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# Work done by isothermic process

Isothermic and adiabatic processes. Calculating the work done by an isothermic process and seeing that it is the same as the heat added. Created by Sal Khan.

Video transcript

Let's start with our classic
system that I keep using over and over again. And that's because it tends
to be very useful for instruction. It also tends to be the system
that is most covered in classrooms. So hopefully it'll
be productive for you and your school work. So I have this container. It has a movable piston
on top, or kind of a movable ceiling. Well, of course, inside of my
system I have a bunch of molecules or atoms bouncing
around, creating some type of pressure on the system. So let's say it has
some pressure, P1. This volume right here,
let's call that V1. And let's say it also has
some temperature that it starts off with. Everything is in equilibrium. Remember these are
macro states. The only way I can even tell
you what the volume, or the pressure, or the temperature
is, is if the system is in equilibrium, if everything
in it is uniform. The temperature is consistent
throughout. Fair enough. And in order to keep it placed
down, I have to put some rocks on top. And I've done this in multiple
processes so far. And, of course, I'm doing these
little pebbles, because I'm going to remove
them slowly. Because I want to approximate
a quasi-static process. Or I want to approximate a
system that's always close enough to equilibrium that I'm
cool with defining our macro states, our pressure, our
temperature, or our volume. Let me write V for volume. Now, in this video, I'm going
to study what's called an isothermic process. And really what it just means
is, I'm going to keep the temperature the same. Iso- just means the same. You probably remember when we
studied the periodic table. Isotopes, those are the same
element just with different mass numbers. So this is the same temperature
we're going to run our process. So my question is, how
can we do that? Because as I remove pebbles,
what's going to happen? If I just did this, without
any-- if it was completely isolated from the world. And actually I'll add in
a word right here. If it was an adiabatic
process-- If it was adiabatic-- fancy word. All that means is, completely
isolated from the world. So no heat is going into
or out of this system. If this was the case, what would
happen as I released or I took away some of these
little particles? Let me copy and paste it. Well, let me just redraw
it actually. So I have my one wall. I have another wall. I have another wall. As I release a couple of
pebbles, one at a time, my volume is going to increase. So I'm going to have a slightly
higher volume. My volume's going to go up. I have fewer pebbles here now. And since I have the same number
of molecules, they're going to bump into this less. So pressure is going
to go down. Volume is going to go up. And if I was adiabatic, if I had
no extra heat being added to the system, what do I know
is going to happen to the temperature? Well, think about it this way. Some work was done, right? Our old ceiling was maybe
some place around here. We pushed it up with some
force for some distance. So we did work. And so we changed some kinetic
energy, or we transferred some kinetic energy, out
of the system. That's essentially what
the work did. That kinetic energy was
turned into work. And temperature is just a
macro measure of average kinetic energy. In fact, we just-- well,
I won't go into it. But in the last video, the kind
of proof you want-- if you didn't watch it because you
didn't want to go through the math, which is completely
fair enough because it normally wouldn't be done in an
intro chemistry class-- I showed that the internal energy
is equal to the total kinetic energy, which was equal
to 3/2 times the number of moles, times R, times
temperature. So temperature is just, by some
scaling factor, a measure of kinetic energy. Now, when I do some work, it's
essentially a transfer of kinetic energy. And I can't replace that energy
with some heat because it's adiabatic. There's no heat going into
or out of the system. So in that situation, the
kinetic energy of the system went down. The average kinetic energy
of the system went down. So the temperature would've
also gone down. And actually, just as a bonus
point, what happened to the internal energy? Well, the internal energy is
the total kinetic energy of the system. And I could even right down
the original formula. Change in internal energy is
equal to change-- let me not do that, because I said I
shouldn't-- is equal to heat added to the system, minus
work done by the system. This is work done
by the system. That's why we're
subtracting it. Now, it's adiabatic,
so there's no heat added to the system. So the change in internal energy
is equal to the minus the work done by the system. Well, in this situation,
the system did do work. It pushed this piston up by some
distance with some force. So, your delta U is negative. It's less than 0. So U went down, and
that makes sense. If temperature changed,
then the internal energy is going to change. And for our simple system,
where internal energy is represented by the kinetic
energy of these molecules, that's always going
to be the case. If temperature doesn't change,
internal energy won't change. If temperature goes up, internal
energy goes up. If temperature goes down,
internal energy goes down. And, of course, they're not
the same thing, though. The difference between
internal energy and temperature is the scaling
factor, 3/2 times the number of molecules, times our
ideal gas constant. So fair enough. I went through this whole
