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Enthalpy _ Thermodynamics _ Chemistry _ Khan Academy

WEBVTT
Kind: captions
Language: en

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I'll draw a favorite pV diagram.
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On this axis is the pressure and on this axis is the volume.
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So, I have pressure and volume.
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A few videos ago I showed that if we start from some condition
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here on the pV chart, and then change
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volume and pressure to reach other states,
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and this happens in a quasi-static process, in principle all the time
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i am close to equilibrium, my variables are clearly defined
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and I have some path to reach the new state.
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Here is my path.
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I go from this state to this state.
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And I showed that in this case the work done by the system,
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is equal to the area under the curve.
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And then if I go back to my original state,
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going some way, some random way,
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then the work done on the system,
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is equal to the area under this light blue curve.
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In the end, the work done by the system is equal
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of the area closed between the two curves.
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I'll do it in a different color.
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The net work is the area between the two curves,
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moving in a clockwise direction.
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This is the net work done by the system.
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And we also know that if we are at some point on the pV chart,
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the state is identical to the previous state at this point.
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So if we go all the way and then come back,
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our macrostatic variables will be no different.
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The pressure is the same as before.
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The volume is the same as before because we have passed
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all the way back to that point on the pV chart.
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And internal energy is the same point as before,
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so the change in internal energy this time,
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we have different internal energy here and here,
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but when we go all the way
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and back to the starting point,
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the change in internal energy is zero.
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We know that the change in internal energy is defined as
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and this follows from the first law of thermodynamics,
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heat absorbed by the system minus the work done by the system.
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If we follow one such circular path in the pV diagram,
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what is the change in internal energy?
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It is zero.
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We have zero change in internal energy because
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we are in the same state which is equal to the heat absorbed
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minus the work done or ... I've done this exercise many times.
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I think I do it for the fourth or fifth time.
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We have that heat absorbed by the system if we add it
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to both sides, is equal to the work done by the system.
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So the face between the two curves, I already said,
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is the work done by the system and if you can't remember
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where it comes from, remember it is
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volume pressure by volume change is the change of work
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and that connects it to that area.
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We've done it many times already.
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We won't do it again.
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And if you have any area here, some heat is added
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in the system, right?
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Some heat was added here and some heat was probably given off
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but we have net heat added to the system.
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I use this to make it clear why the heat is not good
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macrostatic variable.
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Because ... and I have a whole video about it ...
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if I define a macrostatic variable, such as heat.
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Let me define a macrostatic variable amount of heat.
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And I can say that the change in the amount of heat
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is equal to the change in heat.
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This is my definition.
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When I add heat to the system,
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the amount of heat increases.
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But the problem is that the macrostatic variable amount of heat
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is this here, i can say that the amount of heat is 5.
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And I just showed you that if we went some way
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and we come back, and there is some area under this road,
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we have added some heat.
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Say that area here, this is q,
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equal to the work done by the system equal to 2.
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Each time I start with an amount of heat equal to 5,
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it's an arbitrary number, and if I went that whole way
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the amount of heat would be 7.
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And then when I come back, the amount of heat would be 9.
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And it would change by 2 whenever I passed this time.
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I will change with the amount of area the road travels.
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So heat is not a macrostatic variable,
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because it depends on the path that has been traveled.
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Static variable, remember this,
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to be a static variable if you are at this point,
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it should have the same value.
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If the internal energy here is 10, and you go this time and come back,
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the internal energy will be 10 again.
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Therefore, internal energy is a valid static variable.
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It depends only on the condition.
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If the entropy here is 50 when you come back here after
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all the goofy stuff, you come back to this point
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and entropy is again 50.
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If the pressure here is 5 atmospheres when you come back here,
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it will still be 5 atmospheres.
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The static variable does not change because of the path that has been traveled.
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And if you're in a certain state, that's all that matters to the static variable.
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Therefore, the amount of heat does not work and
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in several videos i split it into t
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and we got entropy, which is an interesting species.
