WEBVTT chemistry/physical-chemistry/hovasapian
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Hello and welcome back to www.educator.com and welcome back to Thermodynamics in Physical Chemistry.
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We have come to the very important parts in these lessons.
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We have come to the beginning of the example problems.
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We have gone through a lot of theory.
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We have done a fair number of derivations that is a lot of mathematics.
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As I promised you in the previous lessons, we re going to start on the example problems and we are going to do a lot of them.
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When I say a lot of them, I mean a lot.
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It is very important that we have good graphs of what is happening particularly with the first law.
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With energy, with the work, and heat, that is going to lay the foundations.
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We do not just handle them mathematically, that is true.
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Of course, we want to because we have to do the problems but we want to understand what is going on.
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That will carry us forward.
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With that, let us go ahead and just jump right on in.
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A lot of you are going to approach this particular class with memorizing a bunch of equations.
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Let me go ahead and tell you right off the bat that is not going to work.
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Memorizing equations will work for a specific set of problems or for a specific type of problem, this is higher science,
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this is more sophisticated science or sophisticated mathematics.
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The idea is to have a reasonable degree of understanding to have a handful of equations at your disposal and let the problem at hand,
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However, it is worded because it is going to be worded and the same thing is going to be worded in 10 different ways.
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You need to be able to extract that information, find out what the problem is asking,
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and use the handful equations that you have at your disposal to derive what you need.
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You are not going to be asked to do a complex derivation that is just for your scientific literacy, the derivations that we went through.
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But as long as you understand what is happening, keep the equation that you learn to a minimum,
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just the fundamental set of equations and that is what I'm going to give you.
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In order to solve all these problems, I'm going to give you the basics of equations that you need to know in order to do all these problems.
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Our fundamental equations, the ones that you want to know are the following.
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Fundamental equations and you can use these for every single problem.
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You can use a subset of them for every single problem but these are the ones that you have to know.
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Fundamental equations.
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Work, we have to know what work is and that is the following.
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I'm going to give almost all of these equations the differential form, the finite version is just a δ in front of it for the state functions, for the path functions.
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Do not stick any δ in front of it.
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The work is equal to the external pressure × change in volume that is the definition of work.
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Now what about energy in the first law?
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Energy in the first law says that change in energy of the system is equal to the heat withdrawn from the surroundings - the work produced in the surrounding.
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That is DQ =- DW.
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The definition of enthalpy is very important, it is equal the energy of the system + the pressure of the system × the volume of the system.
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This is not external pressure, the pressure of the system.
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The definition of the heat capacities is very important.
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The heat capacity definitions we have two of them.
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We have a constant volume heat capacity, the constant volume heat capacity is defined as
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the heat withdrawn from the surroundings divided by the change in temperature of the system.
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When I said heat withdrawn from the surroundings under constant volume, under constant volume
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the heat withdrawn from the surroundings is that heat that goes into the system.
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It is just a questionable point of view.
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It is really important be able to switch back point of views because that means you understand what is happening thermodynamically.
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Qualitatively what is going on is which moving and in which direction.
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That was identified with this partial derivative DU DT.
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If you have a function, if you have the energy and it is a function of temperature and volume,
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the partial derivative of that function with respect the temperature is that heat capacity under constant volume.
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And of course, we have the constant pressure heat capacity.
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The constant pressure heat capacity = DQ DU DT,
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the heat withdrawn from the surroundings under conditions of constant pressure ÷ the change in temperature.
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Again, this is the same that definition of the heat capacity is the same.
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It is heat ÷ temperature.
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This happens to be idea, these three lines for definition.
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This happens to be identified with the enthalpy.
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This is the change in enthalpy per change in temperature.
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Under constant pressure conditions that is how best defined.
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The fundamental equations, you have this one.
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It would do this in red.
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This one you have to know, this one you have to know, and you have to know these two.
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When I say you have to know, you have them as a piece of paper right next to you.
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Eventually, you are going to do enough problems so we are going to memorize them.
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These are the ones that you will memorize not the others.
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Let us go back to blue which you wan to know as far as the mathematics.
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In mathematics we have two relations of a mathematics that you have to know.
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DU = CV DT + DU DV DV this is the general equation.
