is right there and very convenient.
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This is equal to the integral from the first volume and the second volume and notice we have the first volume and we have the second volume of an² / V² DV.
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An² comes out, we have V1 / V2 and then we have DV / V.
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This is what we have to integrate in order to solve this.
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On notice we started off with this.
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This is our basic, one of the basic equations that we need to know and we let the problem dmde what sort of constraints, what is going to go to 0.
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If they said that this under constant volume this would go to 0.
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This is what is going on here.
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Once we do that, we end up with the following.
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This is going to be an² × 1/ V1 - 1/ V2 and when I put these numbers in, that is final.
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An², a is 0.366, I’m going to go ahead and leave off the units.
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I hope you guys do not mind.
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I will leave the units for this first problem it might be nice to see.
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Pascal m⁶/ mol² × 1 mol², because we are dealing with 1mol of the gas and of course, these are expressed in dm³.
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We have to express this because this is in meters we have to express the volume in meters.
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We are dealing with Joules and Pascal so the volume has to be in meters.
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We have to make the conversion.
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Volume 1 it is just going to be 25 × 10⁻³.
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1 dm³ is 10⁻³ m³, basic conversion factor -1/ 65 × 10⁻³.
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When we do this calculation, we end up with 9.0 J.
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This is not a strong point but I hope that you actually confirm the arithmetic.
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The arithmetic is secondary.
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What is important is this process, getting to this point, this is what is important.
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Often in a test, for some of the higher end courses you might be asked to calculate and derive.
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You can stop at the equation.
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You do not have to do the numbers.
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9 J that takes care of the change in energy.
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The change in energy of the system is 9 J.
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Let us go ahead and see what we can do about the work.
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Our equation for work is DW = P external DV, again, another one of the basic equations that you have to know.
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This is the definition for work it is the pressure × change in volume.
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If this is a reversible process reversible implies that the P external = P.
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We can write DW = the pressure of the system × the change in volume.
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This is a Vandrual’s gas, the pressure = nrt /V - nb - an² / V².
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We put this into here and we integrate, our total work is going to equal the integral from volume 1 to volume 2 of this expression,
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the P which is nrt / V - nb - an²/ V².
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Do not let this intimidate you, this is not a difficult and it is a very simple integral.
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It was just a logarithmic integral.
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It just looks complicated.
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Do not to let how something looks intimidate you to stop and look at it.
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See what it is, it is very simple integration.
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You can do it by hand or if you like this use mathematical software either mathematical or maple.
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DV that is it.
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When you do this integration, you will end up with following.
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You end up with work = nrt × log V2 - nb/ V1 - nb + an² × 1/ V2 -1 / V1.
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Notice, in the last problem when we integrated I had 1/ V -1 – 1/ V2.
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I did that because the integral itself ended up coming out as -1/ V.
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I just use that negative sign to flip the order.
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In this case, I took the negative sign out and I dmded to go ahead and make this a positive.
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It is not a mistake from previous, it just depends on how you want to express it.
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Again, a particular mathematical equation, at this level you can simplify to any degree you want.
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The simplification is irrelevant.
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What is relevant is that you understand what is going on.
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If you want to leave this as a minus sign, I might put 1/ V1 – 1/ V2.
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Just look at the same answer is not a problem.
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When we put all the numbers in, I’m not going to go ahead and write all the numbers in.
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It is actually a lot.
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Maybe I should, it is not a problem, we should see everything.
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1 × 8.314 the temperature is 298 K × log,
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We have to express the volume in cubic meters.
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This is going to be 65 × 10⁻³ -nb -1 × 0.064 × 10⁻³.
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B was expressed in terms of dm³/ mol, we have to express it in terms of dm³/ mol.
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Thermodynamics, physical chemistry in general, it is a conversion factor that ends up being tedious and toilsome.
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So 25 × 10⁻³ - 1 × 0.064 × 10⁻³ we have that.
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It is going to be +a which is 0.366 that is not a problem × 1² × 1/ V2 which is 65 × 10⁻³ -1/ 25 × 10⁻³.
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When these numbers are floating around, I do hope that you actually confirm my arithmetic.
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What you will end up with is 2371.3 + -9.0 and you will end up with the work is going to be 2362.3 J.
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The system is expanding, the gas is expanding isothermally and reversibly.
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It is doing work on the surroundings which is why the work is positive.
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Work done on the surroundings is positive that is why the work is positive.
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Let us go ahead and see about the next phase here.
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We are looking for Q, it looks like we did δ and we did W, let us do Q.
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We know that δ U= Q – W.
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Therefore, Q = δ U + W.
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Therefore, the heat of this reaction = 9 J which is the δ U, the change in energy + 2362.3 J.
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Therefore, the heat is 2371.3 J.
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I apologize sometimes I write Joules and sometimes I just put J.
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Let us go ahead and see what we can do about the,
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Let us see, with Q let us find δ H.
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Δ H here is how you want to put it.
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The definition of H is U + PV.
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Again, when the basic equation that you start with in this problem is only a handful of equation that you need to know.
