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### 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: Volume ratios in a Carnot cycle

Proof of the volume ratios in a Carnot cycle. Created by Sal Khan.

## Want to join the conversation?

- At2:07time, it was glazed over that P is considered to be constant in the quasi-static process. Why is obvious that P needs to be kept constant? What about the system makes it clear that pressure has be considered constant, except to allow the proof to work out?

I am noting that when adiabatic systems were originally discussed in an earlier video, pressure was shown to vary, thus these approximations have been difficult to conjure for the purposes of making thermodynamic proofs. Much thanks.(25 votes)- Although in an infinitesimal change P is 'almost constant', when Sal substituted

P as a function of V and T he was not keeping P constant. So, when integrating the equation the functionality of P is brought in through P = nRt/V. P is not constant in the total adiabatic process.

Over an infinitesimal interval deltaP approximately equals (nRT/(V+deltaV)-nRT/V) which is much smaller than deltaV(10 votes)

- How can you assume that P is constant?(5 votes)
- As we are taking the process as quasistatic . So for a a very small change we can assume pressure to be constant .(7 votes)

- At7:43, Sal integrated the entire equation. The right hand side of equation is zero. The integration of zero must be a constant as derivative of a constant is zero. But, instead after integrating the equation was written still equal to zero.

Can anyone explain? It will be very helpful. Thanks.(8 votes)- As Andrew said, it's a constant, but it's the same constant at the beginning and end of the integral, so it subtracts to zero.(1 vote)

- What is significance of using only adiabatic processes in the relation ignoring isothermal processes occurring in the the carnot's cycle?(7 votes)
- This proof is for a Carnot cycle, which has two adiabatic segments and two isothermal segments. He used the adiabatic processes to do the math, but since V is a state variable, the V ratio of the "four corners" holds true for the CYCLE regardless of how you derive it. It may be possible to derive it using the isothermal segments, but I have not tried. It would have to yield the same ratio, obviously, since it's still the same overall cycle. My guess is that the math is more complicated with the isothermals, or else Sal would have proved it with the isothermals. Not sure. You could try that proof yourself and see if it is easier.(2 votes)

- At13:50, how is (T2/T1)^3/2 the same as (T1/T2)^3/2(3 votes)
- You should probably watch this again — Sal never says those are equal!

It may be helpful to you to write out the manipulations he is showing on paper and think about what is happening at each step ...(7 votes)

- At 2.35 Sal wrote delta U + P delta V = 0 which is true for adiabatic expansion only, not for adiabatic compression. Using this equation he obtained another expression at10:44which should again be true only for adiabatic expansion. Then at12:54how could it be applied for process D to A which is adiabatic compression ?(3 votes)
- Why is it true for expansion but not compression? The only difference between the two is the sign of delta v.(3 votes)

- at11:16why is it (Tf/Ts)^2/3(Vf/Vs)=1????

why is it 1???(0 votes)- To start with, ln [(Tf/Ts)^3/2(Vf/Vs)]=0. Now, ln (natural log) is logarithm base e (a constant), so log e [(Tf/Ts)^3/2(Vf/Vs)] = 0. From this, and by the property of logarithm, e to the power of 0 (e^0) = [(Tf/Ts)^3/2(Vf/Vs)]. Property of logarithm says whatever number to the 0th power equals 1. So since (e^0) = [(Tf/Ts)^3/2(Vf/Vs)], this whole [(Tf/Ts)^3/2(Vf/Vs)] expression equals 1.(7 votes)

- what is an initial volume and how do you find it on a graph?(2 votes)
- Depending on what your axis are it is usually when x=(0) assuming y is volume(2 votes)

- Could someone explain to me why at5:56Sal is able to divide the 'work done' term by nRT, specifically by T? If it were an isothermal process then I would understand how since T isn't varying we can just cancel it but in an adiabatic process such as this, T is a variable so it is not erroneous to divide through by it in this way ? (He has written pressure as a function of volume AND temperature NOT just volume, as would be the case when dealing with work done in an isothermal process).(2 votes)
- I believe he can divide throughout by a constant or variable, as long as he does it equally on both sides of the equation (which he does)

