WEBVTT chemistry/physical-chemistry/hovasapian
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Hello, welcome back to www.educator.com and welcome back to Physical Chemistry.
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Today, we are going to start talking about, or to continue our discussion of the first law and energy and work and heat and things like that.
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We are going to start talking about changes in energy and state.
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In this particular lecture, we are going to be talking about constant volume processes.
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We are going to start putting some constraints either to constant volume, constant pressure, constant temperature, things like that,
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to see if we can learn some things about what is going on with the first law and with the energy.
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Let us jump right on in.
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Let us recall, this will go with black today.
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U energy is a state function or a state property.
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It does not depend on the path that you take.
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It just depends on your beginning state and your ending state.
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If I start at 0 feet sea level and ago up to 500 feet above sea level, the difference is 500 feet.
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It does not matter whether I drop down to negative 300 and go up to 800, then come back to 500.
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All that matters is the beginning and ending state.
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Work and heat are not state functions, they are path functions.
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W is a path function or path property.
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Heat is also a path function.
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A path function, its value, the value of work or heat depends on the path that we take.
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I’m going from here to here, this symbols less work than this or this.
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The amount of heat from here to here is going to be less than the amount of heat like that.
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It depends on the particular path that we take.
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The first law is this, it says that the change in energy, the total change in energy of the system is equal to
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the heat gained by the system - the work loss by the system.
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That is all that is.
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We will often use this form.
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This is the fully integrated form, we will often use the form, the differential form because
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we will be leading much more on the mathematics and more sophisticated mathematics.
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So du= dq – dw, the differential change in energy = the differential change in heat -the differential change in work.
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A quick word about this, U is a state function.
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W and Q are path functions.
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The symbolism that I have used is a DDD and tends to imply that they are the same.
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They are not the same.
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Many books will differentiate notationally a state function and the path function.
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They used this differential notation to standard, when you are used to the d for a state function.
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They will use some variation of that for a path function.
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You often see something like dw, that is Greek δ and dq or you will see d with a line through it.
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For dw and dq, this let us know that these are path functions.
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Most specifically, for mathematical point of view they are an exact differentials.
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A state function is an exact differential and a path function is an inexact differential.
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And we will be talking more about what we mean by exact and inexact and the properties of exact and inexact differentials.
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I, myself, I do not like to differentiate because I believe, I mean physical chemistry and
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Thermodynamics, there is already just an abundance of symbolism as it is.
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To introduce new symbolism, it just confuses me.
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To me, personally, it is just easier to remember that work and heat are path functions and energy is a state function.
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Other than that, it should not cause you any problem.
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I’m going to go and use the same, but they are not the same.
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This is an exact differential, these are inexact differentials.
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Let us go ahead and talk about integral from an initial state to a final state of this equals U2 – U1.
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This is the fundamental property of a state function but it integrates the way you are used to according to the fundamental theory of calculus.
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It is equal to δ U, this is an exact differential.
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Exact differentials integrate like this.
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You take the final state - the initial state of the integral of the particular function.
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Exact differential, this is a state function.
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When we integrated heat, dq, we do not get Q2 - Q1, we just get whatever the value of Q is.
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This is not δ Q, this is an inexact differential.
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Inexact differentials cannot integrate the same way but exact differential do.
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You do not follow this final – initial.
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It depends on the path so you just get a quantity.
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dQ and dQ, it does not make any sense because a system does not possess heat.
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Heat is something that shows up during a change of state.
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Heat is actually a process, when there is a temperature differential of two of the system in the surroundings,
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or two systems whenever it is, heat is the transfer.
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Heat happens at the boundary.
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That is what is going on, there is a system does not possess heat.
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The system has a certain temperature.
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Temperature and heat are not the same thing.
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Heat is what happens during the change of state.
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It makes no sense to say that there is a certain amount of heat at the beginning, a certain amount of heat in the end,
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and change in heat is the difference between the two.
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That does not make sense the system does not possess heat.
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Granted, we tend to be little loose with our language usage when we talk about heat as if the system does possess it.
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That is just part and parcel of the historical discussion of thermodynamics.
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That sort of looseness in language that existed, signs in general, it is not a problem as long as you remember these things.
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δQ makes no sense because a system does not possess heat.
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And the same thing with work, when we integrate the differential of the work from one state to another,
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all we are adding up all the work done along the particular path, we get the final work its equal to W.
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It is not equal to dW because the system does not possess work.
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It does not make any sense.
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Work inexact path function.
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The dW makes no sense for the same reason, a system does not possess work.
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Here is what is interesting, the Q and W are path functions their difference is a state function.
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That is extraordinary, that is truly amazing.
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Q and W, they are path functions but their difference or sum in a different convention, the chemists convention usually write Q + W.
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Our convention is Q – W but their difference, I will put sum here, is a state function, that is amazing.
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State function, energy.
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Where does this come from?
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The question where does δ U = Q –W come from?
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Or more often we use this with a differential = dq - du come from.
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Let us talk about a cyclic process.
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A cyclic processes is where you start with some initial state, you have a change of state,
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you go to state 2 and then from state 2, you come back to state 1.
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Exactly, it sounds like it is a cyclic process.
