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## Physics library

### Course: Physics library > Unit 10

Lesson 3: Laws of thermodynamics- Macrostates and microstates
- Quasistatic and reversible processes
- First law of thermodynamics / internal energy
- More on internal energy
- What is the first law of thermodynamics?
- Work from expansion
- PV-diagrams and expansion work
- What are PV diagrams?
- Proof: U = (3/2)PV or U = (3/2)nRT
- Work done by isothermic process
- Carnot cycle and Carnot engine
- Proof: Volume ratios in a Carnot cycle
- Proof: S (or entropy) is a valid state variable
- Thermodynamic entropy definition clarification
- Reconciling thermodynamic and state definitions of entropy
- Entropy intuition
- Maxwell's demon
- More on entropy
- Efficiency of a Carnot engine
- Carnot efficiency 2: Reversing the cycle
- Carnot efficiency 3: Proving that it is the most efficient

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# Reconciling thermodynamic and state definitions of entropy

Long video explaining why entropy is a measure of the number of states a system can take on (mathy, but mind-blowing). Created by Sal Khan.

## Want to join the conversation?

- aren't there an infinite number of states in a volume since there are an infinite places a particle can be in the volume?(33 votes)
- In classical mechanics, the position and momentum variables of a particle CAN vary continuously, so the number of possible microstates is actually infinite. Integrals are therefore used. For more, see the following wikipedia article on the partition function: http://en.wikipedia.org/wiki/Partition_function_%28statistical_mechanics%29(14 votes)

- if entropy is a state variable then what is the entropy of the gas at a point where pressure is P volume V and temperature T and internal energy U ?? if we can't say from these how does it becomes a state variable as per def S=Q/T where Q is the heat required to reach there which can be done in many ways so from these it seems that S depends on the process it is done ???(21 votes)
- Good question. We only see delta S = Q / T, i.e., the INCREMENTAL entropy from adding heat at a certain temperature. At absolute zero, S is also 0 and the stuff is a crystal. When you now start heating, the crystal will start vibrating, then it will crack into pieces, individual molecules will come off, etc. The heat capacity will change over these different phases. When you stop heating, the stuff will go into an equilibrium state at a certain temperature. And it won't matter how it got there, i.e., if first molecule 1 broke off and then m 2. I would believe that S not only depends on P, V, T, and U but also on the type of gas we are talking about here, i.e., what type of crystals it forms at very low temparature. So to answer your question: the reason we cannot compute the absolute value of S from macrostate variables is not that it would depend on how we reached that state. It is that the calculation would be extremely complex and would require information (e.g., about crystal properties) that we might not have. Hopefully somebody else can comment on this as well.(19 votes)

- Is change in entropy in adiabatic process zero?? as Q is zero??(14 votes)
- The change in entropy in an adiabatic change will be zero
*only*if the adiabatic process is reversible (quasi-static). The expansion of the gas in a vacuum is not reversible, so the change in entropy must be positive and not zero, even if Q is zero.(14 votes)

- At1:40an equation is started that is going to list all possible combination's for the location and momentum of a system with N particles. Sal says that X is equal to all possible combination's of those two variables for each particle, and to figure out how many possible combination's for all particles you take X^N. Does that then imply that two particles can share the same location? Just curious not in need of explanation for a test or anything.

Thank you Sal!

-Tyler(16 votes)- YES: the equation under discussion tacitly assumes that particles CAN inhabit the same point in space. This seems like a bizarre assumption, but it's actually not that bad!

In these videos, Sal makes repeated use of the ideal gas law, PV=nRT or PV=NkT. Now an ideal gas is the simplest kind of gas so it makes sense to start here; but an important approximation for the ideal gas is that the particles of the gas do not interact with one another, in which case there is no reason for two particles to avoid each other and they can sit in the very same spot.

A different way of looking at it is to say that the particles of an ideal gas occupy zero volume.

