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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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# Thermodynamic entropy definition clarification

Clarifying that the thermodynamic definition of Entropy requires a reversible system. Created by Sal Khan.

## Want to join the conversation?

- why he took friction as heat added to the system with positive sign. shouldn it be minus becaouse while system is expanding systems itself is gonna do this work and loose energy for friction forces?(20 votes)
- Friction always adds heat to a system because you are creating energy by the friction process. During the expansion the work is being done as the piston moves up, in a 'non-friction piston' there would be no Q_f because we assume the expansion is able to do this with no loss in energy. In a 'friction piston' we take accountability of the Q_f because the piston is rubbing against the side creating heat and works against the expansion process so the piston wouldn't reach the same height in a 'non-friction piston' when the same amount of pebbles are removed.

Since the equations relate to internal energy the piston friction is adding energy to the system (friction against the expansion), where the work is losing energy to the system (raising the piston).(3 votes)

- Why does the W cancel out at11:30? If friction was there when the system expanded how can the work done on the system be the same as the work it did on the surrounding before?(9 votes)
- It appears to me that he is looking at the same isotherm, so even if you go back in the other direction ( from A to B, or from B to A) the area under the curve, or the work, will be the same, so they will cancel out.(4 votes)

- at10:00he says the system generates heat and so it will be given to surrounding...so it must be -Qf right...(7 votes)
- Ya I thought of the same thing..

Shouldn't it be dU=Q-(W=Qf) ?

i.e. Q goes into doing displacement work + frictional work or heat(2 votes)

- Has entropy been proven?(4 votes)
- Entropy in an isolated system increases or remains constant. Energy tends to degrade into its lowest state -- heat. Yes, entropy is real. Heat is relative to absolute zero (0ºK) -- the absence of heat. Entropy tends to be linear and never decreases; in this way we can see it as time (particles from the creation of the universe or cosmic background radiation). Therefore, the universe tends to move toward disorder --entropy. The heat your body is producing from cellular respiration proves it.(6 votes)

- In13:12, Sal divided all the terms by the same T. However, the first isothermal process should have a different process from the other (T1, then T2, where T1>T2), based on what I know from his previous videos. So why did he state that the temperature was constant the whole time (although i can understand graphically that he did go from state 2 to 1 using the same path he used when he went from state 1 to 2, and hence, the temperature should stay the same, but why did he draw it that way?)(5 votes)
- what is the difference between quasi-static process & reversible process?(4 votes)
- A reversible process equals a quasistatic proces where you have no change in entropy.

So you can have reversible quasistatic processes and irreversible quasistatic processes.(1 vote)

- I keep hearing that entropy is always increasing that is because Q is increasing in dS=dQ/T. But suppose when I cool something off isn't it releasing heat? And thus isn't entropy decreasing? Because then dQ would be negative(3 votes)
- Entropy is always increasing in the universe. If you cool off something it is just releasing that heat/energy.(3 votes)

- At 5.00, he says when adding the pebble the piston won't go back down to the initial level because friction is always resisting movement. But wasn't friction also resisting movement when we removed the pebble? Shouldn't those effects cancel out?(3 votes)
- No, friction doesn't cancel out that way, it does the opposite: it accumulates. it resists on the way up, and then it resist again on the way down. Each time you go up and down, the range of motion will decrease. Just like a ball eventually stops bouncing, or a toy car eventually stops rolling.(4 votes)

- At13:20Sal showns that the change in entropy after the cycle is (2Q_f)/T. The left side of the equation is a subtraction, not an addition like in the last video. In the last video, he showed that the change in entropy is equal to Q_1/T_1 + Q_2/T_2. What's going on here? These are clearly not the same. This doesn't make any sense.(4 votes)
- In this video Sal talks about irreversible system. That's where (2Q_f)/T comes from: it's a heat generated by friction.

About the signs of Q: in the previous video Sal puts a plus sign in front of the Q, but actual value of Q is negative (he actually mentioned that "we'll see its actually negative" somewhere in the video). In this video Sal assumes the value of Q to be positive (as he mentions at about10:20, so he puts a negative in front.(2 votes)

- if in process 'Q' heat is added and also temperature is increased from T(1) to T(2) then what is change in 'S' is it Q/T(1) or Q/T(2) or any other with a reason(3 votes)
- If the temperature change is caused by the addition of said 'Q' heat, then we use the initial temperature or the temperature at which the heat was added.0:09(3 votes)

