1 00:00:00 --> 00:00:01 2 00:00:01 --> 00:00:02 The following content is provided under a Creative 3 00:00:02 --> 00:00:03 Commons license. 4 00:00:03 --> 00:00:07 Your support will help MIT OpenCourseWare continue to 5 00:00:07 --> 00:00:10 offer high-quality educational resources for free. 6 00:00:10 --> 00:00:13 To make a donation or view additional materials from 7 00:00:13 --> 00:00:16 hundreds of MIT courses, visit MIT OpenCourseWare 8 00:00:16 --> 00:00:20 at ocw.mit.edu. 9 00:00:20 --> 00:00:25 OK, so we ended up last time, we talked about Joule-Thomson 10 00:00:25 --> 00:00:28 expansion, which is an irreversible expansion through 11 00:00:28 --> 00:00:31 a nozzle, through a porous plug, constant 12 00:00:31 --> 00:00:33 enthalpy expansion. 13 00:00:33 --> 00:00:37 And I want to make sure everybody figured out 14 00:00:37 --> 00:00:39 that it really was an irreversible expansion. 15 00:00:39 --> 00:00:42 Anybody have any questions on that, because when we voted, 16 00:00:42 --> 00:00:46 the majority of the class thought it was reversible Yes. 17 00:00:46 --> 00:00:49 STUDENT: Just a quick question on some of the constraints, 18 00:00:49 --> 00:00:51 like isothermal, isobaric, isovolumetric expansion. 19 00:00:51 --> 00:00:57 For isothermal expansion, that means that delta u 20 00:00:57 --> 00:01:02 does not change, but delta q is equal to delta w? 21 00:01:02 --> 00:01:04 PROFESSOR BAWENDI: So the question was, for an isothermal 22 00:01:04 --> 00:01:08 expansion, delta u does not change, therefore, 23 00:01:08 --> 00:01:09 dq is equal to dw. 24 00:01:09 --> 00:01:13 The answer is that's true only for an ideal gas. 25 00:01:13 --> 00:01:16 For a real gas, that won't be true. 26 00:01:16 --> 00:01:19 Ideal gas only depends on the temperature, the energy only 27 00:01:19 --> 00:01:20 depends on the temperature. 28 00:01:20 --> 00:01:24 Therefore, if it's isothermal, the energy of the ideal 29 00:01:24 --> 00:01:25 gas doesn't change. 30 00:01:25 --> 00:01:29 For a real gas it depends on more than the temperature 31 00:01:29 --> 00:01:30 STUDENT: Are there any other constraints similar 32 00:01:30 --> 00:01:37 to that [INAUDIBLE]. 33 00:01:37 --> 00:01:40 PROFESSOR BAWENDI: So, for an ideal gas, the isothermal is 34 00:01:40 --> 00:01:43 the easy one because the energy doesn't change. 35 00:01:43 --> 00:01:45 If you have an isobaric you're going to have to calculate 36 00:01:45 --> 00:01:48 where the energy changes, and that's a calculation that's 37 00:01:48 --> 00:01:51 likely on the homework and very like on an exam as well, too. 38 00:01:51 --> 00:01:53 In fact, we're going to do some of that today, 39 00:01:53 --> 00:01:55 OK, calculate delta u. 40 00:01:55 --> 00:01:57 Any other questions? 41 00:01:57 --> 00:01:59 Good question. 42 00:01:59 --> 00:02:06 OK, so we ended up last time showing that for an ideal gas, 43 00:02:06 --> 00:02:14 that there was was a relationship between the heat 44 00:02:14 --> 00:02:17 capacity under constant pressure, and this is the molar 45 00:02:17 --> 00:02:20 heat capacity, that's why we have the bar on top, and the 46 00:02:20 --> 00:02:24 molar heat capacity for constant volume path. 47 00:02:24 --> 00:02:25 They're related through the gas constant. 48 00:02:25 --> 00:02:27 OK, this is only true for an ideal gas, and we went through 49 00:02:27 --> 00:02:34 that mathematically where the, with a chain rule. 50 00:02:34 --> 00:02:36 And I wanted to do it a different way which is a little 51 00:02:36 --> 00:02:39 bit convoluted, but it introduces the idea of a 52 00:02:39 --> 00:02:42 thermodynamic cycle, and it's something that we're going to 53 00:02:42 --> 00:02:44 use a lot in the class. 54 00:02:44 --> 00:02:47 Because anytime you're working with the function of state, and 55 00:02:47 --> 00:02:50 you're trying to connect two points together in a p-V or a 56 00:02:50 --> 00:02:54 T-V diagram, then it doesn't matter which way you're 57 00:02:54 --> 00:02:56 connecting those two points, as long as you're dealing with 58 00:02:56 --> 00:02:58 state functions it doesn't matter. 59 00:02:58 --> 00:02:59 It only matters if you're looking at the work 60 00:02:59 --> 00:03:01 or the heat right. 61 00:03:01 --> 00:03:05 And so sometimes, when you want to calculate what the energy is 62 00:03:05 --> 00:03:09 at the endpoint, the path of the experiment that you're 63 00:03:09 --> 00:03:11 doing may not be an easy path to calculate. 64 00:03:11 --> 00:03:14 You may have to find a different path which is an 65 00:03:14 --> 00:03:17 easier one to calculate, and it doesn't matter which path you 66 00:03:17 --> 00:03:19 take, because in the end all you care is what the 67 00:03:19 --> 00:03:20 energy is at the end. 68 00:03:20 --> 00:03:24 so we're going to use this concept of the path to go 69 00:03:24 --> 00:03:27 from the initial point to the end point. 70 00:03:27 --> 00:03:30 As long as you're dealing with state functions it doesn't 71 00:03:30 --> 00:03:33 matter to get at the same results here. 72 00:03:33 --> 00:03:39 So we're going to start with a mole of gas, at some pressure, 73 00:03:39 --> 00:03:44 some volume, temperature and some mole so V, doing it per 74 00:03:44 --> 00:03:46 mole, and we're going to do two paths here. 75 00:03:46 --> 00:03:48 We're going to take a constant volume path. 76 00:03:48 --> 00:03:53 So V is constant and we're going to change the pressure 77 00:03:53 --> 00:03:58 to some, a little change in pressure, V is constant here, 78 00:03:58 --> 00:04:00 and then temperatures is also going to change. 79 00:04:00 --> 00:04:05 So this is a path, which is either a temperature change, 80 00:04:05 --> 00:04:08 with the pressure changes at the same time, going up or 81 00:04:08 --> 00:04:11 down, constant volume. 82 00:04:11 --> 00:04:13 And since the temperature is changing, clearly Cv is going 83 00:04:13 --> 00:04:15 to have something to do with this. 84 00:04:15 --> 00:04:18 So constant volume path, where the temperature is changing. 85 00:04:18 --> 00:04:24 So somehow this is going to allow us to get Cv. 86 00:04:24 --> 00:04:27 Cv is going to be related to this path. 87 00:04:27 --> 00:04:32 And if I draw a diagram on a T-V diagram of what I'm 88 00:04:32 --> 00:04:36 doing here, so there's temperature axis. 89 00:04:36 --> 00:04:38 There's the volume axis here. 90 00:04:38 --> 00:04:42 I'm starting with some V1 here. 91 00:04:42 --> 00:04:47 And this is going to be here V plus dV. 92 00:04:47 --> 00:04:48 So I'm starting right here. 93 00:04:48 --> 00:04:51 This is my initial point. 94 00:04:51 --> 00:04:53 And the path that I'm describing then, let's assume 95 00:04:53 --> 00:04:56 that we're raising the temperature up is this 96 00:04:56 --> 00:04:58 path right here. 97 00:04:58 --> 00:05:07 OK, this is my constant volume path going from T to T plus dT. 98 00:05:07 --> 00:05:10 OK, we'll call this 1 here. 99 00:05:10 --> 00:05:14 All right, now I want to make a path where out of that path, 100 00:05:14 --> 00:05:16 C sub p is going to fall out. 101 00:05:16 --> 00:05:18 So that's going to have to be a constant pressure path. 102 00:05:18 --> 00:05:20 Temperature is going to change but pressure is 103 00:05:20 --> 00:05:21 going to be constant. 