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 --> 00:00:06 Your support will help MIT OpenCourseWare continue to 5 00:00:06 --> 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 --> 00:00:17 hundreds of MIT courses visit MIT OpenCourseWare 8 00:00:17 --> 00:00:20 at ocw.mit.edu. 9 00:00:20 --> 00:00:22 PROFESSOR NELSON: Over the last couple lectures through our 10 00:00:22 --> 00:00:27 consideration of a number of somewhat difficult topics, 11 00:00:27 --> 00:00:31 including these Carnot cycles and specific Carnot engine to 12 00:00:31 --> 00:00:36 represent them, we've tried to define and understand a little 13 00:00:36 --> 00:00:44 bit about this very special function, this entropy. 14 00:00:44 --> 00:00:45 Where we've got our dS. 15 00:00:45 --> 00:00:53 It's dq reversible over T. 16 00:00:53 --> 00:00:59 And our treatment of reversible and irreversible cyclic 17 00:00:59 --> 00:01:07 processes, what we saw is that if we go around a cycle and 18 00:01:07 --> 00:01:18 look at dq reversible over T, then this is zero. 19 00:01:18 --> 00:01:20 Now go all the way around a cycle, of course, this is 20 00:01:20 --> 00:01:26 just delta S for the cycle. 21 00:01:26 --> 00:01:28 And that's consistent with the idea that entropy 22 00:01:28 --> 00:01:29 is a state functions. 23 00:01:29 --> 00:01:31 So if we go around cycle the cycle, we have the same ending 24 00:01:31 --> 00:01:35 point, the starting point, same state, then the change in the 25 00:01:35 --> 00:01:39 state function has to be zero. 26 00:01:39 --> 00:01:48 But we saw that for an irreversible path around a 27 00:01:48 --> 00:01:57 cycle, then we have that, this gives something less than zero. 28 00:01:57 --> 00:02:01 And so that was expressed in Clausius inequality that 29 00:02:01 --> 00:02:08 includes both these cases. 30 00:02:08 --> 00:02:10 And now I just want to go through a number of 31 00:02:10 --> 00:02:12 calculations of entropy. 32 00:02:12 --> 00:02:16 How entropy changes for a number processes, both to just 33 00:02:16 --> 00:02:19 learn more about it in general terms and also just to see how 34 00:02:19 --> 00:02:23 it's calculated and what sort of changes in undergoes for 35 00:02:23 --> 00:02:25 different sorts of changes in state. 36 00:02:25 --> 00:02:31 So the first very general example that I want to 37 00:02:31 --> 00:02:33 show is the following. 38 00:02:33 --> 00:02:46 Let's say we've got an isolated system, and it undergoes 39 00:02:46 --> 00:02:57 some irreversible change. 40 00:02:57 --> 00:03:08 So we'll start at one, and we'll go irreversibly 41 00:03:08 --> 00:03:19 to some new state, two. 42 00:03:19 --> 00:03:28 OK, now we can always return the system back to the 43 00:03:28 --> 00:03:36 initial state reversibly. 44 00:03:36 --> 00:03:40 If we do that, then we can't keep the system isolated. 45 00:03:40 --> 00:03:45 So it won't be isolated anymore. 46 00:03:45 --> 00:03:48 In other words, in order to arrange things so that you 47 00:03:48 --> 00:03:51 could have a return along a reversible path, it 48 00:03:51 --> 00:03:53 can't be isolated. 49 00:03:53 --> 00:03:56 In some sense you know that, because when it was isolated 50 00:03:56 --> 00:03:59 it spontaneously irreversibly went from one to two. 51 00:03:59 --> 00:04:03 And example would be ice melting at some temperature 52 00:04:03 --> 00:04:06 that might be higher than the freezing point. 53 00:04:06 --> 00:04:09 So you've got solid ice, but the temperature is above 54 00:04:09 --> 00:04:12 zero degrees Celsius, so irreversibly it'll melt. 55 00:04:12 --> 00:04:14 Systems isolated, it'll melt. 56 00:04:14 --> 00:04:17 Then you could put the system in contact with a cold 57 00:04:17 --> 00:04:23 reservoir reversibly re-freeze the water to form ice again. 58 00:04:23 --> 00:04:26 But of course to do that you couldn't keep it isolated. 59 00:04:26 --> 00:04:30 You need to have it in contact with a low temperature bath. 60 00:04:30 --> 00:04:34 But again, this is just completely general 61 00:04:34 --> 00:04:35 at this point. 62 00:04:35 --> 00:04:37 It's some irreversible process returned then 63 00:04:37 --> 00:04:40 along a reversible path. 64 00:04:40 --> 00:04:43 So let's just look at what happens. 65 00:04:43 --> 00:04:48 So here's our path B. 66 00:04:48 --> 00:04:56 Along path A, well it's an isolated system. 67 00:04:56 --> 00:05:01 So the heat exchanged is zero. 68 00:05:01 --> 00:05:10 And then, for the cycle based on the Clausius inequality 69 00:05:10 --> 00:05:18 expressed here, the integral around the cycle of dq over 70 00:05:18 --> 00:05:24 T in general, it's less than or equal to zero. 71 00:05:24 --> 00:05:28 If there was an irreversible process involved, then in fact, 72 00:05:28 --> 00:05:31 it'll be less than zero. 73 00:05:31 --> 00:05:34 I'll write it here in the more general case, but we know what 74 00:05:34 --> 00:05:40 it's really going to be with the irreversible step. 75 00:05:40 --> 00:05:49 Well, so we can write this as the integral of dq irreversible 76 00:05:49 --> 00:05:58 over T, going from state one to state two. 77 00:05:58 --> 00:06:04 Oh, let me rewrite this down here. 78 00:06:04 --> 00:06:12 So around the cycle, dq over T, it's integral from state one to 79 00:06:12 --> 00:06:20 two of dq irreversible over T, plus the integral from two back 80 00:06:20 --> 00:06:27 to one of dq reversible over T. 81 00:06:27 --> 00:06:30 And of course, we know that that's less than 82 00:06:30 --> 00:06:32 or equal to zero. 83 00:06:32 --> 00:06:36 But this, we know is zero, right, because this 84 00:06:36 --> 00:06:38 irreversible step is taken in isolation. 85 00:06:38 --> 00:06:40 The system is isolated. 