As I said, this is okay-ish because many gases we encounter are low density. If you added up the volumes of all the gazillions of air molecules in the room, the result would be tiny compared to the volume of the room itself. So each molecule can be said to take up approximately zero space.

If you want to go deeper, I suggest you take a look at the "van der Waals gas", which tries to account for finite-volume effects (as well as long-range interactions between molecules).(10 votes)

- Did not the particles in the first box expend heat expanding into the doubled volume? It appears similar to an abstract piston. I may be wrong in assuming that the expansion happens because particles push each other around.(10 votes)
- The expansion is in a vacuum. The gas pushes against nothing, so no energy is spent by the gas during the expansion. The expansion happens not because the particles push against each other but because there is no wall that prevents a particle from moving into the available empty volume.(13 votes)

- At about3:40Sal introduced the natural logarithm to describe the entropy. I didn't really get why a logarithm has to used. Can someone give me an explanation why the natural logarithm has to be used in this exact example?(8 votes)
- In the beginning of the video, he introduces us to the idea of the state of the system, S. As you saw, the number of possible states for the total number of particles can be an exceedingly large number. In such cases, we often resort to the use of a logarithm to reduce the scale of the problem. Specifically, while the natural logarithm of 5 is about 1.6, the natural logarithm of 50 million is only about 18. In this way, large numbers (such as the number of molecules in a volume of gas) can more easily be handled.

The other motivation that we see in in employing the laws of logarithm. These include clearing variable exponents, etc.

Hope that helps.(18 votes)

- Sal's used "ln 2" throughout, but can that be generalized to "ln (v2/v1)", as in natural log of the ratio of the ending volume to the beginning volume?(8 votes)
- Yes. He acquired the two because his volume was doubled. You see something similar in the integral of 1/V from V1 to 2V1. If his original assumption had been 3 times the volume we would've integrated to 3V1 instead. But ultimately the 2 is just the ratio of volumes.(6 votes)

- When Sal blew away the partition wall at5:10, he went on to say that the temperature of the system stayed the same. This makes sense seeing as the particles didn't have to knock against the container to expand it, that is to say they didn't use any of their kinetic energy to increase the volume as in previous examples. I'm happy with that, but when we look at the equation he finishes with: S=Q/T , if T has remained the same but S has changed (there are more possible states so S increases, that's easy) , Q must change... But this isn't intuitive, why should the heat increase just because there's more room for the particles to bounce around in!?(6 votes)
- In the example at5:10, there is no work or heat, so he solves for S by molecular positioning (no mention of Q).

In the "nearly identical but more realistic" example at17:21, the system does work (and receives heat) so he solves for S by Q/T.

He then shows these are the same, and S is a state variable regardless of the path taken. As we saw in the previous video, solving for S can be done using the most convenient manner possible.(4 votes)

- @02:13,

shouldn't it be x(x-1)(x-2)(x-3).....

because a already ocupied 1 state so b has only x-1 states to go to?(6 votes)- But the states are changing as the molecules have velocity. If they wouldn't have had a velocity then your proposition would have been correct(4 votes)

- Watching this video, I realized I do not know how a vacuum works! I get the "big idea" in that the natural tendency of disparate pressures when exposed to each other is to equalize, but how does this work on the atomic level?

Sal implies that once the wall in this video gets eliminated, molecules that were to bounce off of it continue on filling the void, but other molecules not currently headed towards the wall don't "know" about the new volume. It doesn't seem right that it just happens that molecules bounce around enough to eventually equal out. If you poke a hole only large enough for a few molecules to pass through, the chance that they would happen to bounce through is slim, but intuitively there would be a constant stream. Are the molecules somehow dragging along others through?(4 votes)- No, they do not drag each other. What happens is that at the start, all molecules are in one side of the hole. If P is the probability of a molecule to go through the hole, the number of molecules that pass to the other side is n (number of molecules) * P. When some of them did already pass to the other side, n decreases, so there will be less molecules going through. Furthermore, there is also a chance of the molecules that already passed to come back. This eventually balances out when the pressure is the same on both sides.