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But this is not yet satisfactory.
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I want to get something that
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be a static variable but measure the heat.
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We will have to make some compromises,
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because we used the controversial amount of heat here,
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which changes every time you go this route.
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This is not a valid static variable.
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Let's see if we can find one.
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Let's define it.
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Let's call this a new variable
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approximate heat h, for example
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we can call it enthalpy.
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00:06:10.720 --> 00:06:15.420
Let's define it as internal energy
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plus the volume pressure.
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Then what will be the change in enthalpy?
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The change in enthalpy will be the change of these elements.
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This is a change in internal energy
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plus the change in volume pressure.
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This is interesting.
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And I want to emphasize something here.
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This is by definition a valid static variable.
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Why?
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Because it is a sum of other static variables.
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At any point in the pV chart, but this also applies to other charts,
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such as entropy and temperature, or something else,
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involving static variables at each point in the diagram,
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u will be the same no matter how we got there,
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p will by definition be the same.
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That is why it is at this point.
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v will definitely be the same point.
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And if I just collect them, it's a valid static variable,
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because it is simply the sum of other valid static variables.
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Let's see if we can connect with
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other variables we already know to be static.
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By our definition, this job is just a sum of
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fully valid static variables.
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Let's see if we can make a connection with the heat.
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We know how much is Δu.
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And if we look at internal energy
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or the change in internal energy, I don't need others
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chemical potentials and others, this equals
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the heat applied to the system, minus the work done by the system, right?
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Let me put everything else here.
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The enthalpy change is equal to the heat applied to the system,
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minus the work done by the system ... this is a change in internal energy,
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plus ΔpV.
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This follows from the definition of enthalpy.
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This is starting to look interesting.
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What is the work done by the system?
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I can write the change in h, or enthalpy, is equal to
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the heat applied to the system ... and what is it
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the work done by the system?
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If we have any system, and there is a piston,
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and the process is quasi-static, with the classic small pebbles,
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that we talked about in several videos.
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When I apply some heat or say pebbles,
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I have different equilibrium states, but what happens?
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When is work done?
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We have some pressure here,
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and the plunger will rise, the volume will increase,
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A few videos ago I showed that the work done by the system,
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can be considered as
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work to increase the volume equal to
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the volume change pressure.
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And now let's add the rest.
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This is the change in internal energy.
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There are several videos in which I show it.
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Now let's add the other part of the equation.
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Enthalpy, our change in enthalpy,
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can be defined as such. Something interesting is happening here.
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I wanted to define a static variable because I wanted to
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somehow measure the heat content.
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The enthalpy change will be equal to the heat added to the system,
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if these two members are destroyed.
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If I somehow manage to cut them,
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then the change in enthalpy will be equal to
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if these are equal to each other.
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When are they equal?
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Under what conditions the change in volume pressure
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equals the volume change pressure?
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When is this happening?
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If I can make that claim, then these two articles
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equals the enthalpy change
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will be equal to the heat added.
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I can only make such a statement if the pressure is constant.
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Why is that so?
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Let's think mathematically.
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If this is a constant and I change ...
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if it's 5, 5 on changing something
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is equal to changing 5 on this thing,
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this is mathematically true.
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Otherwise, if this is a constant,
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we can just take it to the brackets, right?
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If I say that the change of 5 by x is equal to
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5 x x minus 5 x x initial.
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And you can tell it's simple
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5 (x extreme - x initial).
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That's just 5 on changing x.
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It's too obvious to explain.
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I think sometimes, when things are explained too much,
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they just get more confused.
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I use 5 simply as an analogy to a constant.
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If the pressure is a constant, then this equation is true.
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If pressure is a constant, this is the key assumption,
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if we apply heat at constant pressure in the system
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... we can write it like this.
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I'll write it a few times because it's key.
