WEBVTT chemistry/physical-chemistry/hovasapian
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Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.
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Today, we are going to start our example problem sets for free energy.
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let us just jump right on in.
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Our first example it says given that G of TP = G super 0 T + nRT + LN T for an ideal gas,
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use the fundamental equations and the definitions of the composite functions to derive functions for S, V, H, and U.
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This is a really important problem.
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It is a more important problem theoretically than is practically but I think it just depends on the particular situation you run across.
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Remember, we said when we are talking about free energy in the last lesson that if you know the functional form of the free energy expression,
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If you know what the function looks like, in this case we have this,
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that we can derive all the other thermodynamic functions from this using the basic equations that we had at our disposal.
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From this expression for free energy, we can derive the entropy, the volume, the enthalpy, and the actual energy of the system.
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Again, this is really great way to play with the mathematics to get a sense of what is going on and just become comfortable with it.
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A lot of the problems in the problem sets for free energy are going to involve a lot of mathematical manipulation.
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The manipulations themselves are not difficult but they are mathematical manipulations.
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Again, this is just the way of getting you comfortable with handling the equations
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not just seeing them on a page but actually using them to derive other relations.
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None of them are going to be very long, it is just you have to be very careful because of all mathematics there is going to be a lot of symbolism on the page.
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It is just a question of busyness that is going to be the biggest problem here.
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In any case, let us go ahead and jump right on in.
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Let us go ahead and rewrite this.
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We have G of TP = Z0 of T + nRT nat log of P.
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This G with a superscript 0, this is just a standard free energy.
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My free energy at a given temperature and pressure is going to equal the standard free energy + nRT ×
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log of the new pressure whatever the pressure happens to be that is raised to 2 from the 1 atm.
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Using the fundamental equations of the definitions of a composite function,
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our fundamental equations for G is this.
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We had DG = - S DT + V DP so we know that DG DT,
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That made us slow down a little bit just to make sure we have all of our variables correctly, is equal to –S.
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We know this already from that or we can just read it out from here, this is the total differential expression.
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This is going to be the DG DT holding this variable constant.
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From the equation above, from this equation, if I actually take DG DT holding P constant,
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If I take the derivative of this directly I’m going to end up with DG DT.
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This is just a function of temperature so it is a regular derivative not a partial derivative +
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the derivative of this with respect to T holding P constant is going to be nR LN P.
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From the fundamental equation I have this, from taking a derivative of this equation which is the equation they gave us I have this.
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I’m going to set these equal to each other.
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I have - S = D 0 DT + nR LN P but this, that, this thing right here this DG DT is just - S for the standard conditions.
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DG DT under constant P = - S so DG standard DT is just - S standard, that is all it is.
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We get - S = - S standard + nR LN P.
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I can go ahead and I can leave it like that or I just can go ahead and rearrange things, multiply by -1.
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S = S standard - nR LN P, there we go.
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From this equation which is the equation for the free energy of the system,
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I was able to do derive an equation for the entropy of the system.
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It is equal to the standard entropy - nR LN P, that is what we are doing here.
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That takes care of the S.
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Let us go ahead and let us see what is next.
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Once again, let me go ahead and write the equation DG = - S DT + V DP.
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I know that DG DP under constant T = volume.
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Let us see from the equation of the equation that they gave us, the functional form of the equation which is this one,
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+ nRT LN P, if I take the derivative directly, let me see if I do the DG DP under constant T.
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If I take the derivative of this holding temperature constant and take it with respect to pressure,
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I get nRT × the derivative of LN P which is 1/ P.
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I get nRT/ P.
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I have this equal to nRT/ P from the fundamental equation I have the same thing is equal to V.
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I will just go ahead and set them equal to each other so I get V = nRT/ P which we knew already.
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We actually ended deriving it from the functional form of the free energy as a function of temperature and pressure V= nRT/ P.
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This is just PV = nRT, this is the ideal gas law, this confirms it.
