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

## 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 verbal 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.