WEBVTT chemistry/physical-chemistry/hovasapian
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Hello and welcome back to www.educator.com and welcome back to Physical Chemistry.
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Today, we are going to continue our entropy example problems, let us jump right on in.
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Our first example for today says that 1 mol of an ideal gas with a constant volume heat capacity of 3/2 Rn is in the initial following state 350°K and 1.5 atm.
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This is just a continuation of the example problems from the previous section.
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We have the same initial state, we are just subjecting this initial state to various transformations or transformations and the various circumstances.
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This time the gas expands isothermally, we have a nice isothermal expansion against 0 external pressure.
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This time it is going to be a free expansion.
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A free expansion is when there is nothing on the other end keeping it from expanding.
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0 external pressure of the gas is 0.75 atm, calculate Q, W, δ U, δ H, δ S and again we want compare the δ S with Q/ T.
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0 external pressure free expansion until the pressure goes from 1.5 atm to 0.75 atm.
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Let me go ahead and do this in red since the problems are concerned.
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Let us go ahead and start with a definition of work.
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The definition of work is DW = P external × DV.
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The P external is 0 external pressure so this is 0.
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Therefore, DW =0 which means that our work for this process = 0.
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In a free expansion no work is done because we are not pushing against anything.
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There is no force resisting our push, our expansion.
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This is an ideal gas which makes our job a little bit easier.
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We are dealing with something that is an isothermal process.
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Isothermal implies that δ U=0.
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The whole bunch of things might actually end up being 0 here.
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δ U = Q - W so Q = δ U + W.
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Q = 0 + 0 so it looks like the heat in this is also going to be 0.
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Let us take a look at δ H so DH = let us go ahead and use our equation just so we can actually practice writing it down.
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CP DT + DH DP DP, this is an isothermal process so this is 0.
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This is an ideal gas so this is 0 so we have δ H = 0.
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It looks like everything is going to be 0 here.
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Let us see what happens with δ S so DS = CV.
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It is going to be CV we are using pressure here so this is going to be CP/ T DT - nR/ P DP.
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This is an isothermal process so the change in temperature is 0 so this term goes to 0.
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Therefore, we have just DS = - nR/ P DP and then when we integrate this differential equation
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we are going to get the equation δ S = -n × R × log of P2/ P1 the nat log.
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Let us go ahead and put some numbers in so we get - there is 1 mol that is going to be 8.314 nat log that we have
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0.75 is the second temperature and 1.50 is the initial temperature.
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We get δ S = 5.76 J/°K.
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We end up with 5.76, the initial state the 350°K and 1.5 atm is the final state 350°K, 0.75 atm.
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You remember the previous lesson, the last two examples, these were the initial and
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final states because entropy is a state variable, the state function, state property.
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The path does not matter how you do it.
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It does not matter what you do, a reversible expansion against a particular pressure.
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A constant pressure expansion or a free expansion, the entropy ends up being the same because the initial final states are the same.
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Let us go ahead and do our comparison.
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We have got δ S = 5.
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We are slowing down in writing the numbers out properly 5.76 J/°K and then we have Q/ T is going to equal 0 J/ 350°K = 0 J/°K.
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In this particular case because that Q was 0, our Q/ T = 0 and they do not match.
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Let us go ahead and take a look at the next example.
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1 mol of an ideal gas with this particular constant volume heat capacity is initially in the following state.
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Again, same initial state this time the gas expands adiabatically and reversibly.
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It is no longer isothermal so now are expanding it adiabatically and we are expanding it reversibly until the pressure of the gas is 0.75 atm.
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That part stays the same.
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Calculate all of these variables and compare δ S with Q/ T.
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This time we are doing it adiabatically and reversibly.
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Let us see here, let us see what this looks like if you remember this is the P and this is the V.
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We have this which is an isotherm and we have this which is an adiabat.
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We might have an initial state here, state 2.
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If we follow the isotherm keeping the temperature constant, it is going to be this.
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If we follow the adiabat, in other words an adiabatic condition is where there is no heat allowed to transfer anywhere.
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You are going to end up with a huge temperature drop and reversibly just means we are actually following that path.
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We are not going this way to that state.
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We have an isotherm and we have an adiabat.
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Adiabat that just means that Q = 0, DQ = 0.
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Adiabatic we automatically know what Q is.
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Adiabatic implies that Q = 0 so we have taken care of Q.
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If we look back at reversible adiabatic expansion of an ideal gas, we find the following.
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Let us go ahead and start with the definition.
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This is going to be a little bit of a review of adiabatic processes.