There is no physical significance to what he is doing, it is simply a mathematical way of rearranging the equation to make the integeration possible (seperating V and T terms)

Make sense OK??(2 votes)

- I know delta means difference between two specific states. But, what is only V and T (i mean variables without delta) standing for? A very specific state between two states or any state between them?(2 votes)

## Video transcript

The goal in this video is to
essentially prove a pretty simple result. And that's that the ratio
between the volumes-- let me write this down-- that the ratio
between the volume at state B and the volume at state
A-- so the ratio of that volume to that volume-- is equal
to, in our Carnot cycle, is equal to the ratio between
the volume at state C. So this volume and
that volume. So volume at C to
the volume at D. So this is what I'm about
to embark to prove. A fairly simple result, that
maybe is even, if you look at this, that looks about right. So if you're happy just knowing
that, you don't have to watch the rest
of the video. But if you are curious how we
get there, I encourage you to watch it, although it gets
a little bit math-y. But I think the fun part
about it, even, it will one, satisfy you. That this is true. But the other thing is, we'll
be able to study adiabatic processes a little bit more. So just kind of launching off
of that, the whole proof revolves around to this step
right here and this step right here. when we go from D to A. So by definition, an adiabatic
process is one where there's no transfer of heat. So our heat transfer in an
adiabatic process is 0. So if we go back to our original
definitions-- let me show you that here. Right here, at the step
and this step, we have no transfer of heat. So if we go back to there. Adiabatic-- we're completely
isolated from the rest of the world. So there's nothing to transfer
heat to or from. So if we go to our definition,
almost, or our first law of thermodynamics, we know that the
change in internal energy is equal to the heat applied to
the system minus the work done by the system. And the work done by the
system is equal to the pressure of the system times
some change in volume. At least, maybe it's a very
small change in volume, while the pressure is constant. But if we're doing a
quasi-static process, we can write this. Pressure, you can view it as
kind of constant for that very small change in volume. So that's what we have there. Now, if it's adiabatic, we
know that this is 0. And if that's 0, we can add P
delta V to both sides of this equation, and we will get that--
this is only true if it were adiabatic-- that delta U,
our change in internal energy, plus our pressure times
our change in volume, is equal to 0. And let's see if we can do
this somehow, we can do something with this equation to
get to that result that I'm trying to get to. So a few videos ago, I proved
to you that U, the internal energy at any point in time--
let me write it here. The internal energy at any point
in time is equal to 3/2 times n times R times T. Which is also equal
to 3/2 times PV. Now, if I have a change in
internal energy, what can change on this side? Something must have changed. Well, 3/2 can't change. n can't change. We're not going to change the
number of molecules we have. The universal gas constant
can't change. So the temperature
must change. So there you have it. You have delta U could be
rewritten as delta-- let me do it in a different color-- delta
U could be written as 3/2 n times R times our
change in temperature. And that's why I keep saying
in this-- especially when we're dealing with the situation
where all of the internal energy is essentially
kinetic energy-- that if you don't have a change in
temperature, you're not going to have a change in
internal energy. Likewise, if you don't have a
change in internal energy, you're not going to have a
change in temperature. So let me put this
aside right here. I'm going to substitute
it back there. But let's see if we can do
something with this P here. Well, we'll just resort to
our ideal gas equation. Because we're dealing with an
ideal gas, we might as well. PV is equal to nRT. This should be emblazoned in
your mind, at this point. So if we want to solve
for P, we get O is equal to nRT over V. Fair enough. So let's put both of these
things aside, and substitute them into this formula. So delta U is equal
to this thing. So that means that 3/2 nR delta
T plus P-- P is this thing-- plus nRT over V times
delta V is equal to 0. Interesting. So what can we do
further here? And I'll kind of tell you where
I'm going with this. So that tells me, my change in
internal energy over a very small delta T-- this tells me
my work done by the system over a very small delta V. And we're saying that, you know,
