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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In the previous couple of lessons, we talked about the changes of energy and state and we did them under constant volume.
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We are going to hold something else constant.
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We are going to see what happens to the energy of the system when we hold the pressure constant.
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Let us jump right on in.
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Most law procedures are conducted under constant pressure conditions.
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Most law procedures are carried out under conditions of constant pressure, since it is a lot easier that way.
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One of the experiments that you do end up benched up, you have the constant pressure and the atmospheric pressure.
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The reaction runs and the system changes pressure.
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It accommodates that pressure inside the system.
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It is pretty much going to be the same as the pressure outside.
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There is always an equilibrium with that.
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Constant pressure conditions are very simple experimentally.
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Most law procedures are carried out under constant pressure conditions, under constant P conditions which is atmospheric pressure.
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The pressure which happens to be the pressure of the system, happens to equal the external pressure, that is all that means.
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Let us start, we are just playing with mathematics, that is what we are doing.
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We are hoping that manipulating the mathematical expressions will show us something, will give us something that we recognize.
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Maybe we will see something that we can move this around.
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Let us start with, DU =DQ – DW.
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DW = P external × DV.
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DU =DQ and again we are working under constant pressure conditions - P DV.
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Since P is constant, this is just the first law of thermodynamics.
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It is constant and upon integrating this, we get the following.
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We get integral from an initial state to a final state of DU = integration from an initial to a final state of DQ sub P - P × the integral from initial state of DV.
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Pressure is constant, we can pull it out from the integral.
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We end up with the following.
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This is going to be, instead of using the δ symbol, I’m not going to write it explicitly.
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This U as a state functions, so this becomes U2 – U1.
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This is a path function so it just becomes Q sub P, the heat withdrawn from the system
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at constant pressure is going to be - P × volume is a state functions so this is going to be V2 - V1.
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When I distribute this and move it over to the left, I get the following.
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I get U2 – U1 + P V2 - PV 1 = QP, so far so good.
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We arrange this and put the U and P together and 2’s together.
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I get U2 + PV 2 - U1 + PV 1, this is a –U1 not this one and - this one, that is equal to QP.
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Since P =P1 = P2, the pressure is constant, it does not matter.
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The initial pressure is the same as the final pressure so we can just call it just P.
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That is what P is.
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We can go ahead and put wherever we see a P as associate with the 2, we can put the P2
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and wherever there is a P associated with the subscript 1, we can go ahead and put 1.
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We get the following.
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We get U2 + P2V2 – U1 + P1V1 = QP.
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U is a state function, it is just a function.
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Pressure is a state function, volume is a state function, so P is a state function, volume is a state function.
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Therefore, the sum of the state function, the product state function is a state function.
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This whole thing and this whole thing, they are state functions.
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This U + PV is a state function.
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We are going to define a new state function H = U + PV, we call this the enthalpy of the system.
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The enthalpy of the system is equal to the energy of the system + the pressure of the system × the volume of the system.
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Because of what we just did mathematically, since U2 + P2V2 - U1 + P1V1 = QP.
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This is H2, this is H1, what we end up with is δ H =Q sub P.
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The heat that is transferred from the surroundings in a process that is done under constant pressure = change in enthalpy of the system.
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This is different from when we did constant volume number that the heat transfer during transformation
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under conditions of constant volume = the change in energy of the system under conditions of constant pressure.
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The heat transfer from the surroundings, the heat lost by the surroundings = the change in enthalpy of the system.
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Enthalpy does the same thing that energy does.
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Enthalpy does under constant pressure what energy does under constant volume.
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It is behaving the same way.
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We had, under constant volume, condition that is equal to change in energy.
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Here, the heat transfer is equal the change in enthalpy, it is different.
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Enthalpy is the energy + a little extra.
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That is what is going on, or when I think about energy = enthalpy - the pressure volume work that the system does.
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At constant pressure, what is says is at constant pressure the heat withdrawn from the surroundings, in other words the heat it is going into the system.
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Surroundings = the change in enthalpy of the system.
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We have δ H = QP, δ U + PV = QP, the δ operator is a linear operate, it distributes.
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δ U + δ PZ = Q sub P.
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Since P is constant we can pull it out of there, what we will get is δ U + P δ V = QP.
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The way of looking at it.
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Heat effects are usually measured at constant pressure.
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Again, it is just very easy.
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Constant pressure is just a natural way things, it was constant pressure, atm pressure.
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These effects are changes in enthalpy not changes in energy.
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If we perform a process under constant volume then any heat transfer = change in energy of the system.
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If we do a process under constant pressure, if we measure some heat transfer, heat change, heat effect under conditions of constant pressure.
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As the change in the enthalpy of the system, enthalpy is equal the energy + the pressure volume work that is done by the system.
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It is a little less, if we want a change in energy we have to take the heat and we have to subtract any pressure volume work that is done.