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We have DU = DQ - DW this is the basic equation.
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Since Q was 0, adiabatic DQ is 0.
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What you end up with is DU = - DW.
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For an ideal gas DU = CV DT and the definition of work DU is just P external × the change in volume.
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What you end up with is CV DT = - P external × DV so this is the equation.
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They also said that it is going reversibly and we know what reversible means.
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Reversible means that the external pressure is actually equal to the external pressure on the system = the pressure in the system.
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They are always in equilibrium, that is what reversible means.
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That means we are following this path closely in terms of stair stepping.
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Therefore, we can replace P external with P and what we get is CV DT =- P DV.
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This is an ideal gas, P = nRT/ V and we did this derivation back when we discussed adiabatic processes but I thought it would be nice to do it again.
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We have CV DT =- nRT/ V DV.
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Let us go ahead and divide both sides by this variable T and we would end up with CV/ T DT = -nR/ V DV.
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When we go ahead and integrate this equation, we are going to end up with the following.
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We are going to end up with CV × the nat log of T2/ T1 = - nR × the nat log of V2/ V1.
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This is our fundamental equation for an adiabatic reversible expansion or a compression.
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This is the relationship between the constant volume heat capacity are the volume and the temperature for an ideal gas.
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This is the basic equation.
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This was the equation that we derived when we discuss this, when we talked about the energies.
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This is the basic equation for the reversible adiabatic expansion of an ideal gas.
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Let us go ahead and continue a little bit more.
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Let me write the equation here on this page so we have it.
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We have CV × nat log of T2/ T1 is going to equal - n × R × the nat log of V2/ V1.
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We are going to use this thing.
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Do you remember the ratio of the constant pressure heat capacity divided by the constant volume heat capacity we called it γ.
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We are going to go ahead and use this.
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Let me make this V a little more clear and we are going to do a little bit more mathematical manipulation on this using PV = nRT to derive some equations here.
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I’m not going to go ahead and go through the mathematical manipulation.
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If you want you can check your books, it is usually in every single thermodynamics book or you can go back to the previous lessons.
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I think that I did it there.
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We got the following relations, we got T1 V1 ^γ = T2 V2 ^γ.
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I will go ahead and call this equation 1.
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It expresses a relationship between temperature and volume.
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This was also a relationship between temperature and pressure, it is T1 ^γ P1¹ - γ = T2 ^γ × P2¹ - γ.
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I will call this equation 2 and there is one for relationship between pressure and volume.
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We have P1 V1, I’m sorry this is γ -1.
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The temperature and volume is γ -1, P1 V1 γ = P2 V2 γ.
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I will go ahead and call this equation 3.
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Using this CP/ CV calling it γ and then using the ideal gas PV = nRT, I can actually manipulate this fundamental equation
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to derive these three relationships temperature volume, temperature pressure, and pressure volume, based on the heat capacity.
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This is for an adiabatic process.
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We are going to start, manipulating this.
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The first thing I'm going to do is I’m going to start with equation number 3.
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It just happens to be the one that I picked, so let us just jump right on in.
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Let us go ahead and do this.
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First of all, let us go ahead and see what γ is, so CP/ CV = γ so CP =5/2 Rn and CV is going to be 3/2 Rn.
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γ is going to equal 5/3.
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Let us go ahead and list what it is that we actually have.
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We know our first pressure is 1.5 atm, we have our first volume which we can get from nR T1/ P1.
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Our first volume = nR T1/ P1.
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I’m just trying to get as many of the variables as possible.
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I have the first pressure and I have the second pressure.
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I can calculate the first volume but I want to get the second volume.
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I have the first temperature I want to get the second temperature.
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In other words, for this thing I have temperature 1 which was 350°K.
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We are going to be able to find what the temperature 2 is.
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I want to be able to find this and that so I have some variables to work with this that is why I'm doing this.
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I have 1 mol, in this case R, we are using the ideal gas law so I have to use 0.08206.
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Be very careful with that.
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Temperature is going to be 350°K and the initial pressure is going to be 1.5 atm.
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V1 is actually equal to 19.15 L.
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I know what P2 is, P2 is 0.75 atm.
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I want to know what volume 2 is so I have P1 V1, I have P2, I want to know what volume 2 is.
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I’m going to go ahead and go to this equation right here and I know what γ is 5/3 so I get the following.
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Pressure 1 is 1.5 atm and volume 1 is 19.15 L, 5/3 = 0.75 atm and I'm looking for volume 2 is going to be 5/3.
