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 talk about the properties of the Helmholtz and Gibbs energies.
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Let us jump right on in.
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For the Helmholtz energy, one of the fundamental equations that we have is the following.
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We have DA = - S DT – P DV.
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In this particular case, the temperature and the volume are the natural variables for the Helmholtz energy.
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Essentially, what we are saying is that the Helmholtz energy is a function of temperature and volume.
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We can go ahead and express the total differential this way.
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DA is the general total differential, it is going to be DA DT we already know this, holding the other variable constant not pressure,
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this is volume, DT + the partial derivative with respect to the other variable DA DV holding that one constant DV.
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This is a general total differential.
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This is the one that we actually extracted from the fundamental equation of thermodynamics.
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This is one of the 4 fundamental equations of thermodynamics.
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If we equate this and this, and this and this, here is what we get.
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Equating the differential coefficients, we get DA DT with respect to V = - S and DA DV holding T constant = -P.
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Basically, what this says is that for every unit change in temperature,
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a change in the Helmholtz energy of that system is going to equal negative of the entropy.
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What these negative signs tell us, this is a rate of change, the rate at which
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the Helmholtz energy changes as you change the temperature is going to be a negative of the entropy at that moment.
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That is all this is saying.
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If we make a change in the volume of the system, the Helmholtz energy of that system
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is going to equal the negative of the pressure of the system at that point.
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We are holding V constant, here we are holding T constant.
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Because of these negative signs, this is what it means.
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Because of the negative signs these equations they say the following.
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Number 1, an increase in the temperature implies a decrease in the Helmholtz energy.
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That is what the negative sign means, if you increase this, this is dependent on this.
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If this goes up, this negative sign means that this goes down.
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An increase in temperature implies a decrease in the Helmholtz energy.
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The higher the entropy, the higher this value is, the more quickly the Helmholtz energy decreases.
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It is important when you see equations like this, you are equating some thing, some value.
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You are equating it with a rate, this is a rate.
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It is very easy to forget that we are talking about a rate of change.
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An increase in temperature implies a decrease in A, the higher the entropy, in other words, if S itself is a bigger number,
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the higher the entropy of the substance the greater the rate of change at which it is changing when you change T.
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Let us say I’m at a particular temperature and let us say that my system has an entropy of 50 units.
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If I raise the temperature by a certain number of degrees, the Helmholtz energy of the system is going to decrease at a rate of 50 J/°K.
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At that point, if my entropy was actually higher, if it was 100 then at that point it is going to decrease by 100 J for every °K.
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That is what this means.
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The rate, the higher S is, the fact that there is something going to drop the negative sign.
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The second one says, exactly what you think based on what we did for the first one.
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An increase in the volume implies a decrease in the Helmholtz energy.
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Increase in V implies a decrease in A.
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I’m just making sure that I'm keeping all of my variables clear here.
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What this says, the higher the pressure at that moment, the higher the P the greater the rate of change.
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In other words, the greater the rate of change or decrease.
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If I’m at a particular temperature and if I'm at a particular pressure,
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if I change the temperature the Helmholtz energy is going to change by whatever the pressure happens to be.
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It is going to be a rate of change.
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If the pressure is higher, the rate of change is going to be bigger, the decrease is going to be bigger if I'm increasing the temperature
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and it is the other way around.
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If I’m decreasing the temperature it is going to go the other correction, the Helmholtz energy is going to increase.
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It is very important.
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We are talking about, what we want to do here is we want to cover every single base possible.
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As far as chemistry is concerned, we are going to concern ourselves with Gibbs energy,
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we are not going to concern ourselves with the Helmholtz energy but what we need to know that it is there.
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If for any reason we decide to run an experiment where we are actually holding temperature and
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volume constant, then we have to use the Helmholtz energy and not that Gibbs energy,
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because the Gibbs energy is for conditions of constant temperature and pressure.
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Constant temperature and pressure is what 99% of all chemical reactions are run under so
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that is the standard laboratory conditions for chemistry, not for other things.
