Proof: S (or Entropy) is a valid state variable

Physics

Thermodynamics

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Thermodynamics (part 5)
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Work from Expansion
PV-diagrams and Expansion Work
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
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Efficiency of a Carnot Engine
Carnot Efficiency 2: Reversing the Cycle
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Empirical and Molecular Formulas from Stoichiometry
Example of Finding Reactant Empirical Formula
Stoichiometry of a Reaction in Solution
Another Stoichiometry Example in a Solution
Molecular and Empirical Forumlas from Percent Composition
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Proof: S (or Entropy) is a valid state variable Prroof that S (or entropy) is a valid state variable.

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Proof: S (or Entropy) is a valid state variable

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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.

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