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When I solve for volume 2, I get volume 2 = 29.03 L is my volume 2.
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Let us see what I can do with that.
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Since I have volume 2, now I want to find temperature 2.
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Let us find temperature 2.
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Temperature 2 is just nR T2.
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Again, PV = nRT therefore T2 = P2 V2/ n × R.
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The pressure 2 = 0.75 atm, volume 2 = 29.03 L, n is 1, and R is 0.08206.
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If you use the ideal gas make sure you use the proper form of R and we end up with temperature 2 equal to 265.3°K.
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This is cooler, we started at 350 and the temperature 1 =350°K.
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Temperature 2 = 265.3, the gas is expanding.
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It is expanding adiabatically, we want it to be cooler so our number matches, it is in the proper direction.
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Cooler which makes sense, an adiabatic expansion gives you the largest temperature drop.
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We can go ahead and actually work out the problem.
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We just needed to find what these other values were.
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We needed to find what V2 was, we need to find what T2 was, so we can use these in our equations.
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That is what all of this led to.
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Let us go ahead and start doing this.
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DU = CV DT which means that δ U = CV × δ T which means 3/2 Rn δ T.
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Therefore, δ U = 3/2 × 8.314 × 1 and δ T that is going to be the final temperature which is going to be 265.3 -350 because this is the final - the initial.
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You end up with δ U = -1056 J.
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In the process of expanding adiabatically and reversibly, the system ends up losing 1056 J of energy.
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This is an adiabatic process so we said that δ U = - work = - δ U.
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Let me go ahead and keep writing this – δ U, - - so 1056 J.
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The gas is expanding therefore, it is doing work on the surroundings.
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Work is positive from the surroundings point of view.
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It is expanding so the expanding gas does 1056 J of work on the surroundings.
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Let us go ahead and take a look at DH, DH = CP DT.
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Therefore, δ H = CP × δ T this is what we want, we want δ T.
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We wanted to find that first temperature so we can actually do the problems that we needed to do because DU is CV DT.
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CV δ T this is CP δ T, that is why we went through the initial process.
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Remember, we were only given the initial temperature, the initial and final pressures.
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I use those relationships in order to find the second volume and the second temperature so that I can do these problems.
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That = 5/2 Rn δ T and again CP if the 3/2 Rn is the constant volume capacity, this is an ideal gas at constant pressure heat capacity of 5/2 Rn.
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This is just the 3/2 + 1.
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Based on this relation, from ideal gas the difference between the CP and the CV= R.
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If I’m given this I can find this.
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Therefore, we have δ H = 5/2 0.314 × 1 × 265.3 -350.
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Therefore, our δ H = - 1760 J.
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Let us go ahead and do δ S.
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What shall we do?
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δ S for an ideal gas, I’m not going to go through the entire equation, I’m going to go through the final version of the equation.
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It is equal to the constant pressure heat capacity × LN of T2/ T1 -nR × LN of P2/ P1.
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That = 5/2 × 8.314 × 1 × the nat log of 265.3/ 350 -1 × 8.314 × the nat log of 0.75/ 1.5.
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When I actually do that I end up with δ S= 0.
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If I do Q/ T this is actually equal to Q reversible/ T.
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Q reversible is adiabatic 0/ the temperature which is 350 so we end up with 0.
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You see they match.
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Reversible processes the entropy δ S and Q reversible/ T they will match.
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Let us see what else have we got.
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This one, the entropy of water is 69.95 J/°K at 25°C.
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This is the entropy not that change in entropy, this is S not δ S, 25°C and 1 atm, 1 atm so it is standard.
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This is S standard 298 for water is 69.95.
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In other words, the general sense of disorder of liquid water and 25°C 1 atm pressure is 69.95 J/°K.
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Given the following, calculate the entropy of water vapor at 180°C and 0.75 atm.
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Let us see what we have got.
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They give us the constant pressure heat capacity of the liquid water which is 75.29 J/ mol °K.
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They give us the constant pressure heat capacity of the gas which is 33.58 J/°K.
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They give us the heat of vaporization of water 40.656 kJ have to be injected into 1 mol of water that much energy has to be put into water to convert it from liquid to gas.
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That is what this means.
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Assume ideal behavior for the water vapor accounts, we can assume that it is an ideal gas.
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Let us see what we can do here.
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The basic equation that you are going to work with here is the following.
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Two things that are happening, you are going for 25°C where water is a liquid.
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You are going to a 180°C where water is a gas.
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You have to calculate the entropy of the phase change, the entropy of the temperature rise, and there is one other thing going on here.
