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 off with our example problems for this particular entropy sections.
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We have done a lot of discussion about entropy, a lot of mathematical derivations and now we just want to become familiar with doing problems.
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We are going to do a lot of them just like we did for energy.
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Let us go ahead and get started.
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Again, before I jump into the problems, I want you to do the same thing that I did with energy which was stop and
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take a look at what it is that we want to bring to the table as far as our tools and what we need to know.
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I will discuss a little bit of that in the last few lessons, specially when we did a summary of entropy but I would like to do it again
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just so we have a list of equations that we want to refer to immediately instead of pouring through a whole bunch of derivations and things.
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What we need to know in order to solve entropy problems?
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One of the things that you need to know is the fundamental equation of thermodynamics.
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In general, the fundamental equation of thermodynamics is not necessarily going to be used for problem solving but it is a nice thing to be able to know.
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The fundamental equation of thermodynamics is DS = 1/ T DU + P RT PV that is the fundamental equation of thermodynamics.
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It is derived from the first and second laws.
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First law being DU = DQ - DW and the second law which is the definition of entropy DS = DQ reversible/ T.
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Definitely that equation you want to know.
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For entropy as a function of temperature and volume, this with what you want to know.
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You want to know the total differential expression which is always easily derived.
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You just take the derivative so it is going to be DS/ DT constant V DT + DS DV under constant T DV.
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The actual equation that you want to bring to the table is DS = CV/ T DT + A/ K DV this is the equation that you want to know just like for energy.
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Absolutely, you have to know this most problems begin with this equation or the next one that I’m going to write.
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In the case of dealing with entropy where entropy is going to be a function of temperature and pressure,
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If pressure is mentioned in the problem we have DS = DS DT under constant pressure DT + DS DP under constant temperature DP.
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The actual derivation that we did was the following DS = CP/ T DT - volume × the coefficient of thermal expansion × DP.
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This is the equation, this is the other one that you want to bring to the table.
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Of course, A and K we want to know those definitions.
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A that was equal to 1/ V × , DV DT under constant pressure and K which is the coefficient of compressibility is -1/ V and
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it is a change in volume per unit change in pressure under conditions of constant temperature.
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These are some of the things that we want to bring to the table when we start thinking about these problems.
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We also want to remember the relationship for an ideal gas, the relationship between the constant pressure heat capacity and
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the constant volume heat capacity which was CP - CV = Rn this is for an ideal gas.
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For phase changes, for problems involving phase changes.
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We have a change in entropy for the process of vaporization = the change in enthalpy for the process of vaporization divided by
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the boiling temperature at which the vaporization takes place.
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The change in entropy of fusion or melting, or the back and forth with solid liquid, liquid solid,
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that is equal to the δ H of fusion divided by the temperature of melting.
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For phase change, solid liquid, liquid to gas, or the other way, coming back the other way gas to liquid, liquid to solid,
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these are the equations that we use when we talk about entropy.
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In going from a solid to a liquid the entropy increases because it is becoming more disorder.
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There is more ways for the molecules to move around.
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In going from a liquid to a gas the entropy rises again, it rises a lot because that now you are going from a condensed state liquid to a gas state.
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There is a whole bunch of room for gas molecules to bounce around it so the entropy rises a lot.
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In going from gas to a liquid, entropy decreases.
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In going from liquid to a solid, the entropy decrease, not a lot but it does decrease.
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Because again you are going from a liquid state which is more disordered to a solid state which is more orderly.
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For an ideal gas, when we are dealing with problems that have an ideal gas you can use the previous equations or
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you can use this set of equations, it is totally up to you.
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For S = S a function of temperature and volume we had DS = CV / T DT + nR/ V DV which implies that DS/ DV under constant T = nR/ V which is positive.
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For entropy of an ideal gas as a function of temperature and pressure we have a differential expression DS = CP/ T DT - nR/ P DP
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which implies that DS/ DP under constant temperature is going to equal - nR/ P.