104 00:05:21 --> 00:05:24 So what I can do is I can take my initial point here and 105 00:05:24 --> 00:05:28 create a new, different path, where, which is going to be 106 00:05:28 --> 00:05:30 a constant pressure path. 107 00:05:30 --> 00:05:33 So at the end of this path the pressure is constant. 108 00:05:33 --> 00:05:38 The volume is going to change though, to some new volume dV 109 00:05:38 --> 00:05:40 and the temperature is going to change also and I'm going to 110 00:05:40 --> 00:05:42 make it the same endpoint temperature as the endpoint 111 00:05:42 --> 00:05:44 temperature here. 112 00:05:44 --> 00:05:47 And out of this path, since the pressure is constant but the 113 00:05:47 --> 00:05:50 temperature is changing, I'm hoping that C sub p will fall 114 00:05:50 --> 00:05:53 out of that path of that calculation. 115 00:05:53 --> 00:05:55 All right? 116 00:05:55 --> 00:05:58 This should give us something about C sub p. 117 00:05:58 --> 00:06:00 So this will give us something about C sub v. 118 00:06:00 --> 00:06:02 This will give us something about C sub p. 119 00:06:02 --> 00:06:05 Now these two endpoints here are different. 120 00:06:05 --> 00:06:07 They have a different pressure, a different volume. 121 00:06:07 --> 00:06:09 They do have the same temperature though. 122 00:06:09 --> 00:06:13 So the connection between this endpoint here, 123 00:06:13 --> 00:06:15 and that one here. 124 00:06:15 --> 00:06:16 Same temperature. 125 00:06:16 --> 00:06:18 It's going to be an isothermal path that's 126 00:06:18 --> 00:06:20 going to connect them. 127 00:06:20 --> 00:06:24 So I'm going to connect them in an isothermal way, and as, 128 00:06:24 --> 00:06:27 since I'm dealing with an ideal gas, and as we saw from the 129 00:06:27 --> 00:06:30 first question here, isothermal always means delta 130 00:06:30 --> 00:06:31 u equals zero. 131 00:06:31 --> 00:06:34 So I'm going to write that immediately here, delta u equal 132 00:06:34 --> 00:06:37 zero, because I know the answer to that path here. 133 00:06:37 --> 00:06:39 All right, so let's go again what our paths are. 134 00:06:39 --> 00:06:42 Path number 1 I'm going straight up in the V-T diagram. 135 00:06:42 --> 00:06:48 Path number 2 on my diagram it's a reversible, this path 136 00:06:48 --> 00:06:51 number 2, it's a reversible constant pressure path. 137 00:06:51 --> 00:06:55 So I'm going to be keeping the pressure constant, but my 138 00:06:55 --> 00:07:02 volume is going to go up. 139 00:07:02 --> 00:07:03 My temperature is going to go up. 140 00:07:03 --> 00:07:11 Basically I'm putting heat in this system. 141 00:07:11 --> 00:07:14 That's going to be number 2 right here. 142 00:07:14 --> 00:07:16 It's going to be the same temperature as before but the 143 00:07:16 --> 00:07:19 volume is V plus dV now. 144 00:07:19 --> 00:07:23 That's 2, and then, so I'm heating up the system in this 145 00:07:23 --> 00:07:28 path here, and then to connect the 2 endpoints here, I'm going 146 00:07:28 --> 00:07:31 to connect them through a constant temperature path. 147 00:07:31 --> 00:07:34 Basically I'm going to cool down the system so that the -- 148 00:07:34 --> 00:07:37 no, I'm not going to cool down, I'm going to compresses the 149 00:07:37 --> 00:07:42 system back to the smaller volume V1 and the constant 150 00:07:42 --> 00:07:46 temperature conditions. 151 00:07:46 --> 00:07:48 So that's path number 3 here. 152 00:07:48 --> 00:07:51 And I'm only going to worry about energy in this case here. 153 00:07:51 --> 00:07:53 Because I know energy doesn't care about the path. 154 00:07:53 --> 00:07:57 Energy only cares about where I am on the diagram. 155 00:07:57 --> 00:08:00 And I know energy is related to Cv through Cv dT etcetera. 156 00:08:00 --> 00:08:04 So if I worry about energy I have a pretty good chance of 157 00:08:04 --> 00:08:09 extracting out these heat capacities, right, and I don't 158 00:08:09 --> 00:08:11 have to worry about exactly which path and I can 159 00:08:11 --> 00:08:14 really mix things up. 160 00:08:14 --> 00:08:21 So let's start the process. 161 00:08:21 --> 00:08:28 So the first path then, the first path, constant volume 162 00:08:28 --> 00:08:32 constant V, so I'm going to, again, let's just 163 00:08:32 --> 00:08:33 worry about energy. 164 00:08:33 --> 00:08:39 So du for this first path here, du for path 1, and write dq 165 00:08:39 --> 00:08:42 plus dw what we usually write. 166 00:08:42 --> 00:08:45 Now it's constant volume when the volume is constant. 167 00:08:45 --> 00:08:48 There's no work being done so I can immediately 168 00:08:48 --> 00:08:50 ignore this dw here. 169 00:08:50 --> 00:08:51 Constant volume dq = Cv dT. 170 00:08:51 --> 00:08:55 171 00:08:55 --> 00:08:56 Now it's an ideal gas. 172 00:08:56 --> 00:08:58 So in some sense, I did too much work by writing 173 00:08:58 --> 00:09:02 this, because I knew already du = Cv dT. 174 00:09:02 --> 00:09:04 I could have written that down immediately, but I just went 175 00:09:04 --> 00:09:08 through the process of writing the first law, what's dw, 176 00:09:08 --> 00:09:10 what dq, Cv dT, etcetera. 177 00:09:10 --> 00:09:13 I'm just writing what we already know basically. 178 00:09:13 --> 00:09:15 Let's look at path number 2. 179 00:09:15 --> 00:09:16 Path number 2. 180 00:09:16 --> 00:09:22 OK, let's look at the dw, du again. du for path number 2. dq 181 00:09:22 --> 00:09:27 like the first law down, dq plus dw, then -- what's dq? 182 00:09:27 --> 00:09:31 Well it's a constant pressure path. dq is related to the 183 00:09:31 --> 00:09:33 temperature, the change in temperature. dT through the 184 00:09:33 --> 00:09:37 heat capacity, happens to be constant pressure, so this 185 00:09:37 --> 00:09:38 is exactly what we want. 186 00:09:38 --> 00:09:41 Cp, I forgot to put my little bar on top here because 187 00:09:41 --> 00:09:46 it's per mole Cp dT that's my dq here. 188 00:09:46 --> 00:09:47 The path is constant pressure. 189 00:09:47 --> 00:09:48 And then the dw. 190 00:09:48 --> 00:09:51 Well there's a change now in volume and pressure. 191 00:09:51 --> 00:09:53 So I need, well the pressure is constant, but there's 192 00:09:53 --> 00:09:54 a change in volume. 193 00:09:54 --> 00:09:57 Minus p dV is the change, is the work, right? 194 00:09:57 --> 00:10:07 So I can write that down minus p dV for that path there. 195 00:10:07 --> 00:10:12 OK, now but it's an ideal gas, so I know something about the 196 00:10:12 --> 00:10:15 relationship between dV and dT, because I've got 197 00:10:15 --> 00:10:16 dT here the dV. 198 00:10:16 --> 00:10:17 I don't want to have so many variables around. 199 00:10:17 --> 00:10:19 I've got three variables, T, p and V. 200 00:10:19 --> 00:10:23 I know I only need 2, so I can relate dV to dp 201 00:10:23 --> 00:10:24 through the ideal gas law. 202 00:10:24 --> 00:10:28 P dV is equal to R dT. 203 00:10:28 --> 00:10:34 pV = RT for 1 mole, so I just take dV here. dT here because 204 00:10:34 --> 00:10:43 the pressure is constant, so dV is equal to R over p dT. 205 00:10:43 --> 00:10:44 Then pressure is constant. 