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 on free energy.
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Let us jump right on in.
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I have to warn you, in this particular example problem set, this is going to be heavily mathematical.
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Not heavily mathematical in terms of difficulty, a lot of it is just using equations at our disposal in order to come up with new relations,
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that is really what it is all about.
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Between energy, entropy, and the free energy, we have collected a fair number of equations.
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We sort of close the circle on this discussion of thermodynamics.
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But now we want to do is we want to take all of these things and put them together.
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This is the part where it is going to be just strictly mathematical.
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Most of it is manipulation, what is going to make it seem intimidating is just the symbolism on a page.
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It is going to be a lot of symbols on the page and there is one of the lot equations that we are going to have to recall.
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This is actually an excellent exercise in putting all of this together.
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In any case, let us just jump right on in and see what we can do.
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The first one, using the appropriate Maxwell relation and the cyclic relation among volume, temperature,
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and entropy, demonstrate that DS DP sub V = Κ C sub V/ A T.
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The rate of change of entropy with respect to pressure holding volume constant is equal to the,
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This is the coefficient of thermal expansion, this is the coefficient of compressibility,
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this is the constant volume heat capacity, and this is the temperature.
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We want to find ways of expressing these derivatives, these relations in terms of easily measurable things.
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Like kappa and α, I can look those up.
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Constant volume heat capacity, I can either look it up or measure it directly.
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The temperature, I can just measure it, that is what we want.
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Let us see what we can do.
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When I look at the collection of Maxwell relations that I have, the only relation that involves TV and S and P V and S is the following.
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Let me actually go ahead and do this and blue.
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It is this one, DT DV sub S = - DP DS sub V.
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Well, what we want is DS DP sub V.
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Basically, we want the reciprocal of this.
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It is not a problem, let us take the reciprocal of that.
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This is what we want, we have this relationship.
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We recognize the PS and V, PDS and V, we just want to switch the P and S.
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We want the P on top.
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We want the rate of change of pressure with respect to change in entropy at constant volume.
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We go ahead and we reciprocate this expression.
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Therefore, DS DP at constant V = - DV DT of S.
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Let us use the cyclic relation.
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The cyclic relation between V T and S is the following.
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DV DT S, VTS let me actually show you how I come up with this, in fact.
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I have V T S, let me write those down V T and S.
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Underneath them, I'm going to just choose and I’m going to go ahead and put T here.
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I'm going to put S here, and I’m going to have V here.
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I’m going to do that.
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These VT, therefore this is an S, this is a TS, this is a V SV, this is a T.
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That is how you derive the cyclical relation.
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A cyclic relation among three variables in this case, volume, temperature, and entropy, I can do it that way.
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List any three of them then underneath it just at random, list the other three, and then what you are left out with,
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let us just go ahead and put the particular property that is held constant on the outside.
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The TS TSV SVT and = -1.
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We have this relationship and of course these are partial derivatives.
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You can go ahead and put the partial derivative signs in.
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That is the relationship that exists.
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With this particular relations, we have this and let me go ahead and rewrite it so I can see a little bit better.
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DV DT constant S, DT DS constant V, and DS DV constant S V T = -1.
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This right here is just the reciprocal of DS DT at constant V.
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The rate of change of entropy with respect to temperature at constant volume.
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We know that is equal to the constant volume/ T from back in the entropy.
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This one right here, this DS DV sub T, let us go ahead and bring this down here.
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This is just DP DT sub V from another of the Maxwell relations.
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We are just looking for what is equal to what.
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It is like all of these things, all of these equations at our disposal we are going to pick
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and choose to see how we can arrange this puzzle to get what it is that we want.
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This thing is equal to that from another one of Maxwell's relations.
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Furthermore, DP DT sub V happens to equal Α/ K that came from the coefficient of thermal expansion and compressibility.
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You recall the coefficient of thermal expansion A =1/ V DV DT that and Κ =-1/ V DV DT that.