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Integrating these two, we get δ S = CV Ln T2/ T1 + nR LN of V2/ V1 δ S = CP × LN of T2/ T1 - nR × log of P2/ P1.
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This equation and this equation if you like these two are the differential expressions.
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Third law in entropy, the third law of entropy, the entropy at a given temperature is equal to the integral from 0 to T CP/ T DT.
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This is the entropy of a solid at temperature T and pressure P.
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Let us go ahead in the statement of the third law.
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Let us go ahead and write that down again real quickly.
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It says the entropy of a pure perfectly crystalline solid substance at 0°K is 0.
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The entropy of a pure crystal as 0°K is 0.
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If I want to know what the entropy is at any other temperature, I just take the integral from 0 to that particular °K temperature
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of the constant pressure heat capacity divided by the temperature DT.
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This is the definition of the third law of entropy.
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In computing the entropy of the liquid state of this particular solid, the liquid state of the substance about its melting point,
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in other words I have taken a solid at a particular entropy, it is melted and it is a little higher temperature.
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The entropy of that will include the entropy of the solid, the melting process and then the rise in temperature as a liquid.
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It was entropy at temperature T = the entropy of the solid which is this thing CP/ T DT this standard just means 1 atm pressure.
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When you see the circle there, you are used to thinking the standard temperature pressure , it is not standard temperature,
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this circle is just standard pressure which is 1 atm.
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That is really what it means because now we have come to the point where it is more sophisticated.
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We do not have to stick with 298 K or 25°C but generally we stick with 1 atm.
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These third law of entropy, these are the values that you actually see in your tables of thermodynamic data.
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They are calculated with this particular integral.
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It is the entropy of the solid + the entropy of the change of the melting which is δ H of fusion divided by the melting temperature + from going
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from the melting temperature to another temperature above that, it is going to be integral of the melting temperature to T ×
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the heat capacity of the liquid phase/ T DT, this is the solid phase.
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I’m just accounting for all the phase changes.
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The entropy of the solid that is given from here, the rise of entropy that come from melting, and the rise in entropy that come from taking it
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from the melting temperature to a higher temperature which is not quite yet its boiling temperature.
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For the gas if it actually boils, for the gas at its boiling temperature is the exact same thing,
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just a couple of more terms for the gas above its boiling temperature.
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The total entropy for the gas = 0 to the melting temperature of the constant pressure heat capacity of the solid/ T DT + the entropy change
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that accompanies the melting + the temperature change from the melting temperature
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to the boiling temperature of the liquid phase constant pressure heat capacity divided by T DT.
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This is just the expression of this depending on the temperature range + the entropy change that accompanies the vaporization
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as the liquid goes from liquid to the gaseous state which is going to be isothermal process +
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any extra rise in temperature of the boiling temperature up to my final temperature.
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This time I'm going to use the heat capacity of the gaseous phase / T DT this is it.
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If I want to calculate the third law entropy of the solid, if I want to calculate the entropy of liquid,
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if I want to calculate the entropy of the gas of that substance, this is what I use.
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It implies that I have to know what the heat capacities are, I have to know what the δ H are, what the boiling temperature is,
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what the melting temperature is, all of these things are readily available in tables all over the place.
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It is very easy to get all this information for a solid, for the liquid phase, for the gas phase.
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Let me make this clear, this says gas so this is the G.
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For entropy changes in chemical reactions.
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For problems that involve entropy changes in chemical reaction the δ S = the sum of the entropy, the standard entropy of the products
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including these coefficients - the sum of the standard entropies for the reactants.
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We are looking this up in a table of thermodynamic data, these are tables of third law entropies.
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Just look them up in a table of the data and just take the sum of the products - the sum of the reactants and you have the δ S of that process.
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When we go ahead and write including isometric coefficients and this is nothing that you have done a 1000 times in general chemistry.
