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 discuss entropy changes for an ideal gas.
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The equations that we developed so far, as we said earlier they apply to any every situation solid, liquid, gas, real gas and ideal gas.
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For all practical purposes we do not really need to take those equations and somehow fiddle around with them
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in order to apply them to the case of an ideal gas because they apply as is.
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However, it is nice to do this because most of the time when we are dealing with gases we are doing really specialized work.
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We always going to treat gas as an ideal gas so it is nice to have some equations specifically for ideal gases
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that you can just turn to when you know you are dealing with an ideal gas.
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You do not have to if you do not want to.
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Again, all the other equations are absolutely valid.
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If you want to do the derivations for yourself, it is always nice mathematically.
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Personally, it gives you a nice sense of accomplishment.
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I think it would be nice to go ahead and actually discuss it specifically for an ideal gas because it does tend to be the system that we talk about the most.
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Let us jump right on in.
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What is nice about the equation for ideal gas is that they tend to be really simple.
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Let us go ahead and start off first of all we will do a little bit of a mathematical derivation, very quick one though.
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In general, you remember that we had for energy with respect to temperature and volume,
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the differential change in energy was equal to CV=CV DT + Du DV under constant T DV.
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But because we are dealing with an ideal gas, what was nice about it, you also remember hopefully.
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For an ideal gas, this term the DU DV T that is equal to 0, that was Joule’s law.
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For an ideal gas, this change in energy with respect to a change in volume under constant temperature actually equal 0.
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This was Joule’s law.
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When you change the volume of an ideal gas, you do not change the energy of the system.
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Basically, what you have when this term goes to 0, what you are left with is just that.
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We have DU = CV DT.
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I will go ahead and put that aside for a second.
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Let us deal with fundamental theorem of thermodynamics.
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The fundamental theorem of thermodynamics, it says the following.
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It said that DS = 1/ T DU + P/ T DV.
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We had DU under conditions of an ideal gas so we will just go ahead and put this into here.
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What we end up with is DS = 1/ T CV DT + P / T DV.
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I will go ahead and write this as DS = CV/ T DT + P / T DV.
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This particular differential, it looks familiar.
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It does not really look altogether that different than what we just did with the other questions, it is the same equation.
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Notice, this is a function of temperature and volume but notice we have pressure here.
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The ideal gas law, again, we are dealing with an ideal gas so it make things simpler.
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We have an ideal gas we have a relationship between pressure, volume, and temperature.
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Because this is temperature and volume every single variable in here of this side has to be expressed in terms of temperature and volume.
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As pressure, we have to express it in terms of temperature and volume.
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We can do that it is right here P = nRT / V.
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I take this and I put in there.
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Here is what I get so DS = CV/ T + nRT / V / T DV.
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The T cancels the T and I'm left with the following equation DS = CV/ T + nR/ V DV.
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When we consider entropy as a function of,
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I’m sorry I forgot my differential here DT.
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When we consider entropy as a function of temperature and volume for an ideal gas this is what
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the general equation takes the form as it is CD / T DT that part is the same + nR/ V DV.
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In other words the number of moles × the gas constant divided by the volume of the system.
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This is the differential coefficient for the volume component of the entropy.
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Let us go ahead and call this equation 1 and for an ideal gas this is the equation that you want to memorize if you want to.
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It is pretty easily derived so it is not a problem but there it is.
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We recall the basic total differential expression for S = the function of T and V.
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What you have is DS = DS DT constant V DT + DS DV under constant T DV.
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I will call this one equation 2.
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When we compare equations 1 and 2, the DT DV DT DV that means this and this and this and this are identified with each other.
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We already know the DS DT under constant V is just a constant volume capacity divided by temperature.
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Nothing is new there.
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However, the change in entropy with respect to volume under conditions of constant temperature for ideal gas is equal to nR/ V.
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That is really nice.
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What we have is DV T =nR / V.
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Notice, this is always positive.
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Therefore, because it is always positive that means an increase in the volume implies an increase in the entropy of the system.
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In other words, if I have a smaller volume, if I have a particular ideal gas and if I make the volume bigger now the gas has more room to bounce around in.
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It becomes more chaotic, it becomes more disorder.
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It is entropy rises.
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T is positive S has to be positive.
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This is positive because this is positive.