206 00:10:44 --> 00:10:46 So I can instead of the dV here, I can insert 207 00:10:46 --> 00:10:48 this R over p dT. 208 00:10:48 --> 00:11:00 The p's cancel out, and now I have Cp dT minus R dT. 209 00:11:00 --> 00:11:03 This is equal to Cp minus R dT. 210 00:11:03 --> 00:11:07 211 00:11:07 --> 00:11:10 All right, let's complete the cycle now, the 212 00:11:10 --> 00:11:15 path number 3 here. 213 00:11:15 --> 00:11:18 Path number 3 is a constant temperature path, and I 214 00:11:18 --> 00:11:19 already wrote the answer. 215 00:11:19 --> 00:11:21 Constant temperature isothermal delta u is zero. 216 00:11:21 --> 00:11:23 There's no thinking involved. 217 00:11:23 --> 00:11:26 It's one of these things like I mentioned last time, if I tell 218 00:11:26 --> 00:11:30 you delta H in the middle of the night what you say, q 219 00:11:30 --> 00:11:33 Cp for reversible process. 220 00:11:33 --> 00:11:35 Same thing here. isothermal process, what do you say? 221 00:11:35 --> 00:11:37 Delta u is zero, right? 222 00:11:37 --> 00:11:39 It should be completely second nature. 223 00:11:39 --> 00:11:41 So delta u is zero here. 224 00:11:41 --> 00:11:45 Delta u sub 3 is, for an ideal gas. 225 00:11:45 --> 00:11:47 I gotta put that constraint in there for an ideal gas. 226 00:11:47 --> 00:11:49 Delta u is equal to zero. 227 00:11:49 --> 00:11:52 So now I'm back at the same place. 228 00:11:52 --> 00:11:54 I'm back here right. 229 00:11:54 --> 00:11:58 So path number 1 went from i, let's call this path up here. 230 00:11:58 --> 00:12:01 went to f, and this is how much energy change there was. 231 00:12:01 --> 00:12:05 The sum of path number 2 and path number 3 get me to the 232 00:12:05 --> 00:12:10 same place, so the energy change by going through this 233 00:12:10 --> 00:12:14 time path, this intermediate point here back all the way 234 00:12:14 --> 00:12:18 to final state should be the same the red path. 235 00:12:18 --> 00:12:24 So what I'm saying is that du through path 1 should be equal 236 00:12:24 --> 00:12:30 to du through path 2 plus du going through path 3. 237 00:12:30 --> 00:12:31 All right, because energy doesn't care about the path. 238 00:12:31 --> 00:12:34 It only cares about the end points. 239 00:12:34 --> 00:12:36 OK, so now let's plug what is du through path 1? 240 00:12:36 --> 00:12:47 It's Cp dT and the du through path 2 is Cp minus R dT and 241 00:12:47 --> 00:12:48 then du through path 3 is zero. 242 00:12:48 --> 00:12:50 It's the isothermal process. 243 00:12:50 --> 00:12:56 The dT is cancelled out and voila I have my answer. 244 00:12:56 --> 00:13:04 Cv is equal, oh Cv plus R is equal to Cp whichever way 245 00:13:04 --> 00:13:07 you want to do it, it's a relationship that we had up 246 00:13:07 --> 00:13:10 here that we wanted to prove. 247 00:13:10 --> 00:13:18 Now, I wanted to go to through this just to go through one 248 00:13:18 --> 00:13:20 cycle quickly because we're going to be doing these all the 249 00:13:20 --> 00:13:23 time, and the importance of the fact that the path doesn't 250 00:13:23 --> 00:13:26 matter, and you can always connect things together in 251 00:13:26 --> 00:13:28 a way, whatever you want. 252 00:13:28 --> 00:13:30 Whatever is the easiest. 253 00:13:30 --> 00:13:31 If you want to calculate a change in delta u. 254 00:13:31 --> 00:13:34 Now if you look at my derivation here, there's one 255 00:13:34 --> 00:13:38 spot where I could have just stopped and proven my point 256 00:13:38 --> 00:13:40 without going through the whole thing. 257 00:13:40 --> 00:13:43 So if you look at path number 2 here, the constant pressure 258 00:13:43 --> 00:13:52 path du is Cp minus R dT, du is Cp minus r dT for that path. 259 00:13:52 --> 00:13:56 What is du for an ideal gas? 260 00:13:56 --> 00:14:00 It's -- what else do we know that du is always equal 261 00:14:00 --> 00:14:02 to for an ideal gas? 262 00:14:02 --> 00:14:03 Cv dT right. 263 00:14:03 --> 00:14:08 So I could have written right here immediately equals Cv dT, 264 00:14:08 --> 00:14:10 and that was the end of my derivation. 265 00:14:10 --> 00:14:10 I could have done that. 266 00:14:10 --> 00:14:13 It would've been easier, but I wanted to go through 267 00:14:13 --> 00:14:16 the whole path right? 268 00:14:16 --> 00:14:23 OK, any questions? 269 00:14:23 --> 00:14:24 Cool. 270 00:14:24 --> 00:14:27 All right, so we're going to be doing more 271 00:14:27 --> 00:14:28 thermodynamics processes. 272 00:14:28 --> 00:14:32 We're building up to entropy and to engines, Carnot 273 00:14:32 --> 00:14:34 cycles, etcetera. 274 00:14:34 --> 00:14:37 And so we need to build our repertoire of knowledge in 275 00:14:37 --> 00:14:41 manipulating these cycles like the one I drew on the board 276 00:14:41 --> 00:14:44 that's hidden now You've already seen the simple 277 00:14:44 --> 00:14:47 process, which is the isothermal process, and you've 278 00:14:47 --> 00:14:52 seen how to functionalize, or what the function is 279 00:14:52 --> 00:14:55 that describes the isothermal expansion. 280 00:14:55 --> 00:15:02 Let's say we start from some V1 and p1 here, so high pressure, 281 00:15:02 --> 00:15:05 small volume and we end up with a high volume low 282 00:15:05 --> 00:15:10 pressure, under constant temperature condition. 283 00:15:10 --> 00:15:19 So there's some path, isothermal, so T is constant. 284 00:15:19 --> 00:15:25 And we know the functional form for this path, it's the ideal 285 00:15:25 --> 00:15:28 gas law, pV = RT, pV = nRT. 286 00:15:28 --> 00:15:30 287 00:15:30 --> 00:15:35 T is constant, so the functional form here is pV 288 00:15:35 --> 00:15:38 is equal to a constant. 289 00:15:38 --> 00:15:45 All right, or p is equal to a constant divided by volume. 290 00:15:45 --> 00:15:47 If you want to write a function that describes this line here, 291 00:15:47 --> 00:15:51 it's pressure as a function of volume related to each 292 00:15:51 --> 00:15:53 other with this constant. 293 00:15:53 --> 00:15:54 So we know how to do this. 294 00:15:54 --> 00:15:57 So this allows us to calculate all sorts of things to get 295 00:15:57 --> 00:15:59 these functional forms. 296 00:15:59 --> 00:16:01 That's the isotherm here. 297 00:16:01 --> 00:16:02 We also know how to calculate the work. 298 00:16:02 --> 00:16:07 If it's a reversible process, we know the work is the area 299 00:16:07 --> 00:16:13 under the curve. w reversible. in this case here let's see 300 00:16:13 --> 00:16:17 this is it's a pressure is going down. 301 00:16:17 --> 00:16:22 It's an expansion, so we're doing work against the 302 00:16:22 --> 00:16:23 environment, or to the environment. 303 00:16:23 --> 00:16:26 So work is leaving the system. 304 00:16:26 --> 00:16:32 So minus w reversible is going to be a positive number. 305 00:16:32 --> 00:16:37 V1 to V2 and RT -- you've done this already last time or a 306 00:16:37 --> 00:16:42 couple of times ago and you end up with minus nRT 307 00:16:42 --> 00:16:47 log p2 over p1, OK? 308 00:16:47 --> 00:16:50 So minus w reversible, we're doing work to the environment. 309 00:16:50 --> 00:16:54 This should be a positive number. p2 is less than p1, so 310 00:16:54 --> 00:16:56 p2 over p1 is less than one, log of something less than 1 311 00:16:56 --> 00:16:58 is negative times negative. 