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You go to some final state, you come back, and end up where you started.
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The work of the cyclic process is equal to the integral of all the work done along the particular path that you take.
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The integral dW, we will often put a circle around it to let us know that it is a cyclic process.
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This is sort of an older notation, you probably do not see all that much anymore in modern books.
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Cyclic just means cyclic, that is all it is, nothing strange about it.
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In the Q, the heat lost or gained during a cyclic process that is equal to all the integrals of all of the dQ.
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All of the differential heat elements along that.
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Now these are not usually 0, in other words the cyclic work for a processes is not usually 0.
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The cyclic heat is not usually 0.
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These integrals are usually not equal to 0.
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If they are, it is strictly a coincidence that is it.
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In general, the cyclic integrals of path functions, the cyclic integrals of inexact differentials are not equal to 0.
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In general, the cyclic integral of path functions / inexact differentials does not equal 0.
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Now the cyclic integral of a state function does equal 0.
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The cyclic integral of any state function does equal 0, as it must.
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A state function depends on its initial and final state.
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In a cyclic process, the final state and initial state are the same.
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State - state you get 0.
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In a cyclic involve any state function and the exact differential is equal to 0, as it must.
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Initial, final, you go from initial to final and then you come back.
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The cyclic integral of some dz = 0, where z is a state function.
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Our empirical experience of the first law is the following.
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This is what this says, this is the science, this is thermodynamics, it is all empirical.
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Our empirical experience of the first law is this, if the system is subjected to a cyclic process or cyclic transformation,
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either one, the work transferred to the surroundings, in other words,
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the work gained by the surroundings to the surroundings = the heat transferred from the surroundings.
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Now mathematically, this says this.
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Our convention is we said that heat and work, these are effects that manifested in the surroundings.
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We never measured something directly in the system.
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We know what is happening in the system based on what we observed in the surroundings.
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And we can take the systems point of view but what we are directly measuring is, is what is happening in the surroundings.
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It is not a big deal as long as you think of the surroundings.
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And if you want to talk about the system, you just switch your point of view.
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When we talk about heat gain by work gained by the surroundings, we are talking about work done by the system.
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Work that is lost by the system.
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You just have to switch around.
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Mathematically, it says this, a cyclic process dQ = dW.
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This is our empirical experience of the first law.
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Now let us see what happens here.
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This is our mathematical or empirical experience of the first law.
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What is going to happen is the following.
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This is the same as, let me just move one of the integrals over to the other side which you end up with is this.
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The cyclic integral of dQ - the cyclic interval of dW is equal to 0.
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This is just the same as cyclic integral of dQ -dW = 0 because integration is a linear operator.
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It is linear so I can just pull out the linear and the sort distributes.
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We will talk about that little bit in calculus.
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Now, I have a cyclic integral of something right here, this in the grand underneath the integral sign, the cyclic integral of it is equal to 0.
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I know that the cyclic integral of any state function equal 0.
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The cyclic integral of any exact differential = 0.
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Therefore, dQ - dW must be some state function.
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We call the state function energy, that is where this comes from.
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We drop the definition on you but this is where it comes from becomes it comes from the empirical experience.
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Empirically, this is what happens.
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We want to give this a name we call it the energy of the system.
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Any cyclic integral that = 0 is some state function.
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We need to get the state function a name.
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We call this state function which is dQ - dW or Q – W, we call this state function energy, the energy of the system.
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This is where it comes from.
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dU= dQ – dW, the differential change in energy of the system comes from
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the differential change in heat - the differential change in work or the integrated form δ U= Q – W.
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Remember, δ Q and δ W they make no sense.
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Differential form and integrated form, we will be called the finite form.
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Only δ U is defined.
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This is the differential form, when we integrate this, when we integrate both sides, we end up getting is δ U = Q- W.
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We do not get U, we get Q and W because they are path functions but du is a state function, we get the δ.
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Only the δ U is defined, we do not measure energies of the system.
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We measure changes in energy of the system, that is what we do.
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Science best measures changes in things.
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Only dU is defined, not U.
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Let us see here, we have that δ U = Q – W.
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If we measure the heat loss by the surroundings then subtract from it the work gained by the surroundings, both of which are easily measurable.
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We get δ U, the energy change of the system.
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If we measure the heat lost by the surroundings, the heat of the surroundings loses.
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Once it loses something that goes into the system.
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Then subtract from it the work gained by the surroundings, we get the change in energy of the system.
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Basically, it always says that if you have a system and if you have the surroundings, if heat goes into the system,
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if energy goes into the system as heat and a certain amount of energy to leave the system as work,
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what you get is just the net change in energy.
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It is just simple arithmetic is what it is.
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I give you $100, take back $25, you are left with $75.
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That is all this is, this is a simple accounting of energy in terms of work and heat, in terms of paper or coin money.
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The change in state means changes in properties of the state of the system such as temperature, pressure, and volume.
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The things that are easily measurable and they are properties of the state.
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Also, temperature, pressure, and volume, these things are state functions also just like energy.
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They do not depend on the particular path that you take to get there.
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If you start with a system at 2 atm and then you increase it to 10 atm, and I take it down to 0.2 atm,
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and take it back to 5 atm, the difference is from 2 to 5.
