WEBVTT chemistry/physical-chemistry/hovasapian
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Hello and welcome to www.educator.comn and welcome back to Physical Chemistry.
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Today, we are going to start our discussion of spontaneity and equilibrium.
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We are coming to the end of thermodynamics.
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We are going to close out that circle and here is where everything absolutely starts to come together
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and we start to seriously apply these things to chemical reactions.
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Let us jump right on in.
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The following terms mean the same thing as irreversible.
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We will often refer to a irreversible transformation as a real transformation, natural transformation, spontaneous transformation.
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Let us go ahead and use the blue in here, this is going to be the word that we used most often.
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When we talk about reactions you remember from General Chemistry,
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we talk about spontaneous reaction has nothing to do with speed, it is just a thermodynamic quality and we will talk about a little bit more.
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Irreversible, real, natural, or spontaneous.
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Here is what we want to do, we ask ourselves what differentiates irreversible or an ideal transformation.
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Ideal transformations are ways of dealing with things that we have been dealing with but they are not real,
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they are ideal in the sense that they provide an alternate way of looking at things
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may help us put together some sort of a mathematical structure for it, but it is not real transformation.
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The real world does not work that way but gives us a standard, a reference.
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What differentiates a reversible transformation from an irreversible transformation or real one?
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We consider the relation between the heat that flows and the entropy change that accompanies that heat flow in a given process.
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During reversible process a system deviates only infinitesimally from equilibrium.
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We have talked about that a lot especially when we talked about energy.
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The system does undergo a transformation but it is essentially at equilibrium during the whole step
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because you are only deviating slightly from where you are so essentially you are at equilibrium.
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Therefore, the condition of reversibility is the condition of equilibrium.
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The defining equation for equilibrium is this one right here TDS = DQ reversible.
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All I have done is just the rearrangement of the definition of entropy.
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We had DS = DQ reversible/ T.
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All I have done is multiply both sides by T so that I could write it in this form without any fractions if you will.
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The defining condition for a system to be at equilibrium is TDS = DQ reversible.
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Basically that means that when the system is an equilibrium that differential change in entropy that
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the system experiences × its temperature = the differential change in heat.
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Not the differential change in heat.
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The little bit of heat that flows along the reversible path.
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This is the defining equation for equilibrium.
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The condition for irreversibility, for a real transformation is actually the Clausius inequality.
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The Clausius inequality, the DS is not equal to DQ/ T or DQ reversible/ T.
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DS ≥ DQ/ T.
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We have gotten rid of the irreversible part and now we have this greater than operator.
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Upon rearranging we get the defining equation for irreversible and spontaneity.
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This is important right here.
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Once I had multiplied by that I get that this is the defining equation for a spontaneous process.
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This is the equation that we are going to play with and manipulate.
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It starts right here, a spontaneous process, a natural process, a real process,
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one that happens on its own without any external help these this is the condition for T × DS ≥ DQ.
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We can combine these two equations by writing this.
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Basically, all of them is greater than or equal to, where we take the quality sign to mean a reversible value of the DQ.
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If the heat exchange happens to be done reversibly that is all the says.
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It just include all possibilities, our concern is going to be this one, greater than.
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It is always going to be like that because every real process is this right here so this is the important equation.
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What does this say?
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It says the transfer of heat during a spontaneous process is less than the entropy change
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for the process × the temperature at which the process takes place.
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Once again, for a particular process, a spontaneous process the heat that transfer during the process is going to be less than
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the entropy change for the process × the temperature which the process takes place.
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That is the defining equation for spontaneity and that is the one we are going to manipulate.
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Let us go ahead and get started.
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We have this TDS ≥ DQ.
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Let us go ahead and recall the first law of thermodynamics, our energy.
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We know the DU = DQ – DW.
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Let us go ahead and rearrange that so we end up in solving for DQ.
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We get DQ = DU + DW and we can go ahead and put this expression for DQ into here.
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We end up with the following, we end up with TDS ≥ DU + DW.
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Let us go ahead and rearrange this.
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I'm going to bring both of these over to the left hand side and leave 0 out on the right hand side.
