 So, we understand a fair amount at this point about gases, their PV behavior, how they work when they get compressed or they expand, in particular ideal gases. We've seen equations that describe their PV behavior when we compress them reversibly and isothermally, or when we compress them reversibly and adiabatically, two different types of compression. So, our next step is to combine those two things, talk about both those types of compression and expansion at the same time, and it turns out we'll discover something fairly remarkable when we do both of those things. So, let's start with the diagram. Pressure volume diagram, we'll start with a gas that sets some initial pressure, P1, V1. And what I'm going to do is I'm going to compress that gas reversibly and isothermally down to some, no I'm sorry, expand, let the gas expand to a higher volume, V2 is larger than V1, so larger volume, so it expands to this larger volume and it's at a lower pressure. I'm doing that reversibly and isothermally, so there's an isotherm here, all the points on that curve are at the same temperature, so that's an isotherm, I'm just moving from one point on this isotherm to a different point at that isotherm. So that's the first step in what's going to be a multiple step process, so let me get a little more specific about what's going on in that step. So in this step when we go from P1, V1 to P2 and V2, that's going to be our reversible isothermal expansion. We can say how much heat is involved in that process, how much work is involved in that process, how much internal energy change is involved in that process. Because we have thought about reversible isothermal expansions already for an ideal gas, we can write down what those are. So reversible isothermal expansions, the work associated with that process is minus Nrt log of V final over V initial, so in this case that's V2 divided by V1 inside of this logarithm. It's an isothermal process, so for an ideal gas, delta U is zero whenever the temperature doesn't change, so I know delta U is zero, the work is this, so the heat has to be the opposite of this, so that's going to be positive Nrt natural log of V2 over V1. So that's just repeating what we know about work, heat and internal energy change for reversible isothermal expansion of an ideal gas. Next what we're going to do is after we've expanded the gas to P2 and V2, still on this isotherm, we're going to expand it a little more, but now instead of expanding isothermally, instead of remaining on this isotherm, we're going to expand adiabatically. So I'll draw that curve in a different color, and that curve is going to look something like this because it's going to expand to some new P3 V3. Then I go from step two to step three, that's going to be a reversible and adiabatic expansion. I know to draw the curve in this way because as we know from talking about adiabatic PV changes, when a gas expands adiabatically, its temperature drops, so it's no longer on the same isotherm, there's a different isotherm that it lives on. So here's a set of PV conditions, they're all at the same temperature of the gas, here's another set of PV conditions that are a different temperature. So the first curve, I'll say that's TH for T-hot, this one is TC for T-cold. The upper isotherm is at a hotter temperature than the lower isotherm. So I've done this adiabatic expansion down to a colder temperature. What we know about adiabatic expansions, first of all, it's adiabatic and reversible, so the heat is zero. Secondly, probably the easiest way to think about this is the temperature has dropped for an ideal gas, if the temperature changes, the change in the internal energy is going to be moles times the molar heat capacity times the change in temperature. In fact, let me not say CV because we're not at constant volume. Yes, I do want to say CV. So final temperature T-cold minus initial temperature T-hot, the change in the internal energy is NCV times this change in temperature. And again, because heat and work have to add up to the internal energy, the work must also be the same value. So those are the heat, work, and internal energy change when I do this adiabatic expansion. Now, the next step in this process, we're going to eventually get back to where we started. So the next step in this process is to climb back up this lower temperature isotherm. So now I'm going to do a compression to lower volumes, and it's going to be an isothermal compression. So step from three to four, I'm going to climb back up the isotherm to some point that I'll call P4V4 that's at this cold temperature. And now that I've labeled these temperatures hot and cold, let me remember that the first isothermal expansion took place at not temperature T, but temperature T-hot, I renamed it. This second isothermal process, the compression, is happening at T-cold. So I'm using the same equations for heat, work, internal energy of a reversible isothermal process for an ideal gas. This one is a reversible isothermal compression. The work is still negative nRT log v final over v initial. Let's label these carefully. The temperature at which I'm doing this process is the cold temperature, Tc. V final over v initial is v4 over v3, and that's what I've got for my work. Isothermal, so there's no internal energy change. Q and W have to add up to zero. So heat is positive nRT-cold log of v4 over v3. Before we go any further, I'll point out that we actually know something about this ratio of v4 and v3. Specifically, if I tell you that the next step in the process is going to be another adiabatic expansion up to where I started. So this third step where I go from P3v3 up to P4v4, I don't want to stop just anywhere on the size of the room. I want to stop at the specific place where if I then adiabatically compress the gas, I'm going to get right back to where I started. So there's a particular point on this curve where an adiabatic compression will lead me back to my initial conditions. That's exactly where I want to stop this reversible isothermal compression. That particular point, we know something about this v4. In fact, for the adiabatic expansion step, when the gas cooled down during the second step when I adiabatically expanded it, we know that from what we determined when we thought more deeply about adiabatic expansions, the ratio of temperatures, T final over T initial is the opposite ratio of the volumes, volume initial over v final raised to this power r over Cv bar. So labeling my finals and initials carefully, my final temperature in the expansion was the cold temperature. So T cold over T hot is equal to v initial over v final. So that's v2 over v3 