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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We have introduced this notion of enthalpy and today we would talk about something called the Joule-Thompson experiment.
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Let us just dive right on in.
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Let us recall the following facts, enthalpy H is a state function and that implies a state function is an exact differential.
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Mathematically speaking it is an exact differential.
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If we let, which we did, H = some function of T and P, we are saying that enthalpy is a function of temperature and pressure we have the following.
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We have already done this.
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We had that DH = a partial of H with respect to T at constant P × DT + a partial of H with respect to P at constant T DP.
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We have is that exact differential state function and we can write it in this form depending on the variables that we choose for our function.
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We already identified the DH DT at constant pressure with the heat capacity,
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the constant pressure heat capacity DQDT which was defined as the heat capacity at constant pressure.
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This is the definition of the heat withdrawn from the system or the heat withdrawn from the surroundings.
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We are always looking for this one, the heat withdrawn from the surroundings divided by the change in temperature of the system.
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The heat withdrawn from the surroundings is the heat that goes into the system.
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It is just a question of point of view.
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That is the heat capacity, we identified with its partial derivative.
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It is a change in enthalpy with respect to a change in temperature.
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And this is an easily measure quantity.
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The question is what about that?
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What about the change in enthalpy with respect to pressure at constant temperature?
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Let us see what we can do with this.
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In other words, how does the enthalpy of the system change because the total change in the enthalpy is the sum of the two.
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If we happen to hold one constant it does not matter but if we are not going to hold the temperature of the pressure constant,
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the total change in enthalpy is going to be the sum of the two changes.
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How does enthalpy of the system change upon a unit change in pressure?
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That is all it is, simple straight derivative, rate of change.
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Let us begin with the definition of enthalpy, we have the H=U + PV.
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The enthalpy of the system, the energy of the system + the pressure × the volume of the system, let us go ahead and differentiate that.
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This is a huge differential here.
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This is a product so we can use the product rules, this × the derivative of that + that × the derivative of this.
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We have a differential form DH = DU + this × the differential of that P DV + this × the differential of that + V DP.
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We are going to start with that.
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We have the following.
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We have the DH = DH DT sub P DT + DH DP at constant temperature × DP.
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What else we have the following.
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We also have the differential for U as a state function is exact differential.
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Let me go ahead and write that you already identified this.
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Let me write this as the constant pressure heat capacity.
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This thing is that thing.
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I’m going to write this as C sub P and DU we had a constant volume heat capacity × DT + the other thing that we talked about.
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Remember when we talk about the Joules experiment, DUDV T DV.
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We are going to go ahead and substitute this expression here, this expression here, but rearrange some things and play with this mathematically.
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We have the following.
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We have CP DT + DH DP sub T × DP = the constant volume heat capacity × DT +
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Let me make sure my T's are correct here and let me have a lot of symbols floating around.
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Let me make sure that they are correct and legible.
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+ DU DV sub T × DV + P DV + V DP.
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We have this very long equation but let us see what we can do with this.
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We see that DV and DV is common here so let us go ahead and put these terms together.
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I will work in blue, I hope you do not mind.
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We have got CP DT + DH DP, this is just mathematical manipulation, that is all what is going on here.
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DP =CV DT + DU DV T + P.
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I’m going to put these two together × DV + V DP.
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Holding T constant, it implies that DT = 0, the change in temperature is 0 if we hold the temperature constant.
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The things at T they go to 0 so that becomes 0 and that becomes 0.
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What we are left with is the following.
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We are left with DH DP sub T × DP = this.
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Let us see here we have DU DV, I should call it a little bit slower just to make sure that I'm getting all the symbols done correctly + P DV.
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Mistakes are very easily to make when we have this many symbols floating around + V DP.
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Let us go ahead and divide by DP.
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Divide everything by DP and what you end up with is DH DP sub T = this stays,
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DV ÷ DP that is just DV DP, I can write that as a partial derivative.
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We are holding temperature constant so this is that and V DP divided by DV DP is cancel.
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We are just left with V.
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We were able to extract an expression for this DH DP, this is a general equation.
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This DU DV is exactly what it means.
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I do not think it make a lot of sense physically but at the very least you want to assign some physical meaning to the partial derivatives.
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This is just the rate of change in energy with respect to volume.
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This is the pressure, this is the rate of change of volume with respect to pressure, that is all it is.