exercise just to show you that if I was completely isolated,
and if I removed a couple of these pebbles, that
my temperature is going to go down. Now, I told you already
that I want to do an isothermic process. So I want to do this process
while keeping the temperature the same. So how can I do that? Well, what I'm going to do is
I'm going to place my system on top of what we'll
call, a reservoir. So a reservoir, you can kind of
view as an infinitely large amount of something that is
the temperature that we started off with. So this reservoir is T1. So even though if, I took
two relative things of comparable size. That says, temperature A. This is temperature B. And I've put them next
to each other. They're going to average out to
A plus B over 2, whatever their temperatures are. But if B is massive-- if A is
just a speck of particle-- let's say it's iron dust-- while
B is the Eiffel Tower, then essentially B's
temperature will not change a lot. A will just become
B's temperature. Now, a reservoir is theoretically infinitely large. It's an infinitely
large object. So if something is next to a
reservoir and is given enough time, it'll always assume the
heat of the reservoir, or the temperature of the reservoir. So what's going to happen? So this is adiabatic, but now
I'm actually putting it next to a reservoir. So this isn't going to happen. The adiabatic situation
isn't going to happen. Now, I'm going to have a
situation where I'm going to stay the same temperature. So what's that going to look
like on the PV diagram? So let me draw the PV diagram. This is my pressure. This is my volume. So this is my starting
point right here. And what I'm saying is, if I'm
doing an isothermic process, so I just keep removing
these pebbles. So I start at this
state right here. Let me copy and paste it
since I've done so much art already here. So I'm going from there to
here where I'm removing a couple of the particles. So let's say I've removed a
couple of them over here. And because of that, I've
increased the volume. So let's say the volume,
it's not there anymore. Let's say it's a little
bit higher. Let's say the volume is--
just for the sake of our discussion-- let's say the
volume has expanded a little bit, because I've remove some
particles, the little pebbles on the top keeping it down. So it's like the adiabatic
process, but instead of the temperature going down, my
temperature stays at T1. My temperature's at T1 the
entire time, because I'm next to this theoretical thing
called a reservoir. So because of that, I will
travel along what we'll call an isotherm. So this is my first state. When I'm done, I might end
up some place over here. And so this is state 2. So this is state 2,
this is state 1. What I'm claiming is that my
path along this is going to be on some type of a rectangular
hyperbola, or at least part of it. If I were to add rocks to it and
compress it, I claim that my PV diagram would
go like this. If I were to keep removing rocks
from this diagram, I claim that my PV diagram would
keep going like that. And so what's the intuition? That if I keep the temperature
constant, that I'm essentially moving along this hyperbola. Well, let's just take out
the ideal gas formula. Let me box off all this
stuff over here. If I just take the ideal gas
formula, PV is equal to nRT. If T is constant-- we know that
R is a constant, it's the ideal gas constant. We know that we're not changing
the number of moles of particles. Then that means that PV is
equal to some constant. This whole thing is equal
to some constant. And then, if we wanted to write
P as a function of V, we would just write P is
equal to K over V. Now, this might not look 100%
familiar to you, but if I wrote it in algebraic terms,
if I told you to graph y is equal to 1 over x, what
does that look like? That's a rectangular
hyperbola. That looks like this. And this is the y-axis, that's
the x-axis, at least in this quadrant it looks like this. It also looks like that in the
third quadrant, but we won't worry about that too much. So whenever you hold temperature
constant, you're on some rectangular hyperbola
like this, like an isotherm. Now, if the temperature was a
different temperature, if it was a lower temperature, you'd
be on a different isotherm. So you would be on an isotherm
that looked like this, maybe over here. It would also be a rectangular
hyperbola but at a lower state. Why is that? Because, if you're at a lower
temperature, for any volume, you should have a lower pressure
and that works out. That's why this is some
temperature T2, that is lower than T1. So I want to do a couple of
things in this video. I inadvertently explained to you
what an adiabatic process is, and why the temperature
would naturally go down on its own if you didn't have
this reservoir here. But the whole reason why I even
thought about doing this video is because I wanted you
get comfortable with this idea of, one, that a reservoir will
keep you in kind of an isothermic state. It will keep the temperature
the same. And that if you keep the
temperature the same, that you will travel along this isotherm,
these rectangular hyperbolas. And that each temperature has associated with it an isotherm. So if you take that, let's
just do one more step. And let's think about the
actual work we did by traveling from this state
to this state. Or if you just want to think
of it in visual terms, from removing our pebbles slowly and
slowly with this reservoir down here the whole time, from
this state to this state. Where our volume has increased,
our pressure has-- So our volume has increased. Our pressure has gone down. but
our temperature has stayed the same the entire time. So, several videos ago, we