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If pressure is a constant, then our definition,
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that we invented, that enthalpy that
00:11:47.330 --> 00:11:52.900
we defined as internal energy plus pressure and volume,
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at constant system pressure, the enthalpy change
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we just showed that it is equal to the heat added to the system,
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since all of these things are equal at constant pressure.
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This is true only when heat is added to
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constant pressure system.
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What is the connection of this
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with the pv diagram?
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What happens in the system at constant pressure?
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Let me draw our pV diagram.
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This is p, this is V.
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What happens at constant pressure?
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We have some pressure here.
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Constant pressure means that we are only moving along these lines.
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We can go from here to here, then we can go back.
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We can go from here to here and come back again.
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But what impresses us?
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Do we have an area under the curve?
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There is no curve here because we have constant pressure.
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We kind of cracked this chart.
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The way back and forth is the same.
00:12:59.580 --> 00:13:04.420
So we have no static issues,
00:13:04.420 --> 00:13:08.490
because the net heat is added to the system as we go
00:13:08.490 --> 00:13:12.600
from this point to this and then back to this point.
00:13:12.780 --> 00:13:14.840
So you can see
00:13:14.840 --> 00:13:17.090
that enthalpy at constant pressure when the pressure
00:13:17.090 --> 00:13:20.220
it does not move up and down, then the enthalpy is equal to the heat added.
00:13:20.400 --> 00:13:21.780
You can say here: Hey Sal, this is some kind of compromise,
00:13:21.780 --> 00:13:25.340
constant pressure, this is a serious assumption.
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What can it serve us for?
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This is useful because with most chemical reactions,
00:13:30.530 --> 00:13:33.530
especially those in open containers,
00:13:33.530 --> 00:13:36.755
are done at sea level, and this is an important fact
00:13:36.755 --> 00:13:38.250
they are performed at constant pressure.
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Imagine I'm on the beach,
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I take out a flask or something
00:13:45.290 --> 00:13:48.790
and I put some stuff inside and watch the reaction,
00:13:48.790 --> 00:13:50.450
this is a constant pressure system.
00:13:50.450 --> 00:13:52.070
This is atmospheric pressure.
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1 atmosphere.
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I am at sea level.
00:13:54.310 --> 00:14:00.740
This is a very useful concept for everyday chemical expressions.
00:14:00.900 --> 00:14:02.980
It may not be so useful for engines,
00:14:02.980 --> 00:14:05.480
at which the pressure is constantly changing, but
00:14:05.480 --> 00:14:10.120
about chemistry is because most reactions are there
00:14:10.130 --> 00:14:13.910
are carried out at constant pressure.
00:14:13.910 --> 00:14:16.520
So we can look at this enthalpy
00:14:16.520 --> 00:14:20.280
such as the heat content when the pressure is constant.
00:14:20.440 --> 00:14:22.080
It is precisely the content of heat
00:14:22.080 --> 00:14:24.200
at constant pressure.
00:14:24.200 --> 00:14:27.960
And somehow ... actually I showed you how,
00:14:27.960 --> 00:14:29.930
we could derive this definition which
00:14:29.930 --> 00:14:32.350
is by definition a static variable,
00:14:32.350 --> 00:14:35.710
because it is a sum of other static variables,
00:14:35.710 --> 00:14:39.270
with this constant pressure specification,
00:14:39.270 --> 00:14:43.420
it has become the amount of heat of this system.
00:14:43.420 --> 00:14:46.290
In the future, we will talk more about how enthalpy is measured,
00:14:46.290 --> 00:14:49.310
but now you can just tell if the pressure is constant,
00:14:49.310 --> 00:14:53.220
enthalpy is ... and we can only use it at constant pressure.
00:14:53.380 --> 00:14:56.340
But if we have constant pressure, then enthalpy is
00:14:56.340 --> 00:14:57.655
the amount of heat.
00:14:57.655 --> 00:15:00.900
And it is very useful in determining whether chemical reactions
00:15:00.900 --> 00:15:05.060
need or release heat.
00:15:05.200 --> 00:15:06.450
See you soon!

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