## Video transcript

In the video where I first
introduced the concept of entropy, I just tried
something out. I defined my change in entropy
as being equal to the heat added to a system, divided by
the temperature at which it was added to the system. And then I tested to
see if this was a valid state variable. And when I did that, I looked
at the Carnot cycle. And this is a bit of a review. Never hurts to review. Let me draw the PV
diagram here. We saw that we start at this
state here, and then we proceed isothermically. We removed little pebbles
off the piston. So we increased the volume
and lowered the pressure. Then we proceed adiabatically,
where we isolated things and we moved like that. That was adiabatically. Then at this other isotherm,
we added the pebbles back. And then we isolated
the system again. So adiabatically, we continued
to add more pebbles, and we got back to our original
state. And I did a couple of videos
where I show that if you take the heat added here-- so this is
all being done at some high temperature, T1. This is being done at some
low temperature, T2. There's some heat being added
here, Q1, and that there's some heat being released
here, Q2. And since these are adiabatic,
there's no transfer of heat to and from the system. And when I looked at this, and
when I looked at the Carnot cycle, and I used this
definition of entropy, I saw that the total change in S, when
I go from this point all the way around and got back, the
change in S, was equal to Q1 over T1 plus Q2 over T2. And then I actually showed you
that this was equal to 0, which is exactly the result
that I wanted to see. Because in order for this to be
a state variable, in order for S to be a state variable, it
should not be dependent on how I got there. It should only be dependent
on my state variables. So even if I go on some crazy
path, at the end of the day, it should get back to 0. But I did something, I guess,
a little bit-- what I did wasn't a proof that this is
always a valid state variable. It was only a proof that it's
a valid state variable if we look at the Carnot cycle. But it turns out that it was
only valid because the Carnot cycle was reversible. And this is a subtle but super
important point, and I really should've clarified this
on the first video. I guess I was too caught up
showing the proof of the Carnot cycle to put the
reversibility there. And before I even show you why
it has to be reversable, let me just review what
reversibility means. Now, we know that in order to
even define a path here, the system has to be pretty close to
equilibrium the whole time. That's the whole reason why
throughout these videos, I've been drawing this piston, you
know, have the gases down here, and then always-- instead
of having one big weight on top that I took off
or took on, because it would throw the system out of
equilibrium-- I did it in really small increments. I just moved grains of sand, so
that the system was always really close to equilibrium. And that's called quasistatic. and I've defined that before. And that means that you're
always in kind of a quasi-equilibrium. So your state variables
are always defined. But that, by itself, does not
give you reversability. You have to be quasistatic and
frictionless in order to be reversible. Now, what do we mean
by frictionless? Well, I think you know what
frictionless means. Is that like you see in this
system right here, if I make this piston a little bit bigger,
that when this piston rubs against the side of this
wall, in kind of our real world, there's always a little
bit of friction. Those molecules start bumping
against each other, and then they start making them vibrate,
so they transfer some kinetic energy. From just by rubbing into each
other, they start generating some kinetic energy,
or some heat. So you normally have some heat
generated from friction. Now, if you have some heat
generated from friction, when I remove a pebble-- first all,
when I remove that first pebble, it might not
even do anything. Because it might not even
overcome-- you can kind of view it as the force
of friction. But let's say I remove some
pebbles, and this thing moves up a little bit. But because some of the, I guess
you could say, the force differential, the pressure
differential between the pebbles and the gas inside, and
the pressure of the gas, was used to generate heat as
opposed to work, when I add the pebbles back, if I have
friction, I'm not going to get back to the same point
that I was before. Because friction is always
resisting the movement. So in order for something to be
reversible, when I remove a couple of pebbles, if I removed
ten pebbles and add the ten pebbles back, I should
be at the exact same state. But as, you know, you can just
do the thought experiment. If there's friction, I won't
be at the exact same state. My piston won't move as much as
you would expect if it was frictionless. So this is a key assumption
for reversability. Now, the Carnot cycle, by
definition, is reversible. And that's why no one could
actually really implement an engine that fully does
the Carnot cycle. And we even showed that, that
the Carnot cycle is the most efficient potential engine. That if anyone made a more
efficient engine, you could have a perpetual motion
machine, or a perpetual energy machine. And the reason why the Carnot
engine is the most efficient engine-- and there's no secret
here-- it's because it's frictionless. Any engineer who makes engines
could tell you, wow, if I could just remove all of the
friction from my system, I would get a lot more
efficient. Now, with that said-- so I've
told you that look. The definition doesn't have to
be-- it happened to work, Q divided by T, because
I was dealing with a reversible system. And just to hit the point home,
let me show you that it would not have worked if I had
just defined it Q divided by T on an irreversible system. So let's say that I have-- let
me draw another PV diagram. And I'm going to do almost
a very simple thought experiment. Now I'm going to have an
irreversible system. And I start here at some
point on my PV diagram. And you know, this could be some
type of cylinder, and it has a piston on top, and I have
my rocks, like always. But this time, there's a
little bit of friction. When this thing moves, a
little bit of heat is generated in this. And when it moves in
either direction, some heat is generated. So we can call that the
heat from friction. When it moves either up
or when it moves down. So let's do something. Let's stick this on a big
reservoir, like we tend to do. So it's an isothermal system. So let's call this T1. And let's just start
removing pebbles. And we'll move along
an isotherm. Maybe to that point there. And then we're going to-- and
I want to make a very important point here. Because this is has friction,
I'm not going to get quite as far along the isotherm than
if I didn't have friction. If this was a frictionless
