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## Class 11 Physics (India)

### Course: Class 11 Physics (India) > Unit 17

Lesson 2: Kinetic molecular theory of gases# Heat capacity at constant volume and pressure

Created by David SantoPietro.

## Want to join the conversation?

- Where did you derive Uint= 3/2PV=3/2NKT=3/2nRT?(10 votes)
- Videos are in the wrong order. Kinetic molecular theory of gases.(22 votes)

- At4:51, it was said Pressure remained constant even the piston moved up due to addition of heat. What about in case of work being done on the gas by applying external force (piston moving down) , meaning would pressure still remains constant while work is being done on the system and in addition there is a supply of heat from the bottom?? Not sure we need to add heat at all in this case?? so confused ...Please help(2 votes)
- U is given to be as 3/2 PV or 3/2nRT.....but at7:47it is given that ∆U =3/2PV =nR∆T. Ok let us say for once that they might have forgotten to put the ∆V(in PV) sign over there however the process was being defined for constant volume therefore ∆V =0 so ∆U should also be zero(2 votes)
- Does a monatomic gas have lower or higher heat capacity than a diatomic, or polyatomic?(2 votes)
- Lower. Molar heat capacity at constant pressure = (f+2)/2 and molar heat capacity at constant volume = f/2.

Where f is the number of degrees of freedom. For a monoatomic gas, f =3 and for a diatomic gas we generally consider f=5.(1 vote)

- So does it mean that at constant pressure more heat is required to increase the temperature than whats required at constant volume?(1 vote)
- yes, because at constant pressure some of the energy from the heat transfer will be use to move the piston up (do work)(2 votes)

- how would you figure out the heat capacity of a liquid or solid or a realistic(instead of ideal) gas?(2 votes)
- Then many external factors come into play, which are well beyond the scope of the video. :)

Also, watch the the beginning1:30minutes of the next video in the series (Kinetic Molecular Theory of gases). It list the necessary assumptions for an ideal gas. Peace :)(0 votes)

- hello, for an approximation of heating up the air of an office would we use Cp or Cv?(1 vote)
- In my opinion, its gonna be Cv since the volume of the office will remain constant.(1 vote)

- If I increase the number of moles of gas because of a chemical reaction, but keep volume constant and I increase the temperature is the work done still 0?(1 vote)
- Since the volume does not change, the work done by the gas is still zero.(1 vote)

- When we measure specific heat at constant volume,pressure is changing. We know that specific heat depends on pressure.Therefore does specific heat also changes?(1 vote)
- what about if the piston is pushing down and Q goes out(1 vote)