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When we arranged these, we ended up with that Α/ Κ.
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You know what, I will go ahead and actually do it real quickly here just so you actually see it.
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This is from the previous lesson but that is not a problem, we can go ahead and do it here.
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I have pressure, volume, and temperature.
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The cyclic relation would look something like this DP DT constant V × DT DV constant P × DV DP constant T = -1.
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I get DP DT constant V × based on this DV DT = VΑ.
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DT DV is 1/ VΑ, same thing here.
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DV DP that is just - VΚ = -1.
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When I move this over to your side, these cancel and I get Κ/ Α.
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When we move it over to the other side I end up with the following.
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I end up with DP DT sub V = Α/ K.
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That is where the Α/ Κ came from.
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Let us go back and finish what we started.
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We have DV DT sub S, DT DS sub V, and DS DV sub T = -1.
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This becomes DV DT sub S × T/ CV × Α/ Κ = -1.
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We get DV, therefore, DV DT sub S = - Κ CV/ T Α.
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This DS DP sub V = -DV DT sub S = - Κ CV/ T Α.
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Therefore, I have DS/ DP sub V = Κ CV/ T Α which is exactly what I wanted.
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I have the rate of change of entropy with respect to pressure holding volume constant = Κ × CV/ T Α.
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What do we just do and why is it important?
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We just threw a bunch of mathematics, why is this significant?
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Recall, that we have the following expressions.
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Recall the following, we had DS = CV/ T DT + Α/ Κ DV, where entropy was a function of temperature and volume.
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We also had DS = CP/ T DT – V A DP.
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In this particular case, this is entropy expressed as a function of temperature and pressure.
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What we have done here is deriving an expression for how entropy changes with respect to pressure under constant volume.
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In other words, we have expressed entropy in terms of pressure and volume.
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What we have done is derive an expression for how entropy changes with pressure under constant V.
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In other words, we have expressed entropy as a function of pressure and volume.
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Later in the lesson, we are going to derive an expression for so what we have done is we found DS DP sub V.
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Later in the lesson, we are going to do DS DV holding that constant.
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That is what we have done here.
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We have expressed entropy as a function of temperature and volume.
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We have expressed it as a function of temperature and pressure.
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We are expressing it as a function of pressure and volume.
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Now, we have all three variables accounted for all constraints.
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In this case, we are holding the pressure constant.
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In this case, we are holding the temperature, pressure, the volume constant.
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In this particular case, we end up holding the temperature constant, that is what we have done.
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We have found a way to express the change in entropy with respect to pressure and volume.
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Before, we had temperature and volume, temperature and pressure, now we have pressure and volume.
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That is what we have done based on things we already knew in addition to the equations that we derived.
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That is why this is significant.
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And this is the theme that is going to run through this particular problems set.
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Let us go ahead and continue on here.
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In a lesson on the thermodynamic equations of state we were able to finally break down the following.
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DU = CV DT + this thing DV.
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Do you remember this was DU DV constant T.
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The Joules law for an ideal gas that is equal 0 but it is not an ideal gas it does not equal 0.
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We have to account for it.
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Given this demonstrate the following DU = this thing DT + DP.
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Why is it significant?
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We express energy in terms of temperature and volume.
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When we are talking about energy for temperature and pressure it was DH,
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it was enthalpy, temperature and pressure or the variables for enthalpy.
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Based on this, we can go ahead and now express a change in energy in terms of temperature and pressure also.
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We do not have to be in enthalpy, we can just deal straight with the energy itself, that is why this is significant.
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Let us go ahead and run through the process here.
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The way we are going to do this, this is DT and this is DV.
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We want DT and DP so I need to expand DV in terms of temperature and pressure.
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Let us expand DV in terms of DT and DP and collect terms.
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We are expressing volume as a function of temperature and pressure so the total differential expression is the following DV = DV,
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Temperature and pressure, I’m sorry.