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The change in energy of the system, if the energy is a function of temperature and volume,
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the differential total change if I change temperature and volume, if I change both it is this.
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And I also have an analogous one for the enthalpy, the DH.
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DH = CP DT + DU DP, constant temperature DP.
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If the enthalpy is a function of temperature and pressure, the change in enthalpy of the system is this + that, if I change both temperature and pressure.
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If they say this is a constant pressure process this goes to 0 because DP = 0.
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If they say it is a constant temperature process DT = 0.
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This goes away.
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This is the general equation.
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You are letting the constraints of the system tell you what is happening, what you need to do these terms.
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Let us go ahead and I will write is the fundamental concepts.
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I will right this on this page.
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Fundamental concept, isothermal, when I say something is isothermal that means it is a constant temperature.
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Mathematically, it means the following.
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It means that the DU is= 0 or finite δ U = 0.
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Isothermal means that the temperature does not change.
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If the temperature does not change, the energy of the system does not change, DU = 0.
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In your equations, you can set DU= 0.
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You can out 0 in for DU.
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Adiabatic means DQ = 0, it means no heat is actually allowed to flow.
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If they say the word insulated that means it is adiabatic DQ = 0 or the finite form δ Q.
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Because heat is a path function, it is not a state function.
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Only state functions have the δ.
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Isobaric, if you happen to see the work that just means constant pressure,
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When I say constant pressure, they will use the word isobaric.
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Mathematically, that means D P = 0.
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If you see isochoric or isometric and the usual you would not use these terms.
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The usually terms for pressure and volume, they usually say constant volume and constant pressure.
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This is constant volume.
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For constant volume, DV = 0, that is the mathematics there.
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Final set, for ideal gases, these are just the basic things that we need to know in order to solve the problems.
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This is what we want to have at our disposal.
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For ideal gases, we know that PV = nr, that is our equation of state from ideal gas.
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We also have that relationship between the constant pressure and constant volume heat capacities that is CP - CV = nr.
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Or in terms of molar, CP mol - CV mol= R.
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Molar just means that you divide everything by n that is all it means.
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From ideal gas, go back to the general equations DU DV under constant temperature = 0 for ideal gas.
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If it is not ideal, it is not going to be 0.
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Actually, in this particular case probably will be 0.
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We want to write a general equation for ideal gas is definitely = 0.
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And for ideal gas is DH DP.
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The change in enthalpy, the change in pressure per unit change in pressure, if I change the pressure by one unit, 1 atm how is the enthalpy going to change?
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0 from ideal gas the enthalpy is not a change of pressure.
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It is not a function of pressure.
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The energy is not a function of volume.
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Energy is a function of temperature.
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The enthalpy is a function of temperature from ideal gas.
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With that, let us just go ahead and jump into this very long process of example problems.
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We are going to do a lot of them like it is very important.
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I want to understand this very well.
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Let us see what we got.
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Our first example is going to be 2 mol of an ideal gas undergo the following changes in state a, b, and c.
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What is the change in temperature for each change in state?
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Molar constant volume heat capacity = 12.5 J/ mol/ K.
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In this case they have given us a molar value.
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They give us that, if we want to keep an eye on units as much as anything else.
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2 mol of ideal gas undergo the following changes in state.
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Let us go ahead and deal with part of a.
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Let us go ahead and recall our conventions before we actually get to the problem.
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Our conventions are the following.
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Heat is positive when heat flows into the system.
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Work is positive when work is produced.
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When you see the work, work is produced that means work is positive.
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Work is produced in the surroundings.
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Work we are looking at something from the point of view.
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In the case of work, we are looking at the surroundings point of view.
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When you see the work produced, it means work is positive in the surroundings.
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If you see work is destroyed that means it is negative.
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That means work is leaving the surroundings.
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Work is going into the system.
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Heat is positive when heat flows into the system.
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That means it is flowing out of the surroundings.
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It is just a question of point of view but this is our convention.
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Let us go ahead and see what we have got.
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We want to know what the temperature changes.
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The equations that we have at our disposal, we are talking about heat and work.
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They want to know change in temperature.
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It looks like we would do something like this.
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We have an ideal gas.
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We know that if we are dealing with an ideal gas we are going to use these equations right here DU =DQ – DW.
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Or we are going to use the finite form because they actually give us the values DU = Q – W.