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I have - DU - DW + TDS ≥ 0 this right here, this is our defining equation.
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All we have done is we have taken that defining condition.
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We have included the first law of thermodynamics so we have broadened the heat and the work and energy and now we have expressed it this way.
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This is the defining equation for spontaneous process.
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The negative of the differential change in energy - the work that is done in that process + the temperature ×
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the differential change in entropy is always going to be ≥ 0.
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This defines a spontaneous process.
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If you have a system and this is satisfied the process is spontaneous.
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If you know a process is spontaneous this will always hold.
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So far we only looked at pressure, volume, work.
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Let me actually write that.
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We have only looked at PV work, this DW right here it includes all work not just pressure, volume, work.
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Maybe there are some electrical works done.
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Maybe there is some other mechanical work done.
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This is a combination of all the work that is done not just expansion work.
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This DW includes all work.
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We are bringing this up because we want to make this discussion as general as possible.
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For our purposes, I will to get to that in a minute.
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We are not going to be concerned with other work.
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We are only to be concerned with pressure, volume, work.
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Sometimes when we start putting constraints on this equation,
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we are not even going to be concerned with pressure, volume, work because of the volume change is 0.
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Under constant volume the pressure volume work is 0 so all of these terms are dropping out
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but we want to give the most general conditions for a spontaneous process and we are going to tighten up by placing some constraints on.
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Do not worry if things start a little too complicated, it will be very simple in a minute.
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So far we are going to look at the pressure, volume, work.
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This DW includes all work so we have this DW = the pressure, volume, work which is the external pressure × the change in volume +
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DW sub other, whatever other work that is, we do not really care all that much.
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When we put this expression in for here, we get the following.
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We get -DW - P external DV - DW other + TDS ≥ 0.
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This is the equation we want to concern ourselves with, this is the one we start with.
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These two equations actually are going to be the same.
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Sometimes we are going to talk about work cumulatively.
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Sometimes we are going to separate work out in to pressure, volume, work, and all the other work but these equations are absolutely equivalent.
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Sometimes we will start with this one, this derivation process for new equations.
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Sometimes we will start with this, they are the same.
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We have now expressed the condition of spontaneity in terms of changes of the properties of the system.
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We have the energy and we have volume.
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We have pressure, we have the work done, we have temperature and we have entropy.
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All of these things we have expressed in terms of all the properties of the system.
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This is the condition of spontaneity.
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We do what we normally do.
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We take this condition, we take this as the basis equation and we start putting constraints on it because in the laboratory it is not just a free flow.
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In the laboratory, they are normally you are going to have conditions of constant temperature and pressure.
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That was going to be the majority of it but sometimes we are going to constrain the volume.
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Sometimes we are going to constrain the pressure or the temperature.
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We take this and place those constraints on it the same way we did for energy, the same way we did for entropy.
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Remember we did a constant temperature constant volume, constant temperature constant pressure.
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We are going to do the same thing.
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We are going to take this equation, constrain it, and derive a new set of equations under laboratory conditions, the things that we normally run across.
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The first one I’m going to talk about is transformations in an isolated system, just for the sake of being complete.
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Transformation in isolated system is when the system is isolated there is no exchange of energy.
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I’m going to go back to blue, I think.
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There is going to be no change of energy so DU = 0.
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There is no work done in an isolated system D = 0 and there is no heat exchange DQ = 0.
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If we have - DU - DW + TDS ≥ 0, I used this particular one with just the generic cumulative work.
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That is 0, that is 0, what I end up with is TDS ≥ 0 and of course I get DS ≥ 0.
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In an isolated system, the entropy of that system will always increase.
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By isolated means completely separated from everything so we have no interaction at all with its surroundings,
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no change of energy, no exchange of matter, no exchange of anything.
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The entropy of an isolated system will always increase.
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A little bit of copy out with respect to this one.
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It is not just the entropy of the system that is going to increase,
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the truth is it is going to be the change in entropy of the universe that is actually has to be ≥ 0.
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The entropy of the universe, the change in entropy of the universe is going to be change in entropy of the system + the change in entropy of the surroundings.
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In this particular case, we are isolated so it is not really going to include the surroundings.