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We have got DV DTP DT + DV DPT DP.
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DV DT sub P that is just Α V, the coefficient of thermal expansion DT.
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DV DP sub T that is just the coefficient of compressibility.
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It ends up being - Κ × V DP.
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I'm going to go ahead and put this thing DV, this expression up here into this.
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I'm going to collect terms.
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Here is what happens so I get DU = CV DT + Α T - Κ P/ Κ × Α V DT - Κ V DP.
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DU = CV DT + I’m going to separate this out.
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This is going to be Α T/ Κ - P × Α V DT – Κ V DP.
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DU = CV DT + this times this, this times that.
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This is going to be Α² TV/ Κ DT and this times that, - Α TV.
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The Κ and the Κ cancel that is going to be DP – P A V DT + P Κ V DP.
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Let us collect terms, here is the DT, here is the DT, here is a DT.
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I’m going to do this one in red, so we see it.
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I have a DT term, I have a DT term, I have a DT term, this was going to be CV + Α² TV/ Κ - P A V × DT.
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I have a DP term and a DP term so it is going to be + P K V - Α TV × DP which I think it is exactly what we wanted if I’m not mistaken.
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It is actually, all we have to do is now factor out the V so we have DU = CV + Α² TV/ Κ – P Α V DT + V × P Κ - Α T DP.
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This is what we wanted.
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Notice how everything is expressed in terms of things that are measurable.
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Heat capacity Α, temperature pressure Κ, pressure Κ Α, temperature volume, everything is there.
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Everything is easy, I found a way.
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I have derived a way to express energy in terms of temperature and pressure.
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It used to be enthalpy but if I want to deal strictly with energy I have this option.
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Let us move on to the next problem here.
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Using the result from the previous problem, demonstrate that near 1 atm of pressure
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at a change in energy with respect to pressure holding the temperature constant is approximately equal to – VT Α,
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Then using the following data for water at 20°C, calculate the actual value of DU DP sub T.
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Here we have an Α for water, we have the Κ for water, we have a smaller mass, and we have its particular density.
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Let us see what we can do.
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Let me go back to blue here.
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We have DU = CV + TV Α ²/ Κ – P V A.
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This is DT + V × P Κ - T Α DP.
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At fixed temperature means that DT equal 0.
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A fixed T implies that DT equal 0, this is isothermal so that goes to 0.
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All I’m left with is that term.
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We have DU= volume × pressure × Κ - temperature × Α × DP.
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Let us go ahead and first of all, let us go ahead and calculate.
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Let us do this in red.
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Let us go ahead and find PK.
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That is fine, I can do it here.
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Let us find this term first so we can compare the two.
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PK = the pressure is 1 atm and the Κ is 45.3 × 10⁻⁶/ atm.
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I end up with 45.3 × 10⁻⁶.
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Let us go ahead and find TA.
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TA the temperature is 293 °K because it is at 20° C and Α is 2.07 × 10⁻⁴.
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I end up with a pure number of 0.0607.
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Notice, 45.3 × 10⁻⁶ is 6.07 × 10⁻².
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In this particular case, the P × K is a lot less than the T Α.
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Because that is the case, we can pretty much ignore that term.
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Therefore, we have what we wanted, the DU = VT A – VT Α approximately equal to,
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I’m using the approximate because we ignore this, not exact but this is what we wanted.
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It was very simple, this goes away then compare these two.
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You can ignore this because it is so much smaller than this number.
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This is what you get, the change in energy at around 1 atmosphere = the volume × the temperature × Α under isothermal process.
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This is really simple nice and easy.
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Let us see what we have got here, this is what we want to prove.
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Let me double check and I think we want to express it this way.
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DU DP under constant T = - VT Α.
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There you go, sorry that should be a DP there.
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This and this are the same thing.
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I can just move the P over here, I'm holding T constants so I can just go ahead and expressive in partial differential notation.