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Everything else should be derived.
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Do not think that you can memorize a bunch of equations and to apply them to a situation.
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This is high level science so the problems are not mechanical.
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It is not just this is what this problem is, you plug the numbers in.
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It is not going to work that way.
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There are ton of equations in thermodynamics but there are a handful of basic equations from which the others come.
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The derivations are not difficult, as far as the problems are concerned.
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Do not memorize the equations because they will just make you crazy if you try to memorize.
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So δ H which is what we are looking for = δ U + PV.
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The δ operator is a linear operator.
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So δ H linear just means you can distribute operators symbolically.
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You can distribute that way, distribute that way, which you will end up with is δ U+ δ PV.
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Well, that is the same as δ H = Δ U.
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Put the δ PV, that is just P2 V2 - P1 V1.
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We have to calculate P2 V2 P1 V1 from Van Der Waals equation.
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Again, the pressure = Van Der Waals equation = nrt / V - nb - an²/ V2.
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This is going to be a bit tedious as far as a calculation is concerned but it is not a problem.
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Let us go ahead and calculate P2 V2 first.
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We will do that one.
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It is going to be nrt, this first one I just want to make sure to do everything.
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Over V - nb – an² / V² that is the V2 part, over V2² × V2.
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When you put all the numbers, you have all the numbers.
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You have n, R, T, 298.
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You have the second volume, you have nb.
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You have all these values.
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When you put these values in, you end up with 2473.6 J.
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You do the same for P1 V1.
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In this case, P1V1 = nrt/ V1 - nb - an² / V1² × V1.
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When you put those values in make sure to convert volume to cubic meters.
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In case of 25 × 10⁻³ m³ is 25 dm³.
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When you do that, you end up with 2467.2 J.
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Therefore, P2 V2 - P1 V1 = this – that, you end up with 6.4 J that is P2 V2 - P1 V1.
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This is the δ PV.
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We can go ahead and calculate our δ H.
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Δ H = Δ U + δ PV.
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Δ H where δ U was 9.0 J + 6.4 J for a total of 15.4 J, that is our δ H.
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This is not the only way to solve this problem but one of the beauties of science itself is the problem they come up with different solutions.
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I usually just go with the first solution that occurs to me.
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It might be a little bit longer or shorter, I do not know.
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By all means, any other solution that you can come up with, you should examine that and you should explore it.
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And you should use it because that is the way you see the problem.
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Arithmetic often get in the way of what is going on so I want to pull back a little bit and just go through what is that we did in this particular problem.
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The first thing that we did is we calculated the energy.
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We started off with DU = CV DT + DU DV sub T × DV.
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They said that this was isothermal so this term goes to 0.
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Again, we start with the basic equation and we derive the relationship that we needed.
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DU = DU / DV sub T DV.
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When we integrate that expression, we get the expression for the full energy for the entire change of state
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which is going to be integral from V1 V2 of this DU DV sub T DV.
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When you integrate you put the numbers in.
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That is how we found the energy and then we dmded to go with the work.
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The definition of work = P external DV.
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This one is reversible, P external = P.
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What we end up with is P = the Van Der Waals gas.
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Nrt/ V - nb – an² / V².
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We put this expression into here and then we integrate.
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When we integrate, the work equals the integral from V 1 to V2 of PDV.
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We put this expression into here and we integrate it and we get the value for the work and we put the numbers in to actually get the work itself.
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Number 3 was actually pretty straightforward.
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We used this basic relationship and the differential version is DQ – DW.
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The integrated version is δ U= Q – W.
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These are not exact differentials.
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An exact differential integrates that δ U.
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These inexact differentials just go to Q and W.
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We are looking for Q so I just move things around.
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You already found this and we just found that.
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Q = Δ U + W and we ended up finding the heat for the particular reaction.
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How do we do δ H?
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Δ H = the change in energy which we have + δ PV.
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Δ PV = P2 V2 - P1 V1.
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We used the Van Der Waals equation to find this and to find that, make the subtraction, use that and put it into here and solve for δ H.
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This was the process that we did and it is all based on a handful equations.
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This equation, right here, that is our first equation.
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It is one of the fundamental equations that we have to know in order to solve these problems.
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The second equation right here is the definition of work.
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This is the total energy of the system.
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The energy is dependent on temperature and on volume.
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For this particular problems, isothermal the DT went to 0 so it only depends on the change in volume.
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This is the definition of work.
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This is the first law of thermodynamics.
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The energy of the system = the heat transfer - the work transferred.
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Of course δ H comes from the definition of enthalpy which is one of the basic equations.
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Again H = U + PV that is the definition of enthalpy.
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Excuse me, let us move on to our next problem here.
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We have a lot of extra pages.
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Example number 2, so we have 1 mol of an ideal gas and it is confined under pressure P external.
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This external should actually be down below, I ended up putting on top.
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P external = P = 250 kl Pascal.
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Basically, what this says is that the external pressure and the internal pressure are the same.
