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Current time:0:00Total duration:19:03

let's start with our classic system that I keep using over and over again and that's because it tends to be very useful for instruction and it also tends to be the system that is most covered in classrooms so hopefully it'll be productive for you and your schoolwork so I have this container it has a movable piston on top or a kind of a movable ceiling that's and on top of that well of course inside of my system I have a bunch of pop molecules or atoms bouncing around creating some type of pressure on the system so let's say it has some pressure p1 this volume right here let's call that v1 and let's say it also has some temperature that it starts off with everything is an equilibrium remember these are macro states the only way that I can even tell you what the volume or the pressure or the temperature is is if this system is in equilibrium if everything in it is uniform the temperature is consistent throughout fair enough in order to keep it placed down I have to put some rocks on top rocks on top and I've done this in multiple processes so far and of course I'm doing these little pebbles because I'm going to remove them slowly because I want to approximate a quasi-static process or I want to approximate a system that's always close enough to equilibrium that I'm cool with defining our macro States or pressure or temperature or our volume let me write V for volume now in this video I'm going to study what's called an isotherm ik process isothermic isothermic really what it just means is i'm going to keep the temperature the same iso just means the same you probably remember when we study the periodic table isotopes those are the same element just with different mass numbers so this is the same temperature we're going to run our process so my question is how can we do that because as I remove pebbles as I remove pebbles what's going to happen if I just did it in a if I just did this without any if it was completely isolated from the world and actually I'll added a word right here if it was an adiabatic process if it was adiabatic adiabatic fancy word all that means is completely isolated from the world so no heat is going into or out of this system if this was the case what would happen as I really as I released or I took away some of these little particles let me copy and paste it let me well let me just redraw it actually so I have my one wall I have another wall I have another wall as I release a couple of pebbles you know one at a time my volume is going to increase so I'm going to have a slightly higher volume my volume is going to go up I have fewer pebbles here now and fewer pebbles and since I have the same number of molecules they're going to bump into this less so pressure is going to go down volume is going to go up and if I was adiabatic if I if I had no extra heat being added to the system what do I know is going to happen to the temperature well think about it this way some work was done right our old our old ceiling was maybe someplace around here we pushed it up with some force for some distance so we did work and so we changed some kinetic energy or we transferred some kinetic energy out of the system that's essentially what the work did that kinetic energy was turned into work and temperature is just a macro measure of average kinetic energy average kinetic energy in fact we just well I want to go into it in the vanilla and the and the in the last video the kind of proof e1 if you didn't watch it because you didn't want to go through the math which is completely fair enough because it normally wouldn't be done an intro chemistry class I showed that the internal the internal energy is equal to the total kinetic energy so total kinetic energy which was equal to three-halves times the number of moles times R times temperature so temperature is just by some scaling factor or a measure of kinetic energy now if some of my when I do some work it's essentially a transfer of kinetic energy right and I can't replace that energy with some heat because it's a DI there's no heat going into or out of the system so in that situation the kinetic energy of the system went down the average kinetic energy of the system went down so the temperature would have also go down so the temperature would also go down and actually just as a bonus point what happened to the internal energy well the internal energy is the total kinetic energy of the system and I can even write down the original formula change in internal energy is equal to change yeah let me do not do that because I said I shouldn't is equal to heat added to the system minus work done by the system this is work done by the system that's why we're subtracting it now it's adiabatic so there's no heat added to the system so the change in internal energy is equal to them is equal to the minus the work done by the system well in this situation the system did do work it pushed this piston up by some distance with some force so your Delta U is negative it's less than zero so you went down and that makes sense if temperature changed then the internal energy is going to change and for our simple system where internal energy is represented by the kinetic energy of these molecules that's always going to be the case that if temperature doesn't change internal energy won't change if temperature goes up internal energy goes up if temperature goes down internal energy goes down and of course they're not the same thing though there's the the differing internal energy and temperature is this scaling factor three-halves times the number of molecules x times our ideal gas constant so fair enough so if I do went through this whole exercise just to show you that if I was completely isolated and if I removed a couple