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Current time:0:00Total duration:15:07

Let me draw a good
old PV diagram. That's my pressure axis,
this is my volume axis. Just like that, I have
pressure and volume. I showed several videos ago that
if we start at some state here in the PV diagram, right
there, and that I change the pressure and the volume to get
to another state, and I do it in a quasistatic way, so
essentially I'm always close to equilibrium, so my state
variables are always defined, I could have some path that
takes me to some other state right there. And this is my path. I'm going from this state
to that state. And we showed that if I just did
this, the work done by the system is the area
under this curve. And then if I were to move back
to the previous state, and then if I were, you know, by
some path, just some random path that I happened to be
drawing, the work done to the system would be the area under
this light blue curve. So the net work done by the
system ended up being the area inside of this path. So this is-- let me do it
in different color. The net work done would be the
area inside of this path when I go in this clockwise
type of direction. So this is the net work
done by the system. And now, we also know that if
we're at some point on this PV diagram, that our state is the
same as it was before. So if we go all the way here,
and then go all the way back, all of our state variables
will not have changed. Our pressure is the same
as it was before. Our volume was the same as it
was before because we went all the way back to that same
point on the PV diagram. And our internal energy is also
the same point as it was before, so our change in
internal energy over this path, you're going to have a
different internal energy here than you had here, but when you
go around the circle and you get back, your change
in internal energy is equal to zero. And we know that our change in
internal energy is defined as, and this is from the first law
of thermodynamics, the heat added to the system minus the
work done by the system. Now, if we go on a closed loop
on our PV diagram, then what's our change in internal energy? It's zero. So we get zero change in
internal energy, because we're at the same state is equal to
the heat applied to the system minus the work done or-- and
I've done this little exercise multiple times. I think is probably the fourth
or fifth time I'm doing it. We get that the heat added to
the system, if we just add w to both sides, is equal to the
work done by the system. So this area inside of this
path, I already said, it's the work done by the system, and
if you don't remember even where that came from, it was,
remember, pressure times volume times change in volume is
a little incremental change in work, and that's why it
relates to the area. But we've done that
multiple times. I won't go there just yet. So if you have any area here,
some heat was added to the system, some net heat, right? Some heat was added here, and
some heat was probably taken out here, but you have
some net heat that's added to the system. And I use that argument
to say why heat isn't a good state variable. Because-- and I had a whole
video on this-- that if I define some state variable,
let's just say, heat content. Let's say I want to
define some state variable heat content. And I would say that the change
in heat content would, of course, be equal to
the change in heat. That's what I'm defining. If I'm adding heat to
the system, my heat content should go up. But the problem with that heat
content state variable was that, let's say over here, I say
that the heat content is equal to 5. Now, I just showed you that if
we go on some path here and we come back, and there's some area
in this little path that I took, that some
heat was added. So let's say that this area
right here, so this is q is equal to the work done by
the system, let's say it's equal to 2. So every time, if I start at
heat content is equal to 5, that's just an arbitrary number,
and I were to do this entire path, when I go back,
the heat content would have to be 7. And then when I go back and do
it again, my heat content would have to be 9. And it would have to increment
by 2 every time I do this exact path. It would have to increment by
the amount of area that this path goes around. So heat content can't be a state
variable, because it's dependent on how
you got there. A state variable-- and
remember this. In order to be a state variable,
if you're at this point, you have to have
the same value. If your internal energy was 10
here, when you do the path and you come back, your internal
energy will be 10 again. That's why internal energy is
a valid state variable. It's dependent only
on your state. If your entropy was 50 here,
when you soon. go back and you do all sorts of crazy things,
and you come back to this point, your entropy
is once again 50. If your pressure here is 5
atmospheres, when you come back here, your pressure
will be 5 atmospheres. Your state variable cannot
change based on what path you took. If you're at a certain state,
that's all that matters to the state variable. Now this heat content didn't
work, and that's why we actually led into some videos
where I divided by t and we got entropy, which was an
interesting variation. But that's still
not satisfying. What if we really wanted to
develop something that could in some way be a state variable,
but at the same time measure heat? So obviously, we're going to
have to make some compromises, because if we just do a very
arbitrary kind of heat content variable, then every time
you go around this, it's going to change. That's not a valid
state variable. So let's see if we
can make up one. So let's just make
up a definition. Let's call my new thing that
I'm going to try to maybe approximate heat, let's call it
h, and just as a little bit of a preview, we're going
to call it enthalpy. And let's just define it. I'm just playing around. Let's just define it as the
internal energy plus my pressure times my volume. So then what would my change
in enthalpy be? So my change in enthalpy will
be, of course, the change of these things. But I could just say, that's
my change in my internal energy plus my change in
pressure times volume. Now, this is interesting. And I want to make
a point here. This, by definition, is a
valid state variable. Why is it? Because it's the addition of
other state variables, right? At any point in my PV diagram,
and it's also true if I did diagrams that were entropy in
temperature or anything that dealt with state variables, at
any point on my diagram, u is going to be the same, no matter
how I got there. p is by definition going
to be the same. That's why it's at that point. v is definitely going to
be the same point. So if I just add them up, this
is a valid state variable because it's just the
sum of a bunch of other valid state variables. So let's see if we can somehow
relate this thing that we've already established as a
valid state variable. From the get-go, from our