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Final thing is the change in entropy of a particular chemical process at a temperature other than 25°C or
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other than the temperature at which you happen to know the change in entropy which for tables of thermodynamic data happened to be 25°C or 290°K.
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In general, it is going to be that particular entropy at some initial temperature + T0 to T of the change that δ of the heat capacities products – reactants.
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Where δ CP is the sum of the heat capacities for the products - the sum of the heat capacities for the reactants
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including the volumetric coefficients because heat capacity is an extensive property.
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It depends on how much is there.
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If I have 1 mol twice that is twice the heat capacity.
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Regarding this, this is valid as long as none of the reactants or products undergoes a phase change in that particular temperature range from T0 to T.
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In this particular temperature range, the reactants and products they cannot change their phase,
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they cannot go from solid to liquid, liquid to gas, things like that, then this is valid.
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If I know the δ S of a particular chemical reaction at a given temperature T 0, I can calculate what the entropy would be at any other temperature.
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Let us do some problems.
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This first set is going to be reasonably simple and straightforward.
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We just want to get comfortable with them.
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That is what we do, we get comfortable with the set of concepts by doing problems.
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That is the only way to do it.
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Let us go ahead and start.
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Let us go ahead and do this in red.
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A temperature of 2 mol and an ideal gas, so I have 2 mol of an ideal gas is raised from 25°C to 225°C, the constant volume heat capacity is 3 Rn/ 2.
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They want us to calculate δ S for this transformation under constant volume so we are dealing with constant volume.
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We are dealing with an ideal gas so that is always nice.
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Let us go ahead and see what we can do here.
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I have an ideal gas, I had two things that I can do.
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I can go ahead and use the equations for the ideal gas which is absolutely fine or I can go back to my original equations,
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the equations that I apply to every system and let the problem tell me how to handle it.
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Me, personally, it is just a habit of mine to always fall back on the basic mathematical equations
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that apply to all systems and let the problem tell me where to go from there.
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I'm going to go ahead and do it that way.
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Part A, I’m going to start off with my basic equation.
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This is the temperature changes constant volume so this is the temperature volume issue.
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DS = CV/ T DT + A/ K DV.
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It is telling me the constant volume so DV is 0 so this whole term goes to 0 because V is constant.
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This is beautiful.
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I’m left with is this so I have DS = CV/ T DT well δ S, if I integrate this expression = the integral from T1 to T2 of CV/ T DT.
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This constant volume heat capacity, I have it is 3 Rn/ 2 and it is constant so I can pull that out.
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It ends up being CV × T1 T2 DT/ T.
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I end up getting δ S = CV × LN of T2/ T1 which I would have gotten any way if I had just used the equation for the ideal gas directly.
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It does not matter, this is just my habit, you do not have to do it this way.
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3/2 Rn/ 2 so 3/2 × 8.314 J/mol-°K × n which is the number of moles which we happen to have 2 mol and 2 mol × the nat log of 225 and 25.
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This is going to be 498°K ÷ 298°K and when I do this I end up with the following δ S = 12.81 J/°K because mol and mol cancel.
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Nice and simple, the entropy of this gas from 25 to 225°C with this particular constant volume heat capacity,
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I have actually increased the entropy, I have increased disorder like 12.81 J/°K.
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That is all that is going on here, temperature increase δ S is positive, everything is good.
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If we end up with a -δ S there would be a problem here so we know that we made some sort of a mistake.
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What I did, take a look at this 498 ÷ 298 and I'm hoping that you guys recognize this
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but sometimes things like this actually slip through the cracks, I do want to mention that.
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°K temperature must be used and here is why, in this case °K must be used and the reason is the following.
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Even though, if I took 225°C/ 25°C you know the °C cancels °C, it is fine.
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225 or 25 is a pure number, you can take the log of a pure number.
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However, 225 or 25 = 9 that 9 does not equal 498/ 298 = 1.67 that is the difference.
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The reason is the following, 225 + 273/ 25 + 273, the 273 still not cancel because they cannot cancel these numbers are not the same.