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We are also going to use the following.
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We are going to use DU = CV DT.
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The second term the DU has this one + the partial derivative, the DU DV this is an ideal gas.
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DU DV = 0 because it = 0 we are left with this equation.
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Or the finite form δ U = CV δ T.
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Will we are looking for changes in temperature so we try to find δ T.
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Therefore, I'm going to solve this for δ T.
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Δ T = DU/ CV.
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I have CV, I just need to find δ U and do the division.
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Let us go ahead and do that.
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Part A, they say that 500 J of heat flows out as heat, that means is flowing out of the system.
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That means it is positive when heat flows into the system.
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If it flows out that means Q is negative so this is -500 J.
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Work they are telling me 150 J of work is destroyed which means work is negative.
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Work = -158 J.
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Δ U = Q - W = -500 -150.
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Therefore, δ U = - 350 J.
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Therefore, δ T is = δ U/ CV.
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This is going to be - 350 J ÷ CV.
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They gave us the molar heat capacity.
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This is 12.5 J/ mol/ K.
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They are 2 mol.
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We do a quick calculation here.
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CV the molar heat capacity it is going to be 12.5,
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Let me just do this on the next page.
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I have 12.5 J/ mol/ K × 2 mol that gives me 25 J/ K.
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Now I can do by problem.
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My δ T in this particular case = -350 J ÷ 25 J/ K.
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J and J cancel and if I get my arithmetic correctly which I’m never quite sure that I do, -40 K.
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That is your answer.
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The system experience is -14°K drop in temperature.
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Let us see here, 500 J of energy flows out of the system and 150 J of energy flows into the system.
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The net loss of energy by the system is 350 J.
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If the system loses energy, the energy of the system is going to drop.
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Part B, part B says that the amount of heat is 400 J, 400 J flows in.
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They also say that 400 J of work is produced, +400.
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Δ U= Q - W = 400 -400 = 0.
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δ T = 0/25 J/ K.
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The change in temperature is 0.
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What happened here is the following.
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You have your system, the boundary, and you have your surroundings 400 J of heat flowed into the system.
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400 J of work was done on the surroundings.
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400 J went this way, 400 J went this way.
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There is no change.
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From the system's point of view 400 came in and 400 left.
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From the surroundings point of view, 400 left and 400 came in.
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Δ U is 0, that is what is happening.
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If δ U is 0 and δ T is 0.
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That is what we want here in this particular case, the temperature change.
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Part C , they are telling me that 0 J flow as heat so Q = 0 and they are telling me that 113 J of work is destroyed.
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Work = -130 J.
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Δ U = Q - W = 0 --130 J.
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Therefore, δ U= 130 J.
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Therefore, δ T =130 J ÷ 2 mol or 12.5 J/ mol/ K gives us 25 J/ K.
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J cancels J and what we are left with is 5.2°K increase.
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System surroundings, no amount of heat flowed.
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Work 113 J was destroyed that means this is negative.
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That means it is flowing out of the surroundings into the system.
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This much work was done.
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If these much work as energy transfers to the system, the energy of the system rises.
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The energy of the system rises, the temperature of the system rises.
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5.2 K change in temperature of the system.
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This is what we are looking at, we are looking at the system.
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Our convention is that heat is positive when it flows into the system.
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Work is positive when it is done on the surroundings or it is negative if it is done on the system.
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It is just a question of perspective.
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Let us look at our next example here.
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Doing a particular change of state of 55 J of work or destroyed and the internal energy of the system increases by 200 J,
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the temperature of the system rises by 12°K.
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What is that heat capacity of the system?
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The heat capacity of the system = DQV / DT or Q / δ T.
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Well they give us the change in temperature that so we have the δ T.
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All we have to do is find Q, this division that will give us that heat capacity of the system.
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We need Q, we know that δ U = Q – W, they tell us that a particular change of state 55 J of work are destroyed.
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They tell us that the internal energy of the system increases by 200 J.
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Δ U = 200 = Q – 55 J of work is destroyed -55.
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Therefore, 200 = Q + 55.
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Our Q is = 145.
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Therefore, let us put that back in there.
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Our constant volume heat capacity is going to be 145 J ÷ 12°K.
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If I did my arithmetic correctly, let us check.