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An increase in the volume implies an increase in entropy, we knew that already but it is always nice to have this corroboration,
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that is what is nice about this.
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For a finite change in state, we just integrate the differential expression.
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Integration of equation 1 which I have to go ahead and write again, it is not a problem.
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DS = CV/ T DT + nR/ V DV.
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When I integrate that, I integrate this, I integrate this, I integrate this, and I get the following.
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I get δ S = the integral from temperature 1 to temperature 2 of CV/ T DT + the integral from volume 1 to volume 2 of nR/ V DV.
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If the constant volume heat capacity happens to be constant over the temperature range of T1 and T2.
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In general, we tend to use constant values for heat capacities.
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Heat capacity is actually a function of temperature so as the temperature of the system rises, if heat capacity changes
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but in general over a reasonably good temperature range, it tends to be pretty constant.
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If CV is constant which in most cases it will be, constant over the particular temperature range, we can pull this out of the integral.
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nR our constant so we could pull those out of the integral then we get, when we pull these out of the integral.
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I will go ahead and write that down, it is not a problem.
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We get DS = CV/ the integral from T1 to T2 of DT/ T + nR × the integral from V1 to V2 of DV/ V.
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These are very simple integrations that we can do and we end up with the following equations.
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CV × log of T2/ T1 + nR × log of V2/ V1.
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This is the equation that we want, the final version for a finite change from state 1 to state 2.
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The change in entropy of the system, if I change its temperature and I change its volume or either one, it is a whole lot of constant,
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I will just knockout, I do not need it.
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It is going to be the constant volume heat capacity × the log of T2/ T1 + n × R × nat log of T2/ T1.
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For an ideal gas this will give me the entropy change, nice and simple.
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For a temperature change and volume change or just a temperature change or just the volume change, that is all, very straightforward.
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This was temperature and volume, let us deal with temperature and pressure like we always do.
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For the δ S of an ideal gas, as a function of temperature and pressure here is what we have got.
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In general, we will go ahead and do the derivation.
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We have DH= CP DT + DH DP under constant T DP this was the general expression for the change in enthalpy of the system.
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The enthalpy is just a different way of looking at energy that includes the pressure, volume, work, that was the definition of enthalpy.
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For ideal gas though, this term is 0.
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I would actually write it down.
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For an ideal gas, this term DH DP T = 0.
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Therefore, we just have DH = CP DT.
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Another form of the fundamental thermodynamics that deals with enthalpy instead of just energy is the following DS = 1/ T DH - D/ T DP.
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We have seen this before, this is just the fundamental equation of thermodynamics except now in terms of energy DU
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we have used enthalpy because under conditions of constant pressure.
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That is all that is going on here.
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We have DH = this, so we go ahead and put this there and we end up with DS = CP/ T DT - V/ T DP.
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Notice, this differential expression is in terms of temperature and pressure which means that every single variable
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on the right has to be expressed in terms of temperature and pressure.
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This is volume not a problem, we are dealing with an ideal gas PV= nRT.
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I will just go and ahead express volume in terms of temperature and pressure nRT / P.
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I take this and I put in there I get the following.
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I get the DS = CP/ T DT - nRT/ P / T DP.
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The T cancel and I'm left with DS = CP / T DT –nR / P.
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Let us correct here R / P DP this is the other equation for the differential change in entropy with respect to temperature and pressure.
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I will go ahead and call this equation 3.
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Recall the basic total differential expression for S as a function of T and P it is DS = DS DT P DT + DS DP T DP.
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T and P these 2 I call this equation 4.
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If I compare these 2 that means I have identified this with this which we know already.
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In this case, I’m identifying this for the case of an ideal gas with this right here.
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What I have got is a change in entropy with respect to a change in pressure under conditions of constant temperature for an ideal gas = -nR/ P.
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In other words, if I have an ideal gas and if I hold a temperature constant and I change the pressure, the entropy change is this right here.
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Notice that it is negative.
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What this says that if I increase the pressure of the system and keep the temperature constant, the entropy has to go down,
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that is what is interesting here which says under isothermal conditions, you know what isothermal means, it means T constant.
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If I increase the pressure of the system that implies that the entropy of the system has to go down.
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That is what this says.
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This change is negative, this is the entropy change per unit change in pressure.
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If I change the pressure that entropy change is negative which means it goes down.