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I have derived a thermodynamic property of volume from the free energy using these other equations that I have.
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Using this equation and this equation.
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Let us see what we can do, we have taken care of the S, we have taken care of the V, let us see if we can do something for the enthalpy.
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The composite functions let us just go ahead and recap our definition for composite functions.
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The enthalpy = the energy + PV.
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Our Helmholtz energy which does not concern us in this particular problem but I will go ahead and write it down anyway as the definition.
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G = DU + PV - TS which is equal to G= H – TS.
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These are the definitions of the composite functions, enthalpy A this.
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Let us go ahead and deal with this one because again they gave us G in the problem and now we want to find H.
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Let us go ahead and take this, bring it over here.
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We will rearrange this and we will write H = G + TS.
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H = G is just G is this thing so we are going to write G of T + nRT LN P + T × S.
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H = G standard of T + nRT LN P + T × we already know what S is, we just found it.
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In the previous page that is just is equal to S standard of T – nR LN P.
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I get H = G standard of T + nRT LN P + T S standard T – nRT LN P.
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That cancels with that and I am left with the enthalpy of my system = the standard free energy of the system + the temperature × the standard entropy.
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This is interesting.
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I have G standard + T S standard.
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You remember the definition of H was just G + TS.
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G standard + T S standard is nothing more than H standard.
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This is my equation, this is what I was looking for.
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I was able to derive the functional value for the enthalpy of the system based strictly on the fact that
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I was given the functional form of the free energy of the system and ends up being the standard free energy +
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the temperature I happen to be at that moment × the standard entropy.
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This just happens to equal the standard enthalpy.
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In other words, the enthalpy of the system under standard conditions.
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Again, because H = G + TS that is the definition of a composite function, we just rearranged.
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H standard = G standard + S standard.
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This is the functional form but it just happens to equal that.
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Let us go ahead and see if we can derive something for U.
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They also want U.
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We have H = U + pressure × volume so U = enthalpy - pressure × volume.
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U = enthalpy – we are dealing with an ideal gas so it is going to be H – nRT.
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That is kind of interesting.
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Actually, we want it, here we go.
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U = H – nRT, we already found H.
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H is this nRT so you already found that.
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Here which happens to equal the energy of the system under standard conditions because U = H – nRT.
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Therefore, U standard = H standard – nRT.
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Therefore, let us see, I got U = H – nRT.
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Under standard conditions, the H happens to equal H standard so this is just H.
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Basically, I have got from here, I can stop here because I already got everything else.
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If I want to continue, U = H – nRT, H = H standard nRT.
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Therefore, U = H standard - nRT but that is equal to U standard.
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Everything is falling into place nicely here.
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U = H which is G standard T + TS standard T – nRT.
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I can use the short version of it or I can use this version of it which is expressed in terms of the actual standard free energy.
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Given the functional form of the standard free energy I was able to derive expressions for the entropy.
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I was able to find the entropy, the change in entropy.
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I was able to find the volume of the system.
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I was able to find the change in enthalpy of the system.
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I was able to find the change in energy of the system strictly from having the functional form of the free energy.
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This procedure is the same, this is the general procedure, it will always work.
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If you are given the functional form for free energy, what we did here will always allow you to derive the entropy change,
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the enthalpy change, the volume and the energy change.
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Using the thermodynamic equation of state and the Van Der Waals equation find DU DV sub T for the Van Der Waals gas.
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DU DV sub T for the Van Der Waals gas, not the ideal gas.
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Let us go ahead and let us write the thermodynamic equation of state.
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In this particular case, we have got P = T × DP DT constant volume - DU DV T.
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I’m going to rearrange this because now I'm looking for that.
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I have got DU DV sub T = T × DP DT under constant volume – P.
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The Van Der Waals equation P = nRT/ V - nV - AN²/ V².
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Therefore, I have got this and I have got this so I'm going to find this DP DT.