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We will see what happens in just the second.
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Sure enough if we want δ U, then δ U + P δ V = Q sub P.
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If we want δ U, move that over, = Q sub P - P δ V.
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If we know what the heat lost by the surroundings is and if we know what the pressure is, we know the pressure is the external pressure,
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the pressure the atm whatever happens to be.
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If we know a change in volume that the system undergoes, you can find a change in energy.
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If we know QP which is the heat lost.
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Sorry if I keep repeating myself, I’m going to do that every time a variable I’m going to say what it is.
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You need to hear it over and over again.
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If we know QP which is the heat lost by the surroundings, take the surroundings point of view, that is how we measure changes.
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Δ V of the system and we can find DU.
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It is very, very simple.
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And we will do a problem like that just a little bit.
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Δ H =QP our basic equation, the differential version is the following.
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It is the integrated version of finite version, the differential version is D not DU, D is enthalpy DH = DQ sub P.
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H is a state function.
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We saw that, it is the combination of state functions.
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The enthalpy is a state functions so DH is an exact differential.
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Now we start dealing with the mathematics.
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We know what exact differential is.
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If we choose temperature and pressure as the variables for enthalpy, enthalpy is a function of temperature and pressure we can express it as follows.
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Let us say choosing T and P as our variables, we have that enthalpy = the function of temperature and pressure.
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Because it is an exact differential, we can express it like this, DH = DH DT under conditions of constant pressure holding the other × DT + DH DP × DP.
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We have our total differential.
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If the enthalpy of the system is a function of the temperature and the pressure, a small change in temperature, enthalpy will change by this much.
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A small change in pressure, the enthalpy will change by this much.
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If I add these 2 changes, I get the total differential change in enthalpy.
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If I integrate this expression, I get the total change in enthalpy.
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This is just mathematical consequences of it being a state function.
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A state function is an exact differential and an exact differential can be expressed like this, if you choose the variables appropriately.
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It is your choice as far as the variables are concerned.
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It is not like when you are doing this a 100 years ago or 150 years ago, it is not like they just knew automatically to check temperature and pressure.
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You could express this as temperature volume.
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You can express it as pressure volume.
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If you want to use an expression as temperature, pressure, and volume, you have three terms here.
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It is just what you are seeing is the most fruitful work that is come about.
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It is not like we knew automatically let us just take T and P, we do not.
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We just start fiddling with it and you see what you get.
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The derivations that bear fruit, those are the one that end up in the textbooks, that is how it works.
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DH = this, since P is constant, this DP the change in pressure is 0.
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Pressure does not change so that DP is 0 that term goes to 0 because that is 0.
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What you are left with is DH = DH DT sub P × DT, but DH = DQ sub P.
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That is the equation right there.
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Here we go again with the heat, DQ sub P = DH DT sub PDT.
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Or if I bring this down DQ sub P /DT not partial = DH DT sub P.
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The heat was drawn from the surroundings is related to the temperature change of the system.
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This is a heat divided by the temperature, this is heat capacity, it is another type of heat capacity.
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It is the constant pressure heat capacity.
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This heat capacity is associated with this partial derivative, the change in enthalpy per unit change in temperature.
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We are associating this derivative with a quantity that we can measure.
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We can measure how much heat the surrounding loses.
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We just stick a thermometer in it and see by how much that temperature actually drops.
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We can find out the change in temperature of the system.
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We can stick a thermometer in the system and see how much the temperature goes up.
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When we divide the two, we get the heat capacity.
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DQ sub P / DT, this ratio is the heat capacity, it is a constant pressure heat capacity of the system.
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We have a constant pressure heat capacity is defined as DQ P DT, it is defined as the amount of heat that is lost
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by the surroundings divided by the temperature change of the system.
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Or if you want to take the systems point of view, the heat gain by the system divided by the temperature change
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temperature rise of the system, if we preferred to stick with the systems point of view.
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Again, when we measure heat exchanges we are not measuring the system but we are measuring the surroundings.
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That is when we say, that happens to be associated with identified with this partial derivative which is the total differential DT sub P.
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This partial DH DT under constant pressure is now identified with and easily measured quantity heat capacity under constant pressure.
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I just pull the pressure constant and I measure heat capacity.
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We now write DH = CP DT + DH DP sub T × DP.
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We write this, which is the total differential except now because we associated it did this with the heat capacity CP instead of that partial derivative.
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Let us see what happens.
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Let me rewrite that again so DH = CPDT + DHDP sub TDP.
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If the transformation happens at constant pressure there is a general equation, we are not holding anything constant yet.
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If the transformation happens at constant pressure the DP = 0.
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This goes to 0, all we are left with is DH = CP DT or if we integrate this we get DH = the integral from
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temperature 1 temperature 2 of the constant pressure heat capacity × DT.