of these pebbles that my temperature is going to go down now I told you already that I want to do an isothermal process I want to do an isothermic process so I want to do this process while keeping the temperature the same so how can I do that what I'm going to do is I'm going to place I'm going to place my system on top of what we'll call a reservoir a reservoir so a reservoir you can kind of view as an infinitely large amount of something that is the temperature that we started off with so this reservoir is T one so even though if you know if I took if I took you know two things two things of comparable size let's say a that's a temperature a this is temperature B and I put them next to each other they're going to average out to a plus B over two whatever their temperatures are but if B is massive if you know if a is is just like a I don't know is a is a it's just a speck of a particle let's say it's an iron dust while B is the Eiffel Tower then essentially B's temperature will not change a lot a will just become B's temperature now reservoir is theoretically infinitely large it's an infinitely large object so if something is next to a reservoir its given enough time it'll always assume the heat of the reservoir or the temperature of the reservoir so what's going to happen so this is adiabatic but now I'm actually putting it next to a reservoir so this isn't going to happen the adiabatic situation isn't going to happen now I'm going to have a situation where I'm going to stay the same temperature so what's that going to look like on the PV diagram so let me let me draw the PV diagram this is my pressure this is my volume so this is my starting point right here and what I'm saying is if I'm doing an isothermal process so I just keep removing I just keep removing these pebbles so I started this state right here and let me copy and paste it since I've done so much art already here so I'm going from there I'm going from there to here where I'm removing a couple of the particles I'm removing a couple of the particles so let's say I've removed a couple of them over here and because of that I've increased the volume so let's say the volume it's not there anymore let's it's a little bit higher let's say the volume is just for the sake of our discussion let's say the volume is expanded a little bit because I've removed some particles the little pebbles on the top keeping it down so it's like the adiabatic process but instead of the temperature going down my temperature stays at t1 my temperature is at t1 the entire time because I'm next to this theoretical thing called a reservoir right called a reservoir so because of that I will travel along what we will call an isotherm so this is my first state when I'm done I might end up someplace over here and so this is state two so this is state two this is state 1 1 2 what I'm claiming is that it's my path along this is going to be on some type of a rectangular hyperbola or at least part of it if I were to add rocks to it if I were to add rocks and compress it I claim that my PV diagram would go like this if I were to keep removing rocks from this diagram I claim that my PV diagram would keep going like that and so what's the intuition that if I keep the temperature constant that I'm essentially moving along this hyperbola well let's just take out the ideal gas formula let me box off all this stuff over here if I just take the ideal gas formula PV is equal to N R T if T is constant we know that R is a constant it's the ideal gas constant we know that we're not changing the number of moles of particles then that means that PV is equal to some constant right this whole thing is equal to some constant and then if we wanted to write P as a function of V we would just write P is equal to K over V now this might not look 100% familiar to you but if i wrote an algebraic terms if I wrote if I told you to graph y is equal to 1 over X what does that look like that's a rectangular hyperbola that looks like this if this is the y-axis that's the x-axis at least in this quadrant it looks like this it also looks like that in the third quadrant but we won't worry about that too much so whenever you hold temperature constant you're on some rectangular hyperbola like this like an isotherm now if the temperature was a different temperature if it was a lower temperature you would be on a different isotherm so you would be on an isotherm that look like this maybe over here it would also be a rectangular hyperbola but at a lower state why is that because if you're at a lower temperature for any for any volume if you're a lower temperature you should have a lower pressure and that works out that's why this is this is some temperature - that is lower than t1 so I want to do a couple of things in this video I in add vertical and why the temperature would naturally go down on its own and if you didn't have this this reservoir here but the whole reason why I even thought about doing this video is because I wanted you to get comfortable with this idea of one that they a reservoir will keep you and in kind of an isothermic state it will keep the temperature the same and then if you keep the temperature the same that you will travel along this isotherm these rectangular hyperbolas and that each temperature has associated with an isotherm so if you take that let's just do one more step and let's think about the actual work we did let's think about the actual work we did by traveling from this state to this state or if you just want to think of it in visual terms from removing our pebbles slowly and slowly with this reservoir down here the whole time from this date to this date where our volume is increased our pressure has volume is increased our pressure has gone down but our temperature has stayed the same the entire time so several