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You are going from 1 atm to .75 atm so you have to account for the entropy change for the pressure change.
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Here is what it looks like.
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The standard entropy at any temperature is going to equal the standard entropy at the temperature that you know which in this case is 298 +
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the entropy in going from this initial temperature all the way to the boiling point for water of the liquid state of water/ T DT.
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This from the third law.
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The third law says that if you want to calculate the entropy from temperature 1 to temperature 2
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it is going to be the constant pressure heat capacity of the particular phase divided by the temperature DT.
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The entropy that we start off with a given temperature now we are going to calculate,
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we are going to add to that the entropy in going from, in this particular case 25 C to 100°C.
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The entropy change that accompanies the vaporization of the water at the boiling temperature,
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the conversion of going from liquid to water vapor, we have more temperature rise.
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The temperature boiling up to the particular temperature that we want and this time, since it is going to be water vapor it is going to be this.
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That takes care of the temperature part, now we have to adjust for the pressure part - this is an ideal gas nR × the nat log of pressure 2/ pressure 1.
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I hope that makes sense.
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Remember that the CP/ T DT - nR/ P DP this accounts for the pressure change in entropy, the rise in entropy.
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This accounts for the change in temperature.
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All I have done is I have accounted for the entropy that I start off with.
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The entropy of the temperature rise going from 25 to 100.
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The change in entropy in going from liquid to the water vapor.
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The change in entropy in going from 100°C to 180°C.
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I account for the entropy decrease in the pressure difference.
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In this particular case, because the pressure is going down 1 atm to 0.75 atm, that pressure decline is going to actually accompany volume increase.
00:30:14.000 --> 00:30:17.900
Our volume increase is going to be actually you can end up getting a positive number here.
00:30:17.900 --> 00:30:20.800
It is going to be slightly more entropy than usual.
00:30:20.800 --> 00:30:24.600
Let us go ahead and put the numbers in.
00:30:24.600 --> 00:30:30.000
This is going to be, we said 180°C.
00:30:30.000 --> 00:30:37.400
It is going to be 453°K.
00:30:37.400 --> 00:30:45.200
Our S and now it can no longer standard pressure 1 atm.
00:30:45.200 --> 00:31:06.300
It is going to be 0.7518 atm and 453°K that is going to equal, the entropy of the temperature that we do know so the S at 298 = 69.95 +
00:31:06.300 --> 00:31:17.900
the integral from 298 to 373 of the constant pressure heat capacity 75.29.
00:31:17.900 --> 00:31:40.600
This is going to be 75.29/ T DT + 40656 J divided by the boiling temperature which is 373°K + this 1.
00:31:40.600 --> 00:31:57.400
This is going to be 373 to 453, this time we are going to use the 33.58 which is the constant pressure heat capacity for water vapor, 33.58/ T DT.
00:31:57.400 --> 00:32:16.600
We are going to subtract so it was going to be 1 mol and this could be 8.314 and this is going to be the nat log of 0.75 atm
00:32:16.600 --> 00:32:22.500
which is P2 or the initial which is 1 atm.
00:32:22.500 --> 00:32:25.000
This is the integral that we solve.
00:32:25.000 --> 00:32:30.400
Let us go ahead and write out a little bit more.
00:32:30.400 --> 00:33:03.600
S at 0.75 atm and 453 = 69.95 + 75.29 × the nat log of 373/ 298 + when I do the 4656 divided by the 373 I get 109.
00:33:03.600 --> 00:33:16.300
I'm going to add to that + 33.58 × the nat log of 453/ 373.
00:33:16.300 --> 00:33:48.500
It is going to be -2.39 and I get the entropy of water vapor at 180°C which is 453°K and 0.75 atm is going to equal 205 J/ mol/°K.
00:33:48.500 --> 00:33:55.800
Here we go, just a basic application of the mathematical portion of the third law of thermodynamics.
00:33:55.800 --> 00:33:58.500
There is nothing strange going on here.
00:33:58.500 --> 00:34:03.700
You start with a particular entropy that you know and you account for the any temperature change.
00:34:03.700 --> 00:34:08.000
If there is a phase change you account for the entropy change for the phase change.
00:34:08.000 --> 00:34:14.000
After the change of phase there is still the temperature increase then you account for that temperature increase.
00:34:14.000 --> 00:34:18.700
If there is a pressure increase or decrease, you account for that pressure increase or decrease.
00:34:18.700 --> 00:34:20.100
That is all that you are doing.
00:34:20.100 --> 00:34:28.100
You are accounting for every single change that takes place in the system.