over each little small increments, they're going
to add up to zero. So let me just go back to
the original graph. So this is over a very small
delta V right there. Let me do it in a more
vibrant color. A very small change,
as we go from there to, let's say, there. We're going to have some
change in our volume. And you don't see the
temperature here. So don't try to even imagine,
when we do the integral, that we should think of it in
some terms of area. But we're going to integrate
over the change in temperature, as well. The temperature changes
a little bit from there to there. So what I want to do--
this is, right here, over a small change. I want to integrate eventually
over all of the changes that occur during our adiabatic
process. So let's see if we can simplify
this before I break out the calculus. So if we divide both sides
by nRT, what do we get? So let's divide it by nRT,
let's divide it by nRT. And we have to do it to both
sides of the equation. nRT. Well, on this term, the n's
cancel out, the R cancels out. Over here, this nRT cancels
out with this nRT. And what are we left with? We're left with 3/2-- we have
this 1 over T left-- times 1 over T delta T plus 1 over V
delta V is equal to-- well, zero divided by anything
is just equal to 0. Now we're going to integrate
over a bunch of really small delta T's and delta V's. So let me just change those to
our calculus terminology. We're going to do an infinite
sum over infinitesimally small changes in delta
T and delta V. So I'll rewrite this as 3/2 1
over T dt plus 1 over V dv is equal to 0. Remember, this just
means a very, very small change in volume. This is a very, very, very,
small change, an infinitesimally small change,
in temperature. And now I want to do the total
change in temperature. I want to integrate over the
total change in temperature and the total change
in volume. So let's do that. So I want to go from always
temperature start to temperature finish. And this will be going
from our volume start to volume finish. Fair enough. Let's do these integrals. This tends to show up a lot
in thermodynamics, these antiderivatives. The antiderivative of 1 over
T is natural log of T. So this is equal to 3/2 times
the natural log of T. We're going to evaluate it at
the final temperature and then the starting temperature, plus
the natural log-- the antiderivative of 1 over V is
just the natural log of V-- plus the natural log of V,
evaluated from our final velocity, and we're going
to subtract out the starting velocity. This is just the
calculus here. And this is going to
be equal to 0. Right? I mean, we could integrate both
sides-- well, if every infinitesimal change is equal to
the sum is equal to 0, the sum of all of the infinitesimal
changes are also going to be equal to 0. So this is still equal to 0. See what we can do here. So we could rewrite this green
part as-- so it's 3/2 times the natural log of TF minus the
natural log of TS, which is just, using our log
properties, the natural log of TF over the natural log of TS. Right? When you evaluate, you get
natural log of TF minus the natural log of TS. That's the same thing as this. Plus, for the same reason, the
natural log of VF over the natural log of VS. When you
evaluate this, it's the natural log of VF minus the
natural log of VS, which can be simplified this to this,
just from our logarithmic properties. So this equals 0. And now we can-- this
coefficient out front, we can use our logarithmic
properties. Instead of putting a 3/2 natural
log of this, we can rewrite this as the natural log
of TF over TS to the 3/2. Now we can keep doing our
logarithm properties. You take the log of something
plus the log of something. That's equal to the log
of their product. So this is equal to-- I'll
switch colors-- The natural log of TF over TS to the 3/2
power, times the natural log of VF over VS. And this
is a fatiguing proof. All right. And all that is going
to be equal to 0. Now what can we say? Well, we're saying that e to the
0 power-- the natural log is log base e-- e to the 0 power
is equal to this thing. So this thing must be 1. E to the 0 power is 1. So we can say-- we're almost
there-- that TF, our final temperature over our starting
temperature to the 3/2 power, times our final volume
over our starting volume is equal to 1. Now let's take this result that
we worked reasonably hard to produce. Remember all of this, we just
said, we're dealing with an adiabatic process, and we
started from the principle of just what the definition
of internal energy is. And then we substitute it with
our PV equals nRT formulas. Although this was kind of PV--
this is internal energy at any point is equal to
3/2 times PV. And then we integrated over all
the changes, and we said, look, this is adiabatic. So the total change-- the sum