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I have P so now I’m going to take DP DT and I’m going to hold volume constant.
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When I do that, when I take the derivative of this expression holding volume constant,
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I'm going to end up with derivative with respect to T, it is just going to be this.
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I'm going to get nR/ V –NB.
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Again, constant and constant, NV is a constant so the derivative ends up going to 0.
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I take this value and I plug it into there so I get DU DV constant T = T × nR/ V - NB – P.
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P is just this, the whole thing, just put back in here - nRT / V - NB – N² A/ V².
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That is going to equal nRT/ V - NB - nRT/ V - NB + N² A/ V².
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That cancels and what I’m left with is DU DV constant T for the Van Der Waals gas = N² × A/ V², this is what we wanted.
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Notice, this does not equal to 0 like the ideal gas.
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Recall, when we had DU = CV DT + DU DV under constant T DV, this is the expression for the differential change in energy of the system.
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It is going to be a function of temperature and the function of volume.
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The DU DV, the Joules law for an ideal gas, this DU DV sub T that was equal to 0.
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Because it was equal to 0, this term went away and the energy just became a function of temperature based on the constant volume heat capacity.
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That was for an ideal gas, in this particular case this is a Van Der Waals gas.
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For the Van Der Waals gas, this is no longer equal to 0.
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When we use this equation for a Van Der Waals gas we actually have to include it.
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When we integrate this, we have to integrate this term and we have to integrate this term, in order to find our δ U, our change in energy.
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δ U is no longer just DU = the integral of CV DT from T1 to T2.
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For a Van Der Waals gas, now you have δ U = the integral of this one which is CV DT from T1 to T2 +
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the integral from volume 1 to volume 2 of this Du DV which in this case is N² A² / V² DV.
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If we have an isothermal process for a Van Der Waals gas, isothermal means that this term can go to 0.
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But now, the change in energy is not equal to 0.
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For an ideal gas and isothermal process involves that there is no change in energy and it implies that δ U = 0 but only for an ideal gas.
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Do not just automatically assume that just because we are talking about an isothermal process, the δ U = 0, it does not.
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For an ideal gas and isothermal process this is 0 and this is 0 so δ U = 0.
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Not for a Van Der Waals gas, for Van Der Waals you have to evaluate this integral right here.
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An isothermal process implies that δ U = 0 only for an ideal gas.
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That thing is, you get so many problems and questions that involve ideal gases.
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A lot of students automatically equate isothermal with δ U = 0, that is not the case.
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It is never the case.
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This is the equation you want to know.
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Let the equation tell you what is what.
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Are you dealing with an ideal gas, solid, liquid, Van Der Waals gas, are you dealing with a different kind of gas, whatever it is.
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Let us go ahead and take a look at example 3.
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Calculate δ U for the isothermal expansion of 1 mol of nitrogen as a Van Der Waals gas from 10 dm³/ mol to 100 dm³/ mol.
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It should be just 1 dm³ to 10 dm³.
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Let us see what we have got, 10 to 100.
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A, in this particular case is 0.141 so let us see what we have.
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We always start with our basic equation, those are the ones that we want to use.
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Those are the one that tell us what we are doing.
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DU = CV DT + DU DV at constant T DV.
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This is an isothermal process so isothermal means the DT is 0 so we can take that to 0.
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We have got DU = this thing, DU now equal this thing, that is what we are going to integrate.
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We got δ U = the integral from v1 to v2 of N² A/ V² DV = N² A × the integral DV/ V from v1 to v2.
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Let us put our v², there we go.
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δ U = 1 mol, this is 1 × A is 0.141 × the integral, 10 × 10⁻³ to 100 × 10⁻³.
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The reason it is 10 dm³ to 100 dm³, A is expressed in Pascal’s.
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When you are working with Pascal’s you have to work in m³.
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Recall, volume must be in m³ when working in Pascal.
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Again, it is pesky and it is annoying but you have to keep track of your units.
