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 discussion of entropy and we are going to talk about entropy as a function of temperature and volume.
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Let us jump right on in.
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I have to give you a little bit of a warning, for this particular lesson and the next lesson,
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and several ones that follow there is going to be a lot of mathematics.
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Most of the derivation, the mathematics in itself is really not difficult, it is mostly just taking derivatives
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and you know equating one equation with another and see which coefficients are the same and identifying one thing with the other.
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The only thing you have to worry about, I’m going to warn you about is there is going to be a lot of symbolism on the page.
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It is really important to not lose the four from the three and things like this.
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At the end of these lessons, I will stop and recap and I will tell you what is actually important as far as equations to remember
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and what you should concentrate on.
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I just wanted to give you fair warning that there is going to be a lot of symbolism on page, a lot of mathematics.
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Let us see what we can do with it.
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Nothing is altogether difficult though.
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We defined entropy as follows, we said that DS = DQ reversible / T.
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This relates a change in entropy to an effect in the surroundings or the system.
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It is just a question of perspective in the surroundings, basically, the quantity of heat withdrawn from the surroundings divided by the temperature.
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It is related to an effect in the surroundings.
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Can we use the equations we have so far from our previous work.
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This is the question, can we use our equations to relate entropy to changes in the state variables of the system which we can actually measure very easily.
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In other words, temperature pressure and volume are easily measurable.
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Not that heat, it is not that it is just that basically if I can find a way to use the equations that we had our disposal to be able
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to express this δ S or the DS in terms of the pressure, the temperature, and the volume and maybe energy, it would be very convenient.
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That is what we are going to set out to do, that is our goal, that is our big picture.
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We want to express entropy in terms of temperature, pressure, and volume.
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Let us see what we can do.
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We know that in a reversible transformation, so I just go ahead and put it over here.
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In a reversible transformation and from our discussion of energy we said that the external pressure is actually equal
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to the pressure of the system because of something reversible the system and the surroundings are always essentially in equilibrium.
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The external pressure we just said equal to the internal pressure.
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Therefore, our definition of work because we are going to bring work back into this, definition was P external × DV.
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When P external = P then DW is just equal to P DV.
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Let us go ahead and start with our first law.
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First law expression, energy = DQ and in this case reversible, - DW.
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DU = DQ reversible, DW is just P DV.
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We have this DS = DQ reversible / T so if I move the T over here, I get the DQ reversible.
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I get that DQ reversible = T DS.
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I’m just going to go ahead and put this into here and I get the following I get DU = T DS – P DV.
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I’m going to switch things around and move this over here and I'm going to put T DS = DU + P DV.
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I’m going to divide by T, everything by T, in order to get DS alone, I’m going to be left with DS =1/ T DU + P / T DV.
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This is a very important equation, this is called a fundamental equation of thermodynamics.
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It is going to be the starting 0.4 much of the work that we do subsequently.
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This is the fundamental equation of thermodynamics and is nothing more than a combination of the first law of thermodynamics which is the definition of energy.
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Instead of the second law of thermodynamics which was just an expression of the definition of entropy.
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First law is energy and second law is entropy, we put those together and we come up with a fundamental equation of thermodynamics.
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You do not need to memorize this because the derivation as you see is very simple.
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Just put the definition of entropy in terms of the T DS into here and just manipulate it.
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That is it, it is very simple.
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I will go ahead and actually go to the next page here and rewrite the equation again.
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We have DS =1/ T DU + P/ T DV.
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This expresses a change in entropy, P2 changes in energy and volume.
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In other words, what this says is I can change the energy, I can change the volume, or I can change both.
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Either one of those or combined that changes the entropy. In other words, I have two independent ways of changing the entropy.
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I can change the energy of the system, I can change the volume of the system, either one of those changes
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that I make in the system will change the entropy of the system, that is all this equation says.
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This expresses a change in entropy that changes in energy and volume as well as the temperature and pressure.
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Notice I have T and I have P, notice that every variable is represented, temperature, pressure, volume, energy, and entropy S U T PV is altogether.
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That is as why it is called a fundamental equation of thermodynamics.
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Some other things to notice, the first thing is the coefficients, the 1/ T and the P/ T.
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I think I will go ahead and do this red.