00:34:28.100 --> 00:34:35.700
Let us see what we have got here.
00:34:35.700 --> 00:34:39.800
Silicon dioxide experiences the following transformation.
00:34:39.800 --> 00:34:49.800
We go from 25°C and 1 atm pressure, we are going to raise the temperature to 225°C and we are going to raise the pressure to 1500 atm.
00:34:49.800 --> 00:34:53.600
There are two things going on, change in temperature and a change in pressure.
00:34:53.600 --> 00:34:58.800
Given the following data, calculate the molar δ S for the transformation.
00:34:58.800 --> 00:35:07.400
The constant pressure heat capacity for the solid is this, notice it is not a constant, it is going to be a function of T.
00:35:07.400 --> 00:35:17.400
The density of silicon is 2.648 g/ cm³ and the coefficient of thermal expansion is 3.530 × 10⁻⁵.
00:35:17.400 --> 00:35:24.400
Sorry I forgot the unit here, for thermal expansion this is going to be per °K.
00:35:24.400 --> 00:35:27.700
Let me do it in black, my apologies.
00:35:27.700 --> 00:35:32.300
It is important that we get these.
00:35:32.300 --> 00:35:37.400
Coefficient of the thermal expansion this is α.
00:35:37.400 --> 00:35:42.600
α = 3.530 × 10⁻⁵ /°K.
00:35:42.600 --> 00:35:54.200
It represents the percentage change in the volume of something when you heated up, relative to how much you started off with.
00:35:54.200 --> 00:35:57.500
Let us go ahead and see what we can do.
00:35:57.500 --> 00:36:04.500
We are changing temperature and we are changing pressure, let me go back to red here.
00:36:04.500 --> 00:36:20.700
We are going to use this equation, is our fundamental equation so the DS = CP/ T DT – V A DP.
00:36:20.700 --> 00:36:23.000
We are dealing with a solid here, this is a general system.
00:36:23.000 --> 00:36:25.400
We are no longer dealing with an ideal gas.
00:36:25.400 --> 00:36:34.300
For the most general equation, the entropy change when you change temperature and pressure this is it right here.
00:36:34.300 --> 00:36:38.700
There is nothing strange going on here.
00:36:38.700 --> 00:37:09.200
δ S when we integrate this so we get δ S = the integral from temperature 1 to temperature 2 of CP/ T DT - V × A × the integral from pressure 1 to pressure 2 of DP.
00:37:09.200 --> 00:37:21.400
We have our final equation of DS = this stays the same, we cannot pull the CP out because CP is now no longer constant.
00:37:21.400 --> 00:37:36.000
It is going to be the integral of CP/ T DT from T1 to T2 , this is just DP so - V A and δ P.
00:37:36.000 --> 00:37:39.500
This is the equation that we are going to use.
00:37:39.500 --> 00:37:44.100
It looks like we are going to need the sub V.
00:37:44.100 --> 00:37:46.500
We are going to need the molar volume.
00:37:46.500 --> 00:37:51.100
We need to know what the volume is of this particular thing.
00:37:51.100 --> 00:37:57.400
They wanted us to calculate the molar δ S, they gave us the density, they did not give us the molar volume.
00:37:57.400 --> 00:38:00.600
The first thing we have to do is we have to calculate V.
00:38:00.600 --> 00:38:07.400
Let me go ahead and do that in blue.
00:38:07.400 --> 00:38:27.700
We have silicon dioxide and that is going to be 60.09 g/ mol, let us start off with that way.
00:38:27.700 --> 00:38:43.100
We have 60.09 g/ mol and we have 1 cm³ is 2.648 g.
00:38:43.100 --> 00:38:59.600
2.648 g that takes care of that so that = 22.69 cm³/ mol so that is the molar volume.
00:38:59.600 --> 00:39:02.300
This is V but it is in cm³.
00:39:02.300 --> 00:39:07.700
I want to express it in dm³ which is the same as the L.
00:39:07.700 --> 00:39:15.500
That is going to equal 22.698 × 10⁻³ cm³.
00:39:15.500 --> 00:39:20.100
I hope you guys are okay with converting from cm to dm to m, things like that.
00:39:20.100 --> 00:39:30.200
Dm/ mol which is the same as 22.69 so dm³ is a L.
00:39:30.200 --> 00:39:40.400
22.69 × 10⁻³ L/ mol this is what I wanted, this is my V right here.
00:39:40.400 --> 00:39:47.200
In 1 mol of this stuff because we are calculating molar, the volume is 22.69 × 10⁻³ L.