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If you have this gas and this piston set up, this with a little weight on top, this is an equilibrium.
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It is not moving anywhere.
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In other words, the pressure this way = the pressure that way, that is all these means.
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So ideal gas is very important.
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We are provided with pressure 250 kl Pascal, 250,000 kl Pascal.
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The temperature is changed from 120 to 25°.
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The temperature of the gas we drop it from 120°C to 25°C.
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The constant volume heat capacity= 3 Rn/ 2, n is 1.
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Here we want to calculate the Q, W, δ H.
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This is a pretty standard problem, the four basic properties of the system.
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Notice, they say nothing about a change in volume here.
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Let us see what we can do first.
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Let us start with work, if it stays in red it is not a problem.
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I will start with a definition of work DW = P external × DV.
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The P external is constant, if it is constant when we integrate this, this comes out of the integral sign.
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When you integrate this, this comes out of the integral sign and what you end up with is work = P external × the change in volume.
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Let us see what we have got.
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We have an ideal gas so we have PV = nrt so V volume = nrt / P.
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We have two different temperatures, we have 120 and 25.
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The pressure is actually constant so pressure stays the same.
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For change in volume, let us do this.
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Let us find, we have the P external we just need a change in volume.
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This is P external × V2 - V1.
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I’m going to start with V1, it is going numerical order here.
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V1 = nr T1/ P which = 1 × 8.314 × temperature 1, the initial temperature is 120.
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We have 393 K/ 250,000 Pascal.
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You end up with a value equal to 0.0 1307 and this is going to be in cubic meters.
00:24:55.300 --> 00:25:02.300
If you are concerned about what is happening here, let us do a quick unit check.
00:25:02.300 --> 00:25:15.600
We have mol × J/ K mol × K/ Pascal.
00:25:15.600 --> 00:25:33.600
This is the same as mol × J is N m/ K mol × K and Pascal is a N/ m².
00:25:33.600 --> 00:25:37.300
Mol cancels mol, K cancels K, N cancels N.
00:25:37.300 --> 00:25:42.500
This comes up to the top and ends up with cubic meters is what this comes from.
00:25:42.500 --> 00:25:44.100
This is cubic meters.
00:25:44.100 --> 00:25:46.600
I’m going to do the same for V2.
00:25:46.600 --> 00:25:51.200
Volume 2 that is equal to nr T2/ P.
00:25:51.200 --> 00:26:00.400
We have 1 × 8.314, 25°C which is 298.
00:26:00.400 --> 00:26:03.500
Try to keep everything straight is going to be the difficult part.
00:26:03.500 --> 00:26:13.300
This is 250,000 and it is very easy to actually think that 298 here, and put the 393 here by mistake.
00:26:13.300 --> 00:26:18.500
Again, it is just that is life, lot of numbers floating around.
00:26:18.500 --> 00:26:26.000
Equals 0.00991 m³.
00:26:26.000 --> 00:27:09.500
Our work is equal to P external × V2 - V1 = 250,000 Pascal × C V2 is 0.0091 cubic meters -0. 01307 m³ and we end up with Pascal m³ which is J.
00:27:09.500 --> 00:27:16.800
Our work is going to end up being -790 J.
00:27:16.800 --> 00:27:21.400
When you drop the temperature of this gas, this gas is going to contract.
00:27:21.400 --> 00:27:23.400
It is going to fall.
00:27:23.400 --> 00:27:27.900
The surroundings are going to do work on the system.
00:27:27.900 --> 00:27:34.400
When the surroundings do work on the system that means work is leading the surroundings.
00:27:34.400 --> 00:27:37.600
Our consigned convention work is -790 J.
00:27:37.600 --> 00:27:45.600
790 J is leaving the surroundings and going into the system, that is what this means.
00:27:45.600 --> 00:27:50.900
Let us go ahead and do our C.
00:27:50.900 --> 00:27:54.200
Let us go ahead and do our DH next.
00:27:54.200 --> 00:27:58.800
We have our basic relationship, one of the equations that we have to know.
00:27:58.800 --> 00:28:16.000
The differential change in enthalpy = the constant pressure heat capacity× the differential temperature + DH DP T DP.
00:28:16.000 --> 00:28:22.800
The pressure is constant that means DP = 0.
00:28:22.800 --> 00:28:27.800
Constant pressure, constant P implies the DP = 0.
00:28:27.800 --> 00:28:40.400
This term goes to 0 so you are left with DH = CP DT.
00:28:40.400 --> 00:28:54.900
When you integrate this, CP is constant so what comes out which we are left with is the following equation DH = CP δ T.
00:28:54.900 --> 00:29:01.400
We have δ T that is just going to be the 120 - 25 so we need CP.
00:29:01.400 --> 00:29:03.900
They did not give us the CP, they gave the CV.
00:29:03.900 --> 00:29:11.600
They gave the CV = 3/2 Rn and we have a relationship, this is an ideal gas.
00:29:11.600 --> 00:29:20.700
An ideal gas there is a relationship between the constant pressure heat capacity and the constant volume heat capacity that is the following.