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We would do the same thing for G.
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We do the same thing for G, the Gibbs energy.
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I think I would go back to blue.
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Once again, we take a look at our fundamental equations for G so we have DG = - S DT.
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This is going to be + V DP.
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In the case of free energy, the Gibbs energy it is the temperature and pressure that are the natural variables.
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The total differential DG is going to be DG DT constant P × DT + DG DP at constant T × DP.
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When we equate the differential coefficients, this and this, we have the following relationships.
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The rate of change of the Gibbs free energy per unit change in temperature holding pressure constant is – S,
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the same thing as we saw before the Helmholtz.
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The other relationship is the rate of change of the free energy as you change the pressure of the system
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while you hold the temperature constant = V.
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This is very interesting.
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What this says, I’m going to write down in just a second.
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I will go ahead and tell you.
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As you change the temperature, if you increase the temperature, the free energy, the Gibbs energy of the system is going to decrease.
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It is going to decrease as a rate, at a rate of whatever entropy that happens to be at that moment.
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Here, if you have a system and if you increase the pressure because this is positive,
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if you increase the pressure you increase the free energy of the system and the rate at which you are increasing the free energy
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is going to equal whatever the volume of the system happens to be.
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That is all these says, these equations are not mysterious, just take them for what they are.
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Completely a phase value.
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As you change the pressure, the free energy of the system is going to change the rate
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at which it changes is going to equal the volume at that particular moment.
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That is all this is saying.
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The first one, it says that an increase in T implies a decrease in G.
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A decrease in G is –S.
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Once again, the higher the entropy, in other words, the higher S is the higher the entropy S, the greater the rate of decrease.
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I will just say the rate of change because it might be an increase and decrease in temperature.
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The second one says, now this is positive so an increase in the pressure implies an increase in G.
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The bigger the volume is at that particular moment, the greater the V or volume the, the greater the rate of change of G.
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That is all that is going on here.
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These are very important relationships and again what is important to know is that,
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it is the 4 fundamental equations of thermodynamics.
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They are not really 4 fundamental equations, there is only 1 fundamental equation of thermodynamics.
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These other ones were derived simply based on the definitions of enthalpy, the Helmholtz energy, and Gibbs energy.
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Those are the composite functions.
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Remember what we did, the only real thermodynamics functions are temperature, energy, and entropy.
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The composite thermodynamic functions, the ones that are accounting devices that take other things into consideration,
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that was the enthalpy, the Helmholtz energy, and the Gibbs energy.
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Again, all of these equations they come from only one fundamental equation.
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These are just different variations of it and it is a way of expressing some particular property of the system, some state property in this case.
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G in terms of other properties of the system that is all we are doing.
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At entropy volume, we are expressing G as a function of temperature and pressure but it is also going to be volume and entropy.
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That is all we are doing, we are expressing one state property in terms of other state properties.
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Things that are reasonably easily measurable.
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Let us go ahead and take this particular equation right here and we can actually integrate this.
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Let us go ahead and go to the next page.
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We can integrate the DG/ DP at constant T = V to obtain an expression for the Gibbs energy of a pure substance at a constant temperature.
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We took an expression for the Gibbs energy of a pure substance at constant temperature
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from standard pressure which is equal to 1 atm to any other pressure P.
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Let us go ahead and do that here.
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We have our DG/ DP at constant T = V temperature is being held constant so essentially what this is, this is going to end up being just DG/ DP.
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When we integrate this, let us go ahead and write this way.
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DG = V DP again, we are holding the temperature constant.
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For all practical purposes, the G is just a function of the pressure so T is held constant.
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That is all it says, the function of two variables.
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If we hold one constant it essentially just becomes a function of one variable.
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In this case, the one variable is P so I can go ahead and use regular differential notation.
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I can go ahead and integrate both of these expressions and end up getting the following.
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I end up with δ G = the integral of V DP from P0 which is a 1 atm to any other value of P.