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Take a look at what units you are actually given in the case of the Van Der Waals gas,
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this particular unit is 0.141 using the Pascal and meters so it has to be in cubic meters.
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The volume was given cubic decimeters so you have to convert to m³.
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10 dm³ is 10 × -3 m³ when working with Pascal.
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It is going to be the DV/ V² so we end up with δ U = 1² × 0.141 × 1/ V.
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It is minus because this is going to be when we integrate this, the negative sign is going to be 1/ V -1/ V.
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I'm just going to bring the negative over here 10 × 10⁻³ to 100 × 10⁻³.
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When I evaluate this, I get δ U = 12.69 J.
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An isothermal expansion of 1 mol of nitrogen as a Van Der Waals gas from 10 dm³ to 100 dm³ isothermally you end up with 12.69 J as your δ U.
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Van Der Waals gas δ U is not equal to 0, it is equal to 12.69.
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Let us see what we have got.
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Derive an expression for the change in entropy with respect to volume under a constant temperature for a Van Der Waals gas.
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Find an expression for δ S for the isothermal expansion of the Van Der Waals gas from v1 to v2.
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Comparing the expression in part B with the analogous expression for the ideal gas, for the same increase in volume,
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will the change in entropy be greater for the Van Der Waals gas or for the ideal gas?
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Let us see what we can do.
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Part A, I’m going to do this in red.
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Part A, the equation for the Van Der Waals gas is nRT/ V – NB.
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Let us make V a little bit better, - N² A/ V².
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From one of Maxwell's relations said that DS DV at T = DP DT under constant V.
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We are looking for DS DV and Maxwell's relations said the DS DV = the change in pressure with respect to temperature holding the volume constant.
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I have the equation, the Van Der Waals equation.
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If I hold the volume constant and if I take the derivative of P with respect to T this is just equal to nR/ V – NB.
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There you go, that is my first part.
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The rate of change of entropy as I change the volume for a Van Der Waals gas is nR/V – NB.
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That is all, nothing strange is happening here.
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It is just mathematics, simple derivatives.
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Now, they want us to find an expression for the δ S.
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Let us start with our basic equation again for DS = CV/ T DT + DS DV under constant T DV.
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They say that this process is isothermal.
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Isothermal so that goes to 0, DT is 0.
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All I have is DS = this thing.
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For the Van Der Waals gas, I just derived this thing, it is nR/ V – NB.
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It was equal to nR/ V - NB × DV.
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I just integrate this and I get δ S = nR × integral from v1 to v2 of DV V – NB.
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I get nR × log of V -NB from V1 to V2.
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I get my expression, I get nR × the LN of V2 – NB.
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They just want the expression over v1 – NB.
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For Van Der Waals gas, my δ S is this based on the fact that the DS DV is this.
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That comes from this, the basic equation.
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They want us to compare this with that for the ideal gas.
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Let us go ahead and do that.
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We can go ahead and just recall what the expression is for the ideal gas or we can go ahead and derive it.
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Let us go ahead and derive it because I think it is always good to recall the things.
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We cannot remember everything that is going on.
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Maxwell's relations says the DS DV under constant T = DP DT under constant V.
00:30:52.900 --> 00:30:58.000
For an ideal gas P = nRT/ V.
00:30:58.000 --> 00:31:12.400
Therefore, this DP DT under constant V this thing is just equal to nR/ V.
00:31:12.400 --> 00:31:26.100
DS = CV/ T DT this is our basic equation + DS DV under constant T DV.
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We are dealing with an isothermal process so that goes to 0.
00:31:29.200 --> 00:31:34.800
DS DV = DP DT, DP DT = nR/ V.
00:31:34.800 --> 00:31:43.500
Therefore, DS = nR/ V DV.
00:31:43.500 --> 00:31:53.800
When we integrate this, we end up with δ S.