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Even the units cancel, it looks like you can use that, you cannot use the Celsius temperature but you have to use the Kelvin temperature.
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This is not linear.
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If I multiply it by something then we cancel.
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In this particular case, temperature is not linear, you are actually adding a term, these terms, this does not happen so you have to use the 498 to 298.
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Let us go ahead and do part B.
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In this particular case, we use constant pressure well DS = CP/ T DT - VA DP.
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In this particular case it is going to be the pressure that is constant so that term goes to 0 because P is constant.
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We have the same thing, we end up with just this expression right here.
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When we integrate that expression we get δ S = the integral from temperature 1 to temperature 2 of CP / T DT.
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Let us go ahead and see what the relationship is.
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They gave us CV they said that CV = 3/2 Rn.
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We know the relationship, this is an ideal gas so we know that CP - CV = Rn.
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Therefore, CP = Rn - CV CP = Rn + 3/2 Rn CP = Rn × 1 + 3/2.
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I get CP = 5/2 Rn there we go, that is what I put in to here.
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Therefore, I get δ S and while I do the integration I get δ S = 5/2 Rn temperature 1 to temperature 2 of DT / T = 5/2 Rn LN of T2/ T1.
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I have δ S = 5/2 × 8.314 J/°K × 2 mol × log of 498/ 298.
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I end up with δ S = 21.35 J/°K.
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If I hold the pressure constant and I change the temperature from 25°C to 225°C, I create more disorder.
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That idea if I held the volume constant.
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That is all this is saying.
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Let us take a look at example 2, we have a monatomic solid is measured to have a constant pressure heat capacity of 3.25 Rn,
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we have 2.5 mol of this solid and it is taken from 298°K to 548°K, calculate δ S for this transformation under isobaric conditions.
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Isobaric means constant pressure.
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You can see all these terms, you can see constant pressure, you can see isobaric, you are going to see CP,
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you are going to see molar heat capacity, heat capacity, you just need to interpret what they mean.
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This is constant pressure so we want to deal with, we have DS = CP/ T DT – VA DP.
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That is the general equation.
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Again, I would like to begin with a general equation.
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Let us see, isobaric conditions constant pressure which means DP= 0 so that falls out.
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We are left with δ S = CP × integral from T1 to T2 of DT/ T which is equal to CP × the LN of T2/ T1.
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This is starting to become very simple I hope, δ S = 3.25 Rn × 8.314 J/ mol °K.
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I have 2.5 mol of the solid , mol cancels mol log.
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I’m taking it from 298 to 548 so it is going to be 548 ÷ 298.
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What I get is δ S =241.15 J/ °K.
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The biggest problem you should have what this is one of arithmetic.
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Increase in temperature implies an increase in entropy because a change in entropy is positive and δ is final – initial.
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Let us see what we have got.
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We have the constant pressure heat capacity per mol for aluminum is 20.7 + 12.4 × 10⁻³ × T.
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In this particular case, the constant pressure heat capacity is not constant, it is actually a function of temperature.
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As the temperature rises the heat capacity rises.
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Here we cannot pull it out from under the integral sign.
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We have to leave it there and do the integration but it is pretty straightforward, it is just polynomials with a degree of 1.
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Calculate δ S if 1 mol of aluminum is taken from 25 to 225 ℃.
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In part B, they say if the standard entropy at 298°K is 20.4 J/mol-°K, calculate the standard entropy for aluminum at 498°K.
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Let us take this + some integral which is actually going to be the integral from part 1 so it is really easy.
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Let us go ahead and do part A, taken from 25°C constant pressure heat capacity.
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I have my basic equation of DS = CP/ T DT - VA DP this is going to go to 0 because we are dealing with constant pressure here.
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I have δ S = it is going to be the integral of CP/ T DT as we go from temperature 1 to temperature 2.
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And because it is not constant, it is going to be 298 to 498, I think that is correct.