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Rising pressure of the system means a decrease in entropy of the system.
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Notice this is isothermal that is what makes this important.
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I will get to that in just a minute, I will say actually a couple of things.
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Let us first talk about a finite change.
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For a finite change, we just integrate the DS expression.
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We have the integral of DS = the integral of CP/ TD T from T1 to T2 - the integral from P1 to P2 and nR/ P DP.
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Again, if the constant pressure heat capacity is constant / the particular temperature range,
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this of course is constant we can pull those out of the integral.
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When we pull these out from under the integral sign we end up with the following.
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Let me actually right them out.
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I get CP × the integral from T1 to T2 of DT/ T - nR × the integral from P1 to P2 of DP/ P.
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I get the change in entropy of the system = CP × LN of T2 / T1 - nR × LN of P2 / P1 this is the entropy change for an ideal gas
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whenever I change the temperature and the pressure or either one.
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Let us go back and talk about what I was going to talk about up here regarding the negative.
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Notice that the change in entropy with a change in pressure under constant temperature is actually negative.
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It is -nR/ P so once again an increase in the pressure of the system implies a decrease in the entropy of the system.
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Here is why it makes sense, you probably think yourself and say if I increase the pressure of the system
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does it mean that the volume is actually going to get bigger?
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If the volume gets bigger does not the entropy increase?
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Yes, but we are holding the temperature constant, an ideal gas PV = nRT.
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If temperature is constant that means this n, this R, this T, they are all constants.
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That means that the right side of the equation is constant.
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If I increase the pressure of the system the volume of the system have to go down in order to retain the fundamental equality.
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When a volume of the system goes down, we already know that when volume goes up entropy goes up
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which means that when volume goes down entropy goes down which makes sense physically.
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If the volume increases now there is less room for the molecules to bounce around, there is less disorder in the system, the entropy goes down.
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Pressure actually alternately comes down to a volume.
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Let us go ahead and finish this off with a little bit of example.
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Again, let us finish with a theoretical discussion, we would be doing a lot of example problem.
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Example problems are going to come in a couple of lessons and there are going to be several of them.
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We are going to do in bulk, all at once.
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We have a chance to reuse these equations over and over again in the context of problems.
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Do not worry there are plenty of examples to come, just like there were for energy.
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2 mol of oxygen gas we will go ahead and treat it as an ideal gas is taken from 25°C.
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I’m going to do this one in red, I think.
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It is taken from 25°C and 1 atm pressure to 100°C and 5 atm pressure.
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Here you are changing the temperature and we are changing the pressure.
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Calculate δ S for this transformation, the molar constant pressure heat capacity = 7R / 2.
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This is nice and simple.
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Since we are dealing with constant pressure and it is an ideal gas, I can go ahead and just fall back on the equation.
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If I do not know it, I can derive it, it is not a problem.
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I have CP × LN of T2 / T1 -n × R × LN of P2 / P1.
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Just plug my values in δ S.
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The biggest problem you run into is arithmetic.
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We have 7/2 of 2 mol, the molar CP is 7 R/ 2, molar CP that is CP/n = 7/2 R.
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The CP = 7/2 Rn that is what it means, molar means divided by the number of moles.
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For actual heat capacity you have just move the n over so you have 7/2 Rn.
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We have 7/2 × 8.314 J/ mol-°K × n which is we have 2 mol of the oxygen gas.
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mol and mol cancel × the log 25° to 100° so 100°C is 373°K, we have to use Kelvin temperature divided by 25 which is 298 - the number of moles which is 2 × R which is 8.314.
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I’m going to keep writing the units, it is not a problem 0.314 × log of 5 atm/ 1 atm.
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Hopefully, I have done the arithmetic correctly and I hope you will go and check this out.
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I know I tend to make arithmetic mistakes but this is what is important, this right here, not the actual arithmetic.
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You end up with - 13.7 J/-°K.
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In this particular case, notice the change in entropy is negative.
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The affect of temperature on the entropy is going to be positive.
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The effect of pressure on the entropy is going to be negative.
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In this particular case, the temperature from 25 to 100°C is overcome by the increasing pressure from 1 atm to 5 atm.
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In this particular case, the effect of pressure, the effect of P exceeds the effect of T which is why it ended up with a negative entropy change.
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There you go.
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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, bye.