00:29:20.700 --> 00:29:25.400
CP - CV = Rn.
00:29:25.400 --> 00:29:32.900
They gave the CV we know Rn and we can find CP and put it in here to find our value.
00:29:32.900 --> 00:29:47.000
CP = CV + Rn that is equal to 3/2 Rn + 1 Rn.
00:29:47.000 --> 00:29:51.600
Therefore, the constant pressure heat capacity= 5/2 Rn.
00:29:51.600 --> 00:29:54.100
And now we can go ahead and put this into here.
00:29:54.100 --> 00:30:01.300
Therefore, our δ H = 5/2 Rn δ T.
00:30:01.300 --> 00:30:05.100
We have all the numbers that we need.
00:30:05.100 --> 00:30:24.300
Δ H = 5/2 × 8.314 × 1 mol × change in temperature final - initial 25°C -120°C.
00:30:24.300 --> 00:30:30.800
The change in temperature increment, I did not change this to K because the δ T for Celsius and K is the same.
00:30:30.800 --> 00:30:40.100
The individual things are not the same 393 and this is 298 but 298 -393 is the same as 25 -120.
00:30:40.100 --> 00:30:47.600
The δ T because all you are doing you are adding 273 so it goes up equally.
00:30:47.600 --> 00:30:58.600
And when you do this you end up with δ H = -1975 J.
00:30:58.600 --> 00:31:14.900
Now under constant pressure δ H is actually equal to QP.
00:31:14.900 --> 00:31:17.900
I will probably write it the other way QP= δ H.
00:31:17.900 --> 00:31:23.700
In other words the heat happens to equal the enthalpy so we went ahead and took care of that.
00:31:23.700 --> 00:31:31.200
The Q for this reaction = -1975 J.
00:31:31.200 --> 00:31:34.500
The only thing we are missing now is δ U.
00:31:34.500 --> 00:31:39.000
We have a relationship δ U= Q – W.
00:31:39.000 --> 00:31:54.700
We just found Q, we found W a little bit earlier it is = -1975 J that is the Q - the W which is -790 J.
00:31:54.700 --> 00:32:05.200
Our change in energy of the system = -1185 J.
00:32:05.200 --> 00:32:08.200
There we go and what does all this mean?
00:32:08.200 --> 00:32:10.200
Here is what is happening.
00:32:10.200 --> 00:32:22.000
We have our system and we have our surroundings Q that was equal to -1975 J.
00:32:22.000 --> 00:32:26.500
1975 J left the system.
00:32:26.500 --> 00:32:38.700
Work = -790 J, the surroundings did 790 J of work on the system.
00:32:38.700 --> 00:32:42.500
In other words, 790 J went into the system.
00:32:42.500 --> 00:32:52.000
Therefore, the total change in energy 1975 from the systems point of view 1975 J left, 790 J came in.
00:32:52.000 --> 00:33:08.000
The total net energy, the net loss by the system is 1185 J net loss by the system.
00:33:08.000 --> 00:33:11.200
That is what is happening here.
00:33:11.200 --> 00:33:14.900
The surroundings did 790 J of work.,
00:33:14.900 --> 00:33:20.200
The system in going from 120 to 25 lost 1975.
00:33:20.200 --> 00:33:27.000
The net loss is the systems.
00:33:27.000 --> 00:33:29.800
Let us see what we have got here.
00:33:29.800 --> 00:33:34.600
Let us go to our next example.
00:33:34.600 --> 00:33:37.600
Example 3, we have 1 mol of ideal gas.
00:33:37.600 --> 00:33:46.000
We have 1 mol, we have an ideal gas is taken from 10°C to 90°C under constant volume.
00:33:46.000 --> 00:33:47.200
It is very important.
00:33:47.200 --> 00:33:50.200
This time, our constrained is under a constant volume.
00:33:50.200 --> 00:33:55.000
Our constant volume heat capacity is 21 J/ mol K.
00:33:55.000 --> 00:33:57.200
This is actually a molar heat capacity.
00:33:57.200 --> 00:34:05.500
21 J/ mol K, calculate Q, W, Δ U, and δ H.
00:34:05.500 --> 00:34:15.700
Let us see what we have got.
00:34:15.700 --> 00:34:27.300
Let us see which one do we do first.
00:34:27.300 --> 00:34:38.500
Constant volume implies that DV = 0.
00:34:38.500 --> 00:34:46.500
If you remember, under constant volume our Q = δ U.
00:34:46.500 --> 00:35:06.300
Our equation DU = CV DT + DU DV.
00:35:06.300 --> 00:35:10.200
Constant volume DV equal 0 but we are dealing with an ideal gas so this is already 0.
00:35:10.200 --> 00:35:13.000
In either case, that one goes away.
00:35:13.000 --> 00:35:23.200
Our DU which = QV = CV DT.
00:35:23.200 --> 00:35:26.700
We just go ahead and integrate this.
00:35:26.700 --> 00:35:28.800
Therefore, when we integrate this we end up with the following.