## Video transcript

- [Voiceover] Imagine you
had a monatomic ideal gas in this cylinder here, an
there was this tightly fitted piston above it that prevented
any gas from getting out. Well we know that the
total internal energy for a monatomic ideal gas is just three halves P times V or three halves NKT, or three halves little nRT, and we know that saying U internal the internal energy is really
just code for the total kinetic energy of the monatomic ideal gas, that these are the same thing. When we talk about internal energy we're talking about how fast
are these particles moving, in other words, what's the
total kinetic energy of all of these particles added up. My question is how do we go about changing the internal energy? Let's say we wanted to increase
the total kinetic energy, what could we do? Well, we could say increase the pressure, or the volume, or the
temperature, yeah but I mean physically, actually, in
the lab, what do we do? And there's basically two ways to change the internal energy. If you want to add internal energy, i.e. get these particles moving faster, we can heat it up so
put this above a flame or on a hot plate, and heat will flow into the gas which will cause these particles to
move faster and faster. That's one way to do it, to add heat. The other way to do it
is to do work on the gas. I could take this piston and push it down, and if you push this down hard enough it will squash this gas together, and those impacts with this
piston while it's moving down will cause them to start
moving faster and faster, that will also add
internal energy to the gas. So if we wanted to write
down a formula that told us how you could get a change
in the internal energy, if I want to change the internal energy, Delta U, which is really just saying changing the kinetic energy, well there's two ways to do it. I can add heat. If I added ten joules of heat I'd add ten joules to the internal energy, but I've also got to take into
account this work being done, and so I can do plus the
work done on the gas, and that's it, this is
actually the first law of thermodynamics, it's the
law of conservation of energy it says there's only
two ways to add energy, internal energy to a gas. Let me talk a little bit more
about this work done though cause getting the sign right is important. If you're doing work on
the gas compressing it you're adding energy to the gas, but if you let the gas
push up on the piston and this gas expands
pushing the piston up, then the gas is doing the work, that's energy leaving the system. So if the gas does work
you have to subtract work done by the gas. If the outside force does work on the gas you add that to the internal energy. So you got to pay attention to which way the energy is flowing. Work done on gas, energy goes in. Work done by the gas, energy goes out, and you'd have to subtract that over here. Let's say the gas did expand. Let's say the gas in here
was under so much pressure that the force it exerted
on this piston was enough to push that piston upward
by a certain amount. So let's say it started right
here and it went up to here so that piston went from here to there no gasses escaping cause
this is tightly fitting, but the gas was able to push
it up a certain distance d, how much work was done? You know the definition of work. Work is defined to be the
force times the distance through which that force was applied. So the work that the
gas did was F times d, but we want this to be in
terms of thermal quantities like pressure, and
volume, and temperature. So what could we do? We can say that this volume, not only did the piston raise up, but there was an extra
volume generated within here that I'm going to call Delta V, and I know that this
Delta V has got to equal the area of the piston times the distance through which that piston moved because this height
times that area gives me this volume right in here. Why am I doing this? Cause look I can write d
as equal to Delta V over A, and I can take this, I can
substitute this formula for d into here, and something magical happens, I'll get work equals F
times Delta V over A, but look F over A, we
know what F over A is, that's pressure so I get
that the work done by the gas is the pressure times Delta V. This is an equation that I
like because it's in terms of thermodynamic quantities
that we're already dealing with. So work done you can figure
out by taking P times Delta V but strictly speaking
this is only true if this pressure remained constant, right? If the pressure was changing, then what am I supposed to plug in here, the initial pressure, the final pressure? If the pressure's staying constant this gives you an exact
way to find the work done. You might object and say wait, how is it possible for a
gas to expand and remain at the same pressure? Well, you basically have to heat it up while the gas expands, that allows the pressure
to remain constant as the gas expands. And now we're finally ready
to talk about heat capacities. So let's get rid of this, and heat capacity is defined to be, imagine you had a certain
amount of heat being added. So a certain amount of heat
gets added to your gas. How much does the temperature increase? That's what the heat capacity tells you. So capital C is heat capacity and it's defined to be the amount of heat that you've added to the gas, divided by the amount of change in the temperature of that gas. And actually, something
you'll hear about often is the molar heat capacity, which is actually divided
by an extra n here. So instead of Q over Delta T it's Q over n the number
of moles times Delta T. Pretty simple but think about it. If we had a piston in here, are we going to allow that piston to move while we add the heat, or are we not going to
allow the piston to move? There's different ways
that this can happen, and because of that there's
different heat capacities. If we don't allow this piston to move, if we weld this thing
shut so it can't move we've got heat capacity
at constant volume, and if we do allow this
piston to move freely while we add the heat so that the pressure inside of here remains constant, we'd have the heat capacity
at constant pressure. And these are similar but different, and they're related, and
we can figure them out. So let's clear this
away, let's get a nice, here we go, two pistons
inside of cylinders. We'll put a piston in here, but I'm going to weld this one shut. This one can't move. We'll have another one over
here, it can move freely. So over on this side, we'll have the definition of heat capacity, regular heat capacity, is the amount of heat you add divided by the change in temperature that you get. So on this side we're
adding heat, let's say heat goes in, but the piston does not move and so the gas in here is stuck, it can't move, no work can be done. Since this piston can't move, external forces can't do work on the gas, and the gas can't do work
and allow energy to leave. Q is the only thing adding
energy into this system, or in other words, we've got heat capacity at constant volume is going to equal, well, remember the first
law of thermodynamics said that Delta U, the only way
to add internal energy, or take it away is that you
can add or subtract heat, and you can do work on the gas. In this case, Q, if I
subtract W from both sides I get Delta U minus W over Delta T, but since we're not
allowing this piston to move the work done has got to be zero. So there's no work done at
all so the heat capacity at constant volume is going to be Delta U over Delta T, what's Delta U? Let's just assume this
is a monatomic ideal gas, if it's monatomic we've
got a formula for this. Delta U is just three halves,
P times V over Delta T. That's not the only way I can write it. Remember I can also write it
as three halves NK Delta T over Delta T, and
something magical happens, check it out the Delta T's go away and you get that this is a constant. That the heat capacity for
any monatomic ideal gas is just going to be three halves, Capital NK, Boltzmann's constant, N is the total number of molecules. Or you could have rewrote
this as little n R Delta T. The T's would still have cancelled and you would have got
three halves, little n, the number of moles,
times R, the gas constant. So the heat capacity at
constant volume for any monatomic ideal gas is
just three halves nR, and if you wanted the molar heat capacity remember that's just divide
by an extra mole here so everything gets divided
by moles everywhere divided by moles, that
just cancels this out, and the molar heat
capacity at constant volume is just three halves R. So that's heat capacity
at constant volume, what about heat capacity
at constant pressure? Now we're going to look at this side. Again, we're going to
allow this gas to have heat enter the cylinder, but
we're going to allow this piston to move up while
it does that so that the pressure inside of
here remains constant and this is going to be the heat capacity at constant pressure. Well, again, we're going to get that it's Q over Delta T, and just
like the first law said, Q has to equal Delta U minus W. So we get Delta U minus W, the work done, over Delta T, this time W is not zero. What's W going to be? Remember W is P times Delta V. So this is a way we can find
the work done by the gas, P times Delta V, so this is going to equal Delta U, we know what that is. If this is again, a monatomic ideal gas, this is going to equal
three halves nR Delta T plus this is P times Delta V, but we have to be careful, in this formula this work is referring to work done on the gas, but in this case, work
is being done by the gas, so I need another negative. Technically the work done on
the gas would be a negative amount of this since energy
is leaving the system. So that negative cancels this negative and I get plus P times Delta V, all of that over Delta T, phew. So what do we get? Three halves nR Delta T plus I want to rewrite P times Delta V, but I know how to do that. The Ideal Gas Law says PV equals nRT, well if that's true, then P times Delta V is going to equal nR Delta T. So can rewrite this as nR
Delta T divided by Delta T, almost there, all of
the Delta T's go away. Look what I'm left with. I'm left with C. Heat capacity at constant
pressure is going to be equal to three halves nR plus nR, that's just five halves nR, and if I wanted the
molar heat capacity again I could divide everything,
everything around here by little n, and that would just give me the molar heat capacity constant pressure would be five halves R. And notice they're almost the same. The heat capacity at constant
volume is three halves nR, and the heat capacity at constant pressure is five halves nR. They just differ by nR. So the difference
between the heat capacity at constant volume which
is three halves nR, and the heat capacity at constant pressure which is five halves nR, is just Cp minus Cv which is nR, just nR, and if you wanted
to take the difference between the molar heat
capacities at constant volume and pressure, it would just be R. The difference would just be R because everything
would get divided by the number of moles. So there's a relationship,
an important relationship. It tells you the difference between the heat capacity at constant pressure and the heat capacity at constant volume.