Main content

### Course: Physical Chemistry (Essentials) - Class 11 > Unit 6

Lesson 3: Applications of first law of thermodynamics# Work from expansion

How a system can do work by expanding. Created by Sal Khan.

## Want to join the conversation?

- In these types of problems, you'll hear the term adiabatic used. What is an adiabatic process and how do you know if the system is adiabatic?(30 votes)
- adiabatic process is a process where no transfer of heat is exchanged with the outside the system is isolated, you will know if the problem tells you,(43 votes)

- while piston moving upwards by removing small weight over the piston, there is change in volume and pressure also. but why the change in pressure is not considered(27 votes)
- Remember, we are talking about very small changes here. In order to be a 'quasi-static' process all changes are small and slow. Therefore the change in pressure is 'infinitesimal'.(14 votes)

- When Volume increases, pressure must decrease, that was in the first lesson on thermodynamics.

P1 x V1 = P2 x V2 (assuming there was no exchange of energy from outside).

Here Volume increases by very small number and Pressure decreases also by very small number as seen in the pictures by Sal.

Here we are assuming (as I have read some comments below), that changes in pressure are so infinitesimally small in quasistatic process that we can take pressure initial to the equation. But hey, I'm very confused now, because it seems illogical to me that volume isn't infinitesimly small also. Either both of them change or none of them changes(in my opinion).

Please, is there any logical explanation?? Otherwise I can't accept this equation, for me this is Work = (change in Pressure)x(change in Volume) .

But if Pressure x Volume haven't change has Internal Energy risen?

Thank you for your kindness and please explain this to me :-)(15 votes)- i guess pv/t is constant

the actual formula is pv = nrt

so, pv/t = nr

n is no of particles , so assuming its constant . and of course r is a constant too.

therefore pv/t is constant.

if volume is increasing. p and t must compensate to make the quantity constant. but that does not mean the numerator pv is constant.

i hope it helps you

Edit: I think I have a better answer to this.

When we do work we are losing heat, or u can say we are decreasing temperature.

Pv / t is constant. V is increasing but t is decreasing too. Hence p does not change.(1 vote)

- Why change in volume is not infinitesimal but change in pressure is ?(12 votes)
- That's a good question. I do think that here the changes in pressure and volume are (or should be) both considered infinitesimal since they are taking place in a quasistatic process if I'm not wrong.(5 votes)

- why do the molecules produce pressure when they bump into a surface(4 votes)
- Take your hand and tap against a door that can swing, did it move? No? Tap it harder, did it move? Yes? Well you exerted pressure by your kinetic energy.

Now tape it a few billion times per second and you have the pressure caused by molecules.(20 votes)

- Is PV work the only kind of work a system can do?(8 votes)
- It's the only type of work that a system such as this (piston-cylinder) can achieve. There are many different forms of Work, such as electrical work (dW=Current*Voltage*dt), rotational work (dW=Torque*Angular Acceleration*dt), and many other situations. Basically, anything that has Power (noted as W_dot, which means P=dW/dt) contains the capacity to do work and must be able to do work over time by it's definition.(6 votes)

- why are heat and work not state functions?(4 votes)
- Because work and heat are concepts that deal with the "transfer" of energy. It is not a property of the macro-states. You cannot say "the work inside this system is..." or "the heat inside this system is...". Rather you may state "the work done by the system unto this object is...and the head added to the system is...". So pretty much work and heat are not properties rather the change in properties.(5 votes)

- At6:10when Sal defines the pressure and force, why does he assume that the force applied on the ceiling of the piston is constant? As the volume increases by the force pushing the ceiling a distance "x", shouldn't the pressure decrease? If the pressure decreases in this process, why doesn't the force also decrease?(5 votes)
- Because x is assumed to be a very small number. And over a very small displacement the pressure is assumed to be as good as constant.

The math of that is explained in the videos on finding the area under a curve in the calculus section (Riemann sums and definite integration)(4 votes)

- The pressure of the air in the system changes. Which value of pressure should I use and why?(4 votes)
- Work = Force x Distance

we want to find the force applied by the gas to do the work.