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Do not get too intimidated by these symbols.
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This white side is a little strange, granted you know it is a little difficult to wrap your mind physically but
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as long as you can assign some meaning for the partial derivatives and know what is going on, at least it makes it a little bit more tractable.
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This is a general equation that we were looking for.
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We cannot do much with this but it is nice to see where it came from.
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This is a general equation and it applies to all systems, gas, liquid, solid, it does not matter.
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This is the general equation for all systems.
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For liquids and solids, the first term, in other words this term, is usually very much smaller than this term.
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For all practical purposes, we can ignore that term altogether.
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For liquid and solid, the first term is usually much smaller than V.
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Then I will just say the second term which is the actual volume of the system.
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Therefore, for liquid and solids, the following approximation is valid DH DP of T = the actual volume of the system.
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If I know the enthalpy of the system, the rate of change in enthalpy per unit change in pressure but
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I’m holding the temperature constant for a liquid and solid, it is often just = to the particular volume of liquid or solid.
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That is all this is saying.
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It gives you some numerical value.
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Remember we are trying to figure out what this says because we have that differential for the energy.
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The enthalpy of the system has to do with the constant pressure heat capacity + any change in pressure.
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If that changes the enthalpy a little bit, here you go for a liquid or solid, you can just take that
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number as partial derivative actually = the volume of the system, the liquid or the solid.
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The molar volume of liquid and solids is usually very small.
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This DH DP molar at constant T can be ignored.
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In other words, that heat capacity, the change in enthalpy that comes from the heat capacity term,
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the CP DT that is going to dominate the change in enthalpy of the system.
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The molar volume of liquid and solid is so small but for all practical purposes, that second term of the differential, remember we had this DH = CP DT + DH DP TDP.
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This term is going to dominate.
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For liquid and solids, the molar volume is so small that for all practical purposes, this can be ignored.
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It really does not matter for liquids and solid, this really not that big deal.
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It is only under conditions of really high pressure that this value, this thing actually becomes significant.
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In general, that is not going to happen.
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When I say high pressure I mean extremely high pressure.
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It is very specialized, you will probably never run across it in any of the work that you do unless
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you are specifically working under high pressure conditions for liquid and solids.
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Let us go ahead and talk about ideal gases.
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For ideal gases, it is easy.
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It is the same thing that we have for the Joules experiment.
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For ideal gases, this value DH DP sub T it just = 0.
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In other words, if you change the pressure, the enthalpy of the system is not going to change.
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If you want to go ahead and see a quick derivation of that, other than that is also important but that is fine, you can do it very quickly, it is pretty short.
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We start with a definition U + PV.
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For ideal gas PV = nRT.
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We have H = U + nRT.
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It is very important to be able to actually make these substitutions when you are dealing with the problem.
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If we go ahead and divide by n and we put molar quantities, we have H = U + RT.
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Any variable that has a line over it just means that you divided by the number of moles so it becomes a molar value.
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This is a molar enthalpy J/ mol, molar energy is J/ mol.
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That is all that is going on.
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Let us divide by M.
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We already know that from ideal gas, the energy is just a function of temperature.
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If this is just a function of temperature and we have R × T, it turns out that from ideal gas the enthalpy itself is only a function of temperature, 1 variable.
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It is a function only of temperature not pressure and volume.
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Therefore, if we go ahead and differentiate with respect to pressure, you end up with DH DP T.
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The differential this with respect to P there is no P involved in U so this is 0.
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This is 0, that is it.
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That is the derivation, this is what is important.
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From ideal gas, that second term, this term right here is 0, it drops off.
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For an ideal gas, the change in enthalpy of the system is gone strictly by this and this.
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You do not have to worry about this term.
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That is what we are talking about here.
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The enthalpy depends on this term and this term.
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For liquids and solids, this is really small.
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For all practical purposes, we can ignore it unless otherwise stated.
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For ideal gases we can ignore it completely because that is 0.
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It is only that it matters.
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If I want to measure the enthalpy change, I go ahead and take the temperature change and multiply by the constant pressure heat capacity,
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that gives me the change in enthalpy.
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This is the differential form.
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Let us talk about real gases.
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For real gases, this DH DP sub T is not 0, not definitely.
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It is small but it is measurable.
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When we did the Joules experiment, you remember we were trying to figure out this thing.
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We are trying to figure out DU DV.
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If I change the volume of the system how does the energy of the system change?