of all of our change in internal energy and work done
by the system has to be 0, then we use the property of
log to get to this result. Now let's do these for both
of these adiabatic processes over here. So the first one we could do is
this one where we go from volume B at T1 to
volume C at T2. Watch the Carnot cycle video,
if you forgot that. This was the VB All of these
things up here were at temperature 1. All of the things down here were
at temperature 2 So we're at temperature 1 up here,
and temperature 2 down here, volume C. So let's look at that. So on that right part, that
right process, our final temperature was temperature 2. So let me write it down. Temperature 2. Our initial temperature was
temperature 1, where we started off at point
B to the 3/2. Times-- what was our
final volume? Our final volume was our volume
at C divided by the volume at B. And that's going to
be equal to 1. Neat. That's the result we got from
this adiabatic process. We got that formula saying, this
is adiabatic, we did a bunch of math, and then we
just substitute for our initial and final volumes
and temperatures. Let's do it the same way, but
let's go from D to A. So when you go from D to A,
what's your final temperature? Don't want to get you
dizzy going up. Well your final temperature,
we're going from D to A. So our final temperature
is T1. and our final volume
is the volume at A. Go back down. So our final temperature
is T1. Our initial temperature is T2. We're going from D to A to the
3/2 power is times-- let me write our form formula there. Our final volume is
the volume at A. That's where we moved to. And we moved from
our volume at D. And this is going to
be equal to 1. OK. We're almost there,
if your eyes are beginning to glaze over. But this is interesting. And if anything, it's a little
bit of fun mathematics to wake you up in the morning. So let's see. We can almost relate
these two things. We could set them equal to each
other, but it's not quite satisfying yet. Let's take the reciprocal of
both sides of this equation right here. So obviously if we take the
reciprocal of this-- and we could just say, this is T2
over T1 to the minus 3/2 power, which is the same
thing as T1 over T2, to the 3/2 power. Right? That's just the reciprocal. And we're taking both sides to
the negative 1 power, so we're going to have to take this
to the negative 1 power. VB over VC. And when you take the reciprocal
of 1, that just equals 1, That still equals 1. Which this also equals, so we
could say, that equals this thing over here. So that is equal to T1 over
T2, to the 3/2 power, times VA over VD. Now, these things are
equal to each other. We can get rid of the 1. These two-- actually, let me
just erase some of this. I don't want to make it
say not equal to. They're equal to each other. They both equal 1. So they both equal each other. This thing and this thing
are the same thing. T1 over T2 to the 3/2, T1
over T2 to the 3/2. So let's just divide
both sides by that. Those cancel out. And what are we left with? I think you can see
the finish line. The finish line is near. We have VB over VC is
equal to VA over VD. Now that's not quite the result
we wanted, but it takes a little bit of simple
arithmetic to get there. Let's just cross-multiply. And you get VB times VD is
equal to VC times VA. Now if we divide both sides by
VBVC-- actually, let's do it the other way. Let's divide both
sides by VDVA. What do we get? These cancel out, and
these cancel out. And we are left with VB over
VA is equal to VC over VD. All that work for a nice and
simple result, but that's better than doing a lot
of work for a hairy and monstrous result. So that's what we set
out to prove. That VB over VA is equal
to VC over VD. And we got it all from the
notion that we're dealing with adiabatic process, that our
change in heat is 0. and we just went to our formula,
or our definition of our change in internal energy,
the first law of thermodynamics, that if we have
any change in internal energy, it must be equal
to the amount of work done by the system. Or at least a negative of the
work done by the system. When you add them up,
you get to 0. Then we use that result from a
few videos ago, where we said the internal energy at any
point is 3/2 times nRT. So the change in internal energy
is that times delta T, because that's the only
thing that can change. We used PV equal nRT. And then we just integrated
along all of the little changes in temperature
and volume, as we moved along this line. As we moved along the line,
we took the integral. We said that had to
be equal to 0. And we ended up with this
formula over here, and then we just applied it to our two
adiabatic processes. And we went from B to C, and
we went from D to A. And we got these results. And we got to our finish line. See you in the next video.