We could find this force from the pressure before the force acted through a distance of X and that is, Force = Pressure x Area

Notice that if we want to find the force exerted to perform that work, you should use the INITIAL PRESSURE.(2 votes)

- In my physics textbook it says that work done by the system is positive and work done on the system is negative, but in my chemistry textbook its the opposite. Why are the conventions different?(3 votes)
- Physicists and chemists do it in opposite ways, basically just because those two fields were very separate until the early 1900's(2 votes)

## Video transcript

I've talked a lot about how our
change in internal energy of a system can be due to some
heat being added to the system, or some work being added
to the system, or being done to the system. And I'm going to write it again
the other way, just because you see it both ways. You could say that the change
in internal energy could be the heat added to the
system minus the work done by the system. So there's two questions
that might naturally spring up in your head. One is, how is heat added to or
taken away from a system? And how is work done, or done
by, or done to a system? The heat, I think, is
fairly intuitive. If I have a-- and we'll be a
little bit more precise in the future of this, but I just want
to give you the sense of what we're talking about-- if I
have some system here, some particles in some type
of a canister. And it's at temperature, I don't
know, let's say it's a temperature T1. I'll even give it a-- say
it's at 300 kelvin. If I want to add heat to this
system, what I can do is I can place another system right
next to it, maybe right next to it. Who knows what size it is. And it's got some
particles there. But its temperature is much,
much, much higher. So this system's temperature,
I'll say temp T2 is equal to, I don't know, let's say
it's 1000 kelvin. I'm just making up numbers. So what's going to happen in
this situation, you're going to have heat transferred from
this second system to the first system. So you're going to have heat
going into the system. Now, heat and, work and even
internal energy, this goes back to our conversation
of macrostates versus microstates. Heat is changing the macrostate
of our systems. This system is going to
lose temperature. This system's going to
gain temperature. But we know what's happening
on a micro level. These molecules are going
to lose kinetic energy. These molecules are going
to gain kinetic energy. How is that actually
happening? Well, we assume that there's
some type of a container here. Maybe it's a solid wall. These molecules are going to
bump into that wall, and are going to make the particles in
that wall vibrate, and then they're going to make the
particles in the green container's walls vibrate. And so when the green
container's molecules touch the wall, they're going to
bounce off with even more kinetic energy, with even more
velocity, because of that vibration in the wall will push
them back even further. So that's essentially how you
get this transfer of kinetic energy, or this transfer
of heat. I think that's fairly
intuitive. If we put this next to a cooler,
a system with lower temperature, we would
lose kinetic energy, or would lose heat. And there's other ways
that we can do it. We could compress the-- well,
I don't want to talk about that just now, because that'll
be touching on work. So, how can we add or subtract
work to a system? And this one's a little
bit more interesting. Let's go back to our
piston example. Let me just draw some
lines here. So I have my container. There you go. It's got a little movable
ceiling to it. That's my piston. And go back to the example. Because what we're going to be
dealing with-- especially once I go into the pressure volume
diagram, the PV diagram that I'm about to go into--
we want to deal with quasi-static processes. Processes that are always close
enough to equilibrium that we feel OK talking
about macrostates like pressure and volume. Remember, that if we just did
something crazy and the whole system is in flux,
those macrostates aren't defined anymore. So we want to do a quasi-static
process. So I'll have pebbles instead
of one big rock. I'll draw the pebbles a little
bigger this time. And I have some pressure. So that's my piston and
it's being kept down by these rocks. It's being kept up by the
pressure of the gas. The gas is bumping into
this ceiling. It's bumping into everything. The pressure at every point in
the container is the same. It's at equilibrium. Now, what happens in that
example where I removed one rock from that? So let me copy and paste that. So if I remove one rock from
this thing right here. Copy and paste. So that's the same thing. Now let me remove a rock. I'll remove this top
one, was removed. What's going to happen? Well I now have less weight
pushing down on the piston, and I have a certain amount
of pressure pushing up. The system, it'll very
temporarily go out of equilibrium, but it'll be a very
small difference in how much we're pressing down on it,
so hopefully it won't be a huge change in our
equilibrium. We'll stay pretty close to it. But we know from the previous
example, instead of this thing flying up, it's going to
shift up a little bit. This is just going to shift
up a little bit. Right when we do it it's going
to be like that, right there. And let me fill in that part
with black, because it's not like the space disappeared. So let me fill that
in right there. So our little piston will move
up a very small amount. And what I claim is, when this
happened, when I removed this little pebble from here, the
system did some work. And let's just think