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The change in pressure is 3 atm.
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It does not matter how you got there, same with volume and the same as temperature.
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These are state functions.
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What we want to do, we are going to find relations between the changes in state,
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the changes in these properties, and the change in energy.
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The first law was work and heat, now we want to get a little bit more direct.
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We want to express it in terms of things that the properties of the system itself, the temperature, the pressured, the volume.
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What the changes in does, the quantities and say about the change in energy.
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Let us find relations between changes in these properties and changes in energy.
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Again, that temperature no b a P, temperature, pressure, and volume are state functions.
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They are exact differentials, dT dP dV.
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These are exact differentials.
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We can start by assuming that energy is a function of temperature and pressure, temperature and volume, pressure and volume.
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We have to start anywhere we like and this is certain how it appears.
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What you are seeing us do here on a page, here is where mathematics starts.
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You are seeing the final result of this.
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You are not seeing the process the sort of let up to it.
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When a scientist sits down on a theoretical, he starts playing with mathematics.
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He does not necessarily know where he is going.
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He just starts playing with derivatives.
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He starts playing with equations and by starting to take derivatives and taking other derivatives, integrating and moving this here.
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You will see something that looks like it is important.
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Using the final result of that, you are not seeing the process.
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If it seems like we are pulling things out of the air, when we are doing these mathematics, we are not pulling things out of the air.
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What you are seeing is the final result of all the work that is been done.
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The truth is when somebody has done it for the first time, it is all over the place.
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You go this way, and this way.
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That is the real nature of science.
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You do not see that in your experience and from your books.
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It just seems like one day you are a scientist, woke up, and goes like that.
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It did not happen like that.
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Do not worry about it, it does not entirely make sense to you all at once.
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Let us start with this.
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We are going to let energy equal, we are going to take two variables, temperature and volume.
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We are going to say that energy is a function of temperature and volume.
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In other words, if I change the temperature, if I change the volume, or if I do both, the energy of the system changes.
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If I can express it this way, you get the following.
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I can express the differential change in energy this way, du = du dt vdt.
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Do not worry, I would explain everything here.
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DU dv, these are partial derivatives under constant V.
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This is constant temperature DV.
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This is called a total differential of this variable.
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You have a function which is a function of two variables.
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It is differential can be written like this, the total change in energy is equal, du = partial du dt × dt + du dv under constant temperature × dv.
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Let us talk about what all this actually means.
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You might want to go ahead and go back to the second lesson of the series, one that discusses partial differentiation
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and actually introduces this thing called the total differential.
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This is called the total differential for du.
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It is based on the presumption that is a function of two variables.
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First of all, we want you to notice that the differential of any state property which is an exact differential.
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Sorry if I keep repeating myself, it is important though.
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The differential can be written like this, it can be written in this form.
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Any state property that you deal with, if you know that state property is a function of two variables,
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you can write that state property the differential like this, always.
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This is a proofing, any state property, any exact differential can be written like this on the right hand side of the equality sign.
00:27:31.600 --> 00:27:33.100
The question is what does this mean?
00:27:33.100 --> 00:27:38.000
Let us examine what this means.
00:27:38.000 --> 00:27:43.400
What does this mean?
00:27:43.400 --> 00:27:47.800
What if you are faced with something like this, what if you are faced with some mathematical equation
00:27:47.800 --> 00:27:53.500
and you definitely want to give it a habit of stopping and examining what every single term means
00:27:53.500 --> 00:27:58.100
and what every single term says and you want to get physical meaning.
00:27:58.100 --> 00:28:04.300
In the case of du dt, this is the rate of change of energy with respect to temperature.
00:28:04.300 --> 00:28:09.200
In other words, as the temperature changes by 1 unit, how much does the energy change?
00:28:09.200 --> 00:28:10.600
This is a rate of change.
00:28:10.600 --> 00:28:13.400
This is a derivative, is it like a derivative like any other derivative.
00:28:13.400 --> 00:28:18.000
This happens to be partial derivative because we are dealing with a function of two variables instead of 1.
00:28:18.000 --> 00:28:20.600
This really is just the same as du dt.
00:28:20.600 --> 00:28:27.100
When you take that rate and you multiply it by the change in temperature, you get the total change in energy for the increment.
00:28:27.100 --> 00:28:32.800
The same thing here, this is the rate of change of energy per unit change of volume.
00:28:32.800 --> 00:28:38.500
Time and change in volume of the system experiences using a total change in energy.
00:28:38.500 --> 00:28:44.500
You knew you want to get in the habit of doing this, a lot of work that we do specially in thermodynamics.
00:28:44.500 --> 00:28:46.700
It is going to be strictly mathematical.
00:28:46.700 --> 00:28:52.400
Actually, throughout all of physical chemistry, quantum mechanics, spectroscopy, you need to be able to assign physical meaning.
00:28:52.400 --> 00:28:58.100
After you have done this a couple of times, pretty soon you are not intimidated by the mathematics anymore.
00:28:58.100 --> 00:29:01.000
It makes sense what is going on.
00:29:01.000 --> 00:29:05.300
There is nothing here that is counterintuitive.