312 00:16:58 --> 00:17:01 So in fact this is indeed a positive number so we didn't 313 00:17:01 --> 00:17:03 make a mistake in our signs. 314 00:17:03 --> 00:17:05 You always want to check your signs at the and because it's 315 00:17:05 --> 00:17:10 so easy to get the sign wrong, and I was hoping that it was 316 00:17:10 --> 00:17:11 right here because I wasn't sure, but it turned 317 00:17:11 --> 00:17:13 out to be right. 318 00:17:13 --> 00:17:16 Always check your sign at the end. 319 00:17:16 --> 00:17:17 All right so we know how to do this. 320 00:17:17 --> 00:17:19 Now we are going to do the same thing with a 321 00:17:19 --> 00:17:19 different constraint. 322 00:17:19 --> 00:17:23 Our constraint is going to be an adiabatic expansion 323 00:17:23 --> 00:17:23 or compression. 324 00:17:23 --> 00:17:26 Adiabatic meaning there's no heat involved, and we're going 325 00:17:26 --> 00:17:29 to see how that differs from the isothermal expansion 326 00:17:29 --> 00:17:31 and compression. 327 00:17:31 --> 00:17:37 OK, so this is the experiment then. 328 00:17:37 --> 00:17:40 We're going to take a piston here which is 329 00:17:40 --> 00:17:42 going to be insulated. 330 00:17:42 --> 00:17:47 It's going to have some volume, temperature to begin with, and 331 00:17:47 --> 00:17:50 then we're going to do something to it. 332 00:17:50 --> 00:17:54 We're going to change the pressure above, right now 333 00:17:54 --> 00:17:59 there's a p external, which is equal to p on the inside. 334 00:17:59 --> 00:18:02 OK, we're going to do this reversibly, which means we're 335 00:18:02 --> 00:18:06 going to slowly change the external pressure very, very 336 00:18:06 --> 00:18:10 slightly at a time, so that at every point we're basically in 337 00:18:10 --> 00:18:13 equilibrium, until the pressure reaches a new 338 00:18:13 --> 00:18:15 smaller pressure p2. 339 00:18:15 --> 00:18:21 So that's going to results in an expansion where the new 340 00:18:21 --> 00:18:25 volume new temperature new pressure and an external 341 00:18:25 --> 00:18:28 pressure, which is p2 which is a smaller pressure. 342 00:18:28 --> 00:18:31 It's really important that this, that you remember that 343 00:18:31 --> 00:18:33 this is the reversible case. 344 00:18:33 --> 00:18:36 We're going to also look at the irreversible case briefly. 345 00:18:36 --> 00:18:41 So on the p-V diagram, then, there's a V1 here a V2 346 00:18:41 --> 00:18:44 here, a p1 here a p2 here. 347 00:18:44 --> 00:18:49 This V2 is going to be different than this one here. 348 00:18:49 --> 00:18:52 A priori we don't know what it is. 349 00:18:52 --> 00:18:53 We don't know if it's bigger or smaller. 350 00:18:53 --> 00:18:55 We kind of feel like it's going to be different because it's 351 00:18:55 --> 00:18:56 a different constraint. 352 00:18:56 --> 00:18:58 This is isothermal. 353 00:18:58 --> 00:19:00 This is adiabatic, there's no heat. 354 00:19:00 --> 00:19:02 So there's going to be a line that's going to connect the 355 00:19:02 --> 00:19:08 initial point to the final point, and that line 356 00:19:08 --> 00:19:12 mathematically is not going to be the same as this one here. 357 00:19:12 --> 00:19:15 And our job is to find out what is the mathematical description 358 00:19:15 --> 00:19:19 of this path, this line in p-V's case that connects 359 00:19:19 --> 00:19:21 these two point. 360 00:19:21 --> 00:19:27 And again, I want to stress this is a reversible path. 361 00:19:27 --> 00:19:32 If it was non-reversible, I would be allowed to put an 362 00:19:32 --> 00:19:35 initial point and a final point, but I wouldn't be 363 00:19:35 --> 00:19:37 allowed to put a path between them like this, 364 00:19:37 --> 00:19:39 connecting them together. 365 00:19:39 --> 00:19:42 Because if it's irreversible, it's very likely that I don't 366 00:19:42 --> 00:19:45 know what the pressure inside the system is doing while 367 00:19:45 --> 00:19:46 this is happening. 368 00:19:46 --> 00:19:47 It could be very chaotic. 369 00:19:47 --> 00:19:49 The pressure could not be, might not even be constant 370 00:19:49 --> 00:19:50 throughout the system. 371 00:19:50 --> 00:19:53 It could have eddies and all sorts of things. 372 00:19:53 --> 00:19:56 So for an irreversible process, I wouldn't really be allowed 373 00:19:56 --> 00:19:57 to put a path there. 374 00:19:57 --> 00:20:00 I can do that because it's reversible, and I can get 375 00:20:00 --> 00:20:03 a functional form out. 376 00:20:03 --> 00:20:09 What we're going to find in this process is that the 377 00:20:09 --> 00:20:12 functional form from this is that p is related to the volume 378 00:20:12 --> 00:20:17 through a constant, but instead of being V to the first power 379 00:20:17 --> 00:20:20 as it was for the isothermal, it's now a V to some 380 00:20:20 --> 00:20:22 constant gamma. 381 00:20:22 --> 00:20:31 What we're going to find is that gamma is greater than 1. 382 00:20:31 --> 00:20:39 And if gamma is greater than 1, and you're changing the 383 00:20:39 --> 00:20:42 pressure to the same final point, what we're going to find 384 00:20:42 --> 00:20:44 then is the volume that you go to is going to be a smaller 385 00:20:44 --> 00:20:47 volume than for the isothermal. 386 00:20:47 --> 00:20:48 We're going to go through these points, so don't 387 00:20:48 --> 00:20:53 worry about getting at all straight right now. 388 00:20:53 --> 00:20:54 All right, so let's do this. 389 00:20:54 --> 00:20:55 Let's try to show this. 390 00:20:55 --> 00:20:57 Let's try to show that in fact this is the right functional 391 00:20:57 --> 00:21:02 form, that pV to the gamma is equal to constant is the right 392 00:21:02 --> 00:21:05 form that describes the path in an adiabatic expansion 393 00:21:05 --> 00:21:08 or compression. 394 00:21:08 --> 00:21:09 So let's do the experiment. 395 00:21:09 --> 00:21:11 Let's write everything that we know down. 396 00:21:11 --> 00:21:15 This is where, the way that we should start every problem 397 00:21:15 --> 00:21:17 set or ever problem. 398 00:21:17 --> 00:21:23 It's writing everything we know and what it means. 399 00:21:23 --> 00:21:27 OK, so adiabatic. 400 00:21:27 --> 00:21:31 All the words mean things mathematically. 401 00:21:31 --> 00:21:32 Adiabatic means dq equals zero. 402 00:21:32 --> 00:21:38 Let's write that down. dq equals zero. 403 00:21:38 --> 00:21:38 Reversible. 404 00:21:38 --> 00:21:42 Reversible means that if I looked at dw it's 405 00:21:42 --> 00:21:46 minus p reversible. 406 00:21:46 --> 00:21:50 It's minus p external dV, that's what it is, but in 407 00:21:50 --> 00:21:56 the reversible process p external is equal to p. 408 00:21:56 --> 00:21:58 That's the only time you can set p external equal to 409 00:21:58 --> 00:22:01 p is if it's reversible. 410 00:22:01 --> 00:22:02 So this is the external pressure. 411 00:22:02 --> 00:22:05 This is the internal pressure of the system. 412 00:22:05 --> 00:22:09 You can only do this because it's reversible. 413 00:22:09 --> 00:22:10 What else do I know? 414 00:22:10 --> 00:22:15 It's an ideal gas. 415 00:22:15 --> 00:22:24 So I can write du is Cv dT and pV is equal to RT. 416 00:22:24 --> 00:22:27 417 00:22:27 --> 00:22:30 OK. 418 00:22:30 --> 00:22:32 So what else can I write? 