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Things to notice, the coefficients 1 / T and P/ T are positive.
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There are two independent ways of changing the entropy of the system.
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The first way is changing the energy, that is one, and the other way is change the volume.
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If you make changes to either or both, you change the entropy of the system.
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Let us go ahead and continue here, add a little bit of room.
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Since 1/ T and P / T are positive, if I hold V constant, and if I change the energy, if I raise the energy in other words if the change in energy is positive,
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if this differential is positive, positive × positive is positive, the change in entropy is also positive.
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If I raise the energy of the system, the entropy of the system is raised.
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If I hold the energy constant and if I raise the volume of the system, this is positive and this is positive that means the entropy goes up.
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If I raise the energy of the system, I raise the entropy of the system.
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If I increase the volume of the system, I increase the entropy of the system and vice versa.
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If I decrease the energy, I decrease the entropy.
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If I decrease the volume, I decrease the entropy, that makes sense.
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If I change the energy of the system, I'm agitating the system.
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I'm putting more energy into it, it is becoming more agitated, it is becoming more disordered.
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Again, that qualitative way of looking at entropy.
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If I change the volume of the system and take it from here to here, the particles of the system
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have much more room in order to move so now there is more chaos available to it.
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This is orderly, this is disorderly, if I raise the volume of the system I actually raise the disorder of the system.
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If I take the volume and make it smaller, I actually create more control.
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I have lowered the disorder of the system so that is what is going on here.
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Since 1/T and P / T are positive , if I fix the volume, fixing V and raising U energy raises S the entropy.
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If I fix U, if I fix the energy and raise the volume then S rises.
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Let me go back to black for this one.
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We do not to control volume in experiments, we do not normally control energy and we certainly do not hold it fixed.
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We do not control U in the laboratory.
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Our next phase is let us see, if we cannot transform this equation, this fundamental equation of thermodynamics to one
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that involves temperature and volume or temperature and pressure.
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We want to be able to not worry about controlling the energy because we do not normally control them in a laboratory.
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In the laboratory, we control temperature, pressure, and volume.
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Can we take this equation, the fundamental equation of thermodynamics and mathematically manipulate it to give us something
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that gives us the same expression but in terms of temperature and volume, and temperature and pressure.
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The answer is yes, we can.
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The first thing we are going to do is we are going to look at entropy as a function of temperature and volume.
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We do this in blue, that is okay I will stick with black.
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Entropy, which of course is the title of this particular session.
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Entropy as a function of temperature and volume.
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We will let S= S(TV) this just says that entropy is a function of temperature and volume.
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We can go ahead and write the total differential of this expression then DS is going to equal DS DT, a partial of S
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with respect to the first variable holding the V constant × DT + DS DV holding T constant DV.
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This is just the total differential expression, we know this already.
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Our task is that these differential coefficients right here, can we identify these differential coefficients with things
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that we can easily measure or things that we can look up in a table, that is what we want.
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We are doing the same exact thing that we did with energy.
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We have energy as a function of temperature and volume.
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Energy as a function of temperature and pressure those involved differential coefficients.
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We related, we identify these differential coefficients with things that are easily measurable.
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We are trying to do the same thing here, that is all we are doing or we are finding a way to express
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this elusive quality entropy in terms of things that we can measure and we begin this way.
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Let us see what we can do.
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Let us go ahead and write the fundamental equation again.
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The fundamental equation we have DS =1/ T DU + P /T DV.
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Notice they are almost the same DS DS, DT here is DU DV and DV.
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Here is the same, we need to start manipulating these set of equations.
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Let us go ahead and recall that we had DU = DU DT V DT + DU DV T DV.
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That was equal to the constant volume DT + DU DV T DV.
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This is just this, we know this already from our work in energy DU = this.
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I’m going to go ahead and take this expression for DU and actually substitute into here into the fundamental equation of thermodynamics.
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Here is what I get when I do that I get DS =1/ T × CV DT + DU DV T DV + P/ T DV multiply this out,
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this is equal to CV/ T DT + 1/ T × DU DV sub T DV + P/ T DV.
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The combined terms, this is a DV and this is a DV, I have T in the denominators so I can go ahead and combine that.
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I’m going to write this as CV / T DT + 1/ T × DU/ DV T + P DV.