00:29:05.300 --> 00:29:09.100
It seems that way simply because you just unaccustomed of the mathematics.
00:29:09.100 --> 00:29:11.300
You can get used to it like anything else.
00:29:11.300 --> 00:29:15.400
This is something new that we are going to be using with some rather sophisticated mathematics.
00:29:15.400 --> 00:29:19.400
But it is nothing that you do not understand, that you have not really been exposed to.
00:29:19.400 --> 00:29:26.400
What you have exposed to, we will go over but do not just pull away from it, assign physical meaning to it.
00:29:26.400 --> 00:29:33.000
Eventually, you will get them very quickly of knowing exactly what is going on and be able to relate this math
00:29:33.000 --> 00:29:40.000
to relate what is happening physically in a system, which way temperature is moving, which way a system as the surrounding.
00:29:40.000 --> 00:29:43.600
It says this way and that way, you will be able the see what is happening physically.
00:29:43.600 --> 00:29:51.800
When you see what is happening physically, that is when you understand what is happening mathematically.
00:29:51.800 --> 00:29:55.800
Let us see what this means.
00:29:55.800 --> 00:30:18.600
The first term du dt, du dt with this little subscript V on it, this is the rate of change of energy per unit change in temperature.
00:30:18.600 --> 00:30:28.700
That is it, it is just a rate, it is just a derivative per unit change in temperature.
00:30:28.700 --> 00:30:32.700
Look at the units, that is where I start.
00:30:32.700 --> 00:30:42.000
I look at the units to help it make physical sense for me when I see a derivative like this.
00:30:42.000 --> 00:30:45.800
We are talking about physical things, this is not just theoretical.
00:30:45.800 --> 00:30:54.400
Well energy is in joules, temperature is in Kelvin, and this is J/K.
00:30:54.400 --> 00:30:59.000
The V subscript that tells us that something is happening at a constant volume.
00:30:59.000 --> 00:31:07.400
In thermodynamics, any subscript let us know that it is happening as we keep that variable fixed.
00:31:07.400 --> 00:31:13.000
The rate of change of energy per unit change in temperature, just look at the units.
00:31:13.000 --> 00:31:29.600
Therefore, if T changes by an amount, if the temperature changes by some differential amount dt,
00:31:29.600 --> 00:31:44.000
and this du dt sub V × dt, it gives the energy change for the particular incremental change in temperature.
00:31:44.000 --> 00:31:57.500
It gives the energy change, holding V constant.
00:31:57.500 --> 00:32:03.900
Again just look at the units, du dt that is J/K.
00:32:03.900 --> 00:32:09.900
If you multiply that by a change in temperature, a temperature is in Kelvin, we are left with joules.
00:32:09.900 --> 00:32:17.100
You already notice this is heat capacity and you will see in a minute that it is.
00:32:17.100 --> 00:32:20.100
We are just interpreting what this is right now.
00:32:20.100 --> 00:32:24.100
We are just getting a sense of the mathematics.
00:32:24.100 --> 00:32:32.200
Attaching meaning to the partial derivatives, what they mean, we do not just want them to be scrolls and scribbles on a paper, that mean nothing to us.
00:32:32.200 --> 00:32:35.000
We know how to manipulate this so we have done calculus.
00:32:35.000 --> 00:32:36.600
We are very good at calculus.
00:32:36.600 --> 00:32:41.800
This is a simple calculus.
00:32:41.800 --> 00:32:51.800
For the next one, du dv sub T well this is exactly what you think it is.
00:32:51.800 --> 00:33:26.200
It is the rate of change of energy per unit change in volume at a constant temperature.
00:33:26.200 --> 00:33:37.100
Therefore, if the volume of the system changes by some differential amount,
00:33:37.100 --> 00:34:02.500
dv changes by an amount dv then this du/ dv sub T × dv.
00:34:02.500 --> 00:34:07.300
It gives the total energy change for the differential energy change, du.
00:34:07.300 --> 00:34:17.100
It gives the incremental energy change.
00:34:17.100 --> 00:34:28.200
Again look at the units, energy is in joules, volume always take deci³, dv deci³.
00:34:28.200 --> 00:34:36.600
Volume cancels volume, leaving you just energy, that is what is happening.
00:34:36.600 --> 00:34:59.000
Therefore, this expression du=du dt sub V dt + du dv sub T dv says the following.
00:34:59.000 --> 00:35:07.900
It is absolutely imperative that you do this, that you assign physical meaning, otherwise this stuff is going to get away from you very quickly.
00:35:07.900 --> 00:35:12.200
Because from this point on, it is going to be essentially mathematical.
00:35:12.200 --> 00:35:20.100
It is physical, we assign physical meaning to it but we are expressing these physical changes mathematically with partial derivatives.
00:35:20.100 --> 00:35:27.400
This expression says the following.
00:35:27.400 --> 00:36:00.700
If T changes by an amount dt and volume changes by an amount dv simultaneously, the amount dv
00:36:00.700 --> 00:36:09.700
then the energy change which is du as the sum of the two.
00:36:09.700 --> 00:36:12.600
This is totally intuitive.