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Well, δ G is just G - G0, the G0 on top is a standard free energy, free energy at 1 atm pressure.
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Remember, this little degree sign it does not mean standard temperature and pressure,
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it only refers to standard pressure of 1 atm because temperature can change.
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It is equal this P0/ P V DP and then when we end up moving this over, we end up with the following.
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G a function of temperature and pressure is going to equal G0 which is just a function of temperature.
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I will explain this in just a minute, 0 to P V DP.
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If you are going to find free energy at a specific pressure, you are fixing the pressure, say we fix the pressure 1 atm that is what P0 is.
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The 0 on top of the degree sign, that is standard conditions, standard condition is 1 atm.
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If you fix the pressure, now the pressure is no longer variable.
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The free energy becomes only a function of the temperature that is why we have G °of T.
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However, over here when we are actually calculating what the new G is at a new temperature and at a new pressure,
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This G that you are looking for is a function of the temperature and pressure.
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That is why this is G (TP) it is a function of temperature and pressure but this one is only a function of temperature because of that degree sign.
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That degree sign says we are fixing the pressure at 1 atm, we are locking it in.
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There is our standard, therefore, I no longer have to consider the pressure.
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If you happen to know what the standard free energy is at 1 atm and
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if you want to know what the free energy is off the system at another pressure, this is the equation that you use.
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Let us go ahead and say a little bit more.
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If the substance is a liquid or a solid, because all of these equations that we are developing they actually apply across the board,
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to any substance, any system, and anytime.
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These are universal, it does not apply just to gases or liquid or solids.
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If the substance is a liquid or a solid then the volume can be treated as a constant.
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We know already that for a liquid and a solid, unless you are applying a lot of pressure and by a lot of pressure I’m talking thousands of atmospheres,
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the volume of liquid or solid is not going to change very much.
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In fact, it is going to be very little.
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You remember K, the compressibility coefficient is very tiny on the order of 10⁻⁵ or 10⁻⁶, things like that.
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For all practical purposes, the volume is constant so it actually ends up becoming independent of the pressure.
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If the substance is a liquid or a solid, then V can be treated as a constant and we can pull that V out from under the integral sign.
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You will end up with the following once you integrate it.
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For a liquid or solid it is still going to be a function of temperature and pressure but you do not have to worry about the integration,
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It is going to be + the volume × just the δ P which we will write as P - P0, something like that.
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Unless, P is really high like the difference between 1 and let us say 3000 atm, P is extremely high.
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This term V × P - P0 is going to be very small which means we can usually ignore it.
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Therefore, our G is actually just going to equal G(T).
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Basically, what we are saying is if you are dealing with a liquid or solid this is the general equation for it right here.
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Because V is essentially constant, you can pull it out from under the integral sign of the equation
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that we just had at the previous slide and you end up with just the integral of TP which is δ P.
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That is what this is right here.
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But unless, the pressure is really high this can pretty much drop out.
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When this drops out, you are just left with the free energy of a system of a liquid or solid,
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it is pretty much just equal to free energy under standard conditions.
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For liquids or solids, the free energy of the system is not going to change that much under pressure, that is all this is saying.
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For an ideal gas, let us go ahead and write our expression again.
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We had G as a function of T and P is going to equal G (T) + the integral of V DP from there to there.
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For an ideal gas, V = nRT/ P, just rearranging the ideal gas law.
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We go ahead and put this in here and we get the following.
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We will get G (TP) = G standard function of T + nRT integral DP/ P when P is 0 to P.
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We get our final equations.
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This is for an ideal gas.
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G0 at T + nRT LN (P)/ P0.
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We said that P0 is 1 atm so this just becomes LN P.
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We will go ahead and use this one as my basic equation.
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That is fine, I will just go ahead and write it down.
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Since P is 0 = 1 atm we end up with G of TP = G0 T + nRT × log of the pressure,
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if you are looking for the free energy of a particular substance, of an ideal gas.