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Again, there is nothing strange happening here, this is a basic algebra is was what I'm doing, it is not in calculus.
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Now we have this equation, I'm going to call this equation 2.
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This is from the fundamental equation of thermodynamics and the expression for energy that I substituted into it.
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Notice I have DT and DV.
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I have this thing and I have this thing.
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Equation 1, let us write the total differential expression which was DS DT DT + DS DV T DV, I will call this equation 1.
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Since both equation 1 and the equation 2 express DS in terms of DT and DV, DS DS DT DT DV DV that means that equation 1 and 2 are identical.
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Once again I have 1 equation that derive one way and I have another equation which is from
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the total differential expression DS DT DV DS DT DV they are equivalent.
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Therefore, this thing is this thing and this thing is this thing, that is what I'm going to do next.
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This is that DS DT holding V constant = CV/ T.
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I have achieved what I wanted, I wanted to find a way to express this in terms of something that I can measure.
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I have identified it with something that is measurable but I can measure the constant volume heat capacity divided by the temperature.
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It is fantastic, that is one of them.
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The other one, this is equivalent to that.
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It looks a little bit more complicated and is more complicated but again we are going to simplify that one too.
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We have DS DV T =1/ T × DU DV T + P.
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The change in entropy per change in temperature, in other words, when I change the temperature how does the entropy change?
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That is how it changes.
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Here if I change the volume how does the entropy change under constant temperature?
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That is how it changes.
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I was able to identify this and this with things that are now easily measurable, this was our task.
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Let us deal with these one at a time.
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At constant volume the rate of change of entropy with respect to temperature = CV/ T.
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Since CV/ T is always positive because the constant volume heat capacity is always positive, the temperature is always positive.
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The reason is because CV is always positive and at constant volume,
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if I hold the volume of the system constant an increase in temperature implies an increase in entropy.
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Increase in entropy that is what this says.
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If I raise the temperature, I raise the entropy of the system.
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The temperature dependents of entropy, in other words the extent to which entropy depends on temperature is very simple,
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it is just the constant volume heat capacity divided by the temperature.
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The temperature dependence of entropy is simple, it is just the constant volume heat capacity CV ÷ T.
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Let us see what we have got, so we have DS = DS DT V DT.
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I’m just writing the total differential expression + DS DV T DV at constant volume.
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Constant volume that means DV is 0.
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This is 0 because these constants, if we happen do that what we end up with is the entropy = this thing which we know is this thing = CV / T DT.
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If I integrate this, I end up with the change in entropy = the integral from temperature 1 to temperature 2 of CV / T DT.
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In a particular system, if I hold a volume of the system constant and just raise the temperature this gives me a way of finding the entropy.
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I just integrate the constant volume heat capacity divided by temperature / the temperature range that I do.
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Let us go ahead and see an example.
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Example 1, I think I have here we go.
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Here is my example and we will go ahead and do this in black.
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1 mol of nitrogen gas is heated at constant volume so we have a constant volume process from 350°K to 600°K what is the δ S for this process?
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The constant volume heat capacity is 3 Rn/ 2 and is the number of moles.
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Nice and simple, we just calculated it.
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When you do these problems you want to start with the most general situation and work your way down.
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You do not want to just memorize equations for a particular set of circumstances, that is not going to work with thermodynamics.
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There are far too many equations to actually memorize and there are far too many variations of the circumstances.
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You are just going to make yourself crazy, you are going to cause yourself a lot of stress and you will make it worse.
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You will actually affect your understanding at a very bad way.
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You want to start with the most general situation and then flow from what the problem says,
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go ahead and derive what you need which is what we did with the energy.
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The one of the reasons that we did all the problems that we did with energy and we are going to be doing the same.
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In these lessons, I’m doing one or two example problems.
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I’m saving all the example problems for the end of this particular unit where we are going to do a huge bulk of problems over and over again.
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Do not worry about just having 1 example, you are going to be doing a lot of problems but let us see.
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The best way to handle this is the way we just handled it.
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We know that the change in entropy in general = DS DT V DT + DS DV T DV this is the most general equation.
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This is the equation that you want to now.
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You also know this, we also identified this with CV/ T DT + DS DV DV.
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The general equations we want to start off with is, this is what you want to memorize, this one right here.