This looks kinda weird at first, because intuitively (at least for me) it looks like that vacuum is sucking everything. It's NOT! Don't forget that the Earth is in the space and it's not exploding :D(3 votes)

## Video transcript

If you followed some of the
mathematics, and some of the thermodynamic principles in the
last several videos, what occurs in this video might
just blow your mind. So not to set expectations
too high, let's just start off with it. So let's say I have
a container. And in that container,
I have gas particles. Inside of that container,
they're bouncing around like gas particles tend to do,
creating some pressure on the container of a certain volume. And let's say I have
n particles. Now, each of these particles
could be in x different states. Let me write that down. What do I mean by state? Well, let's say I
take particle A. Let me make particle A
a different color. Particle A could be down here
in this corner, and it could have some velocity like that. It could also be in that
corner and have a velocity like that. Those would be two
different states. It could be up here, and have
a velocity like that. It could be there and have
a velocity like that. If you were to add up all the
different states, and there would be a gazillion, of
them, you would get x. That blue particle could have
x different states. You don't know. We're just saying, look. I have this container. It's got n particles. So we just know that each
of them could be in x different states. Now, if each of them can be in
x different states, how many total configurations are there
for the system as a whole? Well, particle A could be in x
different states, and then particle B could be in
x different places. So times x. If we just had two particles,
then you would multiply all the different places where
X could be times all the different places where the red
particle could be, then you'd get all the different
configurations for the system. But we don't have just
two particles. We have n particles. So for every particle, you'd
multiply it times the number of states it could have,
and you do that a total of n times. And this is really just
combinatorics here. You do it n times. This system would have
n configurations. For example, if I had two
particles, each particle had three different potential
states, how many different configurations could there be? Well, for every three that one
particle could have, the other one could have three different
states, so you'd have nine different states. You'd multiply them. If you had another particle with
three different states, you'd multiply that by
three, so you have 27 different states. Here we have n particles. Each of them could be in
x different states. So the total number of
configurations we have for our system-- x times itself n times
is just x to the n. So we have x to the n states
in our system. Now, let's say that we like
thinking about how many states a system can have. Certain
states have less-- for example, if I had fewer
particles, I would have fewer potential states. Or maybe if I had a smaller
container, I would also have fewer potential states. There would be fewer potential
places for our little particles to exist. So I want to create some type
of state variable that tells me, well, how many states
can my system be in? So this is kind of a macrostate
variable. It tells me, how many states
can my system be in? And let's call it
s for states. For the first time in
thermodynamics, we're actually using a letter that in some way
is related to what we're actually trying to measure. s for states. And since the states, they can
grow really large, let's say I like to take the logarithm
of the number of states. Now this is just how I'm
defining my state variable. I get to define it. So I get to put a logarithm
out front. So let me just put
a logarithm. So in this case, it would be the
logarithm of my number of states-- so it would be x to the
n, where this is number of potential states. And you know, we need some
kind of scaling factor. Maybe I'll change the
units eventually. So let me put a little
constant out front. Every good formula needs
a constant to get our units right. I'll make that a lowercase k. So that's my definition. I call this my state variable. If you give me a system, I
should, in theory, be able to tell you how many states
the system can take on. Fair enough. So let me close that
box right there. Now let's say that I were to
take my box that I had-- let me copy and paste it. I take that box. And it just so happens
that there was an adjacent box next to it. They share this wall. They're identical in size,
although what I just drew isn't identical in size. But they're close enough. They're identical in size. And what I do, is I blow
away this wall. I just evaporate it,
all of a sudden. It just disappears. So this wall just disappears. Now, what's going to happen? Well, as soon as I blow away
this wall, this is very much not an isostatic process. Right? All hell's going