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The constant pressure heat capacity is this thing right here so I have 20.7 + 12.4 × 10⁻³ T/ T DT.
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What I end up solving this integral which I want to do right now.
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I'm hoping that you can actually take this into separate out, it is going to be 20.7/ T + the integral this thing.
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The T and T cancel so I have two integral.
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And when you solve this, you are going to end up with δ S= 13.11 J/ mol-°K.
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And because it is 1 mol it could be 13.1 J/°K.
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This is just 1 mol, I will just go ahead and leave this mol here.
00:33:21.200 --> 00:33:24.700
That is the δ S for the particular process.
00:33:24.700 --> 00:33:36.100
Part B says, the S at another temperature = the S at a given temperature that I do know +
00:33:36.100 --> 00:33:45.600
the integral from that temperature to the next temperature of the CP/ T DT.
00:33:45.600 --> 00:33:54.500
S of 498= 298, the 298 they tell me is equal to 28.4.
00:33:54.500 --> 00:34:00.700
This integral right here that is exactly what which we calculated, it is the 13.11.
00:34:00.700 --> 00:34:22.100
It equals 41.51 J/mol-°K.
00:34:22.100 --> 00:34:31.000
Let us see example 4, we have carbon tetrachloride and has the boiling point of 77°C and the heat of vaporization of 32.5 kJ/ mol.
00:34:31.000 --> 00:34:36.400
Calculate the entropy change of vaporization for CCL4.
00:34:36.400 --> 00:34:37.500
Nice and easy problem.
00:34:37.500 --> 00:34:53.700
We know that the δ S of vaporization is just equal to the δ H of vaporization divided by the boiling temperature.
00:34:53.700 --> 00:35:05.900
We have 32.5 kJ/mol ÷ 77°C, 350°K.
00:35:05.900 --> 00:35:24.800
I get the δ S of vaporization equal to, I'm going to go ahead and change this to J and ends up being 93 J/mol-°K.
00:35:24.800 --> 00:35:38.600
For every mol of carbon tetrachloride and I take from a liquid to the gas phase, I increased its entropy by 93 J/°K.
00:35:38.600 --> 00:35:47.800
Let us go ahead and see what we can do with this one.
00:35:47.800 --> 00:35:57.800
Part A says, what is the δ S for the transformation of 1 mol of water from 0°C to 100°C at constant pressure?
00:35:57.800 --> 00:36:02.900
The molar constant pressure heat capacity is 75.29 J/mol-°K.
00:36:02.900 --> 00:36:06.300
This is pretty straightforward.
00:36:06.300 --> 00:36:15.300
The second part says the δ H of fusion is 6.01 kJ/mol, the δ H of vaporization is 40.66 kJ/mol,
00:36:15.300 --> 00:36:22.600
the constant pressure heat capacity of steam of water vapor is 37.5 J/mol-°K.
00:36:22.600 --> 00:36:30.700
This is liquid and this is gas, we want to calculate δ S for the following transformation.
00:36:30.700 --> 00:36:42.600
I take ice which is at 0°C and 1 atm pressure and I take it and convert it to steam which is now at 125°C still at 1 atm pressure.
00:36:42.600 --> 00:36:47.700
What I'm doing is I’m taking the solid and I'm converting it into a liquid.
00:36:47.700 --> 00:36:59.900
I'm raising the temperature of the liquid from 0°C to 100°C, to 100°C liquid.
00:36:59.900 --> 00:37:12.600
At 100°C I’m going to take the liquid to a gas and now that gas I’m going to take from 800°C, I'm going to take the gas to 125°C.
00:37:12.600 --> 00:37:17.800
The change in entropy for the system is going to be the change in entropy for each of these.
00:37:17.800 --> 00:37:30.600
The melting, the rise in temperature of a liquid, the vaporization of a liquid and the rise in temperature of the gas.
00:37:30.600 --> 00:37:34.400
I'm just going to add all the δ S so let us go ahead and do that.
00:37:34.400 --> 00:37:36.800
This is going to be constant pressure.