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The problem tells us that the volume is constant.
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The volume is constant these just drops off to 0 so what we are left with is DS = CV/ T DT.
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We want δ S so let us just integrate this.
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Therefore, we have our equation δ S = the integral from temperature 1 to temperature 2 of CV/ T DT.
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We have all the information that we need.
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We have CV so let us go ahead and do it.
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Let us go ahead and move over here.
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Therefore, δ S = the integral of temperature is from 350 to 600°K.
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Our CV is 3 Rn/ 2 so it is 3 Rn/ 2.
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We do it this way, we have 3 Rn/ 2/ T DT.
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3 Rn/ 2 is a constant so what it comes out 3 Rn/ 2.
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The integral from 350 to 600 of DT/ T we can do this, this is very simple.
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This is 3 × R which is 8.314 × n which is the number of moles we have 1 mol here.
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The integral of DT/ T is just Ln.
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It is going to be Ln of 600/ 350 and when I solve this I get 6.72 J/°K.
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There you go, that is my δ S.
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If I have 1 mol of nitrogen gas and I take it from 350°K to 600°K, the entropy change for that is 6.72 J/°K.
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It is all based on this, the general equation that applies to every single system solid, liquid, gas, it does not matter.
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This is the equation that you want to memorize, the same thing that we did for energy.
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Once you have this, the problem will allow you to knock things out.
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From the general and then strained we end up with a really simple integral.
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This way we do not have to memorize this equation for this particular situation, that is not going to work with thermodynamics.
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I promise you do not go down that path it will cause you a lot of heartache and it will affect your grades in a bad way.
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There are just too much information for you to memorize.
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The nice thing about thermodynamics is there is only a handful very few general equations that you have to memorize.
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You remember that from what we did with energy, there was only the definition of work, the definition of energy, and a definition of enthalpy, and the two basic mathematical relations.
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That is only 5 to 6 equations that you have to memorize, that is better than any other class.
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Definitely do it this way.
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We have dealt with the temperature dependence now we have to deal with the other one.
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The other one that was not quite so simple.
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The volume dependence of entropy, in other words the rate of change of entropy might change the volume.
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That is not so simple.
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Let us stick with black here.
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The volume dependence of DS at constant T is not so simple.
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We are going to have to fiddle with that mathematically.
00:32:33.800 --> 00:32:36.500
We have the following, it was this.
00:32:36.500 --> 00:32:54.900
It was DS DV at constant T =1/ T × P + DU/ DV sub T.
00:32:54.900 --> 00:33:03.800
I think I actually reversed them, it does not really matter. I think I have the other one first.
00:33:03.800 --> 00:33:08.700
DU/ DV sub T + P does not matter what order you do it in.
00:33:08.700 --> 00:33:15.500
Let us go ahead and call this equation 3.
00:33:15.500 --> 00:33:24.700
Let us see what we have got.
00:33:24.700 --> 00:33:34.000
Here is what we are going to do, let us go ahead and recall this equation, the one that we just had.
00:33:34.000 --> 00:33:42.700
DS DT the temperature dependents under constant volume that was equal to CV/ T.
00:33:42.700 --> 00:33:57.600
I’m going to differentiate this equation with respect to volume holding the temperature constant.
00:33:57.600 --> 00:34:07.800
We differentiate with respect to volume holding T constant.
00:34:07.800 --> 00:34:11.200
I’m going to differentiate this and it becomes the following.
00:34:11.200 --> 00:34:22.300
Differentiate the whole thing so it becomes, I'm going to take the derivative with respect to volume of this,
00:34:22.300 --> 00:34:31.000
it is going to equal, the T is constant 1/ T D CV DV.
00:34:31.000 --> 00:34:47.700
Write the derivative of this with respect the volume =1/ T DV of DU DT and all it done here
00:34:47.700 --> 00:34:56.300
is I have use the definition of constant volume heat capacity D DV of CV is the D DV of this.
00:34:56.300 --> 00:35:08.000
You remember the constant volume heat capacity is actually equal to the change in energy / the change in temperature that is the definition of heat capacity.
00:35:08.000 --> 00:35:25.500
Equals 1/ T I’m going to put this expression together as D² U DV DT.