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Joules experiment it has 2 balls connected by something.
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The gas expanded under free expansion, we measure the change in temperature of water.
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As it turns out it was sensitive enough.
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The heat capacity of the water was too much and the heat capacity of the gas was too small.
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It was far below any measurable level.
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It is not actually 0 but really it is so small that essentially is 0.
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For Joules experiment that ended up being 0.
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For the Joules constant experiment, they were able to actually do this experiment, a variation of it, and actually measure a change.
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Let us see what it is that they did.
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We said that this was 0 for the Joules experiment so we want to find out what is the change in enthalpy upon a change in pressure?
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Here is the setup.
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Let us go ahead and draw this out here.
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Let me draw a little bit lower.
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Based on what they had, they had an insulated pipe.
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There was a pipe and the pipe there is insulation around it.
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No heat can actually cross the barrier.
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The DQ actually ends up being 0.
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It is insulated so heat can escape and no heat can come in.
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This thing right here, this is a porous membrane.
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In the actual experiment themselves, what they use was just a silk scarf.
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Imagine taking a pipe and just sticking a silk scarf in there.
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Basically, what they are trying to do is this.
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They are going to push some gas through this silk scarf.
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They have a pressure gauge attached here and they also have a temperature gauge attached here.
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You know there is a certain pressure on the side of the silk scarf.
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There is a certain temperature inside that pipe.
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There is a gas that is moving in this direction.
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They also have a pressure gauge here and a temperature gauge here.
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You know from your experience that air is going to pass through the silk scarf but on the other side it is going to be a lot less pressure.
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And that is exactly what happens is there is going to be a certain pressure so there was going to be
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a certain amount of gas here and it is going to be under P1 V1 T1.
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It is going to be a certain temperature, a certain pressure, and a certain volume of gas.
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If we pick a certain volume of gas, let us say 1mol, it is going to have a certain temperature and pressure which are going to be read right off the meter.
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On the other side, that same molar gas, the same thing, I mean it is going to now have a new pressure, a new volume, and the new temperature.
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You know this already.
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You stick something to a pipe, you force something through it, the pressure on this side is going to be higher than the pressure on this side.
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If we measure a change in pressure, measure the change in temperature, we can come up with some numbers and that is what we are doing.
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Let us go ahead and do that.
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We have to deal with this mathematically.
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The pipe is insulated so no heat can actually transfer during this process.
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The pipe is insulated so no heat flows.
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What we are going to do was we are going to basically take 1 mol of gas and we are going to push it through this membrane.
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There is 1 mol of gas at a certain pressure and temperature.
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We are going to push it through the membrane and we are going to measure its pressure and temperature on the other side, the same mol of gas.
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Again, because this is obstruction, there is going to be a pressure drop on the other side.
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You know this already just from your intuition tells you.
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The thermometer, we will measure the temperature difference.
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1 mol of gas has to pass through this obstruction.
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We are going to imagine it in two ways.
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I'm going to imagine that I'm taking this 1 mol of gas and squeezing it all into this one member, that is going to be the pushed.
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From this membrane, that same mol of gas has to go this way, it has to expand.
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All I'm doing is, I'm going to be describing mathematically the flow in two stages.
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I'm going to be imagining it as completely crushing it from volume 1 down to volume 0 and from volume 0 to a new volume.
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I just mathematically, I need to break and break it down to the two stages because this 1 mol of gas has to flow through the membrane.
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I can actually do anyway that I want to.
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We will imagine the flow in two stages.
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This I how I think about it, you do not have to think about it this way.
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Basically, you can think about it as a certain volume of gas that is being pushed this way, there is a certain pressure that is being applied.
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The pressure is measured by this gauge.
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That same gas has to push against the gas that comes after it.
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Therefore, it is going to be pushing against this pressure which is this is the flow of gas but the pressure inside is pushing against the flow.
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What we are going to be doing is we are going to be measuring the work done in pushing it this way and work done in expanding that way.
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That is all we are doing here.
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One, is going to be a compression from volume 1 to 0.
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And two, it is going to be the expansion from 0 to volume 2.
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I’m going to compress it and then it is going to expand.
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I’m just taking it a fixed mass of gas and I chose 1 mol.
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The work done in taking it from here to here and compressing this gas into the porous disk is going to be the integral V1 to 0 of P1 DV.