419 00:22:32 --> 00:22:42 Well from the first law, du is equal to dq plus dw, and I 420 00:22:42 --> 00:22:44 wrote down everything I knew at the beginning here. 421 00:22:44 --> 00:22:47 I wrote dq equals zero I wrote what dw was. 422 00:22:47 --> 00:22:51 So du is just minus p dV. 423 00:22:51 --> 00:22:53 Now from the ideal gas, I have another expression 424 00:22:53 --> 00:22:55 for du, Cv dT. 425 00:22:55 --> 00:22:57 So now I have two expressions for du. 426 00:22:57 --> 00:23:00 That's a good thing, because I can them equal to each other. 427 00:23:00 --> 00:23:03 So if you get these two guys together you get 428 00:23:03 --> 00:23:11 Cv dT is minus p dV. 429 00:23:11 --> 00:23:14 So we're connecting temperature and changes in temperature and 430 00:23:14 --> 00:23:21 changes in volume together. 431 00:23:21 --> 00:23:24 OK, now we don't really want this p here. 432 00:23:24 --> 00:23:27 This is going to be a function, I only need two variables, and 433 00:23:27 --> 00:23:28 here I've got three variables down. 434 00:23:28 --> 00:23:31 So I know p is going to be a function of T and V, and I know 435 00:23:31 --> 00:23:33 how to relate p to T and V because I know the 436 00:23:33 --> 00:23:35 ideal gas law. 437 00:23:35 --> 00:23:42 So instead of p, here I'm going to put nRT over 438 00:23:42 --> 00:23:47 V. or RT over V bar. 439 00:23:47 --> 00:23:56 So now, this expression becomes Cv dT over T is 440 00:23:56 --> 00:24:03 equal to minus R dV over V. 441 00:24:03 --> 00:24:05 And I also did a little bit of rearrangement. 442 00:24:05 --> 00:24:06 Yes? 443 00:24:06 --> 00:24:22 STUDENT: [INAUDIBLE] 444 00:24:22 --> 00:24:25 PROFESSOR BAWENDI: Well, let's see. du is, for an ideal gas, 445 00:24:25 --> 00:24:30 it's always equal to Cv dT. 446 00:24:30 --> 00:24:33 Constant volume process tells you that if you have a constant 447 00:24:33 --> 00:24:37 volume, then dq is Cv dT. 448 00:24:37 --> 00:24:38 This is what the constant volume path tells you, that 449 00:24:38 --> 00:24:40 you can equate Cv dT with dq. 450 00:24:40 --> 00:24:45 The ideal gas tells you that this is always true here. 451 00:24:45 --> 00:24:49 It's an ideal gas in adiabatic expansion dq is equal to zero. 452 00:24:49 --> 00:24:54 Adiabatic expansion means you can't use this. dq is zero. 453 00:24:54 --> 00:24:55 The that is adiabatic. 454 00:24:55 --> 00:24:57 It's not constant volume. 455 00:24:57 --> 00:25:01 You're allowed Cv comes out here for this adiabatic 456 00:25:01 --> 00:25:04 expansion, which is not a constant volume only because 457 00:25:04 --> 00:25:06 this is always true for an ideal gas. 458 00:25:06 --> 00:25:09 We didn't use a constraint that V is constant. 459 00:25:09 --> 00:25:16 Just use this as an ideal gas here. 460 00:25:16 --> 00:25:20 OK, so we've got dT here dV here. 461 00:25:20 --> 00:25:22 We want to get rid of derivatives. 462 00:25:22 --> 00:25:23 We want to integrate. 463 00:25:23 --> 00:25:27 So let's take the integral of both sides, going from the 464 00:25:27 --> 00:25:29 initial point to the final point. 465 00:25:29 --> 00:25:35 We've got Cv integral from T1 to T2, dT over T is equal 466 00:25:35 --> 00:25:43 to minus R from V1 to V2 dV over V. 467 00:25:43 --> 00:25:50 OK, we do take that integral, and that leads us to Cv log of 468 00:25:50 --> 00:25:57 T2 over T1 is equal to minus R log of V2 over V1, and now we 469 00:25:57 --> 00:26:00 know how to rearrange this, because the Cv times a log of 470 00:26:00 --> 00:26:04 T2 over T1 is the same thing as the log of T2 over T1 471 00:26:04 --> 00:26:06 to the Cv power, right. 472 00:26:06 --> 00:26:12 So by rearranging these expression of the logs here, 473 00:26:12 --> 00:26:17 this is the same thing is writing log T2 over T1 to the 474 00:26:17 --> 00:26:25 Cv power is equal to log of V1 over V2 to the R power. 475 00:26:25 --> 00:26:27 And I've reversed the fraction here because of 476 00:26:27 --> 00:26:33 the negative sign here. 477 00:26:33 --> 00:26:43 OK, get rid of the logs, and you get that T2 over T1 is 478 00:26:43 --> 00:26:52 equal to V1 over V2 to the R over Cv by taking this Cv and 479 00:26:52 --> 00:26:56 bringing it on the other side here. 480 00:26:56 --> 00:27:01 OK, so now we've got the initial and final points, a 481 00:27:01 --> 00:27:04 relationship between the temperature and volume for the 482 00:27:04 --> 00:27:08 initial and final points. 483 00:27:08 --> 00:27:09 Eventually that's not what we want. 484 00:27:09 --> 00:27:12 Eventually we want the same relationship in the 485 00:27:12 --> 00:27:13 pressure and volume. 486 00:27:13 --> 00:27:18 We want a relationship in p-V space, not in T-V space. 487 00:27:18 --> 00:27:20 So we're going to have to do something about that. 488 00:27:20 --> 00:27:23 But first, it turns out that now we have this R over Cv. 489 00:27:23 --> 00:27:27 and remember, we have this relationship between R Cv and 490 00:27:27 --> 00:27:30 Cp here, which is going to be interesting. 491 00:27:30 --> 00:27:36 So let's get rid of R in this expression here. 492 00:27:36 --> 00:27:41 So R over Cv, R is Cp minus Cv. 493 00:27:41 --> 00:27:45 We proved that first thing this morning, divided by Cv so this 494 00:27:45 --> 00:27:52 is Cp over Cv minus 1 and because I know the answer of 495 00:27:52 --> 00:27:56 this whole thing here, I'm going to call this thing 496 00:27:56 --> 00:27:59 here, Cp over Cv. 497 00:27:59 --> 00:28:01 I'm going to call it gamma. 498 00:28:01 --> 00:28:02 Because I know the answer, right. 499 00:28:02 --> 00:28:02 I know the answer. 500 00:28:02 --> 00:28:06 I'm going to call this gamma. 501 00:28:06 --> 00:28:11 By definition I'm going to define gamma by Cp 502 00:28:11 --> 00:28:15 over Cv by definition. 503 00:28:15 --> 00:28:19 OK, so that means that this is really instead of R over Cv. 504 00:28:19 --> 00:28:22 it's really gamma minus one. 505 00:28:22 --> 00:28:29 So now I've got this gamma placed in there, gamma minus 506 00:28:29 --> 00:28:30 1, so I've ben feeling a little bit better. 507 00:28:30 --> 00:28:33 I've gotten V to the gamma power almost to the gamma 508 00:28:33 --> 00:28:34 power, it's minus 1 here. 509 00:28:34 --> 00:28:39 But it's almost right and I'm looking for something like this 510 00:28:39 --> 00:28:42 here, V to the gamma power pV to the gamma power is a 511 00:28:42 --> 00:28:45 constant to describe that line. 512 00:28:45 --> 00:28:47 So I just got to get rid of the temperature now. 513 00:28:47 --> 00:28:49 Somehow I've gotta get rid of the temperature and 514 00:28:49 --> 00:28:58 make it pressure instead. 515 00:28:58 --> 00:29:04 OK, so let's -- let me get rid of this here, let's use the 516 00:29:04 --> 00:29:08 ideal gas law to get rid of the temperature. 517 00:29:08 --> 00:29:14 So we have T is pV over R. 518 00:29:14 --> 00:29:20 So now if I look at my V1 over V2 to the gamma minus 519 00:29:20 --> 00:29:24 1, that's T2 over T1. 520 00:29:24 --> 00:29:27 I'll replace the ideal gas law for those two T's. 521 00:29:27 --> 00:29:31 That p2 V2 over R, and then I have p1 V1 over 522 00:29:31 --> 00:29:35 R, the R's cancel out. 523 00:29:35 --> 00:29:39 Well let's see, there's a V1 over V2 to the gamma minus 1. 