 Okay, so what we've seen so far about the Carnot cycle is that if we run a Carnot cycle for an ideal gas, we get a heat engine with an efficiency that's one minus this ratio of the cold temperature, the cold reservoir to the hot temperature, the temperature of the hot reservoir. That's not just specific to the Carnot cycle for ideal gases as it turns out, but for lots of other types of heat engines as well. So first let me explain what I mean by other types of heat engines. The Carnot cycle is a sequence of adiabatic and isothermal expansions. So we do an isothermal, reversible and isothermal expansion, adiabatic expansion, isothermal compression, adiabatic compression. Each one of those steps is reversible and either isothermal or adiabatic. The network, we get out of that process, we can calculate numerically. We don't have to do just isothermal and adiabatic steps though. We can imagine, just for the sake of argument, let's say we did an isothermal expansion from here to here, and then somehow we need to get from our hot temperature to some colder temperature, but we don't have to do an adiabatic expansion to get to the colder temperature. Maybe we just drop straight down to the cold temperature. So on this PV diagram, that would correspond to our second step being not an adiabatic expansion. It's not an expansion at all, our volume's not changing. It's just a pressure drop in order to reduce the temperature. So we could call that an isochoric constant volume decrease in temperature. And then we could, reversibly and isothermally, compress to here and then do another isochoric heating to get back up to the initial conditions. So that's another heat engine. We would end up using heat to do work in that process. So that's not the Carnot cycle, it's a different cycle. So there's a number of these different types of cycles. There's a few that we won't dwell on the details of. The Rankine cycle is a sequence of adiabatic followed by isobaric expansions and compressions. That corresponds to what happens in a steam engine. The Otto cycle corresponds more to what happens in a combustion engine. Those are different steps, it's in fact more than just four steps involved in an Otto cycle. But engineers, chemical engineers care a great deal about the details of specific flavors of heat engines. Chemists are less interested in the details of specific types of heat engines and more interested in the thermodynamics of the process. So we're going to not delve into the details of these different heat engines too much. But the question is, how general is this statement? What can we say about heat engines in general, even if we're doing something that's not a Carnot cycle? So let's go back to, again, thinking about the Carnot cycle, but in a more general way. So again, we had four different steps for the Carnot cycle. And if I just remind you, without as much of the math, we had, at the hot temperature and cold temperature, we had we absorbed heat from the surroundings during the first expansion step, and we gave some heat back to the colder surroundings in the third step, the isothermal compression step. So we had, in step number one, reversible isothermal expansion, absorbed some heat, adiabatic expansion, there was no heat, isothermal compression, we gave some heat back to the surrounding, so these two terms have opposite signs from each other and then no heat in the final adiabatic expansion. Adding those up, there was some net amount of heat from that process. I'm going to skip all the rest of the details of this diagram and just remind you that for a cyclic process, any cyclic process, whether it's a ranking cycle or an auto cycle, any cyclic process is going to have zero net change in the internal energy because it's a state function. And the work done by this process, again, even if it's not a Carnot cycle, even if it's some other type of cycle, the work is the area enclosed by this diagram. So there's some total amount of work. So what this tells us, what the first law tells us is these two amounts of heat plus that amount of work must add up to the net change in energy, zero. So Q cold plus Q hot plus work is zero or if I move the work to the opposite side of the equal sign, negative W is Q hot plus Q cold. That's enough information. I don't really have to delve into the details of this diagram and what they are for an ideal gas or Van der Waals gas or some other type of gas. If I'm interested in the efficiency of this process, remember that's the amount of work I get out compared to the amount of heat I had to put in. Just the first law is enough to tell me that the amount of work I get out is equal to negative W is equal to Q H plus Q C. The net heat for the process divided by Q H. That ratio is the efficiency. Q H over Q H is one. Q C over Q H is Q C over Q H. So written in terms of heat, the efficiency of the process is one plus this ratio of the heats. And remember Q C is a negative number. Q H is a positive number so that ratio that's going to end up decreasing the efficiency below one. We had seen previously that this expression involves temperatures not heats. So let's remind ourselves of the Clausius theorem which says, remember a change in entropy, an infinitesimally small change in entropy, is dQ for the reversible process divided by T. Each one of these steps, as we've described them, was reversible. Reversible isothermal, reversible adiabatic, reversible isothermal, reversible adiabatic. So all our steps are reversible so we can use the Clausius theorem and relate the heats for our steps to entropies. So I can say for this total process for my four-step process going around this full cycle the net change in entropy is going to be heat divided by temperature, so Q H divided by temperature and that happens at the hot temperature. This step had no heat. This step, the heat was Q cold and that happened at the colder temperature, T C and then the fourth step had no heat. So total change in entropy is the sum of these terms of which only two of them contribute anything but entropy is a state function so when I start here I go all the way around the cycle when I get back to where I started the net change in the entropy has to be zero for that state function. So this term plus this term have to add up to zero. Q H over T H must be the negative of Q C over T C or since what I'm interested in is ratios of Q's or ratios of T's, let's rearrange this a little bit and write Q C over Q H so I'm going to take this Q H and bring it down underneath the Q C and then I have to take that T C and put it above the T H so Q C over Q H and there's a negative sign involved is negative of T C over T H. That's what I get by rearranging this equation. So I can use that the Clausius theorem applied to this cyclic process tells me that this ratio of heats is negative the ratio of the temperatures. So if I plug that in here I find that the efficiency is one instead of plus Q C over Q H I've got minus T C over T H. So that's not surprising because we've seen it before. We saw it for the case of an ideal gas for the Carnot cycle. What we've just done here, notice I haven't said anything about ideal gases. There were no nRT one over V's or log V's or anything like that that came from the ideal gas law. Only the first law is essentially the only fact we had to use here. So at least for the Carnot cycle where I have some adiabatic steps we've seen that the efficiency is always going to be one minus this ratio of temperatures whether it's an ideal gas a real gas, any type of material the efficiency can be given by this form. This is in fact called the Carnot theorem and it's even more general than just the Carnot cycle. All types of heat engines, all cyclic processes on this PVA diagram will have an efficiency given by this result regardless of what kind of heat engine they are as long as the processes are reversible. So that's the Carnot theorem. The Carnot theorem says any reversible heat engine has this efficiency. Proving that for something other than Carnot cycle is a little more involved in particular it requires us to make use of the second law of thermodynamics which we haven't covered just yet so I'll postpone that for a little bit but it's useful to be able to use that result now and say regardless of what type of heat engine we have the efficiency is given by this result. One minus the ratio of cold temperature over hot temperature for any reversible heat engine operating between two heat reservoirs at these two different temperatures. So what we'll do next is now that we understand how heat engines are good at converting heat into work it's also sometimes useful to do the reverse process convert some work into heat. So that's the next thing we'll talk about.