WEBVTT chemistry/physical-chemistry/hovasapian
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Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.
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In the last lesson we started our discussion of entropy δS.
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Before we actually talk about entropy and investigate how it behaves, I want to take a little bit of a break and discuss a little bit of mathematics.
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The mathematics that I’m going to discuss here is a continuation of some of the mathematics that we started with, the discussion of partial differentiation.
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This is going to be more technique in partial differentiation and it is a series of techniques that we are going to be using absolutely all the time,
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not only for the rest of thermodynamics but once we get into quantum mechanics and spectroscopy and things like that.
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Let us go ahead and get started.
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Let us just go ahead and start with an example here.
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Let F be a function of xy and we will let it equal x² y³.
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Let us go ahead and form some partial derivatives here.
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δF/δx is going to = 2xy³.
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Over on this side, we will form δF/δy that is going to = 3x² y².
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Let us go ahead and form the next of set of partials, the second partials.
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This is a function of x and y. I can take δF/δx of this and I can take δF/δy of this.
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I can take δF/δx of this and I can take δF/δy of this.
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Let us go ahead and do that.
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Let us start off with δx and I will notate it this way.
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I will do δ/δx of δF/δx which is the same as δ² F δx² that is going to = 2y³.
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I will form y, I will form δ/δy of the δF/δx which = δ² F δy/δx this notation that is going to = 6xy².
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I come over here and I’m going to form δ/δx of the δF/δy which is notated as δ² F δx/δy that is going to = 6xy².
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I will do δ/δy of the δF/δy which is δ ⁺2F δy² and that is going to equal 6x² y.
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Notice, 6xy² 6xy² δ² F δy/δx δ² F δx/δy are mixed partials in different orders they actually equal each other.
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δ² F δy/δx = δ² F δx/δy this is not an accident.
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This is not a coincidence, this is generally true.
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Let F again = RF of x and y begin from the total differential of this.
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The total differential is δF = δF/δx holding y constant × δx + δF/δy holding x constant × δy.
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What we saw above, in other word this thing here is true in general.
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If F(xy) is a function and δF/δx δF/δy δ² F δy/δx and δ² F δx/δy exist, in other words if the derivatives of the function exist and
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are continuous ... which in our case there always going to be continuous, then like we said δ² F δy/δx = δ² F δx/δy.
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In other words, the cross partials or the mixed partials are always going to be equal.
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If you are given a function of two variables, three variables, let us just stick with two variables for the time being.
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If we are given a function F of two variables and it is well defined and continuous and if the derivatives, the first derivatives and the second derivatives,
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if they actually exist or continuous then the mixed partials will always be equal to each other.
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Mixed partials are equal.
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For the total differential that we wrote which is δF = δF/δx holding y constant × δx + δF/δy holding x constant × δy, this means this is δF/δx.
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If I take the derivative of this with respect to y and if I take the derivative of this with respect to x, they are going to be equal.
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For the total differential, the derivative of this with respect to the other variable equals the derivative of this with respect to the other variable.
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This means that δ/δy of δF/δx = δ/δx δF/δy.
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Given a total differential expression derived from F = F(xy) from an actual function, given a function and you write down the total differential,
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from that total differential the partial of the differential coefficients are this thing and this thing.
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The top differential coefficients with respect to the other variable are equal.
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It is up here, this blue, the derivative of this expression with respect to this variable = the derivative of this expression with respect to that variable.
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Straight up, so outside and inside.
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The derivative of this with respect to this variable = the derivative of this with respect to this variable or you can just look over here.
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The derivative of this with respect to y = the derivative of this with respect to x, the other variable.
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That is all that is going on here.
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This is always true.
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Let us go ahead and do an example here.
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A thermodynamic example, we know from the first law that δU = δQ reversible - δW this is the definition of the first law of thermodynamics.
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We also have the definition of entropy δS = δQ reversible/ T.
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I will just multiply by T what I get is T δS = δQ reversible.
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When we take this, we are just here and plug it into here I get the following.
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I get δU = T δS – δW.