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Let me just go ahead and bring this up here and I will talk about it later.
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It is actually possible for a system itself to have a decrease in entropy provided that
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in the surroundings the entropy increase more than compensate for the decrease.
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The total entropy change in the universe is always greater than 0.
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This is universally true.
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This is true for an isolated system.
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Again, we are talking about isolated system then the DS of the system is always going to be ≥ 0
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but in general what we are really going to be concerned with and how often do we do isolated systems, we do not.
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In fact of the matter is we do not.
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There is always going to be some surrounding involved.
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This is just for mathematical completion.
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We just want to use this equation in isolated system, this is what happens.
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Let me just go ahead and mention this but we will get back to it and discuss it in more detail.
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Let us start getting into the real important aspects because we are not going to be dealing with isolated systems.
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Let us go ahead and talk about transformations at constant temperature.
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If we hold the temperature constant, isothermal process, here is what happens.
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Transformation at constant temperature, let us go ahead and start with our equation - DU - DW + TDS ≥ 0.
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We have to find a way to manipulate this.
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Here is what we are going to do, when temperature is constant I can rewrite TDS = D of TS.
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In another words, I can put the TS together.
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The differential operator actually operates on both because T is constant, this T actually comes out which is why we write TDS.
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All I did was rewrite it in its more original form.
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The temperature × the entropy, when I take the differential of it because temperature is constant it comes out.
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Instead of writing TDS, I’m going to write DTS so we have - DU + TDS.
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I’m going to go ahead and move this DW over the other side.
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You do not have to it just different ways of looking at the same equation so the derivation itself you can leave the DW on the left, you can put it on the right.
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We are actually going to do both when I do this summary of conditions.
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In the next lesson I'm going to bring everything to left and leave 0 on the right so ≥ DW.
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This is just moving things around.
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This is going to be - DU + DTS ≥ DW.
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I'm going to go ahead and factor out the differential operator, we can do that.
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- DU - TS ≥ DW we end up getting this thing, - the differential of this thing ≥ DW.
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Energy, temperature, entropy, and work.
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This combination right here, this U - TS it shows up so much.
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We actually give it a special name so combination of U – TS.
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Remember what we did with enthalpy, we said that enthalpy =U + PV.
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Enthalpy is a composite function, it is a composite of the energy of the system + the pressure of the system ×
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the volume of the system and we called it the enthalpy.
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This U- TS it is just a composite.
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We define this thing called A the Helmholtz energy.
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A happens to equal U – TS, it is a composite function.
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It is a combination of the energy, the temperature, and the entropy not altogether different than this.
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A is the Helmholtz energy of the system.
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If I know the energy of the system and if I know the temperature of the entropy of the system, I multiply the temperature by the entropy and
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subtract it from the energy of the system, I get the Helmholtz energy of the system.
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I just put something together, it is an accounting device.
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I do not want to deal with UT and S, so I just say U - TS is this thing called Helmholtz energy so that is what we get.
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A is a state function, the Helmholtz energy is a state function.
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It is a state function because it is a combination of state functions.
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Energy, entropy, temperature, is a state function.
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We have - D of U - TS ≥ DW.
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Let us go ahead and we said that A = U - TS so we can write - D of A ≥ DW.
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Or upon integration, if I integrate this I end up with - δ A ≥ W.
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It is not δ W it is W.
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Under conditions of constant temperature, the work in a given process for a spontaneous isothermal process,
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the work done during that process is going to be less than or equal to the decrease in the Helmholtz energy.
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Let us write this down, what does this say.
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it says in a spontaneous isothermal process the work produced in the surroundings or if I just refer to the work in a process,
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regardless of the particular point of view you are taking, the work produced in the surroundings
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is less than or equal to the decrease in the Helmholtz energy.
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Just follow the math, not a problem.
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Let me go back to blue here.
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The DW we said is going to be equal to this pressure, volume, work, so PDV + other kinds of work.
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It includes all the work.
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In general, we are not going to be concerned with other work.
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We are only going to be concerned with pressure, volume, work.
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In general, we will concern ourselves with the PB work only.
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We have - δ A ≥ W, we have - δ U -TS ≥ W.