524 00:29:39 --> 00:29:42 There's a V1 over V2 here or V2 over V1. 525 00:29:42 --> 00:29:45 That's going to get rid of this minus 1 here. 526 00:29:45 --> 00:29:49 Then I'm going to get, be left with a p2 over p1. 527 00:29:49 --> 00:29:54 So this guy here gets rid of this, and I have my answer as 528 00:29:54 --> 00:30:01 V1 over V2 to the gamma power is equal to p2 over p1. 529 00:30:01 --> 00:30:06 Or I can re-write this as p1, V1 to the gamma is equal 530 00:30:06 --> 00:30:11 to p2 V2 to the gamma. 531 00:30:11 --> 00:30:18 That means that p1 and -- where I am, the point number 1 532 00:30:18 --> 00:30:22 and point number 2 were completely arbitrary. 533 00:30:22 --> 00:30:24 They happen to be on the path. 534 00:30:24 --> 00:30:29 So any point I pick on that path will be equal to 535 00:30:29 --> 00:30:30 p1, V1 to the gamma. 536 00:30:30 --> 00:30:34 So if I take p times V to the gamma, anywhere on the path, 537 00:30:34 --> 00:30:36 it's going to be equal to the same relation from 538 00:30:36 --> 00:30:38 my first point. 539 00:30:38 --> 00:30:40 This is going to be true for any point on the path. 540 00:30:40 --> 00:30:44 As long as I'm on the path, pV to the gamma will be whatever 541 00:30:44 --> 00:30:46 it was for the first point, which is going to be 542 00:30:46 --> 00:30:52 some sort of constant. 543 00:30:52 --> 00:30:53 All right, so I've proven my point. 544 00:30:53 --> 00:30:55 I've proven what I was trying to do, which was to show that 545 00:30:55 --> 00:31:03 on this adiabat, everywhere on the adiabat, if you take a 546 00:31:03 --> 00:31:06 functional form that relates p and V together, it's going to 547 00:31:06 --> 00:31:08 have this relationship with gamma as related 548 00:31:08 --> 00:31:10 to the heat capacities. 549 00:31:10 --> 00:31:18 Gamma is Cp over Cv. 550 00:31:18 --> 00:31:26 OK, now there's more we can say because we know that's Cp is 551 00:31:26 --> 00:31:28 related to Cv through this R here, and R is a 552 00:31:28 --> 00:31:30 positive number. 553 00:31:30 --> 00:31:34 So Cp is always bigger than Cv. 554 00:31:34 --> 00:31:42 In fact, if it's an ideal gas Cv and Cp are well defined 555 00:31:42 --> 00:31:44 numbers, it turns out. 556 00:31:44 --> 00:31:56 If you have a monotomic ideal gas, OK, so an atomic version, 557 00:31:56 --> 00:32:02 not a molecular ideal gas, then it turns out that Cv is equal 558 00:32:02 --> 00:32:09 to 3/2 R, and therefore, Cp which is just R plus this is 559 00:32:09 --> 00:32:13 five halves R, which means that gamma for a mono atomic 560 00:32:13 --> 00:32:17 ideal gas is 5/3. 561 00:32:17 --> 00:32:22 So if you have a gas of argon atoms, you know what gamma is. 562 00:32:22 --> 00:32:26 This is something that you're going to prove in statistical 563 00:32:26 --> 00:32:28 mechanics, and so we're not going to worry about 564 00:32:28 --> 00:32:29 where this comes from. 565 00:32:29 --> 00:32:33 We're just going to take it for granted. 566 00:32:33 --> 00:32:37 All right, so gamma is for ideal gas, is bigger than one. 567 00:32:37 --> 00:32:40 In fact we have a number for it if it's an atomic ideal gas. 568 00:32:40 --> 00:32:47 So what it means then is if I look at if this relationship 569 00:32:47 --> 00:32:51 here, gamma is bigger than 1, so gamma something 570 00:32:51 --> 00:32:52 bigger than 1 minus 1. 571 00:32:52 --> 00:32:56 This is, so this is positive here. 572 00:32:56 --> 00:32:57 It's positive. 573 00:32:57 --> 00:33:05 So if I have an expansion where V2 is greater than V1, so V2 is 574 00:33:05 --> 00:33:17 greater than V1, so this is a number which is less than one, 575 00:33:17 --> 00:33:24 then I expect then T2 is going to be less and 576 00:33:24 --> 00:33:29 T1, T2 is less than T1. 577 00:33:29 --> 00:33:37 OK, I have this number here to a power, which is positive, 578 00:33:37 --> 00:33:39 a positive power. 579 00:33:39 --> 00:33:40 And its number is less than one. 580 00:33:40 --> 00:33:41 This is going to give me something which 581 00:33:41 --> 00:33:43 is less than one. 582 00:33:43 --> 00:33:49 So T2 is less than T1, and I have a cooling of a gas. 583 00:33:49 --> 00:33:54 So whenever you have an adiabatic expansion, the 584 00:33:54 --> 00:33:56 temperature cools, it gets colder. 585 00:33:56 --> 00:34:06 It gets colder, and if I look at my graph here, then if the 586 00:34:06 --> 00:34:12 volume of the expansion, if the temperature of the adiabatic 587 00:34:12 --> 00:34:18 expansion was colder, that means that -- this is 588 00:34:18 --> 00:34:23 the wrong graph here. 589 00:34:23 --> 00:34:27 If I want to compare it with the isothermal expansion, 590 00:34:27 --> 00:34:32 which is sitting here. 591 00:34:32 --> 00:34:41 All right, so gamma, the gas is cooling so V2 is going to be 592 00:34:41 --> 00:34:46 less than it what would be if the temperature kept constant. 593 00:34:46 --> 00:34:48 So finally, on the same graph, I put down what an isothermal 594 00:34:48 --> 00:34:55 expansion would be, if my final pressure is the same pressure, 595 00:34:55 --> 00:35:04 my volume for an isothermal expansion will be bigger than 596 00:35:04 --> 00:35:08 for an adiabatic expansion. 597 00:35:08 --> 00:35:10 So when I expand this gas adiabatically and it cools 598 00:35:10 --> 00:35:16 down, why do you think it might cool down? 599 00:35:16 --> 00:35:19 It cools down because I'm doing work through the environment, 600 00:35:19 --> 00:35:23 but the energy has to go somewhere, right? 601 00:35:23 --> 00:35:26 There's an interplay between the energy inside the gas which 602 00:35:26 --> 00:35:29 is the heat energy which is allowing me to do all that work 603 00:35:29 --> 00:35:34 to be outside, and so I'm using up some of the energy that's 604 00:35:34 --> 00:35:37 inside the gas to do the work on the outside. 605 00:35:37 --> 00:35:42 And here, in an isothermal expansion, The bath is putting 606 00:35:42 --> 00:35:50 back the energy that the gas is expanding or using to expand, 607 00:35:50 --> 00:35:53 and so the energy is flowing back into the gas through 608 00:35:53 --> 00:35:59 the environment in the isothermal expansion. 609 00:35:59 --> 00:36:02 In the opposite case, if you have a compression, then it's 610 00:36:02 --> 00:36:03 the opposite of expansion. 611 00:36:03 --> 00:36:08 Compression you're expected to heat up, right? 612 00:36:08 --> 00:36:12 So in the compression, you're going to have 613 00:36:12 --> 00:36:13 V2 is less than V1. 614 00:36:13 --> 00:36:19 So that T2 is going to be greater than T1. 615 00:36:19 --> 00:36:23 That's going to be your heating up of the gas. 616 00:36:23 --> 00:36:25 And that really is the bicycle pump. 617 00:36:25 --> 00:36:29 That example I gave last time with the bicycle pump was not 618 00:36:29 --> 00:36:30 quite the right example. 619 00:36:30 --> 00:36:32 But this time this really is the bicycle pump. 620 00:36:32 --> 00:36:34 That you're pushing really hard on it. 621 00:36:34 --> 00:36:37 So you start with the bicycle pump at one 622 00:36:37 --> 00:36:39 bar, let's say, right? 