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You remember δW = P × δV so what I get is δU = T δS – P δV.
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You have a total differential expression right there. δU = something × the differential S - something × that.
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U is a function of S and V. So I have a differential expression which means that a partial of this with respect to this variable =
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the partial of this with respect to that variable.
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Now I can go ahead and write that.
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Let me write this again on top of the same page, I have δU = T δS – P δV.
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Therefore, a derivative of this with respect to V… So δT δV holding that constant = the derivative this with respect to that... δP δS,
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holding that constant, this is negative.
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That is that. This is a very interesting relation just from the fact that mixed partials are equal.
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I have the total differential expression, I will automatically know that the differential this with respect to this variable = the differential of this
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with respect to this variable holding the other variable constant.
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So I have something like this.
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This relationship this is an example, this relationship is one of a very important set of other relations, they are called Maxwell's relations.
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This is the same Maxwell from electromagnetism.
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This is not Maxwell’s equations, these are Maxwell thermodynamic relations, he did lot of work in a lot of fields.
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These relations are one of very important set of relations called Maxwell's relations...among the variables of state of the system.
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In this particular case, the variables involved are temperature, volume, pressure, and entropy.
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We will come back to these but for right now this is just an example to show you that this applies to thermodynamics,
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this mathematical theorem, but we will come back to Maxwell's relations later on when we discuss the free energy.
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A function F(xy)…
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So a state function F(xy) generates an exact total differential expression, δF = (δF/δx)y δx + (δF/δy)x δy
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which guarantees that the mix partials are equal.
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That takes care of the first part.
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What is interesting is the converse is also true.
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The converse is also true, converse is also true.
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In other words, if we are given the total differential expression such as let us say δF = P δx + Q δy then if δP δy = δQ δx
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then δF is exact and there exists an actual function F which is a function of the variables x and y.
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If we are given the function, we write the total differential and the mix partials are equal.
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If we are given a total differential expression and if we happen to take the mix partials, if the mixed partials are equal
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then the actual function that gave rise to the differential expression that we wrote actually does exist.
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In other words, first we are going from function we are differentiating down, now we are actually given the differential can we integrate backup.
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That is what this is saying.
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The only thing that I need to check is if the mix partials are equal, the δP δy= δQ δx,
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then theoretically I can actually integrate this function and recover some function of x and y.
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It goes in both ways.
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There exist an actual F.
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δQ and δQ for example, are examples of,
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I will have to say a little bit more here.
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If this δP/ δy if it does not equal the δQ/ δx, in other words if I'm given a total differential expression, I take the mixed partials,
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I take the partials and they do not equal each other then there is no guarantee that such a function F(xy) exists.
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In this particular case, δF is inexact.
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δQ and δW are examples of inexact differentials.
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For exact, we have the following.
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In going from S1 to S2, U is exact δU=δ U which is U2 – U1.
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If I go from S1 to S2 and back to S1, the cyclic of an exact differential = 0.
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for inexact, if I go from S1 state 1 to state 2, the integral for example work it just equal the work.
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It does not = δ work just the work.
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If I do a cyclic S1 to S2 and if I come back to S1, the cyclic integral of an inexact differential does not = 0.
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Let us see what else we can say, now let us talk about something called the cyclic rules.
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Mixed partial is our first technique, now we are going to talk about another thing called the cyclic rule.
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Let me move to black here.
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There is another nice relation among the partial derivatives of a given function.
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This one was actually quite beautiful and it is called the cyclic rule.
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Let us start off, we will let z = z (xy).
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z is a function of x and y, let us go over here.
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For various values of x and y, I simply calculate z just the function of two variables.
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z is basically like the third variable, it is the dependent variable.
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Generally, we do things like F(xy) but we can say z(xy) because when I put an x and y and I did a particular calculation
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it is going to spit out some number, that number we call z.
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I have x and y which are independent variables and z is the dependent variable.
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I have three variables.
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z is a function of x and y so I have a total differential expression.
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I can write δ(z) = δz/δx holding y constant × δx + δz/δy holding x constant × δy.