00:35:25.500 --> 00:35:59.300
Our final expression, I’m going to put this together this is D² S DV DT = 1/ T D² U DV DT this is just mathematical manipulation.
00:35:59.300 --> 00:36:06.800
What I did is I took this one, a differentiated with respect to volume holding T constant
00:36:06.800 --> 00:36:16.000
and I’m going to take this equation and I'm going to differentiate with respect to T holding V constant.
00:36:16.000 --> 00:36:41.400
I’m going to take this equation which I will rewrite so I have got DS DV T = 1/ T × let us go ahead and write it the way that I have here DU DV T.
00:36:41.400 --> 00:36:57.600
Let us go ahead and differentiate this with respect to T and this time we will hold V constant so we end up with the following.
00:36:57.600 --> 00:37:19.000
This is going to be a little long so I got DDT of DS DV T and I'm going to put that together as D² S DT DV
00:37:19.000 --> 00:37:23.000
that is going to equal when I differentiate this I'm going to end up with the following.
00:37:23.000 --> 00:37:33.300
I’m differentiating with respect to T so it is going to be this × the derivative of that + that × the derivative of this.
00:37:33.300 --> 00:37:50.600
It is the product rule so was going to be 1/ T × differentiating this thing with respect to T holding V constant +
00:37:50.600 --> 00:38:10.900
differentiate this with respect to U DV T – 1/ T² × P + DU/ DV sub T.
00:38:10.900 --> 00:38:20.100
I know, do not worry about it.
00:38:20.100 --> 00:38:32.000
S = S (TV) and DS is an exact differential.
00:38:32.000 --> 00:38:51.000
You remember from our previous math lesson that the mixed partials are equal which means that D² S DV DT = D² S DT DV mixed partials.
00:38:51.000 --> 00:38:56.600
I just found the square D ⁺S DV DT and I found D² S DT DV.
00:38:56.600 --> 00:39:00.700
I'm going to set those two equal to each other so I get the following.
00:39:00.700 --> 00:39:29.600
I get 1/ T D² U DV DT = 1/ T DP DT V + 1/ T D² U DT DV – 1/ T²
00:39:29.600 --> 00:39:54.200
I’m just taking this from the previous page, I’m just setting these two things equal to each other -1/ T² × P + DU/ DV sub T.
00:39:54.200 --> 00:39:58.500
1/ T² DV DT, 1/ DT² DT DV.
00:39:58.500 --> 00:40:10.400
The V's cancel because again mixed partials U is a state property, U is a function of temperature and volume.
00:40:10.400 --> 00:40:29.600
Therefore, the mixed partials are equal because this time for U, mixed partials are equal.
00:40:29.600 --> 00:41:01.800
We have 1/ T DP DT sub V =1/ T² I just move this thing over there 1/ T² × P + DU DV T.
00:41:01.800 --> 00:41:09.400
We have this thing, 1/T DP DT sub V = 1/ T².
00:41:09.400 --> 00:41:13.200
We also had this which was one of our equations.
00:41:13.200 --> 00:41:25.100
We had DS DV sub T = 1/ T.
00:41:25.100 --> 00:41:29.500
I’m going to make a little change before I do this one.
00:41:29.500 --> 00:41:48.100
I'm going to go ahead and cancel one of the 1/ T and I’m going to end up with DP DT sub V = 1/ T P + DU / DV T.
00:41:48.100 --> 00:41:54.300
Do not worry you do not have to know this derivation, it is only the final result that we need but
00:41:54.300 --> 00:41:59.200
it is important to see the derivation especially at this level of your study of science.
00:41:59.200 --> 00:42:03.000
You need to know where this is actually coming from.
00:42:03.000 --> 00:42:07.000
You need to be comfortable with recognizing where the mathematics is going.
00:42:07.000 --> 00:42:09.700
It is very important.
00:42:09.700 --> 00:42:28.100
We have this, now we also had from one of the previous equations this thing we had DS DV sub T =1/ T,
00:42:28.100 --> 00:42:36.500
this is 1 of our basic equations + DU DV sub T.
00:42:36.500 --> 00:42:45.700
well this and this are the same which means these two are equal.
00:42:45.700 --> 00:42:53.300
What we end up with is, we go back to black.
00:42:53.300 --> 00:43:06.800
Our differential coefficient DS / DV T is actually equal to DP/ DT V.