00:39:47.200 --> 00:39:55.700
1 mol of silicon dioxide has this volume, now I can go ahead and do my problem.
00:39:55.700 --> 00:40:00.700
Let us go ahead and rewrite what I need.
00:40:00.700 --> 00:40:05.900
I need to go back to red.
00:40:05.900 --> 00:40:21.500
There we go so δ S = the integral from T1 to T2 of CP/ T DT – V A δ P.
00:40:21.500 --> 00:40:55.100
I got my δ S = the integral from 298 to 498 and now I write 46.94 + 34.31 × 10⁻³ × T -11.30 × 10⁵ T⁻²
00:40:55.100 --> 00:41:07.200
that is the constant pressure heat capacity of silica/ T that is the integral part and it is going to be –volume.
00:41:07.200 --> 00:41:33.800
The volume is 22.69 × 10⁻³ L/ mol and A which is the coefficient of thermal expansion they give us at 3.530 × 10⁻⁵/°K.
00:41:33.800 --> 00:41:40.800
The change in pressure δ P is going to be 1500 – 1.
00:41:40.800 --> 00:41:51.900
I’ m going to keep multiplying this by, notice this is going to end up being in L atm.
00:41:51.900 --> 00:41:54.900
I need to put this into J.
00:41:54.900 --> 00:42:11.600
I’m going to multiply it by 8.314 J = .08206 L atm.
00:42:11.600 --> 00:42:17.100
It is important if you are going to be working in L atm, you want to convert this to J.
00:42:17.100 --> 00:42:21.900
This is the conversion factor, it is just the ratio of the 2 R.
00:42:21.900 --> 00:42:28.400
8.314 J = .08206 L atm they are both units of energy.
00:42:28.400 --> 00:42:43.000
When I do that, atm cancels atm, L cancels L , you end up with J/ mol °K.
00:42:43.000 --> 00:42:44.400
Everything works out right.
00:42:44.400 --> 00:42:47.500
When you do all this, let us actually calculate this.
00:42:47.500 --> 00:43:09.900
The integral comes out to be 26.88 J / mol °K and this ends up being 0.12 J/ mol °K.
00:43:09.900 --> 00:43:20.200
Our final δ S is 26.76 J/ mol °K.
00:43:20.200 --> 00:43:27.200
Clearly the change in temperature accounts for 26.88 J/ mol °K change in entropy.
00:43:27.200 --> 00:43:34.500
The change in the pressure from 1 to 1500 atm, 1500 atm is massive, it is huge.
00:43:34.500 --> 00:43:38.000
1500 atm, all that pressure change.
00:43:38.000 --> 00:43:46.700
This is solid and a solid is not going to contract all that much under that much pressure, even that much pressure.
00:43:46.700 --> 00:43:49.900
Its only difference is 0.12.
00:43:49.900 --> 00:44:00.200
Clearly, the change in entropy of the system especially of the solid or liquid, under a variation of pressure is virtually negligible.
00:44:00.200 --> 00:44:06.400
You are not losing anything by just ignoring this 0.12 but we want you to see it because we want to be able to solve the problem.
00:44:06.400 --> 00:44:16.700
This is how you solve the problem, your basic equation.
00:44:16.700 --> 00:44:23.000
Let us see what else have we got here.
00:44:23.000 --> 00:44:30.800
The standard entropy of lead at 25°C is 64.8 J/ mol °K.
00:44:30.800 --> 00:44:39.200
Standard means 1 atm it does not necessarily mean 25°C.
00:44:39.200 --> 00:44:46.400
In general chemistry, when you see this little degree sign on top, it generally means 25°C and 1 atm.
00:44:46.400 --> 00:44:49.200
The degree sign ° really just means the temperature.
00:44:49.200 --> 00:44:55.900
We specify the temperature because in these problems now we are a little bit more sophisticated than we were in General Chemistry.
00:44:55.900 --> 00:45:02.100
We can calculate S, δ H, δ S at different temperatures.
00:45:02.100 --> 00:45:07.800
This degree sign ° represents the temperature, that is what standard means.
00:45:07.800 --> 00:45:12.400
Standard pressure is 1 atm.
00:45:12.400 --> 00:45:20.500
The constant pressure heat capacity is this, you notice it is not at constant, it is a function of the temperature.
00:45:20.500 --> 00:45:28.100
The constant heat capacity to liquid at 32.51 again it is not a constant.
00:45:28.100 --> 00:45:38.400
The melting temperature of lead is 327.4°C, the δ H of fusion of melting is 4.770 kJ/ mol.