86 00:06:40 --> 00:06:42 We've already seen that there's no heat crossing 87 00:06:42 --> 00:06:45 the boundary of the system. 88 00:06:45 --> 00:06:48 So we just have this. 89 00:06:48 --> 00:06:58 So of course this dq reversible over T from two to one, that's, 90 00:06:58 --> 00:07:01 what now we've a reversible path, so that's just the 91 00:07:01 --> 00:07:04 entropy at the delta S going from two to one. 92 00:07:04 --> 00:07:11 It's S1 minus S2 it's delta S, we can write it as delta S 93 00:07:11 --> 00:07:15 backwards, right, we're going back along that 94 00:07:15 --> 00:07:17 reversible path. 95 00:07:17 --> 00:07:23 Which is of course negative delta S forward. 96 00:07:23 --> 00:07:29 Because around the whole cycle delta s has to be zero. 97 00:07:29 --> 00:07:43 So what that says is delta S forward is greater 98 00:07:43 --> 00:07:50 than or equal to zero. 99 00:07:50 --> 00:07:52 That's interesting. 100 00:07:52 --> 00:07:54 What that's saying is -- remember, we didn't 101 00:07:54 --> 00:07:55 specify the process. 102 00:07:55 --> 00:08:00 It's just for any completely general irreversible process. 103 00:08:00 --> 00:08:06 For any such process, for any spontaneous process, this 104 00:08:06 --> 00:08:08 is going to be the case. 105 00:08:08 --> 00:08:13 So in other words, for an isolated system, delta S 106 00:08:13 --> 00:08:18 tells us the direction of spontaneous change. 107 00:08:18 --> 00:08:29 In particular, for the isolated system, delta S is greater 108 00:08:29 --> 00:08:40 than zero for something that's irreversible. 109 00:08:40 --> 00:08:51 Delta S equals zero for something that's reversible, 110 00:08:51 --> 00:09:02 and delta S is never less than zero. 111 00:09:02 --> 00:09:04 In a sense, this is a direct consequence of 112 00:09:04 --> 00:09:11 the Clausius inequality. 113 00:09:11 --> 00:09:17 That's a pretty useful result. 114 00:09:17 --> 00:09:21 So remember when we started this whole discussion, I tried 115 00:09:21 --> 00:09:25 to emphasize that the first law of thermodynamics told us 116 00:09:25 --> 00:09:27 about conservation of energy. 117 00:09:27 --> 00:09:30 But it didn't tell us which way something would go 118 00:09:30 --> 00:09:34 spontaneously if you just left the system to its own devices 119 00:09:34 --> 00:09:36 under certain conditions. 120 00:09:36 --> 00:09:40 This is telling us that. 121 00:09:40 --> 00:09:43 We can tell which way the system will evolve in 122 00:09:43 --> 00:09:46 which direction it'll go. 123 00:09:46 --> 00:09:49 This is for an isolated system and we'll generalize this 124 00:09:49 --> 00:10:00 in subsequent lectures. 125 00:10:00 --> 00:10:08 Now, let's look at what happens if the system isn't isolated. 126 00:10:08 --> 00:10:25 So now let's go from one to two not isolated. 127 00:10:25 --> 00:10:32 So of course, delta S I'm going to specify for the system 128 00:10:32 --> 00:10:37 for reasons that'll soon be obvious is S2 for the system 129 00:10:37 --> 00:10:39 minus S1 for the system. 130 00:10:39 --> 00:10:42 Of course it doesn't depend on the path, even though to 131 00:10:42 --> 00:10:52 calculate it, we need to find a reversible path. 132 00:10:52 --> 00:10:55 Just to add emphasis to what we just did here, of course, 133 00:10:55 --> 00:10:57 it was the same thing. 134 00:10:57 --> 00:11:02 In order to calculate delta S along that irreversible path 135 00:11:02 --> 00:11:05 for that irreversible process we needed to construct 136 00:11:05 --> 00:11:09 some reversible path. 137 00:11:09 --> 00:11:13 And what we did is imagined the reversible path going backwards 138 00:11:13 --> 00:11:16 and then said, OK, fine then delta S for the irreversible 139 00:11:16 --> 00:11:18 process must be the opposite of delta S that we calculated 140 00:11:18 --> 00:11:21 along the reversible bath going backwards. 141 00:11:21 --> 00:11:24 The point is that to calculate delta S in either direction, 142 00:11:24 --> 00:11:27 we needed to construct some reversible path. 143 00:11:27 --> 00:11:31 All right, and of course it'll be the same here. 144 00:11:31 --> 00:11:35 Now, in this case since the system isn't isolated, that 145 00:11:35 --> 00:11:38 means heat's going to flow across the boundary. 146 00:11:38 --> 00:11:43 So that means heat is going between, it's being exchanged 147 00:11:43 --> 00:11:45 between the system and the surroundings. 148 00:11:45 --> 00:11:47 So what's happening to the surroundings? 149 00:11:47 --> 00:11:49 Generally, we've worry a lot about the system, but not 150 00:11:49 --> 00:11:54 much in recent discussions about the surroundings. 151 00:11:54 --> 00:12:01 But delta S surroundings must be something. 152 00:12:01 --> 00:12:04 We could calculate it. 153 00:12:04 --> 00:12:07 And that's depends on the path. 154 00:12:07 --> 00:12:10 Not because it isn't the state function in the surroundings, 155 00:12:10 --> 00:12:15 of course it is, but because depending on the path, the 156 00:12:15 --> 00:12:22 final state of the surroundings will be different. 157 00:12:22 --> 00:12:25 After all, depending on how much heat got exchanged, the 158 00:12:25 --> 00:12:30 surroundings got more or less of that heat. 159 00:12:30 --> 00:12:33 Work might have been done on the surroundings. 160 00:12:33 --> 00:12:37 So the surroundings are going to respond, are going to 161 00:12:37 --> 00:12:39 change also, when the system changes in this way 162 00:12:39 --> 00:12:41 and it's not isolated. 163 00:12:41 --> 00:12:44 And how the surroundings change will depend on 164 00:12:44 --> 00:12:45 what path was taken. 165 00:12:45 --> 00:12:47 And you've seen that before in calculating things like 166 00:12:47 --> 00:12:51 pressure, volume, work along different paths. 