00:31:53.800 --> 00:32:03.500
For the ideal gas = n × R × the log of V2/ V1.
00:32:03.500 --> 00:32:19.200
We just calculated that the δ S of the Van Der Waals gas = nR × the nat log of V2 - NB/ V1 – NB.
00:32:19.200 --> 00:32:35.700
We can compare these two, between these two, the δ S of the Van Der Waals gas is going to be bigger than the δ S of the ideal gas.
00:32:35.700 --> 00:32:43.800
The change in entropy of a Van Der Waals gas is going to be bigger than the change in entropy of the ideal gas.
00:32:43.800 --> 00:32:55.400
When you subtract the same value, V2/ V1 V2 – NB V1 – NB.
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You are subtracting the same value from numerator and denominator.
00:32:58.400 --> 00:33:16.000
When you subtract the same value from numerator and denominator of a ratio, the value of the quotient goes up.
00:33:16.000 --> 00:33:20.600
The value of the quotient actually rises.
00:33:20.600 --> 00:33:22.500
The log of the bigger number is bigger.
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Therefore, the δ S of the Van Der Waals, this expression is greater than this expression.
00:33:31.100 --> 00:33:39.800
The value of the quotient rises as just a basic mathematical fact.
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Let us see if we can make our way through this ocean of mathematics that we happen to be swimming through.
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I’m starting to lose my own way here.
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Example number 5, to a first approximation the compressibility factor Z of the Van Der Waals gas is Z = PV/ RT.
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The compressibility factor of any gas is just the pressure × the volume/ R × T.
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It is equal this thing, from this and the thermodynamic equation of state to demonstrate that the change in molar enthalpy
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with respect to pressure under constant temperature for a Van Der Waals gas = B – 2A / RT.
00:34:20.200 --> 00:34:24.200
Again, this is nothing more than mathematical manipulation.
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Let us see what we can do.
00:34:26.900 --> 00:34:31.800
Let us go ahead and rewrite this again.
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We go ahead and go back to blue.
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I have got PV and again when this line over it just means molar, it just means V is nothing more than this V line, this V bar is nothing more than V/ N.
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That said so, if I just want to multiply everything by N, I end up getting V.
00:34:48.300 --> 00:34:55.000
It is not a problem, this is just the molar volume instead of the volume.
00:34:55.000 --> 00:35:09.400
PV/ RT = 1 + B - A/ RT × P/ RT.
00:35:09.400 --> 00:35:15.600
Let us see here, they want this thing.
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The thermodynamic equation of state is the following.
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Thermodynamic equation of state says that DH/ DP = V - T × DV DT.
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I will put these over, it does not really matter, DT under constant P.
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This is equal to this so I need to find DV DT from this equation and plug it into here and see what I end up with.
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I’m going to rearrange this equation and solve for V.
00:35:56.700 --> 00:36:00.500
Let me go ahead and do that over here.
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I get PV =, I’m going to multiply everything by RT.
00:36:05.000 --> 00:36:12.800
I’m going to get RT + B - A/ RT × P.
00:36:12.800 --> 00:36:28.400
I’m going to go ahead and divide by P so I'm going to get V = RT/ P + B - A/ RT.
00:36:28.400 --> 00:36:46.800
Once again, rewrite the DH DP = V - T × DV DT that is the thermodynamic equation of state.
00:36:46.800 --> 00:36:56.800
Now, I have this so let me go ahead and find DV DT while holding P constant.
00:36:56.800 --> 00:37:25.000
I’m going to find this from this equation, holding P constant I end up with R/ P - -A R/ RT².
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When I take the derivative of this with respect to T, this derivative is 0 and I'm going to get - and – that is going to be +.
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We are good.
00:37:36.200 --> 00:37:42.100
This - and this -, when I take the derivative of this with respect to T, I get this -.
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It is going to equal R/ P + AR/ R² T².
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I’m just going to go ahead and cancel the R so I will just do it now.