definition, this works because it's just the sum of completely valid state variables. So let's see if we can relate
this somehow to heat. So we know what delta u is. If we're dealing with all of
the internal energy or the change in internal energy, and
I'm not going to deal with all the other chemical potentials
and all of that, it's equal to the heat applied to the system
minus the work done by the system, right? Let me put everything
else there. The change in enthalpy is equal
to the heat applied minus the work done-- that's
just the change in internal energy-- plus delta PV. This is just from the definition
of my enthalpy. Now this is starting to
look interesting. What's the work done
by a system? So I could write change in h,
or enthalpy, is equal to the heat applied to the system
minus-- what's the work done by a system? If I have some system here, it's
got some piston on it, you know, if we're doing it in
a quasistatic, I have those classic pebbles that
I've talked about in multiple videos. When I apply heat or let's say
I remove some of these pebbles, so I'm at a different
equilibrium, but what's actually happening? When is the work being done? You have some pressure being
applied up here, and this piston is going to be moving up,
and your volume is going to increase. And we showed multiple videos
ago that the work done by the system can be, and you can
kind of view this as the volume expansion work, it's
equal to pressure times change in volume. And let's add the other part. So this was our change
in internal energy. And I had several videos
where I show this. And now let me add the other
part of the equation. So our enthalpy, our change
in enthalpy, can be defined by this. Now. Something interesting
is going on. I said that I wanted to define
something, because I wanted to somehow measure heat content. My change in enthalpy will be
equal to the heat added to the system, if these last two
terms cancel out. If I can somehow get these last
two terms to cancel out, then my change in enthalpy
will be equal to this, if somehow these are equal
to each other. So under what conditions are
these equal to each other? Or another way, under what
conditions is delta pressure times volume equal to pressure
times delta volume? When does this happen? When can I make this
statement? Because if I can make this
statement, then these two terms are equivalent right here,
and then you my change in enthalpy will be equal
to the heat added. Well, the only way I can make
this statement is if pressure is constant. Now why is that? Let's just think about
it mathematically. If this is a constant, then if
I just change-- you know, if this is just 5, 5 times a change
in something is the same thing as the change in 5
times that thing, so it just mathematically works out. Or if you view it another way,
if this is a constant, you can just factor it out, right? Well, if I said, the change in
5x, that would be equal to 5 times x final minus
5 times x initial. And you could say, well, that's
just equal to 5 times x final minus x initial. Well, that's just equal to
5 times the change in x. It's kind of almost too obvious
for me to explain. I think sometimes when you
overexplain things, it might become more confusing. So this applies-- and the 5 I'm
just doing as the analogy for a constant. So if pressure is constant, then
this equation is true. So if pressure is constant-- so
this is a key assumption-- then if heat is being applied
in a constant pressure system-- so we could
write it this way. I'll write it multiple times,
because this is key. If pressure is constant, then
our definition, our little thing we made up, this enthalpy
thing, which we defined as internal energy plus
pressure and volume, then in a constant pressure system,
our change in enthalpy we just showed is equal to the heat
added to the system because all of these two things become
equivalent under constant pressure, so I should
write that. This is only true when
heat is added in a constant pressure system. So how does this gel
with what we did up here on our PV diagram? What's happening in a constant
pressure system? Let me draw our PV diagram. That's P, that's V. So what's happening in
constant pressure? We're at some pressure
right there. So if we're under constant
pressure, that means we can only move along this line. So we could go from here to
there and back to there, or we could go from there to
there, back to there. So we could go there, all the
way there, and then go back. But what do we see about this? Is there any area
in this curve? I mean, there is no curve to
speak of, because we're staying in a constant
pressure. We've kind of squeezed
out this diagram. We've made the forward path and
the return path the same exact path. So because of this, you don't
have that state problem because no net heat is being
added to the system when you go from this point all the way
to this point and then back to this point. So because of that, you can
kind of see visually that enthalpy in a constant pressure,
when you're not moving up and down in pressure,
is the same thing as heat added. So you might say, hey Sal,
this was a bit of a compromise, constant pressure,
you know, that's a big assumption to make. Why is this useful at all? Well, it's useful, because
most chemical reactions, especially ones that occur in an
open beaker, or that might occur at sea level, and that
should be a big clue, they occur at constant pressure. You know, if I'm sitting at
the beach, and I have my chemistry set, and I have some
beaker of something, and I'm throwing other stuff into it,
and I'm looking for a reaction or something, it's a constant
pressure system. This is going to be atmospheric
pressure. 1 atmosphere. I'm sitting at sea level. So this is actually a very
useful concept for everyday chemical expressions. It might not be so useful for
engines, because engines always have pressure changing,
but it's very useful for actual chemistry, for actually
dealing with what's going to happen to a reaction at
a constant pressure. So what we're going to see is
that this enthalpy, you can kind of view it as
the heat content when pressure is constant. In fact, it is the heat content when pressure is constant. So somehow-- well, not somehow,
I showed you how-- we were able to make this
definition, which by definition was a state variable,
because it was the sum of other state variables,
and if we just make that one assumption of constant pressure,
it all of a sudden reduces to the heat content
of that system. So we'll talk more in the future
of measuring enthalpy, but you just have to say, if
pressure is constant, enthalpy is the same thing as-- and it's
really only useful when we're dealing with a
constant pressure. But if we have a pressure
constant, enthalpy can be imagined as heat content. And it's very useful for
understanding whether chemical reactions need heat to occur or
whether they release heat, so on and so forth. See

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