167 00:12:51 --> 00:12:54 Where you calculated delta u for the system and saw that it 168 00:12:54 --> 00:12:59 was the same for different paths but q and w were not the 169 00:12:59 --> 00:13:01 same for the system, and therefore, of course, not for 170 00:13:01 --> 00:13:07 the surroundings either. 171 00:13:07 --> 00:13:21 So let's just consider an irreversible path. 172 00:13:21 --> 00:13:23 And in order to think about what happens to the 173 00:13:23 --> 00:13:29 surroundings, let's take the entire universe which consists 174 00:13:29 --> 00:13:36 of the system and the surroundings -- that is a big 175 00:13:36 --> 00:13:40 isolated system, because it's everything. 176 00:13:40 --> 00:13:45 So it's not somehow in contact with another additional 177 00:13:45 --> 00:13:48 body or universe. 178 00:13:48 --> 00:13:54 We can treat the entire universe as an isolated system. 179 00:13:54 --> 00:14:05 So, we're going to think about the entire universe. 180 00:14:05 --> 00:14:13 We can think about it as an isolated system. 181 00:14:13 --> 00:14:16 Great. 182 00:14:16 --> 00:14:23 And we know that delta S for an isolated system, for 183 00:14:23 --> 00:14:28 a spontaneous process, is greater than zero. 184 00:14:28 --> 00:14:30 That's right there. 185 00:14:30 --> 00:14:38 So delta S for the universe which is delta S for the 186 00:14:38 --> 00:14:44 system plus delta S for the surroundings, and I'll specify 187 00:14:44 --> 00:14:49 since we're considering an irreversible process, specify 188 00:14:49 --> 00:14:52 that, it's greater than zero. 189 00:14:52 --> 00:14:55 The whole thing happened spontaneously, and the whole 190 00:14:55 --> 00:14:58 universe is an isolated system. 191 00:14:58 --> 00:15:05 So if we wanted to, we can write that delta S irreversible 192 00:15:05 --> 00:15:11 for the surroundings must be bigger than minus delta S for 193 00:15:11 --> 00:15:19 the system in order that there sum be positive. 194 00:15:19 --> 00:15:21 All right? 195 00:15:21 --> 00:15:33 Now, if we consider a reversible process, then of 196 00:15:33 --> 00:15:37 course we already know what happens for an isolated system 197 00:15:37 --> 00:15:43 in a reversible process, delta S is zero. 198 00:15:43 --> 00:15:47 So delta S in this case of the whole universe, which is delta 199 00:15:47 --> 00:15:53 S for the system plus delta S for the surroundings in the 200 00:15:53 --> 00:16:02 reversible case must be equal to zero. 201 00:16:02 --> 00:16:08 In other words, delta S of the surroundings in the reversible 202 00:16:08 --> 00:16:15 case is exactly the opposite of delta S of the system. 203 00:16:15 --> 00:16:18 Delta S of the system is the same in either case because 204 00:16:18 --> 00:16:21 reversible or irreversible, we're specifying the system, 205 00:16:21 --> 00:16:24 goes between the same states, but what happens to the 206 00:16:24 --> 00:16:30 surroundings is different in the two cases. 207 00:16:30 --> 00:16:36 So one way of recasting this statement or the Clausius 208 00:16:36 --> 00:16:41 inequality is the entropy of the whole universe 209 00:16:41 --> 00:16:44 never decreases. 210 00:16:44 --> 00:16:47 Of course, in some sense, that's obvious from this 211 00:16:47 --> 00:16:49 statement, since we're considering the universe 212 00:16:49 --> 00:16:51 an isolated system. 213 00:16:51 --> 00:16:56 So the entropy in any process can, it can, the change in 214 00:16:56 --> 00:17:04 entropy can either be zero or positive and never negative. 215 00:17:04 --> 00:17:10 216 00:17:10 --> 00:17:27 So entropy off the whole universe never decreases. 217 00:17:27 --> 00:17:30 That's actually a really quite profound conclusion. 218 00:17:30 --> 00:17:36 All sorts of philosophical and other consequences of it that 219 00:17:36 --> 00:17:39 are sometimes considered in things like evolutionary 220 00:17:39 --> 00:17:41 biology and other things. 221 00:17:41 --> 00:17:43 And how things can change altogether an entire 222 00:17:43 --> 00:17:48 system of things. 223 00:17:48 --> 00:17:55 Now, I'd like to just go through a few calculations of 224 00:17:55 --> 00:17:59 delta S for common processes. 225 00:17:59 --> 00:18:01 We've seen a few general features of entropy. 226 00:18:01 --> 00:18:05 Let's calculate it for a few things. 227 00:18:05 --> 00:18:15 So calculations of delta S. 228 00:18:15 --> 00:18:17 And we know the general procedure. 229 00:18:17 --> 00:18:20 Whatever the process is, we're going to need to find 230 00:18:20 --> 00:18:24 reversible paths, and along a reversible path then, we'll 231 00:18:24 --> 00:18:28 be able to calculate it. 232 00:18:28 --> 00:18:39 One, let's just take two pieces of material at 233 00:18:39 --> 00:18:43 different temperatures. 234 00:18:43 --> 00:18:56 The entire system is going to be isolated. 235 00:18:56 --> 00:19:01 And then let's connect them. 236 00:19:01 --> 00:19:02 And then let's connect them. 237 00:19:02 --> 00:19:03 Put some material between them. 238 00:19:03 --> 00:19:05 So now heat can flow from one to the other. 239 00:19:05 --> 00:19:07 And you know what's going to happen which is if they start 240 00:19:07 --> 00:19:09 out at unequal temperatures, and you wait a little while, 241 00:19:09 --> 00:19:10 eventually they're going to wind up at the 242 00:19:10 --> 00:19:12 same temperature. 243 00:19:12 --> 00:19:14 In other words, you know the direction of 244 00:19:14 --> 00:19:17 spontaneous change. 245 00:19:17 --> 00:19:23 So let's just see what happens to delta S. 246 00:19:23 --> 00:19:33 So isolated system, so there's no work being done. 247 00:19:33 --> 00:19:35 There's no heat that's being exchanged. 