00:35:28.800 --> 00:35:37.100
You end up with δ U= Q = CV δ T.
00:35:37.100 --> 00:35:39.500
When we integrate this, this is constant so it comes out.
00:35:39.500 --> 00:35:42.500
The integral of DT is δ T.
00:35:42.500 --> 00:35:47.000
We have the CV, we have the δ T, so this is really easy.
00:35:47.000 --> 00:36:15.200
Our DU = CV which we said was 21.0 J/ K and our δ T is 90°C -10°C.
00:36:15.200 --> 00:36:28.000
And when we run this calculation, we end up with 1680 J.
00:36:28.000 --> 00:36:39.700
That is δ U= 1680 J also happens to equal Q= 1680 J.
00:36:39.700 --> 00:36:51.300
We have taken care of the Q and the δ U, let us go ahead and take care of the constant volume.
00:36:51.300 --> 00:36:56.500
We did that.
00:36:56.500 --> 00:36:59.000
Let us go ahead and do δ H next.
00:36:59.000 --> 00:37:14.800
Actually, if we want in this particular case because constant volume, no work is done.
00:37:14.800 --> 00:37:19.800
PV so work = 0.
00:37:19.800 --> 00:37:21.800
What about that one?
00:37:21.800 --> 00:37:24.300
Let us go ahead and do δ H.
00:37:24.300 --> 00:37:43.300
Again, we have δ H = Δ U+ δ PV equal to,
00:37:43.300 --> 00:37:54.100
We have to watch out if a constant volume but V is constant, therefore, let us change this equation a little bit.
00:37:54.100 --> 00:38:00.100
That means δ H is going to equal Δ U V comes out.
00:38:00.100 --> 00:38:07.600
V × δ P just the change in pressure.
00:38:07.600 --> 00:38:09.200
Let me rewrite that again on this page.
00:38:09.200 --> 00:38:15.200
We have δ H or a lot of crazy things floating around.
00:38:15.200 --> 00:38:27.200
We have δ H = δ U+ V × δ P or ideal gas so PV = nrt.
00:38:27.200 --> 00:38:35.200
Therefore, the pressure = nrt / V.
00:38:35.200 --> 00:38:48.500
The δ H = δ U+ V δ P well δ P is P2 - P1.
00:38:48.500 --> 00:38:57.200
P2 = n R T1/ V nr T2/ P.
00:38:57.200 --> 00:39:02.700
I'm sorry for V, I’m getting all white variable here.
00:39:02.700 --> 00:39:04.500
Remember, volume is constant so it stays V.
00:39:04.500 --> 00:39:08.100
It is not V1 or V2, temperature 2.
00:39:08.100 --> 00:39:15.600
Nr here is the V and then P1 = nr T1 / V.
00:39:15.600 --> 00:39:25.100
These cancel out so you get δ H = Δ U+ nr × T2 - T1.
00:39:25.100 --> 00:39:27.800
We have T2 and we have T1.
00:39:27.800 --> 00:39:39.800
Therefore, when we put these values in we get δ H = δ U which was 1680 J + 1 mol.
00:39:39.800 --> 00:39:48.600
R is 8.314 and we have 90°C -10°C.
00:39:48.600 --> 00:39:54.600
Again the difference in temperature, δ T in Celsius is the same as the δ T in Kelvin.
00:39:54.600 --> 00:40:09.500
When we run this calculation, we end up with 2345 J is the δ H.
00:40:09.500 --> 00:40:11.800
Let me actually do these individually.
00:40:11.800 --> 00:40:28.000
Here we have the 1680, this particular value ends up being 665 J.
00:40:28.000 --> 00:40:49.500
We took it from 10°C to 90°C so what is going on here is the 1680 J, that much energy comes from the rise in temperature.
00:40:49.500 --> 00:41:02.000
The 665 J it comes from the rise in pressure.
00:41:02.000 --> 00:41:05.500
The δ H, the enthalpy is an accounting device.
00:41:05.500 --> 00:41:11.200
It accounts for the energy change + any change in pressure, volume, work.
00:41:11.200 --> 00:41:14.700
A change in pressure, change in volume, things like that.
00:41:14.700 --> 00:41:22.700
1680 if it comes from just the rise in temperature, 665 come from the actual rising pressure because we are keeping at a constant volume.
00:41:22.700 --> 00:41:23.300
That is what is happening.
00:41:23.300 --> 00:41:25.200
We keep it at constant volume.
00:41:25.200 --> 00:41:31.700
It cannot do work this way so the pressure increases inside that increases the energy of the system.
00:41:31.700 --> 00:41:39.500
You always want to stop, take a look at the numbers, see what they mean in terms of the direction of flow positive or negative,
00:41:39.500 --> 00:41:44.000
and see if you can actually identify what value is coming from what.
00:41:44.000 --> 00:41:51.800
This process will really help you clarify what is happening physically.
00:41:51.800 --> 00:41:58.300
Let us move on to example 4, let us see what we have here.