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Let us go ahead and do the calculation.
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Volume = 1 cm³ = 0.99821 g and we have 18 g/ mol and we have × 10⁻³ dm³/ cm³.
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That cancels and that cancels, I'm left with a volume equal to 0.018O3 L/ mol because dm³ is a liter.
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Therefore, our change in energy with respect to pressure under constant temperature conditions,
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in this particular case let us be specific, 20°C for H2O = - VT Α = -0.01803 × 293°K × Α which is 2.07 × 10⁻⁴.
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I end up with this being equal to 0.001094, this is going to be L atm.
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I need to convert that to Joules, I do not need to but Joules is pretty standard.
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20°C H2O = 0.111 J/ atm, the unit is J/ atm.
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Joules on top energy, pressure on the bottom, this is not J/ mol but J/ atm.
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At 20°C for water, if I change the pressure by 1 atm, I increase the energy by 0.111 J.
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That is what this says, the rate of change of energy with respect to a unit change in pressure is equal to this under these conditions.
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Nice and simple and straightforward.
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Let us get down to some serious business.
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Example 4, we know the DS = CP/ T DT – V Α DP.
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We know this from our expression for entropy, for more chapters on entropy.
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Given this we want to show that the rate of change of entropy with respect to pressure under constant volume is this.
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The rate of change of entropy with respect to volume under constant pressure is this and then -1/ V ×
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the rate of change in volume with respect to pressure under constant S = this.
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Notice, SPV SPV SPV we are establishing relationships now between entropy, pressure, and volume.
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Let us go ahead and do this given that.
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Let us see what we can do.
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Part A, we want to show that DS DP sub V = Κ CV/ T Α.
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Notice, our variables are SP and V.
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Now we have DS = CP/ T DT – V Α DP.
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In other words, S is a function of temperature and pressure.
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We want entropy to be a function of pressure and volume so we expand the same way we did before in the previous problem.
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In this lesson, we expand DT in terms of DP and DV so we will let temperature be a function of pressure and volume.
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For our total differential, we get DT =DT DP constant V DP + DT DV constant P DV.
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We go ahead and put this expression into here and we collect terms and multiply it.
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We get DS = CP/ T × DT DP sub V DP + DT DV sub P DV – V Α DP.
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DS = CP/ T × DT/ DP sub V DP + CP/ T × DT DV sub P DV – V Α DP.
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Let us see how I want to do this.
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Let us go ahead and collect terms here.
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I will do it on the next page.
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DS = CP/ T DT DP.
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I really hope I'm keeping my variables straight here because there are so many symbols on this page.
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It is enough to make you absolutely crazy.
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I understand if you are feeling crazy about this.
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+ CP/ T DT DV sub P.
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My God this is crazy.
00:33:17.800 --> 00:33:40.200
This is just DS DP holding V constant, this thing the total differential.
00:33:40.200 --> 00:33:58.300
DS DP sub V = CP/ T × Κ / Α - V Α.
00:33:58.300 --> 00:34:05.600
We also have the following relation, we also have the other relation which I will do in red.
00:34:05.600 --> 00:34:31.000
We also have from the lesson on the thermodynamic equations of state, we have CP = CV + TV Α²/ Κ.
00:34:31.000 --> 00:35:04.600
Therefore, we are going to put this expression into here and we are going to get DS DP sub V = CV + TV Α² ÷ Κ/ T × Κ/ Α - VA = CV/ T +,
00:35:04.600 --> 00:35:34.300
I’m going to separate this out + V Α²/ Κ × Κ/ Α – V Α = Κ CV/ T Α + VA – VA VΑ.
00:35:34.300 --> 00:35:48.600
Therefore, I get my final expression DS DP sub V = Κ × CV/ T Α.
00:35:48.600 --> 00:35:50.500
This is what we wanted.
