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ემზადებით გამოცდისთვის? მოემზადეთ ამ 4 გაკვეთილის დახმარებით შემდეგ თემაზე: Thermodynamics

იხილეთ 4 გაკვეთილი

# More rigorous Gibbs free energy / spontaneity relationship

ვიდეოს აღწერა

I've hopefully given you a bit
of a gut feeling behind where the formula of Gibbs Free
Energy comes from. In this video I want to do
something a little bit more rigorous and actually,
I guess you could say derive the formula. So to do that, let's just study
two systems that have the exact same change
in entropy. To depict that, I'll get out
our handy PV diagram. What we're going to do is we're
going to compare two systems. One that's this
perfect, reversible system and one that's irreversible or
a spontaneous system. So they're both going
to start here. They're going to go from this
point on the PV diagram to this point right here. And before we go into exactly
what they're doing, I think it's a good review to kind
of talk about what an irreversible process and
what a reversible process really is. An irreversible process is this
theoretical thing where you have no friction, where
you're always so close to equilibrium that you can
always go backwards. You're always-- you can kind of
view that the reaction is never really proceeding
forward or backwards. Although obviously it is moving,
so it's a very-- it actually doesn't really exist
in nature, but it's a useful theoretical construct. So let me draw a little. So if this is a-- I'll get
the good old pistons out. So let's say this is an--
I'll do it twice. So that's my irreversible
piston-- That'll be my reversible, this is going
to be my irreversible. Let me draw them. OK, and let me label them. So this right here is going
to be reversible. And this right here is going
to be irreversible. Or it could also be considered
spontaneous. Using the key word. That's going to matter for
Gibbs Free Energy. But all spontaneous reactions
are irreversible. So they're going to go from
this state to this state. So in our reversible
world, we have this little cap to our piston. And we have our gas in here,
exerting some pressure. And what we'd have is
a bunch of pebbles. We have a bunch of
pebbles out here. And we slowly remove the
pebbles one by one. And as we remove the pebbles,
our piston-- or this movable ceiling up here-- will
move upwards. It'll move upwards. So the pressure will push up,
but as we move up, we will have lower and lower pressure
because the gases will bounce into the surface less. And we'll have a higher,
higher volume. So as we move each
infinitesimally small grain of sand, and we're going to do it
super slow, that we're always infinitely close
to equilibrium. We're going to move from this
state to this state. And even better, let's view that
this is the first stage of our Carnot cycle. Let's say in either case, we're
on top of a reservoir. So in either case, both of these
systems are on top of this infinite reservoir that
has a temperature T1. And what that does it keeps
the temperature constant. So we're going to travel
along an isotherm. Because normally if we were to
remove these things and if we just allowed it to occur
adiabatically, we would actually lose temperature. We would actually lose
average kinetic energy as work is done. But in this case we have
this reservoir. So heat is just going to be
transferred to my system. Heat is going to
be transferred. So let me call that QR. Right? If the reservoir wasn't here,
our temperature would go down. But since we do have the
reservoir, we will constantly be transferring heat. And we've seen this. This is just the first phase
of the Carnot cycle. We'll move along an isotherm
like this. This is the reversible case. And the only reason why we can
even draw the state at every point here is because reversible
processes are quasistatic. They're always infinitely
close to equilibrium. And when we say reversible,
we're also saying there's no friction between this little
piston in the cylinder. That if we put a grain of sand
back, it'll go exactly to where it was before. And no energy was lost
because there was no heat of friction there. So this is what the graph
of the reversible process would look like. Now what does the graph of the
error irreversible-- Well, actually no. I won't draw the graph, but
let's talk about what the irreversible process is
going to be like. So it's going to look similar. It's going to look like that. It's going to have
its gases there. But for the sake of an argument,
to get from that state to that state, instead of
moving the pebbles one by one, let's say I have
these big blocks. And when I remove one of these
big blocks, I go from that state to that state. But obviously all hell
breaks loose. So I'm not really defined in
this in between state. But I definitely go from that
state to that state once I go back to equilibrium. Now the other key thing in the
irreversible process-- and every real process in our world
is irreversible-- is that you're going to
have friction here. As this moves up, it's going to
rub against the side of the container and generate
heat of friction. So let me call that
heat of friction. So let me ask you a question. If, in this case, Q sub R had
to be added to the system to maintain its temperature, what's
going to be the Q sub irreversible here? How much has to be added to this
system in order to keep it at the constant
temperature T1? Will it have to be more or less
than what was added to the reversible system? Well this guy, as this piston
moves up, he's generating some of his own heat. So if this was an adiabatic
process, he wouldn't lose as much temperature as
this guy would. So he's going to need less heat
from the reservoir in order to maintain
his temperature. At T1. In order to get to this
point on the isotherm. Remember, this irreversible
process, we don't know what happens over here. He might be travelling
on some crazy path. In fact, we can't even define
the path, because it goes out of equilibrium. So it's going to be
some crazy thing. But we know it pops back on the
PV diagram right there. But because it's generating some
of its own heat from the friction, it's going to need
less heat from the reservoir. So let me write that down. The heat absorbed by the
reversible process is going to be greater than the heat
absorbed by the irreversible process. And that's because the
irreversible process is generating friction. Fair enough. Now what is the change in