623 00:36:39 --> 00:36:40 And you compress the air in the bicycle pump. 624 00:36:40 --> 00:36:48 You do it so quickly that the heat flow between the inside of 625 00:36:48 --> 00:36:50 the bicycle pump and the outside is too slow compared to 626 00:36:50 --> 00:36:53 the speed at which you compress. 627 00:36:53 --> 00:36:55 But you don't compress it so quickly that you're not in 628 00:36:55 --> 00:37:01 there reversible process so you compress that if you were to 629 00:37:01 --> 00:37:03 feel the temperature of the air and you can feel it through the 630 00:37:03 --> 00:37:07 nozzle gives you an idea of the temperature of the air 631 00:37:07 --> 00:37:08 inside your bicycle pump. 632 00:37:08 --> 00:37:09 You'll feel that it's warmer. 633 00:37:09 --> 00:37:13 You've just done an adiabatic compression of the ideal gas, 634 00:37:13 --> 00:37:15 you can pretend there is an ideal gas. 635 00:37:15 --> 00:37:19 And that gives rise to the heating that you 636 00:37:19 --> 00:37:22 can actually measure. 637 00:37:22 --> 00:37:36 OK, any questions on this part here. 638 00:37:36 --> 00:37:41 All right, so we've just gone through the reversible 639 00:37:41 --> 00:37:46 adiabatic, and I'm not going to do this in detail, but I want 640 00:37:46 --> 00:37:52 to go through the irreversible adiabatic fairly quickly, 641 00:37:52 --> 00:37:58 and see where the difference is here. 642 00:37:58 --> 00:38:01 OK, so now we're going to do the same kind of experiment, 643 00:38:01 --> 00:38:02 but irreversibly. 644 00:38:02 --> 00:38:09 An irreversible adiabatic. 645 00:38:09 --> 00:38:17 OK, so the experiment is going to be to take my cylinder now, 646 00:38:17 --> 00:38:21 and the external pressure and the internal pressure are not 647 00:38:21 --> 00:38:24 going to be in equilibrium with each other during the process. 648 00:38:24 --> 00:38:25 I'm going to start with my cylinder here, I'm going 649 00:38:25 --> 00:38:29 to put little pegs here so it doesn't fly up. 650 00:38:29 --> 00:38:30 There's going to be some external pressure. 651 00:38:30 --> 00:38:32 I'm going to set that equal to p2. 652 00:38:32 --> 00:38:34 There is going to be an internal pressure where 653 00:38:34 --> 00:38:40 p1 is less than p2 and there's V1 and T1 here. 654 00:38:40 --> 00:38:44 The whole thing is going to be insulated. 655 00:38:44 --> 00:38:48 Then I'm going to release these little pegs here and my piston 656 00:38:48 --> 00:38:52 is going to shoot up now because p2 is less than p1. 657 00:38:52 --> 00:38:57 So p2 is less than p1, the external pressures is less 658 00:38:57 --> 00:38:58 than the internal pressure. 659 00:38:58 --> 00:39:01 So it's going to shoot up until the internal pressure and 660 00:39:01 --> 00:39:03 the external pressure are in equilibrium. 661 00:39:03 --> 00:39:13 So they're both p2. external is p2 and I have p2, V2, 662 00:39:13 --> 00:39:17 T2, on the other side. 663 00:39:17 --> 00:39:28 OK, so in my diagram now, I have p1, V1 for my gas. p2, V2 664 00:39:28 --> 00:39:32 for my gas, so I know that I'm starting here and I know that 665 00:39:32 --> 00:39:35 I'm ending here, but I can't connect this path here. 666 00:39:35 --> 00:39:38 I don't know how to do that. 667 00:39:38 --> 00:39:41 It's an irreversible expansion. 668 00:39:41 --> 00:39:43 I don't know what the pressure is doing in there, 669 00:39:43 --> 00:39:45 doing that expansion. 670 00:39:45 --> 00:39:46 It could be quite chaotic. 671 00:39:46 --> 00:39:51 It could be non-spacially constant. 672 00:39:51 --> 00:39:56 OK, so but you still know a number of things. 673 00:39:56 --> 00:40:00 We're now going to be able to write you know dw is minus p 674 00:40:00 --> 00:40:02 dV, but we're still going to be able to write a 675 00:40:02 --> 00:40:03 bunch of things. 676 00:40:03 --> 00:40:05 We're still going to be able to write that it's an adiabat 677 00:40:05 --> 00:40:11 so that dq equals zero. 678 00:40:11 --> 00:40:18 We're still going to be able to write dw instead of pV where p 679 00:40:18 --> 00:40:20 is the internal pressure. dw now is actually much easier. 680 00:40:20 --> 00:40:22 It's minus p external. 681 00:40:22 --> 00:40:24 Well p external is actually p2, so that makes it actually 682 00:40:24 --> 00:40:29 easier. dV we're still going to write that it's an ideal gas. 683 00:40:29 --> 00:40:35 So du is still going to be equal to Cv dT, and we're still 684 00:40:35 --> 00:40:37 going to be able to use the first law, all these things 685 00:40:37 --> 00:40:40 don't matter where the path is. 686 00:40:40 --> 00:40:44 Still know that du is dq plus dw. 687 00:40:44 --> 00:40:48 dq is zero. dw is minus p2 dV. 688 00:40:48 --> 00:40:51 So this is what we know, just like what we wrote there. 689 00:40:51 --> 00:40:52 We write what we know. 690 00:40:52 --> 00:40:54 I'm not going to turn the crank here. 691 00:40:54 --> 00:40:56 I'm going to give you the answer. 692 00:40:56 --> 00:40:59 OK, you use the ideal gas law, etc., then you get a 693 00:40:59 --> 00:41:05 relationship that connects the pressure and the temperature, 694 00:41:05 --> 00:41:08 like here we got a relationship that connected the temperatures 695 00:41:08 --> 00:41:10 and the volumes together. 696 00:41:10 --> 00:41:15 You can get a relationship that connects the temperature so 697 00:41:15 --> 00:41:26 this is T2 times Cv plus R is equal to T1 on Cv plus 698 00:41:26 --> 00:41:31 p2 over p1 times R. 699 00:41:31 --> 00:41:34 This is going to be the connect, what connects the 700 00:41:34 --> 00:41:36 pressures and the temperature. 701 00:41:36 --> 00:41:41 And p2 is less than p1, so this number right 702 00:41:41 --> 00:41:45 here is less than 1. 703 00:41:45 --> 00:41:50 So and Cv plus R, Cv plus something less than R, so T1 704 00:41:50 --> 00:41:52 better be bigger than T2. 705 00:41:52 --> 00:41:57 OK, so T2 is less than T1. 706 00:41:57 --> 00:41:59 Same qualitative result as before. 707 00:41:59 --> 00:42:01 You have an expansion. 708 00:42:01 --> 00:42:03 It's and adiabatic expansion. 709 00:42:03 --> 00:42:05 You get a cooling of the gas. 710 00:42:05 --> 00:42:07 OK? 711 00:42:07 --> 00:42:13 Now here's a question for you guys, which we're going 712 00:42:13 --> 00:42:18 to vote on, so you better start thinking about it. 713 00:42:18 --> 00:42:26 Is the temperature T2 in this process smaller or larger than 714 00:42:26 --> 00:42:31 if I were to do the process reversibly with the same 715 00:42:31 --> 00:42:35 endpoint pressure. 716 00:42:35 --> 00:42:38 So now, instead of having p external equals to p2 here, 717 00:42:38 --> 00:42:42 I put p1 and I let this whole thing go reversibly. 718 00:42:42 --> 00:42:49 And I compare T2 irreversible to T2 reversible. 719 00:42:49 --> 00:42:51 They're both less than T1. 720 00:42:51 --> 00:42:52 They're supposed, there's a cooling happening 721 00:42:52 --> 00:42:54 in both cases. 722 00:42:54 --> 00:42:57 One is going to be colder than the other, maybe. 723 00:42:57 --> 00:42:58 Maybe they're going to be the same. 724 00:42:58 --> 00:42:58 I don't know. 725 00:42:58 --> 00:43:01 I'm going to let you try to figure that out qualitatively 726 00:43:01 --> 00:43:03 without doing any math. 727 00:43:03 --> 00:43:05 See if you can figure it out. 728 00:43:05 --> 00:43:15 OK, so which is colder? 