00:37:36.800 --> 00:37:41.100
Under conditions of constant pressure I have the following.
00:37:41.100 --> 00:37:54.400
I have my basic equation and I just write it over and over again because it is nice to keep in my mind, equal CP/ T DT - VA DP
00:37:54.400 --> 00:38:00.000
this goes to 0 because P is constant.
00:38:00.000 --> 00:38:21.100
Therefore, I got DS = CP/ T DT which implies that δ S is going to equal the integral from T1 to T2 of CP/ T DT.
00:38:21.100 --> 00:38:22.700
When I do that, I end up with the following.
00:38:22.700 --> 00:38:53.200
When I do this integration, I end up with DS = CP LN T2 / T1 which = 75.29 J/mol-°K × 1 mol × log of 373 divided by 273.
00:38:53.200 --> 00:38:59.000
I end up with δ S of 23.50 J/°K.
00:38:59.000 --> 00:39:06.600
Just taking water at constant pressure, the liquid water from 0°C to 100°C.
00:39:06.600 --> 00:39:11.700
I have increased the entropy by 23.5 J/°K.
00:39:11.700 --> 00:39:28.200
Part B, part B is going to look like this, the δ S of the total is going to equal the δ S of the melting of the ice + the δ S of the liquid water and
00:39:28.200 --> 00:39:45.200
I take it from 0 to 100 + the δ S of the vaporization of that liquid at 100 + the δ S, as I take that gas from 100 to 125.
00:39:45.200 --> 00:39:50.400
All these 4 δ S, I have to calculate them.
00:39:50.400 --> 00:40:05.000
δ S T = the δ H of fusion divided by its melting temperature + the integral from 273 to 373
00:40:05.000 --> 00:40:15.800
that is this one that is the melting of the CP of liquid phase / T DT.
00:40:15.800 --> 00:40:23.600
That is that one, the δ S of vaporization is the δ H of vaporization divided by the boiling temperature and
00:40:23.600 --> 00:40:38.500
this is going to be the integral from 373 to 398 of the constant pressure heat capacity of the gas or T DT.
00:40:38.500 --> 00:40:42.000
Let us go ahead and put the values in.
00:40:42.000 --> 00:41:05.800
δ S total = 6010 convert it to J/ mol divided by 273°K, that is the melting temperature of the ice + the integral 273 to 373
00:41:05.800 --> 00:41:26.100
of the heat capacity of the liquid water which is 75.29/ T DT + 40,660 J/mol divided by the boiling temperature
00:41:26.100 --> 00:41:45.000
which is 373°K + my final integral which is going to be the integral from 373 to 398= 37.50/ T DT.
00:41:45.000 --> 00:42:02.200
I end up with δ S total = 22.01 + let us go ahead and I will not do these integrals.
00:42:02.200 --> 00:42:28.700
Let us go ahead, this is going to be + 75.29 × log of 373/ 273 + 109 + 37.50 × log of 398/ 373.
00:42:28.700 --> 00:42:51.900
I get δ S total = 22.01 + 23.50 + 109 + 2.43, I get the δ S total of 156.94 J/°K.
00:42:51.900 --> 00:42:58.400
There we go, per 1 mole.
00:42:58.400 --> 00:43:05.200
In order to take solid ice from 0 all the way to water vapor which is 125 the entropy change
00:43:05.200 --> 00:43:10.600
for all of that I had to account for the entropy change of the melting of the ice.
00:43:10.600 --> 00:43:18.500
I had to account for the entropy change in raising the temperature of the now melted ice from 0 to 100.
00:43:18.500 --> 00:43:31.500
I have to account for the entropy change in going from liquid to gas + now the entropy change in going from now gas from 100 to 125.
00:43:31.500 --> 00:43:35.800
That is my final answer.
00:43:35.800 --> 00:43:38.200
Thank you so much for joining us here at www.educator.com.
00:43:38.200 --> 00:43:39.000
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