00:35:50.500 --> 00:36:03.800
There we go, long semi painful process all in order to get the change in entropy with respect to a change in pressure under constant volume = Κ CV ÷ T Α.
00:36:03.800 --> 00:36:07.800
That is all we have done here.
00:36:07.800 --> 00:36:19.100
Let us move on to part B, we want to show that DS,
00:36:19.100 --> 00:36:24.000
We want to do a change in entropy with respect to volume holding pressure constant.
00:36:24.000 --> 00:36:29.700
We want to show that is equal to CP/ TV Α.
00:36:29.700 --> 00:36:32.100
For part A, we had the following.
00:36:32.100 --> 00:36:55.400
We had DS = CP/ T × Κ/ Α - V Α DP + CP/ T × DT DV sub P and this is DV.
00:36:55.400 --> 00:37:10.600
This part right here, this is just DS DV sub P, that is what we wanted.
00:37:10.600 --> 00:37:28.400
DS DV sub P = CP/ T × DT DV sub P =C sub P/ T × 1/ V Α.
00:37:28.400 --> 00:37:36.000
This is just the reciprocal of the rearrangement of the coefficient of thermal expansion Α.
00:37:36.000 --> 00:37:38.400
There you go, we have it.
00:37:38.400 --> 00:37:49.500
We have DS DV sub P = CP/ TV Α.
00:37:49.500 --> 00:37:54.900
Part A was a change in entropy with respect to pressure constant volume,
00:37:54.900 --> 00:38:00.600
this is a change in entropy with respect to volume under constant pressure.
00:38:00.600 --> 00:38:05.200
We are expressing entropy in terms of pressure and volume.
00:38:05.200 --> 00:38:10.600
This is really beautiful stuff here.
00:38:10.600 --> 00:38:20.600
Let us stop and consider what it is that we have done here, this is really important.
00:38:20.600 --> 00:38:22.800
From the expression of entropy we have the following.
00:38:22.800 --> 00:38:37.700
We have that entropy is a function of temperature and volume and we had the following total differential DS = CV/ T DT + A/ K DV.
00:38:37.700 --> 00:38:55.800
We also had entropy as a function of temperature and pressure and we had DS = CP/ T DT – V Α DP.
00:38:55.800 --> 00:39:11.200
We have S = a function of now temperature volume, temperature pressure, pressure and volume.
00:39:11.200 --> 00:39:13.100
Same thing we did before for energy.
00:39:13.100 --> 00:39:18.000
We have entropy in terms of pressure and volume.
00:39:18.000 --> 00:39:24.700
For entropy equaling a function of temperature, pressure, and volume, we have the following expression.
00:39:24.700 --> 00:39:40.400
DS = Κ CV/ T Α DP + C sub P/ T V Α DV.
00:39:40.400 --> 00:39:45.600
We now have expressions for entropy under all three constraints.
00:39:45.600 --> 00:39:49.200
This is temperature and volume, we are holding pressure constant.
00:39:49.200 --> 00:39:53.800
In this case, this is temperature and pressure, we are holding volume constant.
00:39:53.800 --> 00:39:59.700
We have an expression for entropy, in terms of pressure and volume, we are holding temperature constant.
00:39:59.700 --> 00:40:03.000
This is what we have done here.
00:40:03.000 --> 00:40:07.600
This is what all the mathematical machinery has brought us to.
00:40:07.600 --> 00:40:12.500
It seems a little random all over the place but this is what it means, this is the big picture.
00:40:12.500 --> 00:40:17.900
We have a way of finding entropy under any constraint with simple temperature, pressure, and volume.
00:40:17.900 --> 00:40:23.300
This is absolutely fantastic and absolutely beautiful.
00:40:23.300 --> 00:40:29.000
Let us go ahead and finish it off with part C and see what it is that we are looking for.
00:40:29.000 --> 00:40:30.700
Let me see should I do it in this page or the next?
00:40:30.700 --> 00:40:34.300
I will go ahead and do on the next page.
