Work Done by Isothermic Process

Physics

Thermodynamics

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Work Done by Isothermic Process
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Work Done by Isothermic Process Isothermic and Adiabatic processes. Calculating the work done by an isothermic process. Seeing that it is the same as the heat added.

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Work Done by Isothermic Process

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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.

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