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I can distribute the δ operator so I get – δ U + T δ S ≥ W.
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Starting to look for a little bit familiar or if I want I can write it as, I go ahead and switch this sign and
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move some negatives around and make the δ U positive.
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Δ U - δ S ≤ -W It is another way of looking at this.
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When Helmholtz energy is taught, it is taught as constant temperature and constant volume.
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It is taught as T constant which is what we have done and V constant.
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Normally, the chances are in your book you are going to see this introduced as if I hold temperature and volume constant what do I get.
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That is usually how we introduce this notion of Helmholtz energy.
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If V is constant that means that this right here, this portion of the work is 0 because if V is constant,
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then DV is 0 so that there is no pressure, volume, work and the only work you have is work other.
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We just said that in general we are not going to concern ourselves with other work.
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If we are going to drop, we have DW = P external DV + DW other.
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If we are going to forget about the other work and if we hold volume constant and this is going to be 0 then this is going to be 0.
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Here is what you end up getting, you end up with - DA ≥ DW that is our basic condition.
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If DW is 0 you get – DA ≥ 0 or if you want DA ≤ 0 or the integrated version δ A≤ 0.
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This is normally what you are going to see in your book.
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Under conditions of constant temperature and volume, the change in Helmholtz energy for a spontaneous process is going to be less than 0.
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It should start to look familiar.
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You are accustomed to the δ G being less than 0 for a spontaneous process.
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δ G happens under conditions of constant temperature and pressure which
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we are actually to be talking about in the next lesson but there is this other energy in the system.
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This energy which is the energy - TS which is the Helmholtz energy and if I hold the temperature and
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the volume constant then this for a spontaneous process it is the change in Helmholtz energy, the δ A ≤ 0.
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Let us go ahead and finish up with writing a couple of different versions of this.
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δ A ≤ 0 and once again in your book you are going to be introduced to Helmholtz energy is constant temperature and volume.
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But in general, the volume does not have to be held constant.
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If the temperature that has to be held constant you can go ahead and leave the volume there and in this work term is not 0.
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If we ignore all the other work, we are still to be concerned with the pressure, volume, work that is done.
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This is the general version of this.
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This is the specialized version that holds T and V constant.
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V does not have to be constant here so I do not think that it has to be.
00:27:04.400 --> 00:27:15.800
This is that, let us go ahead and write δ U - TS ≤ 0.
00:27:15.800 --> 00:27:31.100
This is going to be δ U - T δ S ≤ 0 and of course we have δ A = δ U - T δ S.
00:27:31.100 --> 00:27:33.800
This is the basic relationship that exists.
00:27:33.800 --> 00:27:43.400
This is similar to the relationship that you learn in General Chemistry which we are going to learn again which is just as a little bit of a preview.
00:27:43.400 --> 00:27:50.200
δ G = δ H - T δ S remember that one, this is the Helmholtz version of it.
00:27:50.200 --> 00:27:55.100
Under conditions of constant temperature and volume this is what you get.
00:27:55.100 --> 00:27:57.500
All of these things are equivalent.
00:27:57.500 --> 00:28:04.300
This is the general equation right here, that is the general version of the equation.
00:28:04.300 --> 00:28:11.500
This is the one for temperature and volume, for constant temperature volume and constant temperature and volume.
00:28:11.500 --> 00:28:15.500
Please remember, you do not have to hold volume constant but you have to hold temperature constant.
00:28:15.500 --> 00:28:21.200
If you hold volume constant also, you end up with 0 on the right hand side.
00:28:21.200 --> 00:28:26.600
If you do not, there is some pressure, volume, work, is going to be done and there is going to be a work term.
00:28:26.600 --> 00:28:29.000
Again, we are not going to be concerned with other work.
00:28:29.000 --> 00:28:34.300
If there was other work, electrical or mechanical, that would also be part of the work term.
00:28:34.300 --> 00:28:36.100
I hope that makes sense.
00:28:36.100 --> 00:28:38.200
Thank you so much for joining us here at www.educator.com.
00:28:38.200 --> 00:28:42.000
We will see you next time for a discussion of Gibbs free energy, bye.