00:40:34.300 --> 00:40:54.000
Part C, we want to show that -1/ V DV DP under constant S = Κ CV/ CP.
00:40:54.000 --> 00:41:08.400
By the cyclic rule on volume pressure and entropy we have the following.
00:41:08.400 --> 00:41:25.700
DV DP sub S × DP DS sub V × DS DV sub P = -1.
00:41:25.700 --> 00:41:34.400
Therefore, DV DP sub S =, I'm just going to go ahead and multiply by the reciprocal of these.
00:41:34.400 --> 00:41:54.300
I end up with the following - DS DP sub V × DV DS DP DV DS.
00:41:54.300 --> 00:41:58.100
P DV DS that is correct, we have that.
00:41:58.100 --> 00:42:03.400
We just found this, we found DS DP and we found DS DV in parts A and B.
00:42:03.400 --> 00:42:05.300
Let us just put them in here.
00:42:05.300 --> 00:42:15.000
-Κ CV/ T Α, this DV DS is just the reciprocal of the DS DV.
00:42:15.000 --> 00:42:19.900
That is just going to be TV Α/ CP.
00:42:19.900 --> 00:42:39.600
T cancels T, A cancels Α, we are left with DV DP sub S =- Κ CV × V/ CP.
00:42:39.600 --> 00:42:49.900
I'm going to move the negative, I’m going to divide by V so I get -1/ V × DV DP.
00:42:49.900 --> 00:43:00.700
This is a V not a U, S = Κ CV/ CP which is exactly what we wanted.
00:43:00.700 --> 00:43:05.100
We found DS with respect to pressure.
00:43:05.100 --> 00:43:07.000
We found DS with respect to volume.
00:43:07.000 --> 00:43:11.800
We are expressing the final one, now we are putting the volume and the pressure.
00:43:11.800 --> 00:43:16.700
The rate of change of volume with respect to pressure holding the entropy constant.
00:43:16.700 --> 00:43:19.100
This is absolutely beautiful.
00:43:19.100 --> 00:43:22.900
This thing is also written sometimes this way.
00:43:22.900 --> 00:43:37.900
This is also written -1/ V DV DP sub S = Κ/ Γ.
00:43:37.900 --> 00:43:48.700
Remember that Γ is equal to the ratio of the constant pressure heat capacity to the constant volume heat capacity.
00:43:48.700 --> 00:43:51.200
That is what we have done here.
00:43:51.200 --> 00:43:54.100
We have expressed entropy in terms of temperature and pressure.
00:43:54.100 --> 00:43:56.600
You have expressed it in terms of temperature and volume.
00:43:56.600 --> 00:43:59.800
Now, we have expressed it in terms of temperature and volume,
00:43:59.800 --> 00:44:07.400
we found the rate of change of entropy in terms of pressure, constant volume, that was part A.
00:44:07.400 --> 00:44:15.800
We did the rate of change of entropy with respect to volume under constant pressure, that was part B.
00:44:15.800 --> 00:44:24.700
We went ahead and SPV SVP now we did at the rate of change in volume with respect to pressure under constant entropy,
00:44:24.700 --> 00:44:26.300
that is what we have done here.
00:44:26.300 --> 00:44:33.100
If we want the rate of change of pressure with respect to change in volume, all we have to do is reciprocate this and take the reciprocal here.
00:44:33.100 --> 00:44:36.200
We have closed the circle on all of the equations.
00:44:36.200 --> 00:44:48.100
We are now able to express things like entropy and energy and enthalpy strictly in terms of temperature, pressure, and volume, and this Α and this Κ.
00:44:48.100 --> 00:44:56.000
This is absolutely extraordinary, it is more than a minor miracle that is not even possible but there it is.
00:44:56.000 --> 00:44:58.900
In any case, thank you so much for joining us here at www.educator.com.
00:44:58.900 --> 00:45:00.000
We will see you next time, bye.