00:45:38.400 --> 00:45:43.800
They want us to calculate the standard entropy of liquid lead at 525°C.
00:45:43.800 --> 00:45:49.300
Here it is going to be solid lead at 25, what is the entropy of liquid lead and calculate the δ H
00:45:49.300 --> 00:45:57.700
for this transformation of solid lead from 25°C to liquid lead at 525°C.
00:45:57.700 --> 00:46:03.100
This is going to be exactly like the problems that we did a little bit earlier when we calculated the entropy change for water,
00:46:03.100 --> 00:46:04.800
from liquid water to water vapor.
00:46:04.800 --> 00:46:07.800
We are just going from solid lead to liquid lead.
00:46:07.800 --> 00:46:09.900
Let us see what we can do.
00:46:09.900 --> 00:46:14.000
We have got 25 and 525.
00:46:14.000 --> 00:46:20.500
Let us see what we have.
00:46:20.500 --> 00:46:39.100
We have S at 298 = 64.80 J/°K and we want S.
00:46:39.100 --> 00:46:45.000
Still standard so there is no change in pressure, all we are doing is actually changing the temperature, it is actually melting the lead.
00:46:45.000 --> 00:46:52.100
It is going to be at 823 this is what we want.
00:46:52.100 --> 00:46:53.800
Let us see what we have got.
00:46:53.800 --> 00:47:19.200
We are going to write S of 823 = S of 298 + the integral from 298 to the melting temperature of the solid heat capacity/ T DT.
00:47:19.200 --> 00:47:27.900
The entropy at a given temperature + the new entropy at the new temperature which is melting and we are going to account
00:47:27.900 --> 00:47:37.200
for the entropy of the phase change δ H of fusion/ the temperature of melting, boiling from going to solid to liquid.
00:47:37.200 --> 00:47:39.700
This is the melting temperature.
00:47:39.700 --> 00:47:42.400
We are going to heat that liquid some more.
00:47:42.400 --> 00:47:55.100
We are going to go from the melting temperature all the way to this 823, except this time we are going to use the constant pressure heat capacity of liquid.
00:47:55.100 --> 00:47:57.300
That is the integral that we want.
00:47:57.300 --> 00:48:11.400
Let us go ahead and put in our numbers so that S 823 = 64.80 + the integral from 298 to the temperature of melting
00:48:11.400 --> 00:48:20.000
which is 300 and something, they said would actually ends up being 600.4.
00:48:20.000 --> 00:48:44.600
It is going to be the 22.13 + 0.01172 T + 0.96 × 10⁵ T⁻² / T DT.
00:48:44.600 --> 00:48:57.600
We are going to add to that the 4770 J divided by the melting temperature which was 4850.4 °K, that is going to be the transition.
00:48:57.600 --> 00:49:07.600
We are going to add to that the entropy change in going from 600.4 to 823.
00:49:07.600 --> 00:49:22.800
This is going to be the 32.51 -0.00301 T/ T DT.
00:49:22.800 --> 00:49:26.700
I can go ahead and let you work out the arithmetic here.
00:49:26.700 --> 00:49:29.700
I’m not going to give you the answer, this is the answer right here.
00:49:29.700 --> 00:49:34.300
You can work out the integral and whatever number you get that is the answer.
00:49:34.300 --> 00:49:42.700
I just decided that I did not feel working this particular thing out, this is what is important.
00:49:42.700 --> 00:49:50.500
You are just taking the entropy at any other temperature is the entropy at some temperature that you know.
00:49:50.500 --> 00:49:57.000
For all practical purposes, the entropy of the temperature you know comes from the table of thermodynamic data in the back of your books.
00:49:57.000 --> 00:50:04.600
Just like in the back of your General Chemistry books, there was the enthalpy, the free energy, and the entropy.
00:50:04.600 --> 00:50:16.000
Those are standard third law entropy at 25°C and 1 atm pressure, that is going to be your starting point.
00:50:16.000 --> 00:50:23.800
If you want to calculate the entropy of any other temperature at any other pressure, you have to account for the change.
00:50:23.800 --> 00:50:33.500
25°C this is the entropy, you add to it the entropy that you get from the temperature rise from 298 to 600.
00:50:33.500 --> 00:50:45.600
This is the entropy rising going from the solid to liquid state and this is going to be entropy change in going from 600.4 to 823°K.
00:50:45.600 --> 00:50:49.100
That is all you are doing and it is going to be the same thing for δ H.