248 00:19:35 --> 00:19:42 So delta u is zero. 249 00:19:42 --> 00:19:44 What's dS? 250 00:19:44 --> 00:19:49 Well it's just the sum of the changes in entropy for the two 251 00:19:49 --> 00:19:53 sub-systems the two separate pieces. 252 00:19:53 --> 00:20:03 So it's dq1 over T1 plus dq2 over T2, I think there's a 253 00:20:03 --> 00:20:11 type in your notes. it's plus not minus of course. 254 00:20:11 --> 00:20:16 But in this case, dq1 is just the opposite of dq2. 255 00:20:16 --> 00:20:20 Because whatever heat is flowing into T1, into this 256 00:20:20 --> 00:20:23 block, is coming out of this block, or vice versa, depending 257 00:20:23 --> 00:20:32 on which one is hotter. 258 00:20:32 --> 00:20:39 So this is just equal to dq1 times one over T1 minus one 259 00:20:39 --> 00:20:45 over T2, which is T2 minus T1 over T1 times T2. 260 00:20:45 --> 00:20:49 261 00:20:49 --> 00:20:54 So that's dS, and then we could integrate it to get the delta 262 00:20:54 --> 00:20:57 S, but this is sufficient for what I'd like to illustrate, 263 00:20:57 --> 00:20:59 which is just the sign of things. 264 00:20:59 --> 00:21:09 We know dS must be greater than zero for the spontaneous 265 00:21:09 --> 00:21:14 change that's going to occur. 266 00:21:14 --> 00:21:18 So let's just make sure that common sense prevails here. 267 00:21:18 --> 00:21:23 That says if T2 is greater than T1, right T2 is hotter. 268 00:21:23 --> 00:21:27 Then dq must be positive, so this is positive, this is 269 00:21:27 --> 00:21:30 positive, dS we know has to be positive. 270 00:21:30 --> 00:21:35 So, it says T2 is hotter. 271 00:21:35 --> 00:21:40 This being positive means heat is flowing in from this one to 272 00:21:40 --> 00:21:45 this one, from the hotter to the colder body. 273 00:21:45 --> 00:21:47 So it looks right. 274 00:21:47 --> 00:21:52 If T2 is lower than T1, if this one is colder, then in order 275 00:21:52 --> 00:21:55 for dS to be positive, which it has to be, this 276 00:21:55 --> 00:21:57 better be negative. 277 00:21:57 --> 00:22:00 In other words, again heat is going to flow from the 278 00:22:00 --> 00:22:05 hotter toward the colder, into the colder body. 279 00:22:05 --> 00:22:11 So, as promised, the condition that dS must be greater than 280 00:22:11 --> 00:22:17 zero, guides us and tells us the direction that spontaneous 281 00:22:17 --> 00:22:21 change is going to occur. 282 00:22:21 --> 00:22:23 We can tell just from this condition which way 283 00:22:23 --> 00:22:44 things have to evolve. 284 00:22:44 --> 00:22:48 All right, let's try another pretty simple process. 285 00:22:48 --> 00:22:59 Let's just take a gas in some volume V and over here is 286 00:22:59 --> 00:23:05 going to be vacuum of equal volume, and we're going 287 00:23:05 --> 00:23:07 to remove the barrier. 288 00:23:07 --> 00:23:10 You know what's going to happen spontaneously, right? 289 00:23:10 --> 00:23:14 The gas is going to fill the available volume once 290 00:23:14 --> 00:23:16 it becomes available. 291 00:23:16 --> 00:23:23 So this, we'll make this a Joule expansion, and we'll 292 00:23:23 --> 00:23:29 make it an ideal gas. 293 00:23:29 --> 00:23:37 So the process is one mole of gas at our initial volume and 294 00:23:37 --> 00:23:46 some temperature, and we'll make this adiabatic, and this 295 00:23:46 --> 00:23:53 goes to one mole of gas at a new volume, 2V and at 296 00:23:53 --> 00:23:55 the same temperature. 297 00:23:55 --> 00:24:01 All right, so there's no work done. 298 00:24:01 --> 00:24:02 It's adiabatic. 299 00:24:02 --> 00:24:03 There's no heat exchanged. 300 00:24:03 --> 00:24:08 Delta u is zero. 301 00:24:08 --> 00:24:10 Everything's zero. 302 00:24:10 --> 00:24:13 Almost everything is zero. 303 00:24:13 --> 00:24:16 What isn't zero? 304 00:24:16 --> 00:24:16 STUDENT: Entropy. 305 00:24:16 --> 00:24:17 PROFESSOR NELSON: Entropy. 306 00:24:17 --> 00:24:19 It's going to tell us which way this whole thing goes. 307 00:24:19 --> 00:24:22 So this is an irreversible process. 308 00:24:22 --> 00:24:26 In order to calculate delta S we need to construct a 309 00:24:26 --> 00:24:30 reversible path along which this can go. 310 00:24:30 --> 00:24:34 So here's a way to do it. 311 00:24:34 --> 00:24:38 We could compress it back, isothermally and reversibly. 312 00:24:38 --> 00:24:41 Just like the example I gave, the general example, where I 313 00:24:41 --> 00:24:42 mentioned the ice melting. 314 00:24:42 --> 00:24:44 In that case, it won't be adiabatic. 315 00:24:44 --> 00:24:48 We'd have to put it in contact with a heat 316 00:24:48 --> 00:24:51 source of some sort. 317 00:24:51 --> 00:25:00 So reversible process. 318 00:25:00 --> 00:25:21 We could compress it back to volume V, isothermally, 319 00:25:21 --> 00:25:22 and reversibly. 320 00:25:22 --> 00:25:25 Let me say, there's no -- we wouldn't have to consider 321 00:25:25 --> 00:25:26 this process in reverse. 322 00:25:26 --> 00:25:29 Of course, we could also consider the forward process 323 00:25:29 --> 00:25:34 in the presence of some heat source with isn't isolated and 324 00:25:34 --> 00:25:37 do this, but let's just consider the compression. 325 00:25:37 --> 00:25:46 So, OK, now we'll have the -- well, once 326 00:25:46 --> 00:25:48 again, delta u is zero. 327 00:25:48 --> 00:25:50 It's an ideal gas. 328 00:25:50 --> 00:25:52 There's no temperature change. 329 00:25:52 --> 00:25:55 But this time q and w aren't going to be zero. 330 00:25:55 --> 00:25:57 They're going to be some finite numbers that are 331 00:25:57 --> 00:25:59 opposites of each other. 332 00:25:59 --> 00:26:05 Because we're no longer isolated. 333 00:26:05 --> 00:26:21 So let's just write out delta S to go backward is dq reversible 334 00:26:21 --> 00:26:32 over T, and that's minus dw over T, not zero anymore. 