to break loose. I'm going to blow away this
wall, and you know, the particles that were about to
bounce off the wall are just going to keep going. Right? They're going to keep going
until they can maybe bounce off of that wall. So right when I blow away this
wall, there's no pressure here, because these guys have
nothing to bounce off to. While these guys don't
know anything. They don't know anything until
they come over here and say, oh, no wall. So the pressure is in flux. Even the volume is in flux, as
these guys make their way across the entire expanse
of the new volume. So everything is in flux. Right? And so what's our new volume? If we call this volume,
what's this? This is now 2 times
the volume. Let's think about some of the
other state variables we know. We know that the pressure
is going to go down. We can even relate it, because
we know that our volume is twice it. Is 2 times the volume. What about the temperature? Well, the temperature change. Temperature is average kinetic
energy, right? Or it's a measure of average
kinetic energy. So all of these molecules
here, each of them have kinetic energy. They could be different amounts
of kinetic energy, but temperature is a measure of
average kinetic energy. Now, if I blow away this wall,
does that change the kinetic energy of these molecules? No! It doesn't change it at all. So the temperature
is constant. So if this is T1, then the
temperature of this system here is T1. And you might say, hey, Sal,
wait, that doesn't make sense. In the past, when my cylinder
expanded, my temperature went down. And the reason why the
temperature went down in that case is because your molecules
were doing work. They expanded the container
itself. They pushed up the cylinder. So they expended some of their
kinetic energy to do the work. In this case, I just blew
away that wall. These guys did no work
whatsoever, so they didn't have to expend any of their
kinetic energy to do any work. So their temperature
did not change. So that's interesting. Fair enough. Well, in this new world,
what happens? Eventually I get to a situation
where my molecules fill the container. Right? We know that from
common sense. And if you think about
it on a microlevel, why does that happen? It's not a mystery. You know, on this direction,
things were bouncing and they keep bouncing. But when they go here, there
used to be a wall, and then they'll just keep going,
and then they'll start bouncing here. So when you have gazillion
particles doing a gazillion of these bounces, eventually,
they're just as likely to be here as they are over there. Now. Let's do our computation
again. In our old situation, when we
just looked at this, each particle could be in one of x
places, or in one of x states. Now it could be in twice
as many states, right? Now, each particle could be
in 2x different states. Why do I say 2x? Because I have twice
the area to be in. Now, the states aren't just, you
know, position in space. But everything else-- so, you
know, before here, maybe I had a positions in space times b
positions, or b momentums, you know, where those are all the
different momentums, and that was equal to x. Now I have 2a positions in
volume that I could be in. I have twice the volume
to deal with. So I have 2a positions in volume
I can be at, but my momentum states are going to
still be-- I just have b momentum states-- so this
is equal to 2x. I now can be in 2x different
states, just because I have 2 times the volume to travel
around in, right? So how many states are
there for the system? Well, each particle can
be in 2x states. So this is 2x times
2x times 2x. And I'm going to do
that n times. So my new s-- so this is, you
know, let's call this s initial-- so my s final, my
new way of measuring my states, is going to be equal to
that little constant that I threw in there, times the
natural log of the new number of states. So what is it? It's 2x to the n power. So my question to you is, what
is my change in s when I blew away the wall? You know, there was this room
here the entire time, although these particles really
didn't care because this wall was there. So what is the change in s when
I blew away this wall? And this should be clear. The temperature didn't
change, because no kinetic energy was expended. And this is all in isolation. I should have said. It's adiabatic. There's no transfer of heat. So that's also why the
temperature didn't change. So what is our change in s? Our change in s is equal to
our s final minus our s initial, which is equal to--
what's our s final? It's this expression,
right here. It is k times the natural
log--and we can write this as 2 to the n, x to the n. That's just exponent rules. And from that, we're going to
subtract out our initial s value, which was this. k natural
log of x to the n. Now we can use our logarithm
properties to say, well, you know, you take the logarithm of
a minus the logarithm of b, you can just divide them. So this is equal to k-- you