system, I would've gotten a little bit further along
the isotherm. So the number of rocks isn't
going to be the same as it was frictionless. But let's just say I
move from here to here on the PV diagram. And then we go and add a bunch
of rocks back, and we want to go all the way back. And I'm not even saying whether
we have the same number of rocks or
different rocks. You're probably going to have to
add a few more rocks to get back to this point. But the idea here is, is that
we've gotten back to the same point on the state diagram. So our delta U total should be
equal to 0, which is equal to the delta U of expansion. So the delta U of expansion is
the delta u to go that way, delta U of expansion, plus the
delta U of contraction, which is the delta U of going
back like this. Those have to be equal to
0 by definition, right? Because internal energy is a
state variable, and if we get to that same point, our delta
U has to be equal to 0. So what's our delta
U of expansion? What's our change in internal
energy as we expand? Our delta U of expansion is
equal to the heat added to the system minus the work
done by the system. And we know how much
work was done, this whole area right there. And then plus the heat added
by the friction. There's some heat added
by the friction. Let me do that in brown. What's that? I was on some random website off
the screen, and all of a sudden that cartoon
sound started up. I have no idea what that was. But anyway, where was I? So I said our change in internal
energy from expansion is going to be the heat added
to the system from our reservoir minus the work done
by the system, as we expand, plus the heat added to the
system or generated by the system-- I guess you could say,
it's not being added. The system is creating this heat
itself, as it expands. There's this friction
right there. Fair enough. So this is the one variation. Now that we're not dealing with
the reversible process, we have this friction. Now what's our change in energy
from contraction? So our change in internal energy
from contraction is going to be the heat that leaves
the system, that has to go back into the reservoir
as we contract. Because otherwise, if we didn't
have the reservoir, the temperature would go up. But we want to release heat. So we want to say, heat
released-- and I'm going to do something. Let's just assume that all
of the Q's are positive. So if I'm releasing heat, it's
going to be a minus Q release. Let's just say that this is a
positive number, and if I'm releasing it, it's going to
be a minus right there. And I want to just do that, just
to hopefully make things little bit clearer. Plus the work performed
on the system. I'm assuming that work is always
positive, so if we're doing work, it'll
be minus work. If work is being done to
us, it'll be plus work. But in this situation
as well, we're still adding heat from friction. Or heat from friction is still
being generated in the system. This is still positive. In either direction, when we
move upwards or downwards, the system is generating friction. Now, we always said, we went all
the way here, we went all the way back. So the sum of these has to be
equal to 0, because this is a state variable. So if the sum of all of this
has to be equal to 0, let's sum this up. So this gets us to
Qa minus Qr. So the heat accepted minus
the heat released. The W's cancel out. Plus-- let me see right here--
plus 2 times the heat of friction in either direction. All of that has to
be equal to 0. Let's see. What we can do is, we can
rewrite this as the heat accepted minus the heat released
is equal to minus 2 times the amount of heat
generated from friction. And then if we just switch these
around, we'll get the heat released minus the heat
accepted is equal to-- well, I just wanted to get all positive
numbers-- 2 times the heat of friction. Now why did I do all of this? Because I wanted to do
an experiment with an irreversible system, and this
was a very simple experiment with an irreversible system. Now, we said that delta S,
which a long time ago I defined as Q divided by T-- and
in this video, I said it had to be reversible. And I wanted to show you right
now that what if I didn't make the constraint that this
has to be reversible? Because if this doesn't have to
be reversible, and I just use this definition right here,
you'll see that your delta S here would be-- you just
divide everything by T-- because our temperature was
constant the entire time, we were just on a reservoir--
you'll see that this is going to be your delta s. This is your total change in
the, I guess you could say, your net heat added
to the system. So this is, let me say,
this is the heat added to the system. Let me do it this way. Heat added to the system,
divided by the temperature at which it was added. Which is a positive number. Even though we got to
the exact same place on this date diagram. So in an reversible system,
this wouldn't be a valid state variable. So it's only a valid state
variable if it's reversible. Now, does that mean that you
can only talk about entropy for reversible reactions? No. You can talk about entropy
for anything. But what you do is-- and this
is another important point. So let's say that I have some
irreversible reaction that goes from here to here. And I want to figure out
its change in entropy. Right? And it might have done all
sorts of crazy things. It's an irreversible reaction,
its path might have gone like that. That's assuming it's
quasistatic, if we can even look at its path like that. If we wanted to figure out its
change in entropy, though, we wouldn't worry about the heat
that was added to it and the different temperatures at
which it was added. We wouldn't worry about that. We would just say, OK. What would it have taken for a
reversible system to go from this state to this state? And then maybe a reversible
system would have done something like this. Sorry, I want to make
it a smooth curve. Maybe a reversible system
might have done something like that. And this change, this heat added
by the reversible system divided by the temperature for
the reversible system, would be the change in entropy. And this change in entropy--
we could call this S final, and this is S initial, it's
going to be the same for both systems. It's just, we don't use
the irreversible system to figure out our entropy. We would use the reversible heat
and temperature to figure out the actual change. Hopefully that clarifies
something else. It's, on some level, a subtle
point, but on some level it's super important. Because you can't-- the
thermodynamic definition of entropy has to be this. It has to be heat added to a
reversible system divided by the temperature that
was added. Not just heat to any system. It just happened to work when
I did it, and I should have been clearer about it when I
first explained it, that it worked only because it was
a Carnot cycle, which is reversible.