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I never get my arithmetic correct.
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12.1 J/ K that this is our heat capacity.
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Our molar heat capacity, notice there is no mention of a particular moles or anything like that.
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If I want the molar heat capacity I would divide by the number of moles and that would give me J/ K/ mol.
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This is just J/ K.
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Be very careful of the units.
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That was pretty straightforward, nothing too strange here.
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We are starting off easy working our way up.
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3 mol of an ideal gas expand isothermally.
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Isothermally tell us that the temperature is constant.
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Isothermally against a constant pressure 110 kg Pascals for 25 dm³ to 65 dm³.
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We want you to find Q,W, δ U and δ H for this change of state.
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This is a pretty standard problem.
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Find the heat, find the work, find the change in energy, find the change in enthalpy.
00:24:54.400 --> 00:24:57.500
This is a really common problem.
00:24:57.500 --> 00:25:01.900
Some of the problems last for one of the other but again it is nice to ask for all four because that is the relationship.
00:25:01.900 --> 00:25:04.600
This is concerned with work, this and this.
00:25:04.600 --> 00:25:08.200
There is a relationship among these.
00:25:08.200 --> 00:25:11.900
Let us talk about, fundamental concept.
00:25:11.900 --> 00:25:21.800
Isothermal that means the change in temperature = 0.
00:25:21.800 --> 00:25:31.900
If the change in temperature = 0 that means there is no change in energy so that means that DU = 0 or DU = 0.
00:25:31.900 --> 00:25:35.800
I automatically have my DU=0.
00:25:35.800 --> 00:25:42.000
Isothermal means δ U= 0 ideal gas.
00:25:42.000 --> 00:25:45.300
That was nice.
00:25:45.300 --> 00:25:53.900
They gave us a constant pressure 110 km Pascal and the unit of change in volume 25 dm³ to 65 dm³.
00:25:53.900 --> 00:25:58.800
A pressure × a change in volume =work.
00:25:58.800 --> 00:26:19.900
Work = the external pressure × the change in volume, that = 100 × 10³ Pascal × 40 × 10³ m³.
00:26:19.900 --> 00:26:24.800
This is in dm meters, this is km Pascal or Pascal.
00:26:24.800 --> 00:26:28.700
It is a Pascal km that is a Joule.
00:26:28.700 --> 00:26:31.100
We have to make sure working in the right units.
00:26:31.100 --> 00:26:36.300
1 J = 1 Pascal m³.
00:26:36.300 --> 00:26:38.100
We have to make those conversions.
00:26:38.100 --> 00:26:49.000
In this case, 10³ the -3 cancel, 100 × 40 that leaves us with 4000 J.
00:26:49.000 --> 00:26:51.800
The work is 4000 J.
00:26:51.800 --> 00:26:55.800
It is positive that means 4000J of work is done on the surroundings.
00:26:55.800 --> 00:27:03.300
4000 J of work is produced.
00:27:03.300 --> 00:27:10.600
Δ U= Q – W, δ U is 0, isothermal.
00:27:10.600 --> 00:27:14.300
0 = Q – W.
00:27:14.300 --> 00:27:18.700
Therefore, Q = W.
00:27:18.700 --> 00:27:23.100
W is 4000 so Q = 4000.
00:27:23.100 --> 00:27:24.800
There we go we just found Q.
00:27:24.800 --> 00:27:31.800
We have taken care of work, we have taken care of Q, and we have taken care of δ U.
00:27:31.800 --> 00:27:32.900
Now we are almost done.
00:27:32.900 --> 00:27:34.900
All we have to worry about is our δ H.
00:27:34.900 --> 00:27:40.900
Δ H, our recommendation with δ H we will start with the definition of enthalpy.
00:27:40.900 --> 00:27:46.800
Our definition of enthalpy is H = U + PV.
00:27:46.800 --> 00:27:49.400
We want δ H.
00:27:49.400 --> 00:27:57.000
The Δ operator that = δ U + PV and this operating on this.
00:27:57.000 --> 00:28:05.600
The δ operator is linear operator, distributes just regular distribution, consider the symbols they just mean symbolic distribution.
00:28:05.600 --> 00:28:14.700
That = δ U + δ PV.
00:28:14.700 --> 00:28:18.100
We already know what δ U is, δ U is 0.