could factor that out-- times the logarithm of 2 to the
N-- it's uppercase N, so let me do that. This is uppercase N. I don't want to get confused
with Moles. Uppercase N is the number of
particles we actually have. So it's 2 to upper case N times x
to the uppercase N divided by x to the uppercase N. So these two cancel out. So our change in s is equal to
k times the natural log of 2 to the N-- or, if we wanted to
use our logarithm properties, we could throw that
N out front. And we could say, our change in
the s, whatever this state variable I've defined-- and this
is a different definition than I did in the last video--
is equal to big N, the number of molecules we have, times my
little constant, times the natural log of 2. So by blowing away that wall and
giving my molecules twice as much volume to travel around
in, my change in this little state function
I defined is Nk the natural log of 2. And what really happened? I mean, it clearly
went up, right? I clearly have a positive
value here. Natural log of 2 is
a positive value. N is positive value. It's going to be very large
number than the number of molecules we had. And I'm assuming my constant
I threw on there is a positive value. But what am I really
describing? I'm saying that look, by blowing
away that wall, my system can take on
more states. There's more different
things it can do. And I'll throw a little
word out here. Its entropy has gone up. Well, actually, let's
just define s to be the word entropy. We'll talk more about the
word in the future. Its entropy has gone up, which
means the number of states we have has gone up. I shouldn't use the word entropy
without just saying, entropy I'm defining
as equal to S. But let's just keep it
with s. s for states. The number of states we're
dealing with has gone up, and it's gone up by this factor. Actually, it's gone up by
a factor of 2 to the n. And that's why it becomes
n natural log of 2. Fair enough. Now you're saying, OK. This is nice, Sal. You have this statistical way,
or I guess you could, this combinatoric way of measuring
how many states this system can take on. And you looked at the
actual molecules. You weren't looking at the
kind of macrostates. And you were able to do that. You came up with this macrostate
that says, that's essentially saying, how many
states can I have? But how does that relate to
that s that defined in the previous video? Remember, in the previous video,
I was looking for state function that dealt with heat. And I defined s, or change in
s-- I defined as change in s-- to be equal to the heat added
to the system divided by the temperature that the
heat was added at. So let's see if we can see
whether these things are somehow related. So let's go back to our
system, and go to a PV diagram, and see if we can do
anything useful with that. Alright. OK. So this is pressure,
this is volume. Now. When we started off, before we
blew away the wall, we had some pressure and some volume. So this is V1. And then we blew away the
wall, and we got to-- Actually, let me do it a
little bit differently. I want that to be just
right there. Let me make it right there. So that is our V1. This is our original state
that we're in. So state initial, or
however we want it. That's our initial pressure. And then we blew away
the wall, and our volume doubled, right? So we could call this 2V1. Our volume doubled, our pressure
would have gone down, and we're here. Right? That's our state 2. That's this scenario right
here, after we blew away the wall. Now, what we did was not
a quasistatic process. I can't draw the path here,
because right when I blew away the wall, all hell broke
loose, and things like pressure and volume weren't
well defined. Eventually it got back to an
equilibrium where this filled the container, and nothing
else was in flux. And we could go back to here,
and we could say, OK, now the pressure and the
volume is this. But we don't know what happened
in between that. So if we wanted to figure out
our Q/T, or the heat into the system, we learned in the last
video, the heat added to the system is equal to the work
done by the system. We'd be at a loss, because the
work done by the system is the area under some curve, but
there's no curve to speak of here, because our system wasn't
defined while all the hell had broke loose. So what can we do? Well, remember, this is
a state function. And this is a state function. And I showed that in
the last video. So it shouldn't be dependent
on how we got from there to there. Right? So this change in entropy--
actually, let me be careful with my words. This change in s, so s2 minus
s1, should be independent of the process that got
me from s1 to s2. So this is independent of
whatever crazy path-- I mean, I could have taken some
crazy, quasistatic path like that, right? So any path that goes from this
s1 to this s2 will have the same heat going into the
system, or should have the same-- let me take that-- Any system that goes from s1 to