00:43:06.800 --> 00:43:09.300
This is what we wanted.
00:43:09.300 --> 00:43:13.700
We related one of the differential coefficients to constant volume capacity / T.
00:43:13.700 --> 00:43:19.300
With some mathematical manipulation, the other differential coefficient the one which is volume dependent,
00:43:19.300 --> 00:43:27.400
the change in entropy or the change in volume is actually equal to or identifying it to the change in pressure over
00:43:27.400 --> 00:43:30.800
the change in temperature when we hold the volume constant.
00:43:30.800 --> 00:43:35.300
This is easy to measure, this is not easy to measure, this is easy to measure.
00:43:35.300 --> 00:43:45.800
This is telling me that whatever this numerical value is it happens to be the rate of change of entropy with respect to volume.
00:43:45.800 --> 00:44:01.000
Let us go ahead and rewrite that so we have a DS DV at constant temperature = DP DT at constant volume.
00:44:01.000 --> 00:44:06.100
This is our relation that we wanted.
00:44:06.100 --> 00:44:43.600
Now we have a simple expression or a simpler expression at least than what it was for the volume dependence of entropy at constant temperature.
00:44:43.600 --> 00:44:50.900
This DP/ DT constant V on the right side is very easily measurable and that is what we want.
00:44:50.900 --> 00:44:56.000
We want something that is not easily measurable.
00:44:56.000 --> 00:45:00.300
We do not measure entropy directly but this is my easily measurable.
00:45:00.300 --> 00:45:06.000
We are identifying this with that, that is what we want.
00:45:06.000 --> 00:45:16.200
It is very easily measurable.
00:45:16.200 --> 00:45:43.300
Recall from example 2, in the previous math lesson, we said when we are discussing the cyclic rule we ended deriving this relationship.
00:45:43.300 --> 00:45:51.600
We end up deriving the DP/ DT under constant volume is actually equal to Α / K,
00:45:51.600 --> 00:45:56.000
the coefficient of thermal expansion divided by the coefficient of compressibility.
00:45:56.000 --> 00:46:12.700
Therefore, DP/ DT sub V that is right there so we have DS/ DV T this is the V not U = Α / K, there you go.
00:46:12.700 --> 00:46:19.000
A and K I just love those, they are very simple just using your handbook or I measure it.
00:46:19.000 --> 00:46:29.800
Either this equation or this equation, this expresses the volume dependence of entropy.
00:46:29.800 --> 00:46:35.200
Let us say a few more words here.
00:46:35.200 --> 00:46:42.500
Let us do our summary and recap, in case we have lost our way in all this crazy mathematics.
00:46:42.500 --> 00:46:54.900
Let us start on the next page here, I’m going to go ahead and do this one in blue.
00:46:54.900 --> 00:47:01.400
In case we have lost our way, we have a fundamental equation of thermodynamics.
00:47:01.400 --> 00:47:06.300
I’m just going to call it the fundamental equation of thermal F E T.
00:47:06.300 --> 00:47:20.100
It is said that DS =1/ T DU + P/ T DV.
00:47:20.100 --> 00:47:46.600
We also said that for entropy as a function of temperature and volume our differential expression gives us DS = DS DT V DT + DS DV T DV.
00:47:46.600 --> 00:47:59.800
We want to identify these differential coefficients with things that are easily measurable or things that we can look up in a handbook.
00:47:59.800 --> 00:48:46.000
We wanted to identify the differential coefficients these things with quantities easily measured or already tabulated and we did.
00:48:46.000 --> 00:48:58.800
We had DS DT, the temperature dependence of the entropy at constant volume is just equal to the constant volume heat capacity divided by that.
00:48:58.800 --> 00:49:09.600
The volume dependence of entropy under constant temperature happens to equal DP DT
00:49:09.600 --> 00:49:14.700
under constant volume which also happens to equal Α / K.
00:49:14.700 --> 00:49:17.400
This is the temperature dependence of the entropy.
00:49:17.400 --> 00:49:25.000
This is the volume dependence of the entropy.
00:49:25.000 --> 00:49:41.800
K is always positive, Α is mostly positive, when we say mostly, with the exception of water and a couple of other things.