00:28:18.100 --> 00:28:24.300
Therefore, δ H =δ PV.
00:28:24.300 --> 00:28:28.700
This is not P δ V, this is δ PV.
00:28:28.700 --> 00:28:45.600
It is pressure 2 × volume 2 - pressure 1 × volume 1, that is what δ PV is.
00:28:45.600 --> 00:28:56.300
Δ PV =P2 V2 - P1 V1, this is not P δ V, this is not constant pressure.
00:28:56.300 --> 00:29:01.700
This is the pressure of the system.
00:29:01.700 --> 00:29:03.800
Δ PV is P2V2 – P1 V1.
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This is my ideal gas P1 V1 =P2 V2.
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P1 V1 = P2 V2 because P1/V1 = P2/ V2.
00:29:15.700 --> 00:29:18.500
The temperature is constant so the temperature drops off.
00:29:18.500 --> 00:29:23.800
You are left with this basic equation from general chemistry.
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P2 V2 – P1 V1, if they are equal then δ PV = 0.
00:29:30.700 --> 00:29:33.900
Therefore, δ H = 0.
00:29:33.900 --> 00:29:39.300
Their work is 4000 J, your heat is 4000 J.
00:29:39.300 --> 00:29:41.900
Your change in energy of the system is 0.
00:29:41.900 --> 00:29:46.600
Your change in enthalpy of the system =0.
00:29:46.600 --> 00:29:48.600
That is what is going on here.
00:29:48.600 --> 00:29:58.300
Here is the system, here is the surrounding, 4000 J of work is being produced.
00:29:58.300 --> 00:30:06.300
It is expanding the gas, the system is doing work on the surroundings.
00:30:06.300 --> 00:30:14.700
If energy as work goes this way, if the system does work on the surroundings, the energy of the system is going to decrease.
00:30:14.700 --> 00:30:20.900
The energy of the system decreasing implies that the temperature of the system is going to decrease.
00:30:20.900 --> 00:30:22.500
But this is an isothermal process.
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Isothermal means we are not allowing the temperature to decrease.
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Where is that extra energy coming from in order to make sure that the temperature stays up?
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It is coming from the surroundings.
00:30:33.900 --> 00:30:35.900
Work is being done on the surroundings.
00:30:35.900 --> 00:30:42.600
The surroundings is actually giving back heat to the system, heat is 4000 J.
00:30:42.600 --> 00:30:46.800
Heat is coming into the system that is what is happening here.
00:30:46.800 --> 00:30:50.500
This is what this is saying.
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In this isothermal process, the system does work on the surroundings by pushing against it.
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In the process of pushing against it and doing work its losing energy as work.
00:30:58.600 --> 00:31:03.300
If it loses energy as work, the energy is going to drop but the temperature is going to drop.
00:31:03.300 --> 00:31:08.700
If energy drops, the only way for the temperature to stay the same because we are making sure this happens isothermally,
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it has to pull energy as heat from something else.
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The only other source it has to do that is the surroundings.
00:31:15.200 --> 00:31:22.200
It pulls heat this way so heat goes this way, the temperature stays the same.
00:31:22.200 --> 00:31:24.300
That is what is going on here.
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This is the mathematics and this is what is happening physically.
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I hope that makes sense.
00:31:32.800 --> 00:31:40.100
Let us do some more here.
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We have 3 mol of ideal gas to 30°C expands isothermally and reversibly from 25 dm³ to 65 dm³.
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Find Q, W, δ U, and δ H.
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The expansion is the same, it is going from 25 to 65 dm³.
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The only thing that is different is it is happening isothermally.
00:31:59.000 --> 00:32:02.400
It is an ideal gas but now it is happening reversibly.
00:32:02.400 --> 00:32:07.600
It is not just isothermal, we are not just keeping the temperature constant but we are also doing it reversibly.
00:32:07.600 --> 00:32:10.300
We are moving along the isotherm.
00:32:10.300 --> 00:32:14.400
We are not just going this way.
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Let us go ahead and do this.
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In this particular case, isothermally and reversibly we are going to use this one DW = P external × DV.
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Reversible means that the external pressure = to the pressure of the system.
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The P external = P itself, they are in equilibrium.
