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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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# Proof: S (or entropy) is a valid state variable

Proof that S (or entropy) is a valid state variable. Created by Sal Khan.

## Want to join the conversation?

- I got kind of lost when he started using the integrals, could someone explain what he was trying to prove and how he proved it in simpler terms?(16 votes)
- The integral was just the mathematical way to write the heat, Q, in terms of the volume, V, because that's what he had the proof in terms of from the last video. He couldn't prove that Q1/T1 + Q2/T2 = 0, because he doesn't know what Q1 and Q2 are, but he knows for an isothermal (constant temperature) process that the heat, Q, is equal to the integral of pressure, P, over the change in volume, V. So then he ends up with delta(S) equaling something in terms of volumes, which, form the previous video, he can then use to show that the change in S is equal to zero.(18 votes)

- what is state variable and state function?(8 votes)
- A state variable means it will always be the same AT THAT POINT. So if you choose P1 and V1 as your point for a certain system of ideal gas, no matter how you change P and V, when you get back to P1 and V1, the U will be the same as it was at the beginning (U1).

Think if it this way: On a mountain top, there are always the same Latitude and Longitude, not matter if you hike straight up, or cycle around the mountain, or parachute in. Those coordinates DEFINE your location on the mountain top. Is "tiredness" or "effort" a variable that is always the same for someone on top of the mountain? No. If you crawled up the mountain, you would be more tired than the person next to you who parachuted in. Is your elevation a state variable for a mountain top? Yes. Is your thirstiness? No.

Likewise P, V, T, S, and U are always the same at the same point on a PV diagram, no matter what wacky path gets you to that point. But the amount of W or Q needed depends heavily on what path you took to that PV point (iso-volume is no work, etc).(18 votes)

- It's often emphasized how important it is that the process is reversible. I think I have an idea of what reversible and irreversible processes are, but I can't seem to understand why so many things apply to reversible processes only. Could someone explain that?(10 votes)
- Because reversible processes are ideal, and simplifications of real world problems. In the real world we would have to account for changes in kinetic and potential energy along with friction (and more). That would make these problems much more difficult to solve! So reversible is just an assumption created to make the problems more simple in an introductory course and that's why everything seems to apply just to reversible processes in these videos.(6 votes)

- Has entropy been proven?(0 votes)
- well yes, entropy has been proven. It is part of the second "law" of thermodynamics... laws are rigorously tested and have not been disproven since formulated(1 vote)

- Around1:05Sal says that in the second isothermic process (from C to D) you take away less heat than you needed to add in the first isothermic process (from A to B). Can anyone please help clarify why this is the case and why the heat added and taken are not the same?(3 votes)
- The second isotherm is at a lower temperature and pressure, so it takes less work to compress it, and therefore less heat has to leave the system after the compression to keep the system at the same temperature (on that same lower isotherm).(6 votes)

- I have no idea what's going on, as to be expected. I don't have the background. I am watching these videos, so that when I do take Chemistry, I will have heard the terms before. I know you guys are all taking Chem. now, tell me, will my strategy work?(1 vote)
- I don't Know Calculus so this means can i skip it , Or it is really necessary ?(2 votes)
- better u learn it...as it's an application based concept...it shall help u in ol d subjects..!(6 votes)

- Why is it that we can say this derivation applies in general, and not only in Carnot engines?(4 votes)
- How does one complete this entropy problem using algebra based physics?(3 votes)
- at2:12i dont get why heat isnt a state variable. if we went around the cycle, why heat at A would be 10 + W?(3 votes)
- Posted 4 years ago

because Heat isn't an intrinsic property of the system, therefore you are not able to evaluate the heat without talking of exchange of energy with the enviroment, In fact it's a measure of the flux of energy. So you are able to find the value of heat only if you have transition of energy. So it dipends on the path, think about this example: you have two ways to get to the floor one to the floor two, the elevator or the staircase. The height between the floor two and one is a state variable, so we don't care about how do we get to the second floor ( the path), because the height between these two floors is the same.

Instead what you spend in terms of heat it's different, because you follow two different paths ( the elevator or the staircase).

Ok, now, also if I'm not sure of my answer, I'm going to answer your second question.

heat, in the video, is equal to work because the internal energy of your system doesn't change, so the net heat is equal to the work done by the system. Morover you have to add to this value 10J, because it's the heat content in your system at state A: in fact if you keep going around the carnot cycle and eventually you return at state A, the value of your heat content won't be equal to 10J + w, but 10J + 2w, because Q is equal to 2w and in fact you have to add a value of heat equal to the energy transferred thanks to the work to keep the temperature constant, avoiding its reduction( because it is an isothermic process > and so I have to replace the energy which i take away from the system thanks to work, adding the same value of heat). Furthermore you have to add 10 j to 2w because you start from state A.