s2, regardless of its path, will have the same change
in entropy, or their same change in s. Because their s was something
here, and it's something different over here. And you just take the difference
between the two. So what's a system that we
know that can do that? Well, let's say that we
did an isothermal. And we know that these are all
the same isotherm, right? We know that the temperature
didn't change. I told you that. Because no kinetic energy was
expended, and none of the particles did any work. So we can say, we can think of a
theoretical process in which instead of doing something like
that, we could have had a situation where we started off
with our original container with our molecules in it. We could have put a reservoir
here that's equivalent to the temperature that we're at. And then this could have been
a piston that was maybe, we were pushing on it with some
rocks that are pushing in the left-wards direction. And we slowly and slowly remove
the rocks, so that these gases could push the
piston and do some work, and fill this entire volume,
or twice the volume. And then the temperature would
have been kept constant by this heat reservoir. So this type of a process is
kind of a sideways version of what I've done in the Carnot
diagrams. That would be described like this. You'd go from this state to that
state, and it would be a quasistatic static process
along an isotherm. So it would look like that. So you could have
a curve there. Now, for that process,
what is the area under the curve there? Well, the area under the curve
is just the integral-- and we've done this multiple times--
from our initial volume to our second volume,
which is twice it, of p times our change in volume, right? p is our height, times our
little changes in volume, give us each rectangle. And then the integral is just
the sum along all of these. So that's essentially the work
that this system does. Right? And the work that this system
does, since we are on an isotherm, it is equal to the
heat added to the system. Because our internal energy
didn't change. So what is this? We've done this multiple times,
but I'll redo it. So this is equal to the
integral of V1 to 2V1. PV equals NRT, right? NRT. So P is equal to NRT/V. NRT over V dv. And the t is T1. Now, all of this is happening
along an isotherm, so all of these terms are constant. So this is equal to the integral
from V1 to 2V1 of NRT1 times 1 over V dv. I've done this integral
multiple times. And so this is equal to-- I'll
skip a couple of steps here, because I've done it in several
videos already-- the natural log of 2V1
over V1, right? The antiderivative of this
is the natural log. Take the natural log of that
minus the natural log of that, which is equal to the natural
log of 2V1 over V1. Which is just the same thing
as NRT1 times the natural log of 2. Interesting. Now, let's add one little one
little interesting thing to this to this equation. So this is NRT, but if I wanted
to write in terms of the number of molecules, N
is the number of moles. So I could rewrite N as the
number of molecules we have divided by, 6 times 10
to the 23 power. Right? That's what n could
be written as. So if we do it that way, then
what is our-- remember, all of this, we were trying to
find the amount of work done by our system. Right? But if we do it this way,
this equation will turn into-- so let's see. The work done by our system--
this is our quasistatic processes that's going from this
state to that state, but it's doing it in a quasistatic
way, so that we can get an area under the curve. So the work done by this
system is equal to-- I'll just write it. N times R over 6 times 10
to the 23, times T1 natural log of 2. Fair enough. Let's make this into
some new constant. For convenience, let me
call it a lowercase k. So the work we did is equal to
the number of particles we had, times some new constant--
we'll call that Boltzmann constant, so it's really
just 8 divided by that. Times T1, times the
natural log of 2. Fair enough. Now, that's only in
this situation. The other situation did
no work, right? So I can't talk about this
system doing any work. But this system did
do some work. And since it did it along an
isotherm, delta-- the change in internal energy is equal to
0, so the change in internal energy, which is equal to the
heat applied to the system minus the work done by the
system-- this is going to be equal to 0, since our
temperature didn't change. So the work is going to
be equal to the heat added to the system. So the heat added to the system
by our little reservoir there is going to be-- so the
heat is going to be the number of particles we had times
Boltzmann constant, times our temperature that we're on
the isotherm, times the natural log of 2. And all this is a byproduct
of the fact that we doubled our volume. Now, in the last video, I
defined change in s as equal to Q divided by the heat
added, divided by the temperature at which
I'm adding it. So for this system, this