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If I restrict my choices of x and y such that z never changes and I can do that, I can choose x and y, z is a function of x and y.
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I can choose x and y such as z stays the same, it never changes.
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In other words, δz = 0 then 0 = δz/δx y δx + δz/δy (x δy).
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I’m going to make a little bit of change here because I'm actually setting z as constant δz = 0.
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I can write this as δx holding z constant + δz/δyx and I can write this as δy holding z constant.
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I’m going to go ahead and divide by this term right here, divide by the δy sub z.
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What I end up getting is 0 = δz/δxy δx/δy z + δz/δy x.
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I'm going to multiply by the reciprocal of that term.
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When I multiply by the reciprocal everything I get 0 = δz/δx sub y × δx/δy sub z × δy/δz sub x + 1.
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I’m going to rearrange and I’m going to move this over to the other side and I have the following.
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I have δz/δx holding y constant × δx/δy holding z constant × δy/δz holding x constant = -1.
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This is the cyclic rule.
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If I have some function which is a function of x and y, z is a function of x and y, there is a relationship that exists among the partials.
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The partial of z with respect to x holding y constant × a partial of x with respect to y holding the z constant × a partial of y
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with respect to z holding x constant is always going to = -1.
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Notice the relationship among the variables z x y, x y z, y z x, all three are represented in each case.
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The zx xy yz, that is the best way to think about it.
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Basically, what you can do if you want, you can just write the first variable, the xyz and
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you can write the other variables in any combinations underneath that, that just do not repeat it.
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You can write it as, y z and x and just go ahead and do something like this.
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δx/δy δy/δz δz/δx xyz is constant yz x is constant, zx y is constant and it is always = -1.
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That is the relationship that exists between the x, the y, and z.
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Let us take a look at an example.
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Example 2, let be the three variables be temperature, pressure, and volume be our three variables.
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It make some difference which will be expressed as a function of the other two because T is function of P and V,
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P is a function of T and V, V is a function of P and T.
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It is just we arranging the equation, it is not a big deal.
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We immediately write down the relations so let us do it this way.
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Let us write the T, P, R.
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Let us write P and T, V and I will pick T here, I will pick V here, and I will pick P here, this is PT V, TV P, VP T.
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What I have the following.
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I have δP δT holding V constant × δT δV holding P constant × δV δP holding T constant = -1.
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I have a beautiful relationship.
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The rate of change of pressure with respect to temperature under constant volume × the rate of change of temperature
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with respect to volume under constant pressure × the rate of change in volume with respect to pressure under the change in temperature = -1.
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This relation is valid.
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The cyclic rule tells me that it is valid.
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I have a relationship between T P and V, PV = NRT or PV = RT there is a relationship that exists between these.
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I can go ahead and just flat out just write down the relationship that exists between the partial derivatives because I know the cyclic rule is true.
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Let us go ahead and rearrange to make it a little bit nicer.
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Recall, we said α =1/ V × δV / δT under constant pressure.
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I'm going to go ahead and move the V there so I get δV δT under constant pressure = α V.
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I have kappa the coefficient of compressibility -1/ V × δV δP under constant temperature.
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I’m going to move it there so I get δV/ δP under constant temperature = - kappa × volume.
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This thing right here becomes this δP δT under constant volume.
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δT / δV I have δV / δV δT = α × volume this is just the reciprocal.
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δT this is going to be × 1/ α × volume.
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δV δP= - KV that is = -1.
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I just put this and this into here appropriately and the V cancel and I'm left with δP δT V × - kappa/ α = -1.
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Therefore, δP/ δT the constant V = α/ kappa.
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in other words, if I keep the volume constant for every unit increase in temperature or unit change in temperature,
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the change in pressure is α/ kappa, the coefficient of thermal expansion divided by the coefficient of compressibility.
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I have this amazing relationship and these are tabulated for a lot of things or if they are not, they are easily calculated.
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I have this amazing relationship that exists simply by virtue of the cyclic rule.
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This is just an example of one application of the cyclic rule.
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We will see other examples and other applications.
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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 the discussion back to entropy.
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Thank you, bye.