If I make some mistakes, check them(1 vote)

## Video transcript

I've talked a lot about the
general idea that in order to have a state variable like,
say, U, which is internal energy, at any point in this
PV diagram, that state variable should be that value. So for example, if at this
point, U is equal to 5, and I go do this whole Carnot cycle,
when I come back to state A, U should still be equal to 5. It should not have changed. It's not dependent upon what
we did to get there. So if we did some kind of crazy
path on our PV diagram, we got back there, U should
always be the same. That's what it means to
be a state variable. It's only dependent upon its
position in this PV diagram. It's only dependent on its
state, not how you got there. And because of that, heat is
something that we can't really use as a state variable. For example, if I tried to
define some heat-related state variable, let's say I call it
heat content, and I defined change in heat content as equal
to the amount of heat added to the system. Well, if we go back to our
Carnot cycle here, let's say that my heat content
here was 10. Well, I added some heat here,
in this process here. Nothing happened, because this
was adiabatic from B to C. Then from C to D, I took
out some heat. But I took out less heated
than was added here. And then here, nothing was
done with regard to heat. So I did add some heat
to the system. The net heat that I added to
the system as I went around the cycle-- in this case, Q
would be equal to Q1 minus Q2. And we know that this number
is larger than this. The net amount of heat we added
to the system was the amount of work we did on the
system, because the internal energy didn't change. So if this is 0, then the amount
of heat we add to the system is the amount
of work we did. Our internal energy is
definitely 0 as we go all the way around. We did this shaded portion of
work-- I showed you that several videos ago, that the
area inside of our little cycle is the amount
of work we did. And so that's also the
net amount of heat added to the system. So if we added that amount of
heat to the system, if we started off at the heat of 10
here, or whatever my heat content mystical variable I made
up just now, when we go around, it would then
be 10 plus W. If we go around again,
it would be 10 plus 2W and 10 plus 3W. So it can't be a legitimate
state variable, because it's completely dependent on what
we did to get there. And if we keep going around the
cycle, we can increase it, even though we get to
the same point. So this is an illegitimate state
variable, where I define the change in our little made up
heat content to be equal to the heat added to system. Not a valid state variable. Ignore it all. Now we know that Q1, we added
more heat here than we took away, so there was something
net heat added. But there's something
interesting here. We added it at a higher
temperature. And here we took less heat away
at a lower temperature. So maybe we can define another
state variable that can have the result that when we go
around the cycle, we do get back to our same value. Now let me just-- we're
just experimenting. Although I know where this
experiment will go. I wouldn't have been doing
it if I didn't. So let's say I define a
new state variable S. And I define a change in S. So I say, a change in S-- I'm
just making up a definition-- is equal to the heat added to
the system divided by the temperature at which it was
added to the system. Now, I don't know what
this means just yet. In future videos, maybe we'll
get intuition about what this actually means in kind
of our minds. But let's see if at least this
is a valid state variable. If, as we go around the Carnot
cycle, whether our change in delta S is 0, right? To be a legitimate state
variable, we have some value for S here. Maybe it's 100. I don't know. Once we go back around the
Carnot cycle, it should be 100 again, or our delta
S should be 0. So what's the delta S? So the delta S, as we go around
the whole cycle-- let me write delta S-- let
me do another color. Delta S. As we are go around, I'll say
c for the Carnot cycle. As we go around the Carnot
cycle, is going to be equal to-- well, when we went from A
to B, we were at a constant temperature, and we added Q1. So it's Q1, and we were
at temperature T1. Fair enough. Then when we went from B
to C, it was adiabatic. We added or took away no heat. So this value, Q over
T, would just be 0. So it's plus 0. Then we went from C to D. We were at a new temperature,
we were on a new isotherm. We were at T2. And we took away, or I won't put
the sign here, let's just say we added Q2 heat. We're going to actually
solve for it later. We added Q2 heat. We'lll see that it's actually
a negative value. And then finally, when we
went from D to A, it was adiabatic again. So no transfer of heat. So plus 0, right? The 0's are 0's over, you know,
changing temperature, but this is just 0. So this thing should be equal to
0 in order for this to be a valid state variable. So let's figure out what
this value is. What Q1-- so our change in our
mystical new candidate state variable, S, as we go around the
Carnot cycle, is equal to Q1 over T1, plus Q2 over T2. And we'll see, Q2 is negative. So what is Q1? Can we calculate Q1? Well, as we're on this top
isotherm, our temperature doesn't change, our internal
energy doesn't change. So if your internal energy
does not change, if your internal energy is 0, then the
heat added to the system is equal to the work done
by the system. So its the area under
this curve. Not just the area
in the cycle. It would be the whole area
under the curve. So what's the whole area
under the curve? Well, so let me do a
little aside here. So Q1 is equal to the work done
as we went from A to B. And work, remember, can just be
written as pressure times change in volume. We're going to do a little
calculus here, so I'll write dV for a small change
in volume. And we're going to integrate
it all over the little sums, right? This dV is this little
change in volume right there times pressure. That makes a little rectangle. And then we sum up all the
rectangles from our initial volume, which is VA, to our
final volume, which is VB. And then, what's Q2 going