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Basically, because this pressure, the pressure is what is pushing it into, like this pressure here is the pressure here.
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The work being done, the external pressure that was pushing that gas and compressing it, it is P1 DV.
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Work 2, it is going to go from a 0 volume to V2 and the work that is done here, this gas going from 0,
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imagine all the gas is right there in the porous disk, it needs to expand against an external pressure.
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The external pressure against which it is expanding is pressure 2.
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It is the pressure that is on the other side of this obstruction.
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It has to push against that in order to actually expand.
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So that is equal to pressure 2 × the change in volume.
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The total work that is done in this process is work 1 + work 2 = this V1 to 0 of P1 DV + the integral from 0 to V2 of P2 DV.
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This is equal to the pressure is constant, pressure is constant, so it comes out so what you end up with is P1 × - V1 + P2 × V2.
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This is – so let me flip this around so = P2V2 - P1V1.
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Let us go ahead.
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The insulated pipe, we said that there is no heat flow so that implies that DQ = 0.
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The first law expression for this, the change in energy of the system = Q – W.
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If DQ = 0, Q = 0, that means that δ U = -W.
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Δ U is U2 - U1 = - the work which is P2V2 - P1V1.
00:27:23.100 --> 00:27:34.600
We have U2 - U1 = P1V1 – PQ V2.
00:27:34.600 --> 00:27:37.400
Let us go ahead and rearrange this.
00:27:37.400 --> 00:27:42.600
We have to bring this over there and bring this over there.
00:27:42.600 --> 00:27:52.400
I have U2 + P2 V2 = U1 + P1 V1.
00:27:52.400 --> 00:27:53.900
What is U + PV?
00:27:53.900 --> 00:27:56.100
That is the enthalpy.
00:27:56.100 --> 00:28:08.000
Therefore, the molar enthalpy in the final stage = the molar enthalpy of the first stage.
00:28:08.000 --> 00:28:15.400
This shows that the Joules Thomson experiment or expansion is an isoenthalpic process.
00:28:15.400 --> 00:28:26.500
The enthalpy of the system does not change when the gas passes through that obstruction.
00:28:26.500 --> 00:28:42.900
The Joule Thompson expansion because it is an expansion, the gas is more pressure here, is less pressure here, the gas is going to expand.
00:28:42.900 --> 00:28:46.200
As the pressure goes down the volume goes up.
00:28:46.200 --> 00:28:50.400
Volume 2 is bigger than volume 1.
00:28:50.400 --> 00:29:00.000
Expansion is isoenthalpic.
00:29:00.000 --> 00:29:03.900
In other words, the enthalpy is constant through the process.
00:29:03.900 --> 00:29:23.300
The enthalpy is constant through this change of state.
00:29:23.300 --> 00:29:27.500
We see a change in pressure from one side of the pipe to the other, from one side of the obstruction to the other.
00:29:27.500 --> 00:29:31.500
There is a difference in temperature.
00:29:31.500 --> 00:29:34.900
We also see a change in temperature.
00:29:34.900 --> 00:29:44.200
We see a δ T and let us take the ratio of the two.
00:29:44.200 --> 00:29:49.300
We take the ratio of δ T / δ P.
00:29:49.300 --> 00:30:07.600
The P down in the denominator and we defined something called a Joules Thompson coefficient which is μ sub JT = the limit has this δ P goes to 0 of this ratio.
00:30:07.600 --> 00:30:15.300
This ratio as we take δ P smaller and smaller, we are going to arrive at some number.
00:30:15.300 --> 00:30:36.500
δ T/ δ P that is DT DP constant enthalpy, this is called the Joule constant coefficient.
00:30:36.500 --> 00:30:43.100
And it is a measure of the extent to which the temperature changes upon a change in pressure.
00:30:43.100 --> 00:30:55.300
It is a rate of change at which the temperature of a gas changes upon a change in pressure of that gas that is what this measures,
00:30:55.300 --> 00:30:58.400
that is what the Joule constant coefficient does.
00:30:58.400 --> 00:31:05.500
Again, we were after this, we were after DH DP, a change in enthalpy under a change in pressure.
00:31:05.500 --> 00:31:09.100
Let us go ahead and see if we can find that now given this thing.
00:31:09.100 --> 00:31:23.100
We calculated as Joule constant coefficient which is δ T δ P or DT DP.