quasistatic system, what was the change in s? How much did our s-term,
our s-state, change by? So change in s is equal to
heat added divided by our temperature. Our temperature is T1, so that's
equal to this thing. Nk T1 times the natural log
of 2, all of that over T1. So our delta-- these
cancel out. And our change in our s-quantity
is equal to Nk times the natural log of 2. Now, you should be starting to
experience an a-ha moment. When we defined in the previous
video, we were just playing with thermodynamics, and
we said, gee, we'd really like to have a state variable
that deals with heat, and we just made up this thing right
here that said, change in that state variable is equal to the
heat applied to the system divided by the temperature at
which the heat was applied. And when we use that definition
the change in our s-value from this position
to this position, for a quasi-static process,
ended up being this. Nk natural log of 2. Now, this is a state function. State variable. It's not dependent
on the path. So any process that gets from
here, that gets from this point to that point, has to
have the same change in s. So the delta s for any process
is going to be equal to that same value, which was N, in
this case, k, times the natural log of 2. Any system, by our definition,
right? It's the state variable. I don't care whether it
disappeared, or the path was some crazy path. It's a state. It's only a function of that and
of that, our change in s. So given that, even this
system-- we said that this system that we started the video
out with, it started off at this same V1, and it
got to the same V2. So by the definition of the
previous video, by this definition, its change in s is
going to be the number of molecules times some constant
times the natural log of 2. Now, that's the same exact
result we got when we thought about it from a statistical
point of view, when we were saying, how many more different
states can the system take on? And what's mind-blowing here
is that what we started off with was just kind of a nice,
you know, macrostate in our little Carnot engine world,
that we didn't really know what it meant. But we got the same exact result
that when we try to do it from a measuring the
number of states the system could take on. So all of this has been a long,
two-video-winded version of an introduction to entropy. And in thermodynamics, a change
in entropy-- entropy is s, or I think of it, s for
states-- the thermodynamic, or Carnot cycle, or Carnot engine
world, is defined as the change in entropy is defined
as the heat added to the system divided by the
temperature at which it was added. Now, in our statistical
mechanics world, we can define entropy as some constant-- and
it's especially convenient, this is Boltzmann's constant--
some constant times the natural log of the number of
states we have. Sometimes it's written as omega, sometimes
other things. But this time, it's the number
of states we have. And what just showed in this
video is, these are equivalent definitions. or. At least for that one case
I just showed you. These are equivalent
definitions. When we used the number of
states for this, how much did it increase, we got
this result. And then when we used the
thermodynamic definition of it, we got that same result. And if we assume that this
constant is the same as that constant, if they're both
Boltzmann's constant, both 1.3 times 10 to the minus 23,
then our definitions are equivalent. And so the intuition of
entropy-- in the last one, we were kind of struggling
with that. We just defined it this way,
but we were like, what does that really mean? What change in entropy means,
is just how many more states can the system take on? You know, sometimes when you
learn it in your high school chemistry class, they'll
call it disorder. And it is disorder. But I don't want you to think
that somehow, you know, a messy room has higher entropy
than a clean room, which some people sometimes use
as an example. That's not the case. What you should say is, is that
a stadium full of people has more states than a stadium
without people in it. That has more entropy. Or actually, I should even
be careful there. Let me say, a stadium at a
high temperature has more entropy than the inside
of my refrigerator. That the particles in that
stadium have more potential states than the particles
in my refrigerator. Now I'm going to leave you
there, and we're going to take our definitions here, which I
think are pretty profound-- this and this is the same
definition-- and we're going to apply that to talk about
the second law of thermodynamics. And actually, just
a little aside. I wrote omega here, but in our
example, this was 2 to the N. And so that's why
it's simplified. This was x the first time, and
then the second time, when we double the size of our room, or
our volume, it was x to the N times 2 to N. I just want to make sure you
realize what omega relates to, relative to what I just
went through. Anyway, see you in
the next video.