learned that the work done is the area under this graph. It's the area under
that graph. Or, if we want to do in calculus
terms-- and I'm about to break into calculus, so if
you don't want to see calculus cover your eyes or ears. It would be the integral. And the rest of this video will
be a little bit mathy, and I guess I should make
that statement on the title of the video. But if I want to calculate
what this area is, I can now do this. The isotherm assumption makes
our math a little bit easier. Because we know that PV is equal
to nRT, ideal gas law. Or we could say P, if we divide
both sides by V, is equal to nRT, divided by V. So there we have it. We have P as a function of V. This function right here, this
graph right here, is this. We could write P as a function
of v is equal to nRT over V. So if we want to figure out the
area under the curve, we just integrate this function
from our starting, our V1 to our ending point, to our V2. So what is that going to be? Well, we're going to integrate
from V1 to V2. Actually, that shouldn't
be an equal. The work is going to be the
integral from V1 to V2, times our function, P as a function
of V, times dV. We're summing up all of the
little rectangles here. We did that a couple
of videos ago. So what's P as a
function of V? So work done is equal to,
from V1 to V2, nRT over V, times dV. Now, this is our simplifying
assumption. We said we're sitting on
top of a reservoir. That this reservoir keeps
our temperature the same the whole time. And we're going to learn in a
second, it's doing that by transferring heat
into the system. And we're going to calculate how
much heat is transferred into the system. So, if we look at this right
here, temperature-- since we're assuming we're on an
isotherm, is a constant. n and R are definitely constants. So we can rewrite this integral
as the integral from V1 to V2 of 1 over V, dV. And then we could put
the nRT out here. I should have done that
first. nRT, it's just a constant term. Now, what's the antiderivative
of 1 over V? It's the natural log of V. So our work is equal to nRT
times the natural log-- this is the antiderivative-- of V,
evaluated at V2 minus it evaluated at V1. So that is equal to nRT times,
evaluated at V2, so the natural log of V2. Minus the natural log of V1. Now we know from logarithm
properties is the same thing as nRT times the natural
log of V2 over V1. So there you have
it, we actually calculated the real value. If we know our starting volume
and our finishing volume, we can actually figure out
the work done in this isothermic process. The work done in this isothermic
process is the area under this. And we figured out
what it was. By pushing up that piston,
it was nRT. These are the number of moles
we have, ideal gas constant. Our temperature that
we're sitting on. It would be T1 in this case. And the natural log of our
finishing volume divided by our starting volume. Now, let me ask a follow-up
question. How much heat was put into the
system by this isotherm here? It put in heat to keep the
temperature up, otherwise the temperature would have
gone down, right? Heat was going into the system
the entire time. How much was it? Well, since it's an isotherm,
since the temperature did not change, what do we know about
the internal energy? Did the internal
energy change? The temperature not changing
told us that the kinetic energy didn't change. If the kinetic energy didn't
change, then the internal energy did not change. And we know that the change in
internal energy is equal to the heat put into the
system minus the work done by the system. Now, if this is 0-- so we know
that this didn't change, because the temperature
didn't change. So that means 0 is equal
to Q minus W, or that Q is equal to W. So this is the work done
by the system. You'll end up getting
something in joules. And this is also equal to the
heat put into the system. It's also equal to Q. So when you look at this, if we
were to just draw this part of the curve-- let me redraw it
just to make things neat. I want to give you a little bit
of the convention of what people in the thermodynamics
world tend to do. I'll make a neat drawing here. We started here, at state 1. And we moved along this
rectangular hyperbola, which is an isotherm, to state 2. And now we calculated the area
under this, which is the work done, which was this
value right here. Let me write it there. It's nRT natural log
of V2 over V1. This is V2. This is V1. This whole axis, remember, was
the V-axis, volume axis. This axis here was the
pressure axis. And the convention is that
because we did work, but we were constant temperature, so
our internal energy didn't change, we had to add energy to
the system to make up for the work we did. So some heat must have been
added to the system. And then what they do is they
just put this little downward arrow and they write
a Q right there. So some heat was put into the
system during this isothermic process right there. And the value of that Q is
equivalent to the work we did. We put the exact amount of heat
into the system as the work that was performed. And because of that, our
internal energy didn't change. Or you could say our temperature
didn't change. Or you could go the other way. Because our temperature didn't
change, these two things have to be equal. Anyway, I want to
leave you there. Hopefully I gave you a little
bit more of an intuition of how PV diagrams work, a little
bit more intuition behind what isotherms and adiabatic mean. And the most important thing is,
once we get a little bit mathy, this result can be useful
for coming up with other interesting things about a
lot of these thermal systems that we're dealing with. See you in the next video.