335 00:26:32 --> 00:26:39 And where we're going is from 2V to V, we're compressing, 336 00:26:39 --> 00:26:41 and it's an ideal gas. 337 00:26:41 --> 00:26:59 So this is the same as p dV, but it's p dV over T which 338 00:26:59 --> 00:27:02 is the same thing as going from 2V to V. 339 00:27:02 --> 00:27:05 Now we're going to replace this with RT over V, and 340 00:27:05 --> 00:27:07 the T's will cancel. 341 00:27:07 --> 00:27:15 So it's R dV over V going from 2V to one V. 342 00:27:15 --> 00:27:23 So it's just R times the log of 1/2. 343 00:27:23 --> 00:27:27 So now we've just calculated the value of delta S by 344 00:27:27 --> 00:27:30 constructing this reversible path. 345 00:27:30 --> 00:27:33 Now this is delta S backward. 346 00:27:33 --> 00:27:37 To do the compression, so of course delta S forward is 347 00:27:37 --> 00:27:46 just the opposite of that. 348 00:27:46 --> 00:27:58 So, delta S forward is R log 2. 349 00:27:58 --> 00:28:12 Reassuringly, that's a positive number. 350 00:28:12 --> 00:28:16 So this process that we considered originally, this 351 00:28:16 --> 00:28:20 irreversibly process in this isolated system, remember, for 352 00:28:20 --> 00:28:25 the irreversible expansion we considered the system isolated, 353 00:28:25 --> 00:28:27 Happened spontaneously, and of course you know 354 00:28:27 --> 00:28:30 that's the case. 355 00:28:30 --> 00:28:37 Now, I'll just mention, and later on we'll go into this in 356 00:28:37 --> 00:28:40 considerably more detail, but I'll just mention at this stage 357 00:28:40 --> 00:28:45 through a little bit of sort of foreshadowing that everything 358 00:28:45 --> 00:28:48 we've discussed so far about entropy is with 359 00:28:48 --> 00:28:50 a macroscopic picture. 360 00:28:50 --> 00:28:54 We started with heat engines and Carnot cycles and 361 00:28:54 --> 00:28:56 macroscopic processes, reversible and irreversible 362 00:28:56 --> 00:28:58 processes and so on. 363 00:28:58 --> 00:29:02 But there's a, and that's certainly the way it was 364 00:29:02 --> 00:29:04 all formulated originally. 365 00:29:04 --> 00:29:06 Part of the power of thermodynamics is that it 366 00:29:06 --> 00:29:12 doesn't depend on a specific microscopic model of matter. 367 00:29:12 --> 00:29:15 That said though, we have a pretty good microscopic 368 00:29:15 --> 00:29:16 model of matter. 369 00:29:16 --> 00:29:20 We know about atoms and molecules, and in fact there is 370 00:29:20 --> 00:29:26 an entirely microscopic formulation of entropy that has 371 00:29:26 --> 00:29:31 to do with disorder and the number of available states, 372 00:29:31 --> 00:29:36 microscopic states available to a system. 373 00:29:36 --> 00:29:41 So I'll just state what we'll see in much 374 00:29:41 --> 00:29:44 more detail later on. 375 00:29:44 --> 00:29:57 Which is in microscopic terms, it's going to turn out that the 376 00:29:57 --> 00:30:04 entropy of a system can be given by R over Avogadro's 377 00:30:04 --> 00:30:16 number times the log of the number of distinct 378 00:30:16 --> 00:30:24 microscopic states that are available to a system. 379 00:30:24 --> 00:30:28 Right so in this isolated system, when we double that 380 00:30:28 --> 00:30:33 volume, each molecule suddenly has twice as many 381 00:30:33 --> 00:30:36 states available to it. 382 00:30:36 --> 00:30:38 You could formulate that in a lot of ways. 383 00:30:38 --> 00:30:42 You can sort of divide up the volume into tiny little 384 00:30:42 --> 00:30:46 molecule-sized cubes and realize well, now, there 385 00:30:46 --> 00:30:49 are twice as many of them. 386 00:30:49 --> 00:30:52 Now if I have one molecule, the number of states doubles. 387 00:30:52 --> 00:30:56 If I have n molecules, the number of states goes 388 00:30:56 --> 00:30:57 up by 2 to the n. 389 00:30:57 --> 00:30:59 Because for each one of the molecules, I can select any one 390 00:30:59 --> 00:31:04 of those states, and then the next one I can select any one. 391 00:31:04 --> 00:31:05 So you have this enormous increase in the 392 00:31:05 --> 00:31:08 number of states. 393 00:31:08 --> 00:31:19 So delta S in this case looks like R over NA log of 2 to the 394 00:31:19 --> 00:31:22 power NA, it's Avogadro's number, for a mole 395 00:31:22 --> 00:31:25 of molecules. 396 00:31:25 --> 00:31:30 This can come out and cancel this. 397 00:31:30 --> 00:31:34 There's our R log 2 result. 398 00:31:34 --> 00:31:37 But the point I'm making here is that could actually be 399 00:31:37 --> 00:31:42 derived from entirely microscopic consideration. 400 00:31:42 --> 00:31:44 We haven't gone through that yet. 401 00:31:44 --> 00:31:48 We've done entirely macroscopic thermodynamics so far, but 402 00:31:48 --> 00:31:51 we'll do a little bit of treatment of statistical 403 00:31:51 --> 00:31:56 mechanics, is what is called the whole microscopic 404 00:31:56 --> 00:31:59 formulation of entropy in thermodynamics. 405 00:31:59 --> 00:32:03 And that's a really important fundamental result. 406 00:32:03 --> 00:32:08 And we'll get to it in much more detail in a while. 407 00:32:08 --> 00:32:12 But for now, I just want to continue calculating entropy 408 00:32:12 --> 00:32:31 changes for a few more kinds of processes. 409 00:32:31 --> 00:32:37 So, how about if we have two different substances and we 410 00:32:37 --> 00:32:50 mix them, entropy of mixing. 411 00:32:50 --> 00:33:00 So, we start with some number of moles of substance A in some 412 00:33:00 --> 00:33:04 volume VA, and some other number of moles of substance 413 00:33:04 --> 00:33:09 B in volume VB, separated. 