entropy for both systems? Well they both started here. And they both ended here. And entropy is a
state variable. So the change in entropy for
the reversible process is going to be equal to the change
in entropy for the irreversible process. They're both going from
there to there. And obviously the entropy
has changed. We're going from one state
to another state. And the entropy, well I won't
go too much into it. But let's ask another
question. What is the total change in
entropy of the universe for the reversible process? So for the universe, that's
going to be-- Our universe here is the reservoir
in our system. So let me write here. Reversible. I don't want to run
out of space. Let me see, I'll do it
in a different color. So this is the reversible
process. So the change in entropy of the
universe is equal to the change in entropy of our
reversible process plus the change in entropy of-- oh, I
already used R for reversible, so let's call it the reserv--
well, the first three letters are the same, so let me call
it of our environment. Right? And then the reversible process,
the change in entropy of our reversible process is
the heat added for our reversible process. And we can use this definition
because it is a reversible process. It's that over T1. And then what's the change in
entropy of our environment? Well it's giving away Q sub R. Right? So its heat absorbed
is minus Q sub R. And, of course, it's at a
constant temperature. It is a heat reservoir. It's at T1. So it equals 0. It equals 0. Interesting. So actually I should take
a little side note here. That, the change in entropy of
the universe for a reversible process is 0. And actually that should make
a lot of intuitive sense because the whole point of a
reversal process is you could go in this direction, or you
could go in that direction. It's always so close people to
equilibrium you can move in either direction. And if the entropy was greater
than 0 in one direction, it would have to be less than
0 on the other direction. So it wouldn't be able to go in
the other direction by the second law of thermodynamics. So it actually makes sense
that the entropy of the universe-- the change in entropy
of the universe, not just of the system--
when a reversible process occurs, is 0. Let's see if we can relate
that to the irreversible process So if I wanted to
figure out the change in entropy of the irreversible
process-- What's the change in entropy of the irreversible
process? And then let me subtract from
that the heat that was taken away from the reservoir, from
the irreversible process, Q sub IR, and of course
all this over T1. What is this going to be equal
to in relation to 0? Remember this is an irreversible, spontaneous process. Well, this value, the
irreversible process is starting here and going there. So its change in delta S is the
exact same thing as that. Change in delta S is the same
as the reversible process. So these two things
are equivalent. Now, I just told you that since
there's some heat of friction, this guy has to take
in less heat from the reservoir than this guy. So if this-- and I
wrote it here. I mean, let me clean this
thing up a little bit. I wrote it right here. Q sub IR, the heat that the
irreversible process has to take from the reservoir, because
generating its own heat from friction, is less
than the heat from the reversal process. So this number, right here, is
less than this number here. Or you could view it this way. This number here was
equal to this. So this number here is going
to be less than this. So this has to be greater than
0 for the irreversible, spontaneous process. Now let's just do a little
bit of mathematics. So this is the heat that was
essentially given to the irreversible system. It's a minus here because
this is the term. It's kind of taken away from
the actual reservoir. So let's just do a little bit
of-- let's just multiply all sides of this equation by T1. And we get T1 times delta S of
the irreversible process minus Q of the irreversible process
is greater than 0. Now, well, let's just--
So what is this? How can we say this? But let's actually just multiply
both sides of this by negative 1. And remember, this
is true for any irreversible, spontaneous process. If you multiply both
sides of this by negative 1, you get this. The heat added to the
irreversible process minus T times delta S of the
irreversible process is going to be less than 0. This is true for any irreversible, spontaneous process. And at this point,
this should look reasonably familiar to you. When we wrote the Gibbs Free
Energy formula, we said change in G is equal to delta H
minus T times delta S. And we said that if this is less
than 0, then we have a spontaneous process. And this all makes sense. Because these two are equivalent
to each other. What's the only difference
here? Here we wrote heat added
to the system. Here we wrote change
in enthalpy. And I've done three or four
videos right now, where I say that change in enthalpy is equal
to the heat added to a system, as long as we're
dealing with a constant pressure system. So we get this result just by
comparing a irreversible to a reversible system. And this is true for all irreversible, spontaneous processes. And then if we assume that we're
dealing with a constant pressure system-- so you can
forget a little bit of what I just drew, because we know
that this is true, and we assume constant pressure--
then we get to this. And then we know that if
something is spontaneous, then this right here must
be less than 0. So hopefully you found that
a little bit interesting. Actually I'll do one
more point here. This kind of gels with the
idea that second law of thermodynamics tells us that for
any spontaneous process, delta S is going to be greater
than or equal-- or well for any spontaneous process, is
going to be greater than 0. Because although this right
here isn't the formal definition of entropy, because
we're not dealing with a reversible process here,
you can kind of think of it that way. And so, at least, on an
intuitive level it gets you there, that we have delta
S greater than 0. And I won't fixate on that too
much, because what I did earlier in the video is more
rigorous than what I'm heading to right now. So hopefully that gives you a
sense of where you can get to the Gibbs Free Energy formula,
and how it drives spontaneous reactions from just our basic
understanding of what reversible and irreversible
processes are, and how they relate to entropy and heat
exchange and enthalpy.