00:31:23.100 --> 00:31:41.000
Let us go back to our differential statement, CP DT + DH DP sub T DP, this is just a statement of the change in enthalpy
00:31:41.000 --> 00:31:46.000
with respect to a change in temperature and a change in pressure.
00:31:46.000 --> 00:31:52.400
We can always change the temperature and pressure in such a way that the enthalpy stays constant.
00:31:52.400 --> 00:32:00.000
Experimentally, we can do that.
00:32:00.000 --> 00:32:25.700
It is always possible to vary the temperature and the pressure which are the two variables in such a way that the enthalpy remains constant.
00:32:25.700 --> 00:32:29.800
The Joule constant effect is an isoenthalpic process so we do that.
00:32:29.800 --> 00:32:32.200
We will discuss what enthalpy means.
00:32:32.200 --> 00:32:39.600
The constant enthalpy that means that the change in enthalpy = 0 that is great.
00:32:39.600 --> 00:32:43.600
Or the differential form the DH = 0.
00:32:43.600 --> 00:33:02.400
Therefore, I’m going to put this over here I get 0 = CP DT + DH DP T DT.
00:33:02.400 --> 00:33:04.500
I’m going to divide by the DP.
00:33:04.500 --> 00:33:24.000
When I divide by the DP I get 0 = CP DT DP constant enthalpy + DH DP T.
00:33:24.000 --> 00:33:26.200
I’m going to go ahead and move this over the other side.
00:33:26.200 --> 00:33:28.900
When I do that, I end up with the following.
00:33:28.900 --> 00:33:43.700
I end up with DH DP sub T = - CP × DT DP H.
00:33:43.700 --> 00:33:48.200
Well this is the constant pressure heat capacity.
00:33:48.200 --> 00:33:52.200
This DT DP we just found as the Joule Thompson coefficient.
00:33:52.200 --> 00:33:59.100
It is the change in temperature upon a unit change in pressure of the system at constant enthalpy.
00:33:59.100 --> 00:34:04.700
I am going to rewrite it at the second line.
00:34:04.700 --> 00:34:20.500
DHDP for real gas = - CP × the Joule Thompson coefficient.
00:34:20.500 --> 00:34:26.400
DH DP which is what we wanted for this term, for our real gas.
00:34:26.400 --> 00:34:29.300
It happens to equal from this.
00:34:29.300 --> 00:34:32.700
The constant pressure heat capacity of the system is very easily measured.
00:34:32.700 --> 00:34:37.300
You just take the heat that goes into the system divided by the change in temperature of the system.
00:34:37.300 --> 00:34:40.200
The Joule Thompson coefficient is very easily measured.
00:34:40.200 --> 00:34:45.500
I multiply the two, put a negative sign in front of it, and I end up with this term right here.
00:34:45.500 --> 00:34:51.600
I just take this, put it right here, and if I want find out what the change in enthalpy of the system is,
00:34:51.600 --> 00:34:57.300
upon a change in pressure and a change in temperature, the temperature component is this one.
00:34:57.300 --> 00:34:59.000
The pressure component is this term.
00:34:59.000 --> 00:35:02.700
I add them together and I get the change in enthalpy.
00:35:02.700 --> 00:35:06.800
This is what we are after, right here, this second term.
00:35:06.800 --> 00:35:11.300
We wanted to find a way to identify it with something that we can measure.
00:35:11.300 --> 00:35:15.300
Sure enough, we can measure both of these very easily.
00:35:15.300 --> 00:35:17.900
That is all what is going on here.
00:35:17.900 --> 00:35:21.900
You, yourself do not have to derive this.
00:35:21.900 --> 00:35:24.300
As long as you can follow the derivation, that is all that we ask.
00:35:24.300 --> 00:35:30.100
At your level, that is all that is going to be asked of you, to be able to assign some meaning to this.
00:35:30.100 --> 00:35:32.200
You will understand what these things mean.
00:35:32.200 --> 00:35:41.500
It is like the change in enthalpy per unit change of pressure, constant pressure heat capacity, what the Joule Thompson coefficient is,
00:35:41.500 --> 00:35:44.700
which is actually a change in temperature per unit change in pressure.
00:35:44.700 --> 00:35:48.800
Always assign physical meaning, you do not have to worry about doing the derivation.
00:35:48.800 --> 00:35:53.000
Again like I said before, it is not like somebody sat and said let us do it this way.