to be equal to? Well, Q2 is going to be
essentially the same thing. It's going to be the sum of the
work done by our system, which in this case is going to
be negative, because work was done to our system as we
go from here to here. Right? That's when Q2 was operating,
the heat was being taken out of the system. So we're going to go from--
where was our starting point? VC and we go to VD. Now, how can we evaluate
these integrals? Well, we've done this before
in a previous video. We use both of these
circumstances-- When we go to A to B, and we go from C
to D-- both of these circumstances occur on
isotherms, right? So the only things that
are changing are pressure and volume. Temperature is not changing. And so if we go back to our
ideal gas equation-- PV is equal to nRT-- we can just
rewrite this by dividing both sides by V, as P is equal
to nRT over V. And substitute that back in
for P, in both cases. That is P as a function of V. We now have the equation
of the curve. And we're taking the area
under it in both cases. So Q1 is equal to the integral
from VA to VB of nRT over V, dv. And Q2 is equal to the integral
from VC to VD of nRT over V, dv I'm going to do two
integrals in parallel, just so that you kind of see that we're
essentially solving the same thing. OK. So how can we solve this? Well, we know in both of
these cases, we're moving along an isotherm. That our temperatures
are constant. And actually, we know the
temperatures are. When we're moving from VA to
VB, our temperature is T1. It was kept that way
by our reservoir. When we moved from VC to VD,
our temperature was T2. It was kept that way by
our reservoir, right? T2, when we move from C to D. And T1, we move from A to B
Those were our temperatures. And they're constant. Fair enough. So we can take-- so
n is constant. R is definitely a constant. n is just a number of molecules
we have. And then our temperature is also
constant, so we can take it out of the integral. So we can rewrite Q1 is equal to
nRT1 over the integral from VA to VB times 1/V dv, and Q2 we
can write as nRT2 times the integral from VC
to VD, 1/V dv. All right. Now this integral is fairly
straightforward to evaluate. The antiderivative of 1/V
is the natural log of V. So we get Q1 is equal to nRT1
times the natural log of V evaluated at VB, minus
it evaluated at VA. And Q2-- well, let me just
solve this whole equation right here. So this is equal to what? This is equal to a natural
log of VB minus the natural log of VA. Which is the same thing as the
natural log of VB over VA times nRT1. And all that is equal to Q2. Now, the same logic, Q2, is
going to be equal to what? Q2 is going to be
equal to nRT2. Now the only difference with
this integral is where I had VB, now I have VD. Sorry. So then it becomes natural
log of VD. And where I had VA, now
I have VC, So over VC. All right. Now what was our original
question that we were dealing with? We said, this is a legitimate
state variable. If the change in this-- whatever
this value S as we go around the cycle-- is
equal to 0, that means it didn't change. So these two things, when you
sum them, have to equal 0. Q1 over T1, plus Q2 over T2. So let's add them. So Q1 over T1 is equal
to that over T1. That cancels out. Q2 over T2 is equal
to that over T2. That cancels out. So our change in our mystical
state variable, as we go around the Carnot cycle,
is equal to Q1 over T1, plus Q2 over T2. Which is equal to nR times the
natural log of VB over VA. That's that, right there. And then plus Q2 over T2, which
is just nR times the natural log of VD over V. This is VC. This is a VA here. All right. Now let's see what we can do. This is equal to--
almost there. Home stretch. nR. We can factor out an nR. And then the natural log of A
plus the natural log of B is just the same thing as the
natural log of AB. So this is equal to times the
natural log of VB over VA, times VD over VC. All right. So this is our change in our
S, state variable that we're playing with right now. Now what is this equal to? Let me think of the best
way to say this. So let's divide the numerator
and the denominator by VC. So let me take this expression
here, and divide its numerator and its denominator,
essentially, by VC over VD. Or let me multiply
its numerator and denominator VC over VD. So I can rewrite this as the
natural log of VB over VA, divided by-- right? Instead of multiplying it times
this, I can divide it by this reciprocal, VC over VD. So I just rewrote it. I just did a little bit
of fraction math. That's all it is. Instead of multiplying it times
this, I divided by its reciprocal. Now you see why the previous
video I did was done. What is this equal to? On the previous video, I showed
you that VB over VA is equal to VC over VD. We did that big convoluted hairy
proof to prove this. And now that we proved it, we
can use this to know that this quantity is equal to
this quantity. So if you divide something by
itself, they're equal to each other, this is equal to 1. If that's equal to 1, then
what's the natural log of 1? So our change in our mystical
S state variable is nR times the natural log of 1. What's the natural log of 1? E to the what power
is equal to 1? e to the 0 is equal to 1. n times R times 0-- I don't
care how big or whatever-- this is 0. So it equals 0. So there we have it. We've stumbled upon a legitimate
state variable that deals with heat. If we define change in S is
equal to the heat added to the system, divided by the
temperature at which the heat was added to the system,
this is a legitimate state variable. Now, we don't have much
intuition about what it really means at kind of a micro
state level. But at least we've
stumbled upon some property of something. If S is 10 here, and we go
around here, our change in S will be 0, S is 10 again. If S is, I don't know, let's
say S is 15 here, and we go around some crazy cycle and we
come back here, our change in S is going to be 0 again. Or sorry, it's going
to be 15 again. So we didn't have-- our change
in S will be 0, so our s itself will be 15 again. So S is a legitimate state
variable, but we don't have a good sense of what it
actually means. We'll leave that to
a future video.