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Or if over the temperature range, if CP happens to be constant we can just pull it out of the integral and we get the other version of it,
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a finite version which is δ H = CP δ T, this is the version that you see in your general chemistry classes.
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These equations they apply to any system of fixed mass, any system like gas, liquid, solid.
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Just like the previous equation, in front of a constant volume.
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This is not just for gases, in general we tend to do the examples for gases but they apply to any system.
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These are general equations applied at any system of fixed mass, gas, liquid, or solid, as long as there is no phase changed during the process.
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The equation applied in the system of fixed mass, as long as no phase change takes place or
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no chemical reaction takes place during this process, during the transformation.
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If a chemical reaction takes place from the thermodynamic things happening, that is not just this, this is a chemical reaction.
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This equation holds as long as this is the case.
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Important equations, let us go ahead and do that.
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For this particular lesson, the important equations to keep in mind.
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We will be going through all of these when we do our problems in bulk when we start doing entire lessons of problems.
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Do not worry, we are going to give a lot of them.
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We have the general definition of enthalpy, U + PV.
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It is very important you have to know this definition.
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We have δ H = δ U + P δ V that is just the finite version of it.
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We have δ H =QP or the differential version DH = DQP.
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A state function, exact differential, path function, inexact differential, dealt with the fact that these are the same confuse you.
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We have the basic mathematical relation DH = DH DT under constant pressure DT + DHDP under constant T and DP which is the same as.
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That is the final, let us go ahead write the other version, the DH = the constant pressure heat capacity DT + DH DPTDP.
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At constant pressure, these are the general equations at constant pressure here is what we have.
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We get DQ sub P = DT sub PDT, in other words DQ sub P = constant pressure heat capacity × DT, DH = CPDT because DQP = DH.
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And if we integrate this, we get the change in enthalpy of the system = the integral from one temperature
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to the other temperature of the constant pressure heat capacity × the differential change in temperature.
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Let us do some problems.
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During these individual lessons when we are discussing mostly theory, examples are going to be reasonably simple.
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The more sophisticated or complicated examples, the broader range of them are going to come when we do them in bulk.
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So no worries, I will not leave you hanging across.
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Let us see what we have, sample 1.
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Example 1, the constant pressure molar heat capacity for solid silver is, this in J/ mol K.
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Be very careful when you are reading these questions, every single word matters.
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It is not specific heat capacity but a molar heat capacity.
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If it is a constant volume heat capacity, its constant pressure heat capacity.
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For sold silver is this, 23.4 + 0.0063 T.
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Notice, this is a function of temperature which means as the temperature changes the heat capacity changes which is the real behavior of heat capacity.
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The capacity changes with temperature.
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It is usually not constant over a large range of temperatures.
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If 100 g of the silver is taken from 25° C to 900° C at 1 atm pressure, what is the change in enthalpy of this 100 g of silver that is our system?
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Let us see what we have got.
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Molar heat capacity, first of all let us go ahead and take care of this 100 g.
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Let us change it to mol, so we have 100 g × 1 mol silver is 107.9 g.
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If I did my arithmetic correctly, this will be 0.0927 mol of Ag.
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I'm not pretty big concern with significant figures, I do not care about that much.
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Honestly, I do not like to think that important.
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If 2 or 3 decimal places is usually pretty great.
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We want the δ H, the δ H, the general equation is.
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It is the one that we should always fall back on, the integral from temperature 1 to temperature 2 of the constant pressure heat capacity × DT.
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It is equal to the integral, we are taking it from 25° to 900° C.
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We are doing integration directly.
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This constant pressure heat capacity is not constant, it is a function of temperature.
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Therefore, we cannot just pull it out and use DT.
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We cannot just use 900 – 25.
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We have to integrate from 298 which is 25 C to 98 K, 900 is 1173 K, them we go ahead and
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integrate the 23.4 this is our function, this is our CP 0.4 + 0.0063 TDT.
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I’m not going to do thee valuation here.
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Mathematical software - I use maple or mathcad, whatever it is that you need to do the integration.
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So using the software, I evaluated this and I end up with following value.
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Δ H I got was 24,529 J/ mol, notice it is just the K that canceled so we are still left with J/ mol, this is the molar heat capacity.
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We had a 0.927 mol in this particular sample so we multiply this we end up with 22,733 J.
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In taking this piece of silver from 25° C to 900° C, we have increased the enthalpy of this 100 g of silver by 22,733 J.
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We said that the enthalpy δ H =δ U + P δ V.
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In the case of solid, when you heat up a solid, it does expand a little bit.
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The volume changes, but it changes very little.
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Therefore, this V δp component, the δv because of δ Vis tiny, we pretty much ignore it.
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In this particular case, when it comes to liquid and solids the change in enthalpy, we can usually take it to be the same as the change in energy.