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That is what reversible means, the external pressure is = to the pressure of the system.
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Therefore, DW = PDV or we are going to do the derivation itself instead of memorizing equation.
00:32:57.700 --> 00:33:04.600
We are going to do the derivation because we can, we have a mathematics just a simple calculus, PDV.
00:33:04.600 --> 00:33:16.000
Therefore, work = to the integral from V1 to V2 of PDV.
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It equals the integral from V1 to V2 of nrt / V DV because for ideal gas P = nrt / V.
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I’m just substituting this value in for P.
00:33:32.100 --> 00:33:42.000
Our T is a constant so I will pull it out so I get nrt × the integral from V1 to V2 of DV / V.
00:33:42.000 --> 00:33:50.800
I end up with nrt LN of V2 / V1, that is my work.
00:33:50.800 --> 00:33:52.900
And I just put my values in
00:33:52.900 --> 00:34:05.100
I have 3 mol, R is 8.314 J/ mol K.
00:34:05.100 --> 00:34:07.200
This is happening at 30°C.
00:34:07.200 --> 00:34:10.400
It is isothermal because it stays at 30°C.
00:34:10.400 --> 00:34:12.700
This is 303 K.
00:34:12.700 --> 00:34:24.700
In this case I do not have to worry about changing this to meters because the unit cancels 65/25.
00:34:24.700 --> 00:34:30.400
I think this is supposed to be, I think I did my numbers wrong here.
00:34:30.400 --> 00:34:34.700
I think this is supposed to be 75.
00:34:34.700 --> 00:34:45.000
If I’m not mistaken, I think this is supposed to be 75 but any way you can work out the actual arithmetic itself does not really matter.
00:34:45.000 --> 00:34:47.600
All that matters is the process.
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3 mol × log of, I think I’m going to ahead and keep it as written.
00:34:55.000 --> 00:35:02.100
I’m going to do 65/ 25, the dm³ end up cancelling.
00:35:02.100 --> 00:35:12.000
The ratio is what matters so I ended up with 8303 J.
00:35:12.000 --> 00:35:20.000
I can recall this 8303 come from this being 65 of this come from being 75.
00:35:20.000 --> 00:35:26.800
I may have actually been thinking of 75 when I did this problem and put it to my calculator but this is the correct process.
00:35:26.800 --> 00:35:31.500
The number itself is actually not that relevant.
00:35:31.500 --> 00:35:32.700
This takes care of the work.
00:35:32.700 --> 00:35:39.000
So work = 8303 J.
00:35:39.000 --> 00:35:43.700
Isothermal implies that the temperature is constant.
00:35:43.700 --> 00:35:48.100
If the temperature is constant that means there is no change in energy.
00:35:48.100 --> 00:36:00.000
Therefore, isothermal means that δ U = 0.
00:36:00.000 --> 00:36:17.800
If δ U = 0 that means 0 = Q – W, Q = W.
00:36:17.800 --> 00:36:27.200
Therefore, Q = 8303 J.
00:36:27.200 --> 00:36:32.800
Enthalpy is going to be the same as that problem that we just did.
00:36:32.800 --> 00:36:39.800
Let me go over it, H = U + PV.
00:36:39.800 --> 00:36:51.000
Δ H = δ U + PV = δ U + δ PV.
00:36:51.000 --> 00:37:01.900
Δ U = 0 PV, this is an ideal gas so P1 V1 = P2 V2 so that also = 0.
00:37:01.900 --> 00:37:06.100
Therefore, δ H for this process = 0.
00:37:06.100 --> 00:37:16.100
And again this is the pressure of the system not the external pressure.
00:37:16.100 --> 00:37:25.200
The previous example, in example 3, Q = W = 4000 J.
00:37:25.200 --> 00:37:35.100
In this example Q = W = 8303 J.
00:37:35.100 --> 00:37:41.900
This one was just a isothermal against an external pressure 110 km Pascal.
00:37:41.900 --> 00:37:54.100
This one was not just isothermal but it was also reversible.
00:37:54.100 --> 00:38:06.400
It was also reversible, this is the maximum amount of work that this gas can do in expanding from one volume to another.
00:38:06.400 --> 00:38:18.300
The other path that was taken was 4000 J, 8303 J is the maximum amount of work that this gas can do in expanding.