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I’m sorry this is an ideal gas under a certain temperature and pressure conditions, you find the standard free energy
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which is under 1 atm conditions and you just take the number of moles you are dealing with × R × T × log of the new pressure.
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That is all that is going on here.
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This is just an equation to find the free energy of an ideal gas at a particular temperature and a particular pressure.
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Special note, the free energy is a very special thing.
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If the function G (TP) is known, if you happen to know the functional form, this is free energy is some function of temperature and pressure.
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If you happen to know what that function is, it is known, then all other thermodynamic functions can be derived from it.
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Can be derived not just from it, all to say all the other thermodynamic functions can be derived
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using the relations that we have + the definitions of the composite functions.
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We are actually could be doing one of these problems when we do the whole problem sets for free energy.
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At a specific temperature, they should be obvious from the equation above but not a problem, we will go ahead and write it down anyway.
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At a specific temperature, the G of an ideal gas, the free energy of an ideal gas is a function of pressure only.
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Of course, if you hold one of these constant at a certain temperature, that temperature it is just going to be a function of P.
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Let me go ahead and talk about the temperature dependence of the Gibbs energy.
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We already have a relation for the temperature dependence of the Gibbs energy.
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Depending on a particular problem that you are working on or a problem that you run across,
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we have developed different relations to express the relationship,
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that is temperature dependence of Gibbs energy to what extent does Gibbs energy rely on temperature.
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It is not just one equation, we actually have several equations.
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We will go ahead and derive those now and they are actually quite simple.
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Let us go ahead and start with the first one, the one that we already know.
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We know that DG DT under constant pressure = -S.
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We know this already, as the temperature changes the free energy of the system is going to change according to –S.
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This is the relationship, this expresses the temperature dependents of G on T at constant pressure.
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I’m going to call this equation number 1.
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We know the G = H - TS that is the definition of G.
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Let us go ahead and move this around a little bit.
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G - H = - TS and let us go ahead and divide by T so we get G - H ÷ T = - S.
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-S = G - H/ T - S = DG DT at constant P.
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We are going to go ahead and put this into here and we are going to get our second equation.
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DG DT at constant P = G - H/ T.
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This is just another way of expressing so I will call this equation number 2.
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This is just another way of expressing the temperature dependence of the Gibbs energy.
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For every unit change in temperature the Gibbs energy is going to change according to what the G is
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at the time - the enthalpy that quantity ÷ the temperature of the system.
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That is all we are doing, let us say I do not have the entropy but let us say I have the G at the time, let us say I know the temperature and enthalpy.
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This just gives me another way of expressing the change in the Gibbs energy as a function of temperature.
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Sometimes, in this case I’m going to say do not worry about Y.
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Sometimes, we will want the derivative of the function G/ T with respect to T.
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In this particular case, we get the following.
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D (G/T DT) is equal to, we have a ratio.
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When you take the derivative of a ratio you get this × the derivative of that - that × the derivative of this all over that squared.
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1 - S let me go ahead put P here.
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We will go ahead and substitute the equation.
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We already have the equation DG, always variables floating around.
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T at constant P = - S so this is our first equation, equation number 1.
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We know the DG DT P = - S so we can go ahead and put that - S in for this DG DT P and we end up with the following.
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The partial derivative of the function G/ T with respect to T under constant pressure is going to equal
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- TS - G/ T² that = - S/ T - G/ T² or it also = -ST + G ST + G/ T².
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We know that G = H - TS so H = G + TS, G + TS which is ST + G that is actually equal to H.
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I can go ahead and put that in there and I have my final form which it can end up being equation number 3.
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Let me write it again.
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The derivative of the function G/ T with respect to T constant P is going to equal - H/ T².
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This is equation 3 and this is usually the equation let us call the Gibbs Helmholtz equation.
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This is the form that is called the Gibbs Helmholtz equation.