00:41:58.300 --> 00:42:05.300
Find δ U and δ H for the transformation of 1 mol of ideal gas from 25°C and 1 atm.
00:42:05.300 --> 00:42:11.300
They give us an initial temperature and initial pressure to a final temperature and a final pressure.
00:42:11.300 --> 00:42:17.000
The molar heat capacity of the gas CP.
00:42:17.000 --> 00:42:23.800
This is a CP, CV is 20 + 0.045 J/ mol K.
00:42:23.800 --> 00:42:28.500
I will make one little correction here.
00:42:28.500 --> 00:42:33.100
20 + 0.045 × T, T is the variable.
00:42:33.100 --> 00:42:37.600
In this particular case, the heat capacity is temperature dependent.
00:42:37.600 --> 00:42:43.800
It starts at 20 but as the temperature rises the heat capacity changes.
00:42:43.800 --> 00:42:49.600
This is not constant heat capacity, this is something we are going to have to integrate over a change in temperature.
00:42:49.600 --> 00:42:56.800
I apologize for that mistake so this should be 20 + 0.045 × T.
00:42:56.800 --> 00:43:05.300
Let me write it here CP = 20.0 + 0.045 T.
00:43:05.300 --> 00:43:09.100
T is a variable like x.
00:43:09.100 --> 00:43:12.600
In this particular case you can do Δ U first, you can do δ H first.
00:43:12.600 --> 00:43:14.600
It does not really matter.
00:43:14.600 --> 00:43:17.800
Let us go ahead and start off with a basic equation.
00:43:17.800 --> 00:43:20.100
I’m going to do δ U first.
00:43:20.100 --> 00:43:31.300
Δ U is equal to CV DT + DU DV T DV.
00:43:31.300 --> 00:43:35.600
This term goes to 0 because the gas is ideal.
00:43:35.600 --> 00:43:38.600
An ideal gas the DU DV = 0.
00:43:38.600 --> 00:43:48.100
What I'm left with is DU = CV DT.
00:43:48.100 --> 00:44:02.800
When I integrate that, of course I will have to integrate that, I will be left with δ U= the integral from temperature 1 to temperature 2 of CV DT.
00:44:02.800 --> 00:44:07.500
I do not have CV DT, I have CP DT but it is an ideal gas I know that there is a relationship.
00:44:07.500 --> 00:44:13.000
I know that CP - CV = Rn.
00:44:13.000 --> 00:44:24.200
Therefore, CV = CP – Rn.
00:44:24.200 --> 00:44:42.200
CP = 20.0 + 0.045 T - 8.314 × 1 mol.
00:44:42.200 --> 00:44:55.500
For my constant volume heat capacity I end up with some 20 - 8.314, I end up with 11.686 + 0.045 T.
00:44:55.500 --> 00:44:59.000
This is the value that I put in here.
00:44:59.000 --> 00:45:06.200
Therefore, δ U= the integral from 298 K.
00:45:06.200 --> 00:45:10.200
These cannot be a Celsius, you have to put these in K.
00:45:10.200 --> 00:45:28.300
298 K to 225 which is 498 K of the constant volume heat capacity which was 11.686 + 0.045 T DT.
00:45:28.300 --> 00:45:42.800
I will go ahead and just use math software and if it did it correctly I ended up with 5919 J that is our δ U.
00:45:42.800 --> 00:45:46.000
Let us go ahead and do a δ H for this.
00:45:46.000 --> 00:45:50.200
Δ H again, it is up to you how you want to do this.
00:45:50.200 --> 00:45:55.000
I’m going to do it with a long way with DU and δ PV.
00:45:55.000 --> 00:46:09.500
You already know that DH = CP DT + DH DP T DP.
00:46:09.500 --> 00:46:14.300
This is one of the fundamental equations for enthalpy.
00:46:14.300 --> 00:46:17.300
This is an ideal gas so this goes to 0.
00:46:17.300 --> 00:46:30.100
Our δ H = CP DT from T1 to T2, this is one way you can actually calculate the enthalpy.
00:46:30.100 --> 00:46:39.900
I did it another way, I did it with a basic equation by the definition of enthalpy DU + D PV.
00:46:39.900 --> 00:46:41.900
This is probably the quicker method.
00:46:41.900 --> 00:46:47.400
This is just one that I happen to have gotten used to so I just tend to automatically default to that one.
00:46:47.400 --> 00:46:51.700
But either one is fine however you want to see it.
00:46:51.700 --> 00:47:03.700
This = δ U + P2 V2 - P1 V1 and = Δ U.
00:47:03.700 --> 00:47:08.700
P2 V2 is an ideal gas PV = nrt.
00:47:08.700 --> 00:47:17.400
P2 V2 = nr T2 – nr T1.
00:47:17.400 --> 00:47:24.400
You end up with δ U + nr × T2 - T1.
00:47:24.400 --> 00:47:35.400
We have T2 and T1 so you end up with δ U which was 5919 + 1 mol ×,
00:47:35.400 --> 00:47:50.400
I really need to slow this down + 1 × 8.314 × 225 -25 again δ T.