00:38:18.300 --> 00:38:23.800
That is was happening here.
00:38:23.800 --> 00:38:28.300
Example 5, we have 4 mol of an ideal gas.
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They are compressed this time isothermally from 100 L to 20 L by a constant pressure of 7 atm.
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Find Q, W, δ U and δ H.
00:38:38.700 --> 00:38:39.600
This is pretty straightforward.
00:38:39.600 --> 00:38:41.700
We have to take on it by now.
00:38:41.700 --> 00:38:44.300
We have done a couple of problems just like this.
00:38:44.300 --> 00:38:50.700
We are just going to keep doing a lot of problems that are the same just to get a sense of it and to become comfortable with that.
00:38:50.700 --> 00:38:53.000
That is what we want.
00:38:53.000 --> 00:38:55.200
Let us go ahead and do work first.
00:38:55.200 --> 00:39:02.300
Work = the external pressure × the change in volume.
00:39:02.300 --> 00:39:06.500
They give us the change in volume that is just the 120.
00:39:06.500 --> 00:39:11.800
It is final - initial so 20 -100 so it was actually -80.
00:39:11.800 --> 00:39:30.500
The external pressure is 7 atm and -80 L so we end up with -560 L atm.
00:39:30.500 --> 00:39:33.700
-560 let us go ahead and convert this to J.
00:39:33.700 --> 00:39:53.000
-560 L atm × 8.314 J ÷ 0.08206 L atm, that cancels.
00:39:53.000 --> 00:39:59.500
I’m going to go ahead and express in kJ, -56.72 kJ.
00:39:59.500 --> 00:40:02.100
It is an isothermal process.
00:40:02.100 --> 00:40:08.200
We already know that an isothermal process δ U = 0.
00:40:08.200 --> 00:40:21.100
If δ U = 0 that means 0 = Q - W which means that Q = W so Q = -56.7 kJ.
00:40:21.100 --> 00:40:26.800
That takes care of Q and we already work out δ H part.
00:40:26.800 --> 00:40:39.000
Δ H = δ U + δ PV, we are dealing with an ideal gas and this is 0, this is 0.
00:40:39.000 --> 00:40:42.200
Our δ H is 0.
00:40:42.200 --> 00:40:44.500
In this particular case, what is happening is this.
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Here is the system, here is the surroundings, notice in this case that the other work and heat are negative.
00:40:53.800 --> 00:41:00.700
Work is -5.
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Sorry I’m not writing it properly -56.7.
00:41:07.200 --> 00:41:09.100
This is right, I have the next line.
00:41:09.100 --> 00:41:17.700
Everything is good.
00:41:17.700 --> 00:41:21.100
The work done is actually negative, that means work is destroyed.
00:41:21.100 --> 00:41:30.300
It means work is leaving the surroundings 56.7 kJ is moving from the surroundings to the system as work.
00:41:30.300 --> 00:41:33.400
Heat -56.7 kJ.
00:41:33.400 --> 00:41:40.200
Heat is negative when it s leaving the system.
00:41:40.200 --> 00:41:42.300
This is a compression.
00:41:42.300 --> 00:41:48.200
The surroundings are doing work on the system 56.7 kJ of work.
00:41:48.200 --> 00:41:53.100
56.7 kJ energy transfer as work from the surroundings to the system.
00:41:53.100 --> 00:41:58.100
This is an isothermal process which means the temperature have to stay the same.
00:41:58.100 --> 00:42:04.100
If we put energy in the system, the temperature is the one to rise but we want to keep the temperature down.
00:42:04.100 --> 00:42:10.300
We have to take energy away from the system in order to keep the temperature at the same level.
00:42:10.300 --> 00:42:17.600
Heat moves as energy or energy moves as heat from the system to the surroundings.
00:42:17.600 --> 00:42:21.000
Therefore, you get a total of δ U = 0.
00:42:21.000 --> 00:42:26.200
In other words, there is no energy change for the system.
00:42:26.200 --> 00:42:28.300
That is was going on here.
00:42:28.300 --> 00:42:33.000
Let us go ahead and we will stop it with these 5 example problems.
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Certainly there are more example problems to come, quite a few more.
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Thank you for joining us here at www.educator.com.
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We will see you next time, bye.