about that. So work, according to the
definitions that you learned in first-year physics, and when
you're using classical or dealing with classical
mechanics, you learn that work is equal to force
times distance. So if I'm claiming that when
this piston moved up a little bit, when I removed that pebble,
I'm claiming that this system here did some work. So I'm claiming that it applied
a force to this piston, and it applied
that force to the piston for some distance. So let's figure out what that
is, and if we can somehow relate it to other macro
properties that we know reasonably well. Well we know the pressure
and the volume, right? We know the pressure that's
being exerted on the piston, at least at this
point in time. And what's pressure? Pressure is equal to
force per area. Remember, this piston, you're
just seeing it from the side, but it's a kind of a flat plate
or a flat ceiling on top of this thing. And at what distance
did it move it? You know I could blow
it up a little bit. It moved it some-- I didn't draw
it too big here-- some x, some distance x. So this change, it moved it up
some distance x there, right? So what is the force that
it pushed it up? Well, the force, we know
its pressure, the pressure's force per area. So if we want to know the force,
we have to multiply pressure times area. If we multiply both sides of
this times area, we get force. So we're essentially saying
the area of this little ceiling to this container right
there, you know, it could be, I could draw with some
depth, but I think you know what I'm talking about. It has some area. It's probably the same area as
the base of the container. So we could say that the force
being applied by our system-- let me do it in a new color--
the force is equal to our pressure of the system, times
the area of the ceiling of our container of the piston. Now that's the force. Now what's the distance? The distance is this
x over here. The distance is-- I'll do it
in blue-- it's this change right here. I didn't draw it too big,
but that's that x. Now let's see if we can
relate this somehow. Let me draw it a little
bit bigger. And I'll try to draw in
three dimensions. So let me draw the piston. What color did I do it in? I did it in that brown color. So our piston looks something--
I'll draw it as a elipse-- the piston
looks like that. And it got pushed up. So it got pushed up
some distance x. Let me see how good
I can-- whoops. Let me copy and paste
that same-- So the piston gets pushed
up some distance x. Let me draw that. It got pushed up some
distance x. And we're claiming that our--
oh sorry, this is the force. Sorry, let me be clear. This is the force, and
this is the distance. So work is equal to our force,
which is our pressure times our area, times the distance. I want to be very
clear with that. Because when I wrote this I
said, OK, the force that we're applying is the pressure we're
applying, times the area of our cylinder. This is the area of our
cylinder right here. That's the area of our
cylinder right there. So if you do the pressure
times this area, you get the force. And then we moved it
some distance x. Now, we could rearrange this. We could say that the work is
equal to our pressure times our area, times x. What's this? What's this area, this area
right here, times x? Well that's going to be our
change in volume, right? This area times some height
is some volume. And that's essentially how
much our container has changed in volume. When we pushed this piston up,
the volume of our container has increased. You can see that, even looking
from the side. Our rectangle got a
little bit taller. When you look at it with a
little bit of depth, you see the rectangle also didn't
get taller. We have some surface area. Surface area times
height is volume. So this right here, this
term right here, is a change in volume. So we can write work now in
terms of things that we know. We can write work done
by our system. Work done is equal to pressure
times our change in volume. Now this has a very interesting
repercussion here. So we could-- actually many--
we can rewrite our internal energy formulas. So, for example, we can write
internal change and internal energy is now equal to heat
added to the system, plus the work-- let me say minus the
work done by the system. Well what is the work
done by the system? Well it's the pressure of the
system times how much the system expanded. In this case, the system is
pushing these marbles, or these pieces of sand up. It's doing work. If we were doing it the other
way, if we were adding the sand, and we were pushing down
on our little canister, we would be doing work
to the system. So this is the situation where
I'm doing here, where I'm removing the sand and the piston
goes up, essentially the gas is pushing up on
the piston, the system is doing the work. So if we go back to our little
formula, that internal energy is heat minus the work done by
a system, so done by, then we can write this as, this is equal
to the heat added to the system minus this quantity, the
pressure of the system, times the change in volume. And it's interesting, if the
volume is increasing, then the system is doing work. And this applies-- we're going
to talk a lot more about engines in the future-- but
that's how engines do work. They have a little explosion
that goes on inside of a cylinder that pushes up on the
piston, and then that piston moves a bunch of other stuff
that eventually turns wheels. So the volume increases, you're
actually doing work. So I'm going to leave you
there in this video. In the next video, we're going
to relate this, this new way of writing our internal energy
formula, and we're going to relate it to the PV diagram.