raised to this powers. That's also the same, the difference, this adiabatic cooling from T hot to T cold, that would have been the same as if I adiabatically cooled from P1v1 down to this P4v4 or equivalent adiabatically compressed up to the hotter temperature. So this ratio T cold over T hot is this ratio of volumes raised to this power. That's also this ratio of volumes, v1 over v4 raised to the same power. So if this ratio raised to a power is equal to this ratio raised to a power, it must be true that v2 over v3 is v1 over v4 or since I have this ratio of v4 over v3, if I rearrange this, if I put v4 up here and therefore I have to take v2 over the other side, this ratio v4 over v3 is the same as the ratio of v1 over v2. Let's make sure we understand what that means. This volume relative to that volume, so this value on the v axis relative to this value on the v axis, that ratio is exactly the same as this ratio. If this adiabatic expansion doubled the volume, this adiabatic expansion or compression also doubled the volume. That's perhaps not a surprising fact, but it's one that takes a little bit of thought to convince us that we do in fact know that that's true. So when I write minus nR T cold log of v4 over v3, I could equivalently write v1 over v2, in fact I'll go ahead and do that, minus nR T cold log of v1 over v2 or if I want to exchange the positions of v1 and v2, which I do to make it look more like this one, I can change the sign out front and I'll write that as positive nR T cold log v2 over v1. Any one of these three ways of writing it is exactly the same. I like this last one best because it has a log v2 over v1, which looks a lot like this one. Same thing for heat, so nR T c log v4 over v3, I could write it as v1 over v2 or I could change the sign and write it as minus nR T c log v2 over v1. So this column is our heat, this column is our work. For this third step, the adiabatic compression. So I've expanded, expanded again, compressed now to p4 and v4. My last step is to compress the rest of the way from v4 up to v1. So for this step, the step for 4 going to 1, that's going to be another adiabatic reversible and adiabatic step. This one is a compression. Like any adiabatic process, there's no heat associated with it, any reversible adiabatic process. The work and the delta u as in the adiabatic expansion are going to be the same number. In this case, I've changed, instead of changing from T hot to T cold, I've changed from T cold to T hot. So this is just the same thing in reverse. I've gone, my final temperature is T hot, my initial temperature is T cold. So we're done with our four stage process, expand, expand, compress, compress, get back to where we started. This is a cycle, a cyclic process because I ended up back where I started. It's called specifically the Carnot cycle, the sequence of doing a reversible and then adiabatic expansion followed by isothermal and then adiabatic expansion followed by isothermal and then adiabatic compression is called the Carnot cycle after Sadi Carnot, a French thermodynamicist from the 1800s. What's special about this process in particular that we've done for the adiogas? Let's take a look at what all these numbers add up to. So if I do all four steps in sequence ending back up where I started, the total amount of internal energy change, delta U, zero and zero, NCV times cold minus hot, if I add that to NCV times hot minus cold, those two terms exactly cancel each other. So the net amount of internal energy change for the process is zero. Shouldn't be too surprising because internal energy is a state function. Whatever I do, if I end up back where I started, the internal energy at that state is exactly the same. So if I go away and come back, the change in the internal energy has been zero. Heat and work of course are not state functions. So if I add these up, I won't get zero. I won't necessarily get zero and I don't get zero. In this case, let's see, if I add up some zeros, if I add up this term and this term for the heat. Remember, this is just another way of writing this one. The total amount of heat that's associated with this process is NR. There's a log V2 over V1 that both these terms have in common. This one has a positive TH. This one has a negative TC. So I've got T hot minus T cold. If I think about the work, there is some cancellation that happens. So NCV T cold minus T hot cancels with NCV T hot minus T cold. So that work associated with the second step and the fourth step cancel. Again, remember, I only need one of these if I look at this one and this one. The total amount of work is I've got an N and an R and a log V2 over V1. This term has a minus TH. This term has a positive TC. So I'll write this one as T cold minus T hot. Again, not surprisingly, if I add up the heat and the work, NR log V2 over V1 times TH minus TC, add that to the same thing with a TC minus TH. Those two terms cancel. So this plus this is equal to 0. Q plus W must be equal to delta U. That's still true. The heat is the negative of the work, vice versa. If we think about the signs of these two results, the heat, T hot is bigger than T cold. Hot temperature is larger than the cold temperature. So this term in parentheses is positive. N is positive. R is positive. V2 and V1, that was an expansion step. So V2 is a larger volume than V1. So this ratio is bigger than one. Its log is positive. So this term is positive. The heat associated with this process is positive. The work, therefore, is going to be negative. That happens because T cold is less than T hot. So T cold minus T hot is a negative number. So for this cyclic process, I haven't changed the internal energy of the system. It has required some heat and it is associated with some work. If I think about what those terms mean and what the signs of those terms mean physically, in the process of doing this cyclic process, Q being greater than 0, that means the system absorbs some heat from the surroundings. Some heat was absorbed into the system because Q is positive. Work is negative. So that means work was done by the system. Energy is still conserved. The system absorbed heat from the surroundings and used that amount of energy in the form of heat to do work on the surroundings, specifically to do PV work on the surroundings. So putting all this in a nutshell, this process, this cyclic Carnot process, has converted heat into work. That sounds like a fairly useful thing to be able to do, convert heat into work and get some useful work out of the process. So that's what we'll do next is explore in a little more detail. What are these processes that allow us to convert heat into work?