videos ago we learned that the work done is the area under this graph it's the area under that graph or if we want to do in calculus terms and I'm going to about about to break into calculus so if you don't want to see calculus cover your eyes or ears it would be the integral and this the rest of this video will be a little bit mathy and I guess I should make that statement on the on the the title of the video but if I want to calculate what this area is I can now do this the the isotherm assumption makes our math a little bit easier because we know that PV is equal to n R T ideal gas law or we could say P if we divide both sides by V is equal to n R T divided by V so there we have it we have P is a function of V this this function right here this graph right here is this we could write P as a function of V is equal to NRT over V so if we want to figure out the area under the curve we just integrate this function from our starting our V one two our ending point to our v2 so what is that going to be well we're going to integrate from v1 to v2 actually this isn't that shouldn't be an equal this is the work it's going to be the integral from v1 to v2 times our function P is a function of V times DV right we're summing up all of the little all the little rectangles here we did that a couple of videos ago so what's P is a function of V so work done is equal to from v1 to v2 n RT / / V times DV now this is our simplifying assumption we said where we were sitting on top of a reservoir that this reservoir keeps our temperature the same the whole time and we're going to learn in a second it's doing that by transferring heat into the system and we're going to calculate how much heat is transferred into the system so if we look at this right here temperature since we're assuming is that we're on an isotherm is a constant n an R are definitely constants so we can write rewrite this integral as the integral from v1 to v2 of 1 over V DV and then we could put the NRT out here I should have done that first NRT it's just a constant term now what's the antiderivative of 1 over V it's the natural log of V so our work is equal to n RT times the natural log this is the antiderivative of V evaluated at v2 - it evaluated at v1 so that is equal to n R T times evaluated v2 so natural log of v2 minus the natural log of v1 now we know from logarithm properties is the same thing as n R T times the natural log of v2 over v1 so there you have it we actually calculated the real value if we know our our starting volume and our finishing volume we can actually figure out the work done in this isothermic process the work done in this isothermic process is the area under this and we figured out what it was by pushing up that piston was NRT these are you know the number of moles we have ideal gas constant our temperature that we're sitting on it would be t1 in this case and the natural log of our finishing volume divided by our starting volume now let me ask a follow-up question how much heat was put into the system how much heat was put into the system by this isotherm here how much heat was put into the system it put in heat to keep the temperature up otherwise the temperature would have gone down right he was going into the system the entire time how much was it well since it's an isotherm since the temperature did not change what do we know about the internal energy did the internal energy change the temperature not changing told us that the kinetic energy didn't change if the kinetic energy didn't change then the internal energy did not change and we know we know that the change in internal energy is equal to the heat put into the system minus the work done by the system now if this is zero so we know that this didn't change because the temperature didn't change so that means zero is equal to Q minus W or that Q is equal to W so this is the work done by the system you'll end up getting something in joules and this is also equal to the heat put into the system it's also equal to Q so when you look at this if we were to just draw this part of of the the curve if I were to let me redraw just to make things neat I want to give you a little bit of the convention of what people in the thermodynamics world tend to do make a neat drawing here we started here at state one and we moved along this rectangular hyperbola which is an isotherm to state two to state two and now we could have calculate we calculated the area under this which is the work done which was this value right here let me write it there it's n R T natural log of v2 over v1 this is v2 this is v1 this whole axis remember was the V axis volume axis this axis here was the pressure axis and the convention is is that because we did work but we were constant temperature so our internal energy didn't change we had to add energy to the system to make up for the work we did so some heat must have been added to the system and what they do is they just put this little downward arrow and they write a Q right there so some heat was put into the system during this this isothermic process right there and the value of that Q is equivalent to the work we did we put the exact amount of heat into the system as the work where that was performed and because of that our internal energy didn't change or you could say our temperature didn't change or you could go the other way or because our temperature didn't change these two things have to be equal anyway I want to leave you there hopefully give you a little bit more of an intuition of how PV diagrams work a little bit more intuition behind what isotherms and adiabatic mean and the most important thing is once we get a little bit mathy this result can be useful for coming up with other interesting things about a lot of these thermal systems that we're dealing with see in the next video