[Music] all right 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 a diagram pressure volume diagram we'll start with a gas that's at 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 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 p1v1 to p2 and v2 that's going to be our reversible isothermal expansion and 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 and 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 all right 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 so it's going to expand to some new p3 v3 when i go from step 2 to step 3 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 for the gas here's another set of pv conditions that are a different temperature so the first curve i'll say that's t hot 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 uh 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 cd because we're not at constant volume yes 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 need 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 uh and now that i've labeled these temperatures hot and cold let me remember that the iso 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 p3 v3 up to p4v4 i don't want to stop just anywhere on this isotherm 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 this 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 of the volumes volume initial over v final raised to this power r over c v 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 uh 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 if i adiabatically compressed up to the hotter temperature so this ratio t cold over t hot is this ratio of volumes raised to this power it's also this ratio of volumes v 1 over v 4 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 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 so 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 nrt cold log of v4 over v3 i could equivalently rewrite 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 nrt 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 log v2 over v1 which looks a lot like this one same thing for heat so nrtc log v4 over v3 i could write it as v1 over v2 or i could change the sign and write it as minus nrtc log v2 over v1 all right 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 four going to one 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 all right so we're done with 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 a reverse i'm sorry isothermal and then adiabatic expansion followed by isothermal and then adiabatic compression is called the carnot cycle after saricarno a french thermodynamicist from the 1800s what's special about this process in particular that we've done for the ideal gas 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 0 and 0 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 that 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 it 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 this is just another way of writing this one the total amount of heat that's uh 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 n cv 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 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 c t c add that to the same thing with a t c minus t h those two terms cancel so this plus this is equal to zero 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 negative is 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 zero 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 all right that sounds like a fairly useful thing to be able to do convert heat into work and get some 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