00:36:12.600 --> 00:36:23.200
It is just as sum of the two, if I change the temperature, this amount, and the rate at
00:36:23.200 --> 00:36:30.200
which the energy changes per unit change in temperature, that is going to give me the total energy change for temperature movement.
00:36:30.200 --> 00:36:38.200
And if the volume changes also, that change in volume × the rate of change of the energy of the volume gives the energy change for the volume.
00:36:38.200 --> 00:36:41.200
If I add those two together because the total energy change, that is it.
00:36:41.200 --> 00:36:43.900
That is all this is saying, you know this already.
00:36:43.900 --> 00:36:46.100
When you know this intuitively, you know this since you were a kid.
00:36:46.100 --> 00:36:49.800
There is nothing here that is strange.
00:36:49.800 --> 00:36:56.300
It is just the symbolism that can look a little intimidating.
00:36:56.300 --> 00:37:11.400
Let us see here, if we happen to know what these partial derivatives are, in other words
00:37:11.400 --> 00:37:24.700
if we happen to know what du dt and du dv sub T are, we just integrate the expression.
00:37:24.700 --> 00:37:31.200
We just put into the differential because we just integrate the expression.
00:37:31.200 --> 00:37:58.000
We can just integrate this expression to get the total energy change.
00:37:58.000 --> 00:38:00.200
We have the du, we have the expression on the right.
00:38:00.200 --> 00:38:06.800
If we know what these partial derivatives are, we just put them in and integrate that function that gives us the total energy change.
00:38:06.800 --> 00:38:11.900
Nice and simple.
00:38:11.900 --> 00:38:31.900
Let us see if we can express these partial derivatives in terms of things that we know, in terms of things that we can measure.
00:38:31.900 --> 00:38:33.400
We know what they mean.
00:38:33.400 --> 00:38:37.600
Let us see if we can actually measure them somehow.
00:38:37.600 --> 00:39:02.200
Let us see if we can express these partial derivatives, this and this, in terms of things we can measure.
00:39:02.200 --> 00:39:22.300
Let us see, we have the du is equal to du dt sub v dt + du dv sub T dv.
00:39:22.300 --> 00:39:34.500
We also have the first law which says du=dq – dw which is equal to dq and dw is equal
00:39:34.500 --> 00:39:38.800
to the external pressure × the differential change in volume.
00:39:38.800 --> 00:39:42.600
Right, that is the definition of work, pressure × volume.
00:39:42.600 --> 00:39:54.000
I’m going to go ahead and put this expression over here on the left side so we get dq - P external dv
00:39:54.000 --> 00:40:07.500
is equal to the right side du dt sub v dt + du dv sub T dv.
00:40:07.500 --> 00:40:09.300
I'm going to ask that you actually confirm this.
00:40:09.300 --> 00:40:16.900
There are a lot of symbols that I’m writing on the page and all of the subscripts and letters so it does get to be a little confusing.
00:40:16.900 --> 00:40:23.600
Please make sure I’m actually writing this correctly.
00:40:23.600 --> 00:40:25.400
In any case, this is where we start.
00:40:25.400 --> 00:40:27.400
Let us see what we can do with this.
00:40:27.400 --> 00:40:34.700
We have the first law, we have the expression of the change in energy in terms of the properties temperature, volume,
00:40:34.700 --> 00:40:37.600
we sent them equal to each other and we are left with this.
00:40:37.600 --> 00:40:40.000
Let us see if we can make sense of this.
00:40:40.000 --> 00:40:41.900
This is the equation we start.
00:40:41.900 --> 00:40:47.300
I apply this equation to our changes of state.
00:40:47.300 --> 00:41:01.400
We apply this equation to various changes of state.
00:41:01.400 --> 00:41:08.700
Here is where it begins, various changes of state.
00:41:08.700 --> 00:41:23.800
The first change in state we are going to do is we are going to change of state under conditions of constant volume.
00:41:23.800 --> 00:41:27.700
A system goes from state 1 to state 2 but in that process, the volume stays constant.
00:41:27.700 --> 00:41:38.200
We have dealt with the volume change.
00:41:38.200 --> 00:41:52.900
If V remains constant and then this dv equal 0, V final V initial, the volume stays the same.
00:41:52.900 --> 00:41:57.300
There is no differential change in volume so dv = 0.
00:41:57.300 --> 00:42:20.500
Let me write this expression again, we have dq - P external dv = du dt sub V × dt + du dv sub T × dv.
00:42:20.500 --> 00:42:28.600
If dv = 0 then this goes to 0, that goes to 0, what we are left with is the following.
00:42:28.600 --> 00:42:49.400
Dq, we are going to go ahead and put that V there because we are under constant volume process, = du dt sub V dt.
00:42:49.400 --> 00:43:09.200
Let us see, it gets to U = dqv and we write the other equation also, du = dq - the external dv.
00:43:09.200 --> 00:43:19.200
This goes to 0 so we are also left with du = dqv.
00:43:19.200 --> 00:43:27.600
What we have here is the following, under constant volume, the change in energy of the system = the change in heat.
00:43:27.600 --> 00:43:35.100
It is equal the heat that is a lost by the surroundings or that heat gained by the system, if you want to take the systems point of view.