00:47:50.400 --> 00:48:03.800
225 -25 I can leave that δ H = 7582 J.
00:48:03.800 --> 00:48:11.500
There you go, nothing strange going on here.
00:48:11.500 --> 00:48:17.000
P2 V2 P1 V1.
00:48:17.000 --> 00:48:32.800
This said nothing about volume being constant so despite the fact that this is an ideal gas, P2 V2 - P1 V1 is not 0 because temperature changes.
00:48:32.800 --> 00:48:39.000
That is what is going on here, be very careful because sometimes you can go with P2 V2 – P1 V1.
00:48:39.000 --> 00:48:43.600
If other things are the same, if this were isothermal that would not be a problem.
00:48:43.600 --> 00:48:47.800
If this were isothermal then Q would be constant.
00:48:47.800 --> 00:48:52.100
P2 V2 would equal nrt and P1 V1 equal nrt.
00:48:52.100 --> 00:48:56.600
Therefore, P2 V2 – P1 V1 equal 0.
00:48:56.600 --> 00:49:06.800
But temperature is not constant here so again what the problem says very careful on how you approach it, which is why you do not want to memorize equations.
00:49:06.800 --> 00:49:08.600
This is an excellent example.
00:49:08.600 --> 00:49:15.800
If you just memorize the equation you will just plug numbers in without stopping to save yourself and say this is not isothermal situation.
00:49:15.800 --> 00:49:18.700
If it were an isothermal situation, this term will go to 0.
00:49:18.700 --> 00:49:21.400
If not isothermal, therefore, this is not 0.
00:49:21.400 --> 00:49:23.500
You have to run the calculation.
00:49:23.500 --> 00:49:27.500
You have to be very careful, go slowly.
00:49:27.500 --> 00:49:29.500
That takes care of that one.
00:49:29.500 --> 00:49:35.500
Let us go ahead and go to example number 5, it should be our final example for this particular lesson.
00:49:35.500 --> 00:49:39.000
Let us see.
00:49:39.000 --> 00:49:47.200
When an ideal gas undergoes an expansion such that the relationship pressure × volume raise to the nth power = a constant
00:49:47.200 --> 00:49:54.200
or C is a constant and n is greater than one, it is called a reversible poly tropic expansion.
00:49:54.200 --> 00:50:00.700
We would like to calculate the work for such an expansion of 1 mol of the gas from V1 to V2.
00:50:00.700 --> 00:50:09.700
When the temperature 1 is 350 K and temperature 2 = 200 K, and n=2.
00:50:09.700 --> 00:50:16.400
We would like you, given that the constant volume heat capacity= 5/2 Rn or 5n/ 2.
00:50:16.400 --> 00:50:22.200
We want you to find Q, Δ U, δ H, for this particular expansion.
00:50:22.200 --> 00:50:23.300
Let us see what we have got.
00:50:23.300 --> 00:50:25.300
Here is a lot going on here.
00:50:25.300 --> 00:50:36.300
A new term poly tropic expansion, here we are given a relationship that the pressure × the volume raised to the nth power is actually a constant.
00:50:36.300 --> 00:50:42.600
Again, this is where the things we just have to dive in and see what is going on.
00:50:42.600 --> 00:50:48.100
There is only a handful of equations that you know, you want to start with those and see where it takes you.
00:50:48.100 --> 00:50:51.100
Let us go ahead and start with part A.
00:50:51.100 --> 00:50:54.800
I think I will go ahead and go back to blue for this.
00:50:54.800 --> 00:51:06.700
Part A, telling me that n = 2, I have PV² = a constant.
00:51:06.700 --> 00:51:13.700
I know that the pressure is equal to C/ V², C being a constant.
00:51:13.700 --> 00:51:20.900
My work = I know my basic relationship.
00:51:20.900 --> 00:51:28.700
Our basic relationship by definition is DW = P DV.
00:51:28.700 --> 00:51:37.000
Therefore, my work = the integral from V1 to V2 of P DV.
00:51:37.000 --> 00:51:44.700
I have P here, the relationship is C/ V² so I can just plug this into here and integrate.
00:51:44.700 --> 00:51:55.500
Work = the integral from one volume to another volume of P DV which P = C / V².
00:51:55.500 --> 00:52:06.500
This is C/ V² DV which = C × the integral from V1 to V2 of DV / V².
00:52:06.500 --> 00:52:18.200
And I'm left with C × 1 /V1 -1 / V2 that is great, I have expression for the work = C × 1/ V1.
00:52:18.200 --> 00:52:22.500
That is what they wanted, calculate the work for such an expansion.
00:52:22.500 --> 00:52:34.500
However, they gave us temperatures here so it looks like they want an actual numerical value not just some expression.
00:52:34.500 --> 00:52:41.500
Let us see what we have got.
00:52:41.500 --> 00:52:50.500
V = C/ nrt so let us see if we can play with this a little bit.
00:52:50.500 --> 00:53:01.500
We have work = constant × 1/ V1 -1/ V2.