00:49:41.800 --> 00:49:54.800
For most situations and by most I mean like 99.999% of situations, Α / K is positive.
00:49:54.800 --> 00:49:57.700
Positive what does this mean?
00:49:57.700 --> 00:50:09.100
It means that if I increase the volume, I increase the entropy.
00:50:09.100 --> 00:50:12.200
I knew that already but this just confirms it mathematically.
00:50:12.200 --> 00:50:23.300
It confirms it in another way, that is what is going on here.
00:50:23.300 --> 00:51:03.900
The total differential expression which is what we start off with DS DT V DT + DS DV T DV can now be written DS = CV/ T DT + Α / K DV.
00:51:03.900 --> 00:51:09.200
If I'm dealing with the system and, by the way this applies to every systems gas, liquid, solid,
00:51:09.200 --> 00:51:13.300
under any set of circumstances this is the most general equation.
00:51:13.300 --> 00:51:22.200
If I want to know what the change in entropy of the system is, under conditions of change in temperature, change in volume, this is the equation that I use.
00:51:22.200 --> 00:51:26.800
If volume is held constant in a particular problem this term goes to 0.
00:51:26.800 --> 00:51:30.800
If temperature is held constant in a particular problem this term goes to 0.
00:51:30.800 --> 00:51:31.700
I just do it with this term.
00:51:31.700 --> 00:51:36.000
Α and K, I can easily look up or the problem will give it to me.
00:51:36.000 --> 00:51:42.200
C, V, or T I can either measure or the problem will give it to me, that is what makes this beautiful.
00:51:42.200 --> 00:51:45.000
This is the equation that you want to memorize this and this.
00:51:45.000 --> 00:51:52.100
This is the general version that you want to start off with your problems and this is the version
00:51:52.100 --> 00:51:59.400
that actually involves what it is that we have identified this with.
00:51:59.400 --> 00:52:02.300
This is identified with this, this is identified with this.
00:52:02.300 --> 00:52:10.700
This is what we wanted, we want to express entropy in terms of variables that we do know temperature, pressure, volume, things like that.
00:52:10.700 --> 00:52:16.100
We wanted to identify the differential coefficients once we had the expression for how entropy behaves
00:52:16.100 --> 00:52:19.700
in terms of things that we can actually measure, that is all that we have done here.
00:52:19.700 --> 00:52:24.600
The rest is just mathematical manipulation.
00:52:24.600 --> 00:52:28.600
Please, do not think that is mathematical manipulation or something that was just obvious.
00:52:28.600 --> 00:52:37.000
It took a lot of months, years, hours, for us to figure out the stuff out, for people to figure the stuff out in the 19th century.
00:52:37.000 --> 00:52:38.700
It took a lot of time.
00:52:38.700 --> 00:52:41.100
What we are getting is the result of all that.
00:52:41.100 --> 00:52:46.800
It looks like let us just do this, let us differentiate here, let us identify this with this.
00:52:46.800 --> 00:52:53.600
A lot of paper, a lot of time has been spent trying to put this together.
00:52:53.600 --> 00:53:14.900
Now one final word, for liquids and solids the volume dependence, the second part of the equation
00:53:14.900 --> 00:53:35.500
the volume dependence of entropy is usually so small that it can be ignored.
00:53:35.500 --> 00:53:42.900
In a particular problem, unless they tell you to specifically include this for anytime you are dealing
00:53:42.900 --> 00:53:46.900
with a system which is a liquid or solid because automatically take that to 0.
00:53:46.900 --> 00:53:56.100
You cannot do it with a gas so but you can with a liquid a solid for the most part, unless the problem tells you otherwise.
00:53:56.100 --> 00:54:17.400
Let us go ahead and revisit Α = 1/ V DV DT under constant pressure and K happens to equal -1/ V DV DP under constant temperature.
00:54:17.400 --> 00:54:20.200
This is the coefficient of thermal expansion.
00:54:20.200 --> 00:54:25.300
This is the coefficient of compressibility.
00:54:25.300 --> 00:54:30.200
This is one of the equations and this is pretty much what you want.
00:54:30.200 --> 00:54:34.000
You want to know this equation, you want to start your problems with that equation.
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Thank you so much for joining us here at www.educator.com.
00:54:36.500 --> 00:54:37.000
We will see you next time, bye.