414 00:33:09 --> 00:33:11 And then we remove the barrier, of course we know 415 00:33:11 --> 00:33:14 they're going to mix. 416 00:33:14 --> 00:33:20 So we'll have some total number of moles, nA plus nB, and 417 00:33:20 --> 00:33:23 they'll be filling some total volume, which is the sum of 418 00:33:23 --> 00:33:26 the two original volumes. 419 00:33:26 --> 00:33:30 Certainly this is what we expect to happen spontaneously. 420 00:33:30 --> 00:33:41 So are process is nA moles of A, gas, initial volume VA and 421 00:33:41 --> 00:33:47 T, plus nB of substance B, I think this is miswritten 422 00:33:47 --> 00:33:49 as A there in your notes. 423 00:33:49 --> 00:34:01 Gas VB and T goes to nA moles of A plus nB moles of B. 424 00:34:01 --> 00:34:08 Gas, total volume and T. 425 00:34:08 --> 00:34:15 And we'll have constant temperature and pressure. 426 00:34:15 --> 00:34:19 Well, there was no p v worked done. 427 00:34:19 --> 00:34:22 The temperature didn't change and it's ideal gases, so delta 428 00:34:22 --> 00:34:28 u is zero, but what about delta S? 429 00:34:28 --> 00:34:31 Well again, to calculate it, we need to find a reversible path. 430 00:34:31 --> 00:34:34 This is irreversible if we just remove the barrier 431 00:34:34 --> 00:34:37 and let it all go. 432 00:34:37 --> 00:34:40 But we could at least imagine a reversible path, and we could 433 00:34:40 --> 00:34:44 actually construct this for the right substances. 434 00:34:44 --> 00:34:53 We could imagine that we can find some sort of piston here. 435 00:34:53 --> 00:35:01 And we're going to prepare to push it inward, with a membrane 436 00:35:01 --> 00:35:07 that's permeable only to b. 437 00:35:07 --> 00:35:14 And over here we could set up the exact same thing, but this 438 00:35:14 --> 00:35:23 one's permeable only to a. 439 00:35:23 --> 00:35:27 And then we could just push them to recover this state. 440 00:35:27 --> 00:35:29 And we can do it reversibly. 441 00:35:29 --> 00:35:33 So we could construct a reversible process that 442 00:35:33 --> 00:35:36 does the reverse of this. 443 00:35:36 --> 00:35:40 And in this case, it'll be isothermal still, constant 444 00:35:40 --> 00:35:45 pressure, reversible compression of the two gases. 445 00:35:45 --> 00:36:22 So, reversible compression and demixing. 446 00:36:22 --> 00:36:23 All right, so what happens? 447 00:36:23 --> 00:36:28 Well, of course, our delta S of demixing is going to be 448 00:36:28 --> 00:36:34 minus delta S of mixing. 449 00:36:34 --> 00:36:40 Delta u of demixing is still zero. 450 00:36:40 --> 00:36:47 Temperature didn't change, ideal gases, and there's some 451 00:36:47 --> 00:36:52 reversible work and heat. 452 00:36:52 --> 00:37:04 So dw in this reversible case, is just minus pA dVA minus pB 453 00:37:04 --> 00:37:12 dVB, the infinitesimal amount of work done as these things 454 00:37:12 --> 00:37:15 are gradually moved toward each other. 455 00:37:15 --> 00:37:27 So delta S for demixing, of course it's dq reversible over 456 00:37:27 --> 00:37:30 T, but just like we did right there, of course, in this case 457 00:37:30 --> 00:37:33 that's just negative dw. 458 00:37:33 --> 00:37:40 So it's integral from V to VA, right, were compressing. 459 00:37:40 --> 00:37:43 So for substance a, we're going to go back to there. 460 00:37:43 --> 00:37:45 It was occupying the whole volume, and then it's going 461 00:37:45 --> 00:37:48 to end up in only part of the volume, VA. 462 00:37:48 --> 00:37:54 pA dVA over T. 463 00:37:54 --> 00:38:09 Same thing for B, going to volume B, pB dVB over T. 464 00:38:09 --> 00:38:14 And now we can substitute for pressure, right? pV is nRT, so 465 00:38:14 --> 00:38:25 it's nA times R log of VA over V plus nB of R log 466 00:38:25 --> 00:38:33 of VB over V. 467 00:38:33 --> 00:38:37 Now, so this is a suitable answer, but I'm going to make 468 00:38:37 --> 00:38:42 if a little easier by putting in terms of mole fractions. 469 00:38:42 --> 00:39:05 So, mole fractions, of course, XA is nA over n, and XB is nB 470 00:39:05 --> 00:39:14 over n and XA is also equal to VA over V and XB is equal to VB 471 00:39:14 --> 00:39:17 over V for ideal gases, right? 472 00:39:17 --> 00:39:18 The molar volumes are the same. 473 00:39:18 --> 00:39:21 We're at the same temperature and pressure. 474 00:39:21 --> 00:39:32 So that immediately gives us the delta X of demixing is nR 475 00:39:32 --> 00:39:42 times XA log XA plus XB log XB. 476 00:39:42 --> 00:39:48 And of course delta S of mixing is the opposite of this, minus 477 00:39:48 --> 00:39:58 nR XA log XA plus XB log XB. 478 00:39:58 --> 00:40:00 Now XA and XB are 479 00:40:00 --> 00:40:01 mole fractions. 480 00:40:01 --> 00:40:06 They are between zero and one, So their log rhythms 481 00:40:06 --> 00:40:08 are both negative. 482 00:40:08 --> 00:40:10 There's a negative sign. 483 00:40:10 --> 00:40:13 So delta S of mixing is positive. 484 00:40:13 --> 00:40:15 That's reassuring. 485 00:40:15 --> 00:40:18 That tells us as we expect that the mixing should happen 486 00:40:18 --> 00:40:28 spontaneously and irreversibly when we remove the barrier. 487 00:40:28 --> 00:40:31 Any questions so far about how we're going about these 488 00:40:31 --> 00:40:34 calculations of delta S? 489 00:40:34 --> 00:40:52 OK, let's just do a couple more examples. 490 00:40:52 --> 00:40:54 Here's a really straightforward one. 491 00:40:54 --> 00:40:58 What if we just heat stuff up or cool it down. 492 00:40:58 --> 00:41:02 It happens all the time, but it's pretty important, right? 493 00:41:02 --> 00:41:12 So let's just heat or cool at constant volume. 494 00:41:12 --> 00:41:17 So, we're going to go from substance A at T1 and some 495 00:41:17 --> 00:41:22 volume going to substance A at T2 at some volume. 