00:37:55.400 --> 00:38:03.300
A/ RT², 1 R cancels 1 R leaving just 1 R on the bottom.
00:38:03.300 --> 00:38:30.500
My DH DP T = V - T × DV DT which is this R/ P + A/ RT².
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This is equal to V, V is equal to this whole thing so I’m going to put that back in.
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It is going to be RT/ P + B - A/ RT –,
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I’m going to go ahead and distribute, this is RT/ P - A/ RT².
00:38:59.400 --> 00:39:32.700
RT/ P cancels RT/ P and I get the DH/ DP under constant T = this - that - that = B -2A/ RT².
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What happened here?
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I think I have lost my way in this mathematics.
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Let us see what is going on.
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Let us see if I copy everything correctly here.
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There is our T – T, I'm sorry this is here and I distribute the T.
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I forgot to distribute the T over this one.
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This is T there, sorry about that.
00:39:56.700 --> 00:40:00.500
There we go, let me go ahead and erase this part and let me do this part again.
00:40:00.500 --> 00:40:03.500
This RT/ P cancels this RT/ P.
00:40:03.500 --> 00:40:10.500
We have B - A/ RT and this T cancels that - A/ RT.
00:40:10.500 --> 00:40:16.700
There we go, -2A/ RT.
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There you go, that takes care of the first one.
00:40:24.000 --> 00:40:48.600
The change in molar enthalpy with respect to a change in pressure at constant temperature for a Van Der Waals gas = B – 2A/ RT.
00:40:48.600 --> 00:40:56.900
Example 6, using the same expression from example 5 for the compressibility Z of
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the Van Der Waals gas which was this thing, show that DS DP sub T = this thing.
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By one of Maxwell's relations, we have the following – DS DP sub T which is actually what we want.
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That is actually equal to DV DT sub P.
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This is the same thing that we did before, we are going to take this equation.
00:41:56.600 --> 00:42:03.100
We want to take the derivative of V with respect to T so we are going to rearrange this equation
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just like we did before in the previous example.
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We ended up with the following, we ended up with V = RT/ P + B - A/ RT.
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When we take the derivative of this with respect to T holding P constant so that DV DT holding P constant we end up with R/ P + AR/ RT².
00:42:51.300 --> 00:43:12.900
We go ahead and put this in for here and we are going to end up with DS DP under constant T =...
00:43:12.900 --> 00:43:31.300
Let me just rewrite this again, DT under constant P which is actually equal to – R/ P + AR/ RT².
00:43:31.300 --> 00:43:38.100
This one was not that bad, this was just using Maxwell's relation and taking the derivative of the Van Der Waals equation.
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That was nice and simple.
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Example 7, this one is a little interesting.
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Using the results from the previous two problems for DH DP and DS DP, calculate δ H and δ S
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for the isothermal pressure increase of CO₂ as a Van Der Waals gas from 1 mPa to 15 mPa at 298°K.
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These are the values for the Van Der Waals constants for CO₂ and they ask us to run the same calculation of 598.
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They say compare the values you calculated for 298 to those obtained for an ideal gas.
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We just found expressions for a Van Der Waals gas for the change in enthalpy with respect to pressure,
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the change in entropy with respect to pressure, we could integrate these expressions to get expressions for the actual change in enthalpy and
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the change in entropy and when you compare that to the ideal gas.
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Let us see what we have got.
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We know what δ H is and we want to know what δ S is.
00:44:57.400 --> 00:45:08.900
Let us go ahead and write down what we got.
00:45:08.900 --> 00:45:20.600
Our CO₂ gas, we are taking it from 1 mPa, we are taking it to 15 mPa and we have our value of.
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A = 0.366 m 6 Pa/ mol² and B = 42.9 × 10⁻⁶ m³/ mol.
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From the previous example, we found that DH DP at constant T = B - 2A/ RT.
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Therefore, I have got DH = V -2A / RT DP.