00:35:53.000 --> 00:35:57.000
No, they went this way and they have tried this mathematical path.
00:35:57.000 --> 00:36:02.100
They tried this partial derivative, that partial derivative, until they actually saw some relationships.
00:36:02.100 --> 00:36:08.200
What you are seeing is the final process that actually did give them something fruitful.
00:36:08.200 --> 00:36:15.100
You do not have to do the derivation but as long as you follow the derivation, that is what is important.
00:36:15.100 --> 00:36:16.800
Again, we can measure this and measure this.
00:36:16.800 --> 00:36:21.600
We can calculate this very beautiful.
00:36:21.600 --> 00:36:29.200
Let us go ahead and close this discussion out of the Joule-Thompson experiment, the Joule-Thompson effect.
00:36:29.200 --> 00:36:40.600
The Joule-Thompson coefficient is positive at and below room temperature for all gases except hydrogen and helium.
00:36:40.600 --> 00:36:47.000
Hydrogen and helium are the only two gases that have a negative Joule-Thompson coefficient.
00:36:47.000 --> 00:36:52.100
This means that hydrogen and helium become hotter upon Joule-Thompson expansion.
00:36:52.100 --> 00:36:58.500
All the other gases that pass through that pipe, upon Joule-Thompson expansion, the gas will cool.
00:36:58.500 --> 00:37:04.400
Intuitively notice this just from your own experience with the expansion of gases and what you did in General Chemistry.
00:37:04.400 --> 00:37:08.300
When a gas expands, it tends to cool.
00:37:08.300 --> 00:37:17.300
The Joule-Thompson expansion which is keeping the enthalpy of the system constant, it is strange that hydrogen and helium
00:37:17.300 --> 00:37:20.000
actually end up getting hotter when the volume expands.
00:37:20.000 --> 00:37:23.200
That is very odd.
00:37:23.200 --> 00:37:33.800
Every gas, including hydrogen and helium, every gas has a temperature above which the Joule-Thompson coefficient becomes negative.
00:37:33.800 --> 00:37:38.700
The Joule-Thompson inversion temperature.
00:37:38.700 --> 00:37:47.100
At room temperature, carbon dioxide gas, if you do a Joule-Thompson expansion is going to cool but above a certain temperature,
00:37:47.100 --> 00:37:51.100
what that happens to be for carbon dioxide gas, I do not know.
00:37:51.100 --> 00:37:56.200
Above that temperature, if you do a Joule-Thompson expansion it would actually get hotter.
00:37:56.200 --> 00:38:00.200
It is called the Joule-Thompson inversion temperature.
00:38:00.200 --> 00:38:05.800
For hydrogen gas, the inversion temperature is -80° C.
00:38:05.800 --> 00:38:10.300
Below this temperature, hydrogen actually cools.
00:38:10.300 --> 00:38:21.400
We said above which, in a particular case, under normal circumstances hydrogen has a negative Joule-Thompson coefficient, below this temperature.
00:38:21.400 --> 00:38:27.200
80 is the inversion temperature: -80° C is the inversion temperature for hydrogen gas.
00:38:27.200 --> 00:38:30.900
Below this, it actually cools.
00:38:30.900 --> 00:38:37.000
If I'm doing a Joule-Thompson experiment for hydrogen gas during a Joule-Thompson expansion,
00:38:37.000 --> 00:38:45.200
if I run it below about negative -80…-90° C at that point upon expansion a hydrogen gas cools.
00:38:45.200 --> 00:38:51.300
Again, at or below room temperature.
00:38:51.300 --> 00:38:53.200
It is going to be the most of what we do.
00:38:53.200 --> 00:38:58.200
All gases had positive Joule-Thompson coefficient.
00:38:58.200 --> 00:39:00.800
That means if the pressure drops, the temperature drops.
00:39:00.800 --> 00:39:03.300
If the pressure rises, the temperature rises.
00:39:03.300 --> 00:39:04.500
It is positive, they correlate.
00:39:04.500 --> 00:39:06.400
They have the same sign.
00:39:06.400 --> 00:39:08.500
The δ T and δ P, they have the same sign.
00:39:08.500 --> 00:39:12.000
Only hydrogen and helium are switched.
00:39:12.000 --> 00:39:14.300
Thank you for joining us here at www.educator.com.
00:39:14.300 --> 00:39:15.000
We will see you next time, bye