496 00:41:22 --> 00:41:25 We can do this. 497 00:41:25 --> 00:41:34 Delta S is integral of dq reversible over T. 498 00:41:34 --> 00:41:43 Going from T1 to T2, but we know how to do this, right? 499 00:41:43 --> 00:41:44 It's the heat that we need. 500 00:41:44 --> 00:41:47 It's just given by the constant volume heat capacity times 501 00:41:47 --> 00:41:49 the change in temperature. 502 00:41:49 --> 00:42:00 So this is just T1 to T2, Cv dT over T. 503 00:42:00 --> 00:42:11 And that is just Cv times the log of T2 over T1 if Cv is 504 00:42:11 --> 00:42:16 temperature independent. 505 00:42:16 --> 00:42:18 That's not always the case. 506 00:42:18 --> 00:42:21 Usually if it's not too big an excursion of temperature then 507 00:42:21 --> 00:42:25 it's a reasonable approximation. 508 00:42:25 --> 00:42:33 So that's straightforward. 509 00:42:33 --> 00:42:42 By the way, notice that delta S is greater than zero if 510 00:42:42 --> 00:42:44 T2 is greater than T1. 511 00:42:44 --> 00:42:45 1 512 00:42:45 --> 00:42:50 In other words, if we heated it up, delta S is positive. 513 00:42:50 --> 00:42:54 Notice also delta S is less than zero if 514 00:42:54 --> 00:42:57 T2 is less than T1. 515 00:42:57 --> 00:43:01 Delta S is negative if we cooled it. 516 00:43:01 --> 00:43:05 Now, under certain conditions we've seen that delta 517 00:43:05 --> 00:43:07 S can't be negative. 518 00:43:07 --> 00:43:09 Why can it be negative here? 519 00:43:09 --> 00:43:14 STUDENT: [UNINTELLIGIBLE]. 520 00:43:14 --> 00:43:17 PROFESSOR NELSON: Yes, I heard it. 521 00:43:17 --> 00:43:19 It's because the system isn't isolated, right. 522 00:43:19 --> 00:43:23 It was for the isolated system that delta S is always greater 523 00:43:23 --> 00:43:24 than or equal to zero. 524 00:43:24 --> 00:43:27 And you know, for the whole universe entropy never 525 00:43:27 --> 00:43:30 decreases and so on. 526 00:43:30 --> 00:43:33 In this case, though, you know we're taking a system and we're 527 00:43:33 --> 00:43:36 putting it in contact with a cold bath and cooling it down. 528 00:43:36 --> 00:43:38 It's certainly not isolated. 529 00:43:38 --> 00:43:39 Heat is being exchanged. 530 00:43:39 --> 00:43:43 It's going from the system to the surroundings, to the bath. 531 00:43:43 --> 00:43:44 So it's fine. 532 00:43:44 --> 00:43:48 Delta S can be negative. 533 00:43:48 --> 00:43:53 What would happen if we calculated delta S of a new 534 00:43:53 --> 00:43:55 entire system consisting of the original system 535 00:43:55 --> 00:43:58 plus the heat bath? 536 00:43:58 --> 00:44:00 What would delta S be then? 537 00:44:00 --> 00:44:07 We already kind of did that with those two blocks, T1 and 538 00:44:07 --> 00:44:11 T2 in an isolated systems. 539 00:44:11 --> 00:44:13 What do you think is going to happen? 540 00:44:13 --> 00:44:17 In other words, if we call our system the stuff that is put in 541 00:44:17 --> 00:44:21 contact with the heat bath, well OK, then we've seen delta 542 00:44:21 --> 00:44:24 S can be plus, can be positive or negative. 543 00:44:24 --> 00:44:28 Now let's call our system the original system plus the heat 544 00:44:28 --> 00:44:32 bath, and we'll put them all in some isolating box, that's 545 00:44:32 --> 00:44:34 thermally isolating and so on. 546 00:44:34 --> 00:44:37 And now we'll put our original system, which is now sort of a 547 00:44:37 --> 00:44:40 sub, part of the system, and it's now going to be in 548 00:44:40 --> 00:44:45 contact with the cold bath. 549 00:44:45 --> 00:44:48 The bath that it's in contact with is colder than it is. 550 00:44:48 --> 00:44:50 That's why its delta S is less than zero. 551 00:44:50 --> 00:44:53 But what's delta S to be for the entire new system 552 00:44:53 --> 00:44:58 including the cold bath? 553 00:44:58 --> 00:44:59 STUDENT: Wouldn't it greater than zero? 554 00:44:59 --> 00:45:01 PROFESSOR NELSON: Yes, it's going to be positive, because 555 00:45:01 --> 00:45:04 it's an isolated system and something happened 556 00:45:04 --> 00:45:09 spontaneously in it, namely the original system got cooler and 557 00:45:09 --> 00:45:12 presumably the heat bath got at least a little bit warmer. 558 00:45:12 --> 00:45:15 And it happened spontaneously, which immediately means delta S 559 00:45:15 --> 00:45:20 had to be positive for that entire assembly, of you know 560 00:45:20 --> 00:45:25 the original system plus the bath. 561 00:45:25 --> 00:45:32 Here's another important sort of process. 562 00:45:32 --> 00:45:44 Reversible phase change, we'll melt something or freeze it. 563 00:45:44 --> 00:45:50 How about water -- liquid 100 degrees Celsius 564 00:45:50 --> 00:45:54 one bar goes to water. 565 00:45:54 --> 00:46:01 Gas 100 degrees Celsius one bar. 566 00:46:01 --> 00:46:05 Well in that case, you know what the heat is. 567 00:46:05 --> 00:46:11 It's just the delta H of vaporization, right, 568 00:46:11 --> 00:46:15 at constant pressure. 569 00:46:15 --> 00:46:17 So there's nothing really much to calculate here. 570 00:46:17 --> 00:46:28 Delta S of vaporization for H2O at 100 degrees Celsius is just 571 00:46:28 --> 00:46:36 q of vaporization over the boiling temperature, which is 572 00:46:36 --> 00:46:43 delta H of vaporization over the boiling temperature. 573 00:46:43 --> 00:46:45 So that was easy. 574 00:46:45 --> 00:46:49 One last thing. 575 00:46:49 --> 00:46:53 What if we want to calculate the entropy of melting or 576 00:46:53 --> 00:46:56 some phase transition not at equilibrium? 577 00:46:56 --> 00:46:59 In this case it's at equilibrium because it's 578 00:46:59 --> 00:47:03 water and water vapor at 100 degrees Celsius and 579 00:47:03 --> 00:47:05 one atmosphere or one bar. 580 00:47:05 --> 00:47:08 So it's at its boiling point. 