00:43:35.100 --> 00:43:38.700
Again, we are taking the surroundings point of view for the most part.
00:43:38.700 --> 00:43:46.800
For a constant volume processes, the heat that the surroundings loses happens to equal the change in energy.
00:43:46.800 --> 00:43:52.500
If I want to know what the change in energy is, I just have to measure how much heat the surroundings loses.
00:43:52.500 --> 00:44:01.500
Over here, change in heat, the amount of heat that the surroundings loses which is equal to the change in energy
00:44:01.500 --> 00:44:11.600
is equal to the rate of change of the energy of the temperature × the temperature increment.
00:44:11.600 --> 00:44:28.900
This equation which relates dqv which is the heat withdrawn from the surroundings, the heat lost by the surroundings.
00:44:28.900 --> 00:44:56.800
The heat withdrawn from the surroundings to an increase in temperature of the system which is the dt.
00:44:56.800 --> 00:45:11.600
Well, both of these are easily measurable, the change in temperature and the change in heat.
00:45:11.600 --> 00:45:17.600
We can measure the change in heat lost by this, we can measure the heat lost by the surroundings.
00:45:17.600 --> 00:45:21.700
We can measure the change in temperature of the system.
00:45:21.700 --> 00:45:32.700
The ratio which is dqv/ dt it is called heat capacity.
00:45:32.700 --> 00:45:37.900
The change and heat over a change in temperature, that is the definition of heat capacity.
00:45:37.900 --> 00:45:44.900
In this particular case, because it is a constant volume process, this ratio of the heat of the surroundings loses
00:45:44.900 --> 00:45:54.100
divided by the temperature by which the system, the temperature increase of the system,
00:45:54.100 --> 00:46:00.600
this is defined as the constant volume heat capacity.
00:46:00.600 --> 00:46:06.300
Constant volume heat capacity is defined as heat capacity in this particular case because
00:46:06.300 --> 00:46:15.900
the process is happening under constant volume it is called the constant volume incapacity.
00:46:15.900 --> 00:46:24.200
Dq sub V dt which is defined as a constant volume heat capacity C sub V.
00:46:24.200 --> 00:46:32.300
That happens to be associated, identified, with this partial derivative.
00:46:32.300 --> 00:46:38.800
A partial derivative which is the change in energy or change in temperature at constant volume.
00:46:38.800 --> 00:46:47.000
We were able to associate some partial derivative with something that we can easily measure,
00:46:47.000 --> 00:46:50.800
this P capacity which happens to be under constant volume.
00:46:50.800 --> 00:47:01.900
The heat capacity just happens to be the heat withdrawn from the surroundings divided by the change in temperature of the system.
00:47:01.900 --> 00:47:10.800
You are in chemistry, you are accustomed to thinking about it as a heat gain by the system divided by the temperature increase of the system.
00:47:10.800 --> 00:47:16.000
That heat change of the system divided by the temperature change of the system, that is fine, it is the same thing.
00:47:16.000 --> 00:47:18.500
Again, we are taking the surroundings point of view.
00:47:18.500 --> 00:47:21.200
This is the heat lost by the surroundings.
00:47:21.200 --> 00:47:24.200
It happens to be the heat gained by the system.
00:47:24.200 --> 00:47:27.700
It is just a question of point of view, as long as you know what is happening.
00:47:27.700 --> 00:47:33.400
That is the very definition of heat capacity, that is how we define it.
00:47:33.400 --> 00:47:36.400
Let us just go ahead and do what we do with differentials.
00:47:36.400 --> 00:47:48.800
We can just move this over here so dq sub V = cvdt.
00:47:48.800 --> 00:47:52.300
We just move that there.
00:47:52.300 --> 00:47:59.600
This is the differential form and we know this already.
00:47:59.600 --> 00:48:03.000
The change in heat is equal to the heat capacity × the change in temperature.
00:48:03.000 --> 00:48:08.400
We know this from general chemistry, constant volume.
00:48:08.400 --> 00:48:14.700
Let us go ahead and write the integrated form here.
00:48:14.700 --> 00:48:42.700
Our integrated form is, let me integrate that function, we get the energy = the integral from temperature 1, temperature 2, and CVDT.
00:48:42.700 --> 00:48:52.100
We said that dqv=cvdt, dqv is equal to du.
00:48:52.100 --> 00:48:56.100
That is what we get as the first law tells us when there is no change in volume.
00:48:56.100 --> 00:49:01.500
We just put this over here, we will get this thing and we integrate it.
00:49:01.500 --> 00:49:07.300
In other words, du is equal to cvdt and we integrate both sides.
00:49:07.300 --> 00:49:17.900
The integral of du is δ U, the integral is that integral from temperature 1 to temperature 2.
00:49:17.900 --> 00:49:34.200
From du= dqv, we also get this integrated version which is δ U= QV.
00:49:34.200 --> 00:49:42.100
Q is a path function, so there is no δ Q, the change in energy of the system happens to equal at a constant volume processes,
00:49:42.100 --> 00:49:49.200
the change in energy of the system happens to equal the heat withdrawn from the surroundings.