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When I go ahead and integrate this expression, I’m going to get δ H = B - 2 A/ RT × the integral from P1 to P2 of DP and it is going to = B -2A/ RT × δ P.
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Our δ H is going to equal, again this is molar.
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We have got D – 2A.
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Let us just plug in all of our values so we have B which is going to be 42.9 × 10⁻⁶ -2 × 0.366/ 8.314 × 298 × δ P.
00:47:17.500 --> 00:47:27.000
It is going to be 15 -1, I’m going to do × 10⁶ because this is mPa.
00:47:27.000 --> 00:47:35.400
We are working in Pascal’s in order to make sure that the units match.
00:47:35.400 --> 00:47:41.600
15 -1 is 14 mPa, it is 14 × 10⁶ Pa.
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When I do this, I get δ H of -3536 J/ mol that is all this line means molar.
00:47:54.000 --> 00:48:05.900
There we go, that is the δ H based on what we calculated for DH DP and integrated it.
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For the DS, we work out that DS DP under constant T = - R/ P + RA/ RT².
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Therefore, DS = when I move P over I integrate the expression, I'm going to get from P1 to P2 of this thing.
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This thing R/ P + R A/ RT² DP.
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DS = - R × the integral from P1 to P2 of DP/ P.
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This - comes out so this becomes a - and this becomes a - that distributes over both.
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- A / RT² and I cancel one of the R × the integral P1 to P2 DP.
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Therefore, our DS = - R LN P2/ P1 - A/ RT² × δ P.
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We put the values in, δ S = -8.314 × the nat log of 15/ 1.
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The mPa cancels mPa, the units cancel, I can leave the numbers alone.
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-0.366/ 8.314 × 298² × 15 - 1.
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And again, this one × 10⁶ and there is no cancellation here.
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I’m going to do these separately, - 22.5 - 6.9.
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Therefore, the entropy change is -29.4 J/°K mol.
00:50:47.100 --> 00:50:51.700
That takes care of the δ S.
00:50:51.700 --> 00:51:01.200
Let us see what is next, they ask us to run the same calculations part B under a new temperature.
00:51:01.200 --> 00:51:08.200
It is the same exact process same, same exact thing we just did the only difference is instead of 298 for the T, we are going to put 598 for the T.
00:51:08.200 --> 00:51:16.100
What we end up with are δ H, I will go ahead and put 598.
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It is going to equal- 1460 J/ mol and δ S = - δ S at 598 = -24.22 J/mol °K.
00:51:57.700 --> 00:52:04.700
Let us go ahead and compare the two values.
00:52:04.700 --> 00:52:10.100
Let us see, C wants us to compare the values for the ideal gas and the Van Der Waals gas.
00:52:10.100 --> 00:52:23.900
The δ S for an ideal gas = - R LN P2/ P1 that was the first term.
00:52:23.900 --> 00:52:27.400
We will do it in just a second.
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It is - 22.5 J/°K mol
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The δ H for the ideal gas = 0.
00:52:48.000 --> 00:53:04.900
It is 0 because DH/ DP T = 0 for an ideal gas, remember Joule Thompson coefficient for an ideal gas.
00:53:04.900 --> 00:53:13.300
I will put Joule Thompson.
00:53:13.300 --> 00:53:26.900
In case you are wondering where this came from, right here for the ideal gas, just look back on lesson 21.
00:53:26.900 --> 00:53:37.700
Lesson 21 which is the entropy changes for an ideal gas.
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Notice, this expression right here this was the first half of the expression for the Van Der Waals equation that we just did in part A.
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The Van Der Waals expression, remember I did into -22.5 - another value.
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That -22.5, that takes care of the ideal gas portion of the Van Der Waals gas.
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Because it is a Van Der Waals gas, there is actually an additional term.
00:54:04.200 --> 00:54:06.400
It is no different than the Van Der Waals equation.
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Remember, we had -22.5.
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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.