581 00:47:08 --> 00:47:14 But what if it isn't at its boiling point? 582 00:47:14 --> 00:47:24 So what if instead, let's take H2O, liquid, at minus 583 00:47:24 --> 00:47:29 ten degrees Celsius and one bar so it's cold. 584 00:47:29 --> 00:47:31 It's going to freeze. 585 00:47:31 --> 00:47:36 So it's going to turn into solid water at minus 10 degrees 586 00:47:36 --> 00:47:40 Celsius and one bar, we're going to have it be isothermal. 587 00:47:40 --> 00:47:43 Now you know that's going to happen spontaneously, 588 00:47:43 --> 00:47:52 and it's irreversible. 589 00:47:52 --> 00:47:56 Irreversible, which means we can't just straight away 590 00:47:56 --> 00:48:00 calculate delta S along this path because it's irreversible. 591 00:48:00 --> 00:48:04 But what we can do, just like we've done before for making 592 00:48:04 --> 00:48:07 cycles, we can make some other set of events that 593 00:48:07 --> 00:48:08 are all reversible. 594 00:48:08 --> 00:48:12 We can construct a reversible pass that will get there. 595 00:48:12 --> 00:48:13 How do we do it? 596 00:48:13 --> 00:48:17 Well you know for the phase change, for the freezing to be 597 00:48:17 --> 00:48:20 reversible, it has to happen at zero degrees Celsius right. 598 00:48:20 --> 00:48:23 You know that's where you have reversible, 599 00:48:23 --> 00:48:25 freezing and melting. 600 00:48:25 --> 00:48:28 Reversible equilibrium between liquid and solid water. 601 00:48:28 --> 00:48:33 So surely that's got to get involved here. 602 00:48:33 --> 00:48:39 H2O liquid zero degrees Celsius, one bar, in 603 00:48:39 --> 00:48:44 equilibrium with H2O solid at zero degrees Celsius, one bar. 604 00:48:44 --> 00:48:48 Whatever reversible path we construct that's going to go 605 00:48:48 --> 00:48:52 from liquid to solid water, somewhere along the set of 606 00:48:52 --> 00:48:55 steps, that better be one of them. 607 00:48:55 --> 00:48:58 So now what we need to do is go from liquid at minus 10 degrees 608 00:48:58 --> 00:49:02 Celsius and one bar to liquid at zero degrees 609 00:49:02 --> 00:49:05 Celsius and one bar. 610 00:49:05 --> 00:49:08 That's called heating. 611 00:49:08 --> 00:49:09 We have to heat it. 612 00:49:09 --> 00:49:15 And here we have to cool it. 613 00:49:15 --> 00:49:18 And we've already seen all these three processes, right. 614 00:49:18 --> 00:49:23 So for the phase change, it's reversible now. 615 00:49:23 --> 00:49:26 This is just delta heat of fusion. 616 00:49:26 --> 00:49:30 Great. 617 00:49:30 --> 00:49:33 Because this is now reversible. 618 00:49:33 --> 00:49:35 This is going to be reversible. 619 00:49:35 --> 00:49:37 This is going to be reversible. 620 00:49:37 --> 00:49:38 Great. 621 00:49:38 --> 00:49:43 So dq reversible, it's going to be the heat capacity at 622 00:49:43 --> 00:49:46 constant pressure, the example we did before it was constant 623 00:49:46 --> 00:49:51 volume for the liquid dT. 624 00:49:51 --> 00:49:59 Here, dq reversible is going to be Cp solid, dT. 625 00:49:59 --> 00:50:01 Those won't be the same right? 626 00:50:01 --> 00:50:03 It's one thing to say the heat capacity of something doesn't 627 00:50:03 --> 00:50:07 change over some small excursion in temperature, but 628 00:50:07 --> 00:50:09 it's another thing when the thing actually changes phase. 629 00:50:09 --> 00:50:11 The heat capacity of the solid and the liquid 630 00:50:11 --> 00:50:14 won't be the same. 631 00:50:14 --> 00:50:18 So this we can just finish up in a jiffy, because we're just 632 00:50:18 --> 00:50:23 going to add the three things. 633 00:50:23 --> 00:50:30 So delta S is delta S of heating, minus, I think there's 634 00:50:30 --> 00:50:34 a plus that should be a minus there, minus delta S of fusion 635 00:50:34 --> 00:50:36 because it's going in the direction of freezing the 636 00:50:36 --> 00:50:38 way we've written it. 637 00:50:38 --> 00:50:42 Plus delta S of cooling. 638 00:50:42 --> 00:50:52 So integral from T1 to the melting point of Cp of the 639 00:50:52 --> 00:50:59 liquid dT over T, minus delta H of fusion over 640 00:50:59 --> 00:51:03 the melting temperature. 641 00:51:03 --> 00:51:07 Plus the integral going from the heat of the temperature 642 00:51:07 --> 00:51:15 of melting to T1 of Cp of the solid dT over T. 643 00:51:15 --> 00:51:18 That's it, all right? 644 00:51:18 --> 00:51:24 So delta S is minus delta H of fusion over T. 645 00:51:24 --> 00:51:26 That's what it would be for just the reversible phase 646 00:51:26 --> 00:51:30 change happening at zero degrees Celsius. 647 00:51:30 --> 00:51:34 And then there's this additional part which is, 648 00:51:34 --> 00:51:39 we can write it from T1 to melting point of Cp of 649 00:51:39 --> 00:51:47 the liquid, minus Cp of the solid, dT over T. 650 00:51:47 --> 00:51:50 Now usually we can assume that these will be temperature 651 00:51:50 --> 00:51:55 independent in their own phases. 652 00:51:55 --> 00:52:00 So we can usually write this as minus delta H of fusion over 653 00:52:00 --> 00:52:10 the melting temperature plus Cp of the liquid minus Cp of the 654 00:52:10 --> 00:52:21 solid log of T of fusion over T1. 655 00:52:21 --> 00:52:24 Any questions? 656 00:52:24 --> 00:52:25 What did we do? 657 00:52:25 --> 00:52:28 We constructed a reversible path going from here to here 658 00:52:28 --> 00:52:31 and calculated delta S that way, and of course that has 659 00:52:31 --> 00:52:36 to be the same as delta S in the one irreversible step. 660 00:52:36 --> 00:52:40 All right, more on entropy and its consequences for how to 661 00:52:40 --> 00:52:43 figure out what happened spontaneously next time. 662 00:52:43 --> 00:52:44