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We cannot do that with a gas.
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When you raise the temperature of the gas, the volume goes from this to that.
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The change in volume is huge so the change in energy, the change in enthalpy they are not exactly the same.
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You will see that in a minute.
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In the case of the solid or a liquid, because of the change in volume is so small, the change in enthalpy is almost the change in energy.
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That is what is happening here.
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Enthalpy is a counting device, it accounts for the change in energy of the system but also accounts for any pressure volume work but the system does,
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that is why enthalpy and energy are not the same.
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In a constant volume process because the change in volume is 0, the enthalpy and the energy and the heat are the same.
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Let us try more examples here.
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Consider the following reaction, at 25°C under 1 atm pressure.
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We have the decomposition of ammonia gas to nitrogen gas and to hydrogen gas.
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2 mol of gas is being converted to 3 mol, 2 mol initial 3 mol total.
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The δ H for this reaction is 92 kg J.
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They give you the δ H in this particular problem.
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Δ H is 92 kg and notice that kg J/ mol it is kg J because it is for the whole process, this is not molar.
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Find δ U for this reaction as written.
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What are our equations?
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We know that the δ H = δ U + P δ V.
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We want δ U so let us just go ahead and move this over here and we get δ U = δ H - P δ V.
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If we want δ U, we just have to take the enthalpy and we have to subtract the pressure which is 1 atm, we have it already.
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So we have δ H, we have the pressure, we just need a change in volume.
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We just need to find out the change in volume under 1 atm of 2 mol of gas turning into 4 mol of gas.
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We just the δ V.
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Let us go ahead and do that.
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It is very simple.
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Let us start off with our ideal gas law Pv = nRT.
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The volume = nRT /P.
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Therefore, the volume 1, the initial volume = the initial number of mol × RT / P.
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The initial number of mol is 2 mol.
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We start off with 2 mol, the system starts off with 2 mol of ammonia gas 0.08206 this is going to be L atm/ mol K.
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It is taking place at 25° C so it is 298 K and is happening all under pressure 1 atm.
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Atm atm, mol mol, K K, if I did my arithmetic properly I ended up with something which is 48.91 L.
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So under these conditions I started off with 48.91 L.
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As written, the decomposition produces 4 mol of gas, 4 mol of gases could occupy a greater volume.
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Therefore, let us go and see what the new volume is.
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The change in state is from volume 1 to volume 2, pressure stays constant 1 atm.
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Let us find out what volume 2 is.
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Well, volume 2 is the number of mol for the products × RT / P.
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We have 4 mol × 0.08206 × 298 K, the temperature stays the same it is still 1 atm and you end up with 97.82 L.
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The final state has gone to this much, the δ V = V2 – V1 = 97.82 - 48.91.
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I end up with 48.91 L that is our δ V.
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We are going to calculate the δ U = δ H - P δ V = 92 kg J which is 92000 J - pressure is 1 atm and we have 48.91 L, we end up with 92,000 J - 48.91 L atm.
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I’m going to go ahead and convert that.
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I have got 8.314 J/ 0.8206 L atm and I end up with 92,000 J - 4955 J.
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I end up with 87,045 J this is δ U.
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Let us see what I have got here.
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I find my δ U as 87,045 J.
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Notice my δ H is 92,000 J.
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The system gains, the decomposition the change in enthalpy of the system is 92,000 J.
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The system gains 92,000 J of heat.
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All of those 92,000 J, 4955 J are converted to work.
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That much energy is lost by the system in pushing out from 49 L to 98 L.
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Therefore, what is leftover is the energy of the system 87,045.
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92,000 J of energy go into the system, 4,955 J leave the system that leaves to work.
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Work done in pushing away the atm in order to expand the volume to accommodate the fact that 2 mol is turning into 4 mol.
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When we are dealing with the gas because there is a huge volume difference the 48.91 L, the enthalpy and the energy are not the same.
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For a liquid and a solid, the energy and enthalpy are pretty much the same because the volume changes a little.
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For a gas that is not the case, that is what is happening.
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In other words, it is very important that you understand what is happening physically.
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In other words, δ H = QP the heat that is lost by the surroundings is the change in enthalpy of the system.
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The heat that is gained by the system is the change in enthalpy of the system.
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The system, the reaction the ammonia gains 92 kg J of energy as heat energy but about 5 kg J of this 92
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is used in the volumetric expansion of the system as the system goes from 2 mol to 4 mol.
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Some of the heat is used as work to push away the atm upon the expansion of this gas.
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This leaves 87 kg J of energy in the system.
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Enthalpy and energy are not the same.
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The enthalpy it accounts for any pressure, volume, work done by the system under conditions of constant pressure.
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There you go, hope that makes sense.
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And again, no worries, we will be doing a lot more problems in complete blocks.
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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.