Heat capacity at constant volume and pressure | Physics | Khan Academy

Imagine you had a monatomic ideal gas in the cylinder here, and 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 and Katie or three-halves little n RT.

We know that saying "internal energy" is really just code for the total kinetic energy of the monatomic ideal gas; 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, you could say increase the pressure or the volume or the temperature. Yeah, but I mean physically, actually in the lab, what do we do?

There are 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'll 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, so if I added 10 joules of heat, I'd add 10 joules to the internal energy, but I've also got to take into account this work being done.

So I could do plus the work done on the gas, and that's it. This is actually the first law of thermodynamics; it's a law of conservation of energy. It says there's only two ways to add internal energy to a gas. Let me talk a little bit more about this work done, though, because 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 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've 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 that started right here and it went up to here, so that piston went from here to there. No gas is escaping because 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, that there was an extra volume generated within here that I'm going to call Delta V. 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? Because look, I can write D as equal to Delta V over A, and I can take this and 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 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 or the final pressure? If the pressure is staying constant, this gives you an exact way to find the work done.

You might object and say, "Wait, how's 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 capacity. So let's get rid of this. 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 as the amount of heat that you've added to the gas divided by the amount of change in the temperature of that gas.

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 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 are different ways that this can happen, and because of that, there are 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.

These are similar but different, and they're related, and we can figure them out. So let's clear this away. Let's get up. 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; they can move freely. 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 the system, or in other words, we've got heat capacity at constant volume which 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'd 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 n k ΔT over ΔT, and something magical happens—check it out; the Delta T's go away, and you get that this is a constant.

The heat capacity for any monatomic ideal gas is just going to be three-halves n k, Boltzmann constant, and n is the total number of molecules. Or you could have rewritten this as little n R ΔT—the T's would still have canceled, 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 n R. And if you wanted the molar heat capacity, remember that's just divided by an extra mole here, so everything gets divided by moles everywhere. Divided by moles, that just cancels the n, so 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 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 that if this is again a monatomic ideal gas, this is going to equal three-halves n R 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. So what do we get? Three-halves n R 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 n R Delta T, so I can rewrite this as n R Delta T divided by Delta T. Almost there; all the Delta T's go away and I’m left with, I'm left with C—the heat capacity at constant pressure is going to be equal to three-halves n R plus n R.

That's just five-halves n R, 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 at constant pressure would be five-halves R.

Notice they're almost the same. The heat capacity at constant volume is three-halves n R and the heat capacity at constant pressure is five-halves n R. They just differ by n R. So the difference between the heat capacity at constant volume, which is three-halves n R, and the heat capacity at constant pressure, which is five-halves n R, is just C_P minus C_V, which is n R.

Just n R. 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—that tells you the difference between the heat capacity at constant pressure and the heat capacity at constant volume.