00:49:49.200 --> 00:49:57.000
If we are measuring temperature, the change in energy of the system happens to equal the integral of the change in temperature,
00:49:57.000 --> 00:50:03.300
the integral of the heat capacity of the system × the change in temperature from one temperature to the other.
00:50:03.300 --> 00:50:10.600
These are the equations that are important.
00:50:10.600 --> 00:50:46.700
If this is heat capacity happens to be, if T is constant over the range of temperature increase then we have δ U = CV δ T.
00:50:46.700 --> 00:50:52.600
In other words, over a certain temperature increase, let us say 25 to 50, if the heat capacity does not change.
00:50:52.600 --> 00:50:55.200
We do not need to integrate, we can just take the change in temperature.
00:50:55.200 --> 00:50:57.600
This is how we seen it before in general chemistry.
00:50:57.600 --> 00:51:03.300
Our assumption was that heat capacity is constant over a range of temperatures.
00:51:03.300 --> 00:51:06.000
As it turns out, heat capacities temperature dependent.
00:51:06.000 --> 00:51:09.700
The hotter something gets, the heat capacity changes.
00:51:09.700 --> 00:51:18.000
This is the wheel equation, that right there.
00:51:18.000 --> 00:51:34.500
These two equations, in other words, δ U= the integral from temperature 1 to temperature 2 of the constant volume heat capacity × that.
00:51:34.500 --> 00:51:43.700
The change in energy = the heat withdrawn from the surroundings, these two equations,
00:51:43.700 --> 00:52:02.200
they express the energy change of the system, in terms of measurable quantities.
00:52:02.200 --> 00:52:04.900
That is what we wanted.
00:52:04.900 --> 00:52:15.400
In terms of measurable quantities, that is good, that is what we want.
00:52:15.400 --> 00:52:17.500
In science, we measure things.
00:52:17.500 --> 00:52:19.700
We need to be able to measure things.
00:52:19.700 --> 00:52:26.400
The theory that we got from manipulating to get a partial derivatives, that tells us one thing
00:52:26.400 --> 00:52:30.500
but we need to associate these partial derivatives as something we can measure.
00:52:30.500 --> 00:52:40.000
We have something that dudt V this, we can associate it with the heat capacity.
00:52:40.000 --> 00:52:42.400
Heat capacity is easily measurable.
00:52:42.400 --> 00:52:48.600
We measure the heat lost by the surroundings or the heat gained by the system and we divide by the temperature change in the system.
00:52:48.600 --> 00:52:54.300
We have something mathematical that is associated with something physical, something we could measure.
00:52:54.300 --> 00:52:58.700
This is beautiful, it is what we want.
00:52:58.700 --> 00:53:18.400
Mathematically, δ U = QV.
00:53:18.400 --> 00:53:26.200
Δu QV they have the same sign.
00:53:26.200 --> 00:53:28.900
They have the same sign.
00:53:28.900 --> 00:53:50.800
Now given our particular convention, a positive value of heat means that heat is flowing from the surroundings to the system.
00:53:50.800 --> 00:54:15.600
Heat flows from the surroundings, in other words through the system.
00:54:15.600 --> 00:54:23.200
If heat is positive and change in energy is positive.
00:54:23.200 --> 00:54:32.400
A positive QV implies that δ U of the system, this is what all these means.
00:54:32.400 --> 00:54:37.700
I’m just explicitly writing at this time in general.
00:54:37.700 --> 00:54:43.900
It means that δ U is positive.
00:54:43.900 --> 00:54:55.500
That means the system has increase in the energy of the system.
00:54:55.500 --> 00:55:02.600
A negative QV means that heat is flowing to the surroundings.
00:55:02.600 --> 00:55:04.800
In other words, the heat is flowing from the system.
00:55:04.800 --> 00:55:11.100
The system is losing heat, therefore, the energy of the system decreases.
00:55:11.100 --> 00:55:14.300
Δ U is negative.
00:55:14.300 --> 00:55:17.300
If δ U is Q, Q is positive W is positive.
00:55:17.300 --> 00:55:21.000
If Q is negative δ U is negative.
00:55:21.000 --> 00:55:30.000
This is a decrease in energy.
00:55:30.000 --> 00:55:39.400
The constant of volume heat capacity is always greater than 0.
00:55:39.400 --> 00:55:48.400
A positive temperature change implies that δ U is positive.
00:55:48.400 --> 00:55:58.100
It is an increase in energy, a drop in temperature implies that the δ U is negative.
00:55:58.100 --> 00:56:01.300
If the temperature of the system increases, the energy of the system increases.
00:56:01.300 --> 00:56:05.000
If the temperature of the system decreases, the energy of the system decreases.
00:56:05.000 --> 00:56:07.200
That is what is going on here.
00:56:07.200 --> 00:56:09.300
This is actually really important.
00:56:09.300 --> 00:56:37.900
Therefore, at a constant volume, at constant V the temperature is a direct representation of the energy of the system.
00:56:37.900 --> 00:56:40.300
Let us see what temperature actually is.
00:56:40.300 --> 00:56:47.100
Temperature is a measure of the energy, the average energy of the system.
00:56:47.100 --> 00:56:50.000
OK let us go ahead and do an example.