00:53:01.500 --> 00:53:18.300
We have that P = C/ V², this is PV = nrt.
00:53:18.300 --> 00:53:32.600
I’m going to go ahead and put this P value of P into here and I end up with C/ V² × V = nrt.
00:53:32.600 --> 00:53:42.300
This V cancels that V so I’m left with C/ V = nrt.
00:53:42.300 --> 00:53:51.600
That means V = C/ nrt.
00:53:51.600 --> 00:54:03.800
That means that V1 is equal to C/ nrt 1 and V2 = C/ nr T2.
00:54:03.800 --> 00:54:08.600
I can go ahead and put these values into here and here.
00:54:08.600 --> 00:54:30.400
I got work = C × 1/ C/ nr T1 -1/ C/ nr T2.
00:54:30.400 --> 00:54:37.000
Work = C × nr T1/ C-1.
00:54:37.000 --> 00:54:40.000
This is working out really great.
00:54:40.000 --> 00:54:52.800
C cancels therefore our work = nr × T1 - T2.
00:54:52.800 --> 00:54:55.200
This is fantastic, we have T1 and T2.
00:54:55.200 --> 00:55:14.000
We have n and R so work = 1 mol × 8.314 J/ K mol × temperature 1 was 350 K and temperature 2 was 200 K.
00:55:14.000 --> 00:55:20.700
And therefore, for work for this process is 1247 J.
00:55:20.700 --> 00:55:24.000
That was fantastic.
00:55:24.000 --> 00:55:26.700
Work = 1247 J.
00:55:26.700 --> 00:55:31.500
Let us go ahead and see what we can do with DU.
00:55:31.500 --> 00:55:49.300
DU= CV DT so δ U = CV δ T well that is equal to 5/2 Rn δ T.
00:55:49.300 --> 00:55:51.300
I had all information.
00:55:51.300 --> 00:56:06.300
Therefore, δ U= 5/2 × 8.314 × 1 mol × δ T.
00:56:06.300 --> 00:56:14.300
Δ T is final - initial so we have now 200 -350.
00:56:14.300 --> 00:56:24.300
Therefore, the change in energy = -3118 J.
00:56:24.300 --> 00:56:26.600
That takes care of the energy.
00:56:26.600 --> 00:56:32.100
We have a relationship Δ U= Q – W.
00:56:32.100 --> 00:56:48.600
Therefore, Q = Δ U+ W so Q = δ U which is - 3118 J + 1247 J.
00:56:48.600 --> 00:56:57.100
Therefore, heat is -874 J.
00:56:57.100 --> 00:57:01.100
Let us see what we can do with δ H.
00:57:01.100 --> 00:57:09.400
Δ H = Δ U + δ PV and again this is as my default, I automatically go to this.
00:57:09.400 --> 00:57:23.900
It is equal to δ U + P2 V2 - P1 V1.
00:57:23.900 --> 00:57:33.400
We have PV² = a constant therefore P = C / V².
00:57:33.400 --> 00:57:45.100
Therefore, P2 V2 = C/ V 2² × V2.
00:57:45.100 --> 00:58:07.100
This V2 cancels one of the V2 so I'm left with C / V2 and P1 V1 analogously is C/ V1.
00:58:07.100 --> 00:58:37.400
We have δ U + C/ V2 - C/ V1 = δ U + C / nr T2 –C/ nr T1.
00:58:37.400 --> 00:59:09.400
We end up with δ U+ nr this time T2 - T1 = - 3118 J + 1 × 8.314 × 200 -350 = -3118 J + -1247 J.
00:59:09.400 --> 00:59:24.200
Our final δ H = -4365 J there we go.
00:59:24.200 --> 00:59:30.900
A long process and a little tedious and a little involved.
00:59:30.900 --> 00:59:39.600
But again you are starting off with a basic set of equations and any other information that they gave you that PV ⁺nth power = C.
00:59:39.600 --> 00:59:41.900
They gave you n = 2.
00:59:41.900 --> 00:59:45.900
You are going to use that to start fiddling around with it.
00:59:45.900 --> 00:59:52.900
This relationship holds, this is an ideal gas so this relationship holds.
00:59:52.900 --> 00:59:58.100
You can start fiddling with these and playing around with the equation.
00:59:58.100 --> 01:00:01.000
And again, this is not necessarily the only way to do this problem.
01:00:01.000 --> 01:00:03.300
Maybe there are other ways to do this problem I do not know.
01:00:03.300 --> 01:00:06.000
I just sort of jump right out on in and did it this way.
01:00:06.000 --> 01:00:09.500
And now everything seemed work out okay.
01:00:09.500 --> 01:00:12.300
I will go ahead and leave it like that.
01:00:12.300 --> 01:00:14.600
Thank you for joining us here at www.educator.com.
01:00:14.600 --> 01:00:18.600
We will see you next time for a continuation of some more example problems.
01:00:18.600 --> 01:00:22.100
We want to make sure to get this really nice and tight.
01:00:22.100 --> 01:00:23.000
Take care, bye.