729 00:43:15 --> 00:43:17 I'm going to let you think about it for about 730 00:43:17 --> 00:43:35 thirty seconds or so. 731 00:43:35 --> 00:43:57 You can talk to each other if you don't, can't figure it out. 732 00:43:57 --> 00:43:58 All right, let's take a vote. 733 00:43:58 --> 00:44:01 How many people say that T2 irreversible is colder 734 00:44:01 --> 00:44:04 than T2 reversible? 735 00:44:04 --> 00:44:14 One, two, don't be shy, three four, anybody else, five. 736 00:44:14 --> 00:44:15 OK, I've got five votes here. 737 00:44:15 --> 00:44:15 Professor Nelson? 738 00:44:15 --> 00:44:19 PROFESSOR NELSON: Oh, I'm abstaining. 739 00:44:19 --> 00:44:22 PROFESSOR BAWENDI: You're abstaining, OK, good choice. 740 00:44:22 --> 00:44:27 OK, what about T1 cooler than T2? 741 00:44:27 --> 00:44:29 OK, I've got a lot of votes here. 742 00:44:29 --> 00:44:32 So the majority of the people, so this is 743 00:44:32 --> 00:44:34 much bigger than five. 744 00:44:34 --> 00:44:35 OK. 745 00:44:35 --> 00:44:37 So remember last time the majority was wrong, right? 746 00:44:37 --> 00:44:39 The majority was wrong last time. 747 00:44:39 --> 00:44:41 That doesn't mean that the majority is wrong here. 748 00:44:41 --> 00:44:43 It could be right. 749 00:44:43 --> 00:44:45 Let me give you another piece of information here 750 00:44:45 --> 00:44:46 that you already know. 751 00:44:46 --> 00:44:54 Minus w irreversible, this is the work which is done to the 752 00:44:54 --> 00:44:59 environment by the system, minus w irreversible is always 753 00:44:59 --> 00:45:01 smaller than minus w reversible. 754 00:45:01 --> 00:45:04 You learned that a little while ago. 755 00:45:04 --> 00:45:08 OK, so this is a little hint. 756 00:45:08 --> 00:45:09 Let's vote again. 757 00:45:09 --> 00:45:12 Let's think about it for ten seconds why this could be 758 00:45:12 --> 00:45:19 interesting and relevant, and I want to ask if anybody 759 00:45:19 --> 00:45:22 changed their mind. 760 00:45:22 --> 00:45:24 Anybody change their mind, change their vote? 761 00:45:24 --> 00:45:27 Yes You want to change your vote? 762 00:45:27 --> 00:45:29 To which one? 763 00:45:29 --> 00:45:33 STUDENT: From T2 reversible is greater than T2 irreversible, 764 00:45:33 --> 00:45:36 saying that T2 reversible is [UNINTELLIGIBLE]. 765 00:45:36 --> 00:45:37 PROFESSOR BAWENDI: So the question then, which 766 00:45:37 --> 00:45:40 one is colder? 767 00:45:40 --> 00:45:42 STUDENT: T2 reversible should be colder. 768 00:45:42 --> 00:45:44 PROFESSOR BAWENDI: T2 irreversible should be colder? 769 00:45:44 --> 00:45:47 STUDENT: Yes. 770 00:45:47 --> 00:45:49 PROFESSOR BAWENDI: All right. 771 00:45:49 --> 00:45:50 OK. 772 00:45:50 --> 00:45:52 I'm going to keep you there then, the majority. 773 00:45:52 --> 00:45:52 Yes? 774 00:45:52 --> 00:45:54 STUDENT: I'm switching to [INAUDIBLE]. 775 00:45:54 --> 00:45:56 PROFESSOR BAWENDI: You're switching this way here? 776 00:45:56 --> 00:46:00 So now we have six here. 777 00:46:00 --> 00:46:04 Anybody else want to switch? 778 00:46:04 --> 00:46:08 All right, let's take the example of, the extreme 779 00:46:08 --> 00:46:12 example, let's go to the extreme example where p 780 00:46:12 --> 00:46:17 external is really small. 781 00:46:17 --> 00:46:22 And I have this adiabatic expansion where p external 782 00:46:22 --> 00:46:26 is really small. it's kind of like the Joule 783 00:46:26 --> 00:46:30 expansion, an ideal gas. 784 00:46:30 --> 00:46:32 What happened to the temperature in a Joule 785 00:46:32 --> 00:46:33 expansion in ideal gas? 786 00:46:33 --> 00:46:37 Anybody remember? 787 00:46:37 --> 00:46:39 Anybody remember? 788 00:46:39 --> 00:46:45 Joule expansion, this is where we proved that delta u was only 789 00:46:45 --> 00:46:46 dependent on temperature. 790 00:46:46 --> 00:46:51 Bob Field taught that lecture. 791 00:46:51 --> 00:46:52 Nobody remembers? 792 00:46:52 --> 00:46:56 Well, all right, delta u is equal to zero for that. 793 00:46:56 --> 00:46:59 What happened to the temperature in that expansion? 794 00:46:59 --> 00:47:04 It's an adiabatic expansion. 795 00:47:04 --> 00:47:04 Well, we'll revisit that. 796 00:47:04 --> 00:47:07 Let me ask you another question here. 797 00:47:07 --> 00:47:11 So w, the work is less for the irreversible process than 798 00:47:11 --> 00:47:13 the reversible process. 799 00:47:13 --> 00:47:14 There's less work done to the outside. 800 00:47:14 --> 00:47:19 So there's less energy expanded by the system. 801 00:47:19 --> 00:47:22 The energy expanded by the system is smaller for the 802 00:47:22 --> 00:47:23 irreversible process. 803 00:47:23 --> 00:47:25 It's not doing as much work. 804 00:47:25 --> 00:47:27 It's not pressing against as much pressure. 805 00:47:27 --> 00:47:31 Right? p external equals p2 is less than p external 806 00:47:31 --> 00:47:34 equals p internal. 807 00:47:34 --> 00:47:36 So it doesn't have to do as much work. 808 00:47:36 --> 00:47:39 It doesn't have to expend as much energy, therefore, the 809 00:47:39 --> 00:47:40 majority in this case is right. 810 00:47:40 --> 00:47:43 The temperature is not going to cool as much for the 811 00:47:43 --> 00:47:44 irreversible process. 812 00:47:44 --> 00:47:46 The reversible process is going to be colder because it 813 00:47:46 --> 00:47:48 has to expend more energy. 814 00:47:48 --> 00:47:50 It's going to have a higher pressure. 815 00:47:50 --> 00:47:53 The pressure is going to decrease along the way, 816 00:47:53 --> 00:47:56 but it's going to have to do more work because 817 00:47:56 --> 00:47:57 of this right here. 818 00:47:57 --> 00:47:58 So the majority in this case is right. 819 00:47:58 --> 00:48:01 Now the Joule expansion, the temperature doesn't change, T 820 00:48:01 --> 00:48:08 equals zero. eta sub J, the Joule coefficient for an ideal 821 00:48:08 --> 00:48:17 gas, is equal to zero and eta sub J was dT/dV under 822 00:48:17 --> 00:48:21 constant energy. 823 00:48:21 --> 00:48:23 All right, any questions? 824 00:48:23 --> 00:48:23 Yes? 825 00:48:23 --> 00:48:29 STUDENT: [INAUDIBLE]. 826 00:48:29 --> 00:48:32 PROFESSOR BAWENDI: The only -- a priori, that's true. 827 00:48:32 --> 00:48:36 But it's hard to imagine an irreversible compression the 828 00:48:36 --> 00:48:44 way we can just imagine an irreversible expansion. 829 00:48:44 --> 00:48:47 You have to do it awfully fast to get your irreversible 830 00:48:47 --> 00:48:47 compression. 831 00:48:47 --> 00:48:49 I don't know how to do it really. 832 00:48:49 --> 00:48:54 I think it's, but it's more complicated, but you're 833 00:48:54 --> 00:48:57 basically right, basically right. 834 00:48:57 --> 00:49:00 That's why we always talk about irreversible expansion, because 835 00:49:00 --> 00:49:02 it's easy to write down an irreversible expansion. 836 00:49:02 --> 00:49:04 A compression is a little bit more complicated. 837 00:49:04 --> 00:49:08 Any other questions? 838 00:49:08 --> 00:49:10 OK, great. 839 00:49:10 --> 00:49:15 Monday Professor Nelson who is sitting here will be taking 840 00:49:15 --> 00:49:19 over for a few weeks and then you'll see me again after that. 841 00:49:19 --> 00:49:21