00:56:50.000 --> 00:56:52.500
It is a very simple example.
00:56:52.500 --> 00:56:56.800
A lot of these lectures I’m only going to have a lot of these preliminary lectures.
00:56:56.800 --> 00:57:03.600
I already have 1 or 2 examples, do not worry about that.
00:57:03.600 --> 00:57:09.200
In subsequent lessons and for several lessons, after we get these preliminary theoretical discussions out of the way,
00:57:09.200 --> 00:57:12.000
that is where we are going to do the bulk of our examples.
00:57:12.000 --> 00:57:16.800
It is not going to be a lesson where we have a little bit of discussion and a handful of examples.
00:57:16.800 --> 00:57:26.000
I’m going to set aside complete lessons, several of them at the end of this unit, to do all of the example problems that we need.
00:57:26.000 --> 00:57:29.600
Do not worry there is only 1 or 2 showing up in these particular lectures.
00:57:29.600 --> 00:57:33.900
We are going to do a lot and when I say a lot I mean a lot.
00:57:33.900 --> 00:57:36.100
We definitely need to get familiar with this material.
00:57:36.100 --> 00:57:39.400
Handling the first law, we need to set a good foundation.
00:57:39.400 --> 00:57:44.200
We are going to do a lot of problems, I promise you.
00:57:44.200 --> 00:57:46.300
Let us take a look.
00:57:46.300 --> 00:57:56.800
The first example we have, 2 mol helium gas, they are taken from a temperature of 25° C to a temperature of 55° C.
00:57:56.800 --> 00:58:04.700
The molar heat capacity happens to be 3 ½ R, notice molar heat capacity.
00:58:04.700 --> 00:58:11.400
This is the amount of heat J/ K/ mol.
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Molar heat capacity, if I just said heat capacity it would be J/ K/ C.
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If I say specific heat capacity, we are accustomed to in general chemistry it would be J/ K/g or J/℃/ g,
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because specific heat does talk about mass, lower heat capacity, J/ mol/ K/ mol.
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We want to find the change in energy of the system and we want to find the heat that is lost by the surroundings for the transformation.
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We know that δ U is equal to the integral from temperature 1 to temperature 2 of CVDT, we know that already.
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We also happen to know that δ U is equal to QV.
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We find this and we find that.
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Let us go ahead and work this one out, it is really simple.
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Δ U is equal to the integral of T1 to T2, the CV or constant volume heat capacity is 3 ½ rdt.
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The 3 ½ R is not a function of T, it is a constant so we can pull it out, = 3 ½ R × the integral of T1T2 dt.
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That = 3 ½ R × δ T, notice I have not put the values in.
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I may have to change this to 298 and change the 55 to whatever it is, 55 goes to 73.
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In this particular case, because this is constant I do not to have to evaluate the integral.
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This right here becomes R δ T, temperature is a state function.
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The integral of the state function is δ of the T.
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This is really simple, let us just go ahead and put the values in.
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We get 3 ½ × R, the R value we are going to take is 8.314 J/ mol/ K.
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The change in temperature is going to be the 55 - to 25, so 55° - 25°.
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The δ T, in terms of Kelvin and Celsius is the same.
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A difference of 1° C is the same as 1° K.
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I do not actually have to convert to it to K when I’m doing δ.
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This is going to be 30 K, K and K cancels and I end up with is, if I did my arithmetic correctly which more often that I do not, 374 J/ mol.
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We are left with J/ mol.
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Since we have 2 mol of helium, we multiply that and we get a total of 748 J.
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748 J that is the change in energy.
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So δ U= 748 J.
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QV happens to equal that, QV = 748 J.
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Under these circumstances of constant volume, this 2 mol of helium gas that goes for 25° to 55° C, the energy change was 748 J.
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The energy of the system, the heat lost by the surroundings is 748 J.
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The 748 J went from the surroundings to the system, that is what happened.
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Let us go ahead and close this out.
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The important equations for this particular lesson, we have the heat capacity which is defined as dq/ dt,
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the change in heat over the change in temperature.
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The heat lost by the surroundings divided by the temperature increase of the system or
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the heat gain by the system divided by the temperature increase of the system.
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Totally your choice, as long as you are consistent.
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This happens to equal, it is associated with the derivative, this partial derivative,
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the change in energy or the change in temperature at constant volume.
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Du = CVDT, du = dq, δ U + integral from T1 to T2 of CVDT, this is just the differential form, this is the integrated form.
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Δ U = Q not dq.
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At constant volume, the heat that is lost by the surroundings = the change in energy of the system.
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If I want to know what the change in energy of the system is, all I have to do is measure how much heat is lost by the surroundings.
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And that is what I do, I do not measure directly into the system.
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I measure what is happening in the surroundings which is why we keep saying surroundings.
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I know that in chemistry we are accustomed to thinking about the system but where we can take our measurements
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in order to find what was happening in the system is in the surroundings.
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The change in energy of the system is equal to at under constant volume heat capacity ×
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the differential change in temperature for the particular increment.
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If I integrate all those increments over the temperature change, I get the change in energy of the system.
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Thank you so much for joining us here at www.educator.com.
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We will see you next time for a continuation of discussion, bye.