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Actually, any one of these equations I’m actually going to write one more you can consider as the Gibbs Helmholtz equation.
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This is just variations of the same equation, pretty much this one.
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We just took it and we are expressing the temperature dependence of the free energy in different ways depending on the particular problem we want.
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We can go ahead and stick with this and at a particular problem, on a particular situation,
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we can derive all of these which is how it actually happened.
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What they decided to do was they decided they were talking about free energy so when we present this material to you as students,
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we are just going to present all of these variations of the equation for you.
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That is all we are doing here.
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This is called the Gibbs Helmholtz equation and this is usually the form that you see it in.
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The derivative of G/ T with respect to T holding P constant = -H/ T².
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If you want to know how the free energy ÷ temperature changes as you change the temperature,
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it is equal to the -enthalpy/ the temperature².
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It is just equations.
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A lot of times in thermodynamics, especially when it comes to things like free energy, when we are dealing with G,
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I do not worry about that, I am actually going to explain what G is and what Helmholtz is in relation to the system.
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I’m going to give a physical meaning to it, particularly when I start doing the problems.
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A lot of this is just mathematical so just create it mathematically, sort of pull yourself away a little bit.
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And if you do not completely understand what is happening physically, that is not a problem, that is natural.
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Just go ahead and treat it mathematically.
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Let us go ahead and give one more version of the Gibbs Helmholtz equation.
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Let us go ahead, if I take the derivative with respect to T of 1/ T that is going to equal -1/ T².
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We have D of 1/ T = -1/ T² DT.
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I'm going to move the T² over so I have - T² D of 1/ T = DT.
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I can do this, I can go ahead and create this thing D of G/ T with respect to D of 1/ T.
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I can go ahead and rewrite this as DT = - T² D of 1/ T.
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I have D of GV/ DT, I have this thing.
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I have this, I'm going to go ahead and put this in the denominator so I end up with.
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Let me go ahead write this whole thing out, things are confusing enough as it is.
00:36:32.700 --> 00:36:44.400
We have the D of the G/ T DT = - H / T², that is the Gibbs Helmholtz equation, that is equation number 3.
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I went ahead and I did a little bit more manipulation with this D of 1/ T.
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I expressed it this way so DT = -T² D 1/ T.
00:36:54.100 --> 00:37:02.200
I went ahead and converted to a partial derivative and now wherever I see the DT which is right here, I can go ahead and replace it with this.
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I end up with the following.
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I end up with D of G/ T - T² D of 1/ T = - H / T².
00:37:17.900 --> 00:37:33.600
The - T² cancels and I end up with a partial derivative of G/ T with respect to 1/ T = H.
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And this is going to be equation number 4 for the Helmholtz equation.
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And again, these are all just 4 different variations of the same equation.
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The temperature dependence of the free energy, when I change the temperature how does the free energy change?
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This is just a different way of expressing it.
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Personally, I do not like this form, it is confusing and you probably not going to run across it
00:37:58.600 --> 00:38:01.500
altogether that much except maybe in a couple of problems.
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The most problems are really only going to require some mathematical manipulation.
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The way we put those problems in these problem sets with the free energy section of your books
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simply to get you familiar with the mathematics of the equations.
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You have noticed, in this free energy section when we dropped all of those fundamental equations
00:38:22.100 --> 00:38:25.600
at Maxwell relations there is a lot of mathematics going on.
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One of the problem sets in thermodynamics books for the free energy section
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tend to involve mathematical manipulation a little more so than with energy and entropy.
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That is all that is going on here, it is not altogether important that you totally grasp this.
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Again, if there is one equation that you should probably remember more than any other it is the original one, the DG DT.
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This one that was the equation number 1 that gave rise to the equation number 2 and equation number 3.
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This one is the Gibbs Helmholtz equation.
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They are all Gibbs Helmholtz equations, they are all just variations of this one.
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In any case, I will go ahead and leave it that.
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We will see some more of this when we actually do the problem sets and probably make more sense.
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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.