00:50:49.100 --> 00:51:01.300
You remember when we did that earlier for energy, we are going to be doing it again in just a moment.
00:51:01.300 --> 00:51:26.200
If we want the standard entropy at still higher temperatures, at temperatures above boiling,
00:51:26.200 --> 00:51:31.900
let us say we took this solid lead to liquid lead, that is why we boiled off that lead to turn it into lead vapor
00:51:31.900 --> 00:51:36.100
and we raise the temperatures too, then we just add more terms.
00:51:36.100 --> 00:51:53.100
Temperatures above boiling then we just add the appropriate terms.
00:51:53.100 --> 00:52:03.900
In this particular case, they would add the δ H of vaporization divided by the temperature of boiling, that is the entropy in going from liquid to vapor.
00:52:03.900 --> 00:52:11.200
We add to that if we are going to still raise it above the boiling temperature, temperature of boiling to whatever final temperature and
00:52:11.200 --> 00:52:18.500
this time we are going to use the constant pressure heat capacity of the gas / T DT.
00:52:18.500 --> 00:52:24.700
That is all nice and simple.
00:52:24.700 --> 00:52:36.900
Let us do part B, this time it asks for the δ H.
00:52:36.900 --> 00:52:46.400
I’m going to go ahead and write δ H = CP DT + DH DP T DP.
00:52:46.400 --> 00:52:51.500
The pressure is constant in this particular case so this is 0.
00:52:51.500 --> 00:53:01.200
δ H is just equal to the integral from T1 to T2 of CP DT.
00:53:01.200 --> 00:53:09.100
If there are any phase changes we simply add the δ H for the phase change, same as before.
00:53:09.100 --> 00:53:27.200
Our general equation becomes the δ H at any temperature T that I want is going to equal T0 to T melting of the CP solid DT.
00:53:27.200 --> 00:53:40.200
In this particular case, lead going from 298 to its melting temperature + the δ H of fusion and we are talking about δ H here not δ S.
00:53:40.200 --> 00:53:46.800
Once I have actually melted it, I raise the temperature some more, it is going to be the melting temperatures
00:53:46.800 --> 00:53:50.300
to whatever final temperature that I want.
00:53:50.300 --> 00:53:58.700
This time it is going to be the constant pressure heat capacity of the liquid DT.
00:53:58.700 --> 00:54:34.200
We have δ H at 823 = 298 to 600.4 of the 22.13 + 0.01172 T + 0.96 × 10⁵ T⁻² DT.
00:54:34.200 --> 00:54:51.200
I'm going to add to that the 4770 J and I'm going to add to that the change in enthalpy for going from the melting temperature of 823.
00:54:51.200 --> 00:55:05.000
This time I’m going to use the heat capacity of liquid lead which is 32.51 -0.00301 T DT.
00:55:05.000 --> 00:55:12.300
I'm going to let you work out whenever this number is.
00:55:12.300 --> 00:55:20.100
This is the answer, the rest is just arithmetic and integration and stuff like that.
00:55:20.100 --> 00:55:29.100
If I needed to know the δ H at higher temperature still, let us say I want to take solid lead turn it into liquid lead like I did here and
00:55:29.100 --> 00:55:36.400
then take the liquid lead and turn it into a gaseous lead and still raise the temperature, I would add the two following terms.
00:55:36.400 --> 00:55:45.000
I have to account for the δ H that comes from the vaporization and then if I raise the temperature beyond that,
00:55:45.000 --> 00:56:06.400
I would go from the boiling temperature to whatever temperature I wanted and this time I would use the constant pressure heat capacity of the gas.
00:56:06.400 --> 00:56:25.000
It is solid to liquid, transition from solid to liquid temperature change during the liquid phase, if I needed to,
00:56:25.000 --> 00:56:32.300
I would include the δ H of the transition from liquid to gas and then any other temperature change during the gas.
00:56:32.300 --> 00:56:39.100
This is just normal heat.
00:56:39.100 --> 00:56:46.400
This is the solid phase, the liquid phase, the gas phase, these are constant temperature processes.
00:56:46.400 --> 00:56:50.800
These are the phase changes so these are accounted for by the δ H.
00:56:50.800 --> 00:56:55.900
This is a vaporization, this is the δ H of fusion.
00:56:55.900 --> 00:57:00.000
Here I have to do them in terms of integration.
00:57:00.000 --> 00:57:04.000
That is all that is happening here.
00:57:04.000 --> 00:57:06.000
Thank you so much for joining us here at www.educator.com.
00:57:06.000 --> 00:57:06.000
We will see you next time, bye.