1 00:00:00,030 --> 00:00:02,400 The following content is provided under a creative 2 00:00:02,400 --> 00:00:03,820 commons license. 3 00:00:03,820 --> 00:00:06,840 Your support will help MIT OpenCourseWare continue to 4 00:00:06,840 --> 00:00:10,510 offer high-quality educational resources for free. 5 00:00:10,510 --> 00:00:13,390 To make a donation or view additional materials from 6 00:00:13,390 --> 00:00:17,490 hundreds of MIT courses visit MIT OpenCourseWare at 7 00:00:17,490 --> 00:00:20,360 ocw.mit.edu. 8 00:00:20,360 --> 00:00:21,900 PROFESSOR NELSON: Over the last couple lectures through 9 00:00:21,900 --> 00:00:27,400 our consideration of a number of somewhat difficult topics, 10 00:00:27,400 --> 00:00:31,930 including these Carnot cycles and specific Carnot engine to 11 00:00:31,930 --> 00:00:36,970 represent them, we've tried to define and understand a little 12 00:00:36,970 --> 00:00:44,250 bit about this very special function, this entropy. 13 00:00:44,250 --> 00:00:45,860 Where we've got our dS. 14 00:00:45,860 --> 00:00:56,960 It's dq reversible over T. And our treatment of reversible 15 00:00:56,960 --> 00:01:04,320 and irreversible cyclic processes, what we saw is that 16 00:01:04,320 --> 00:01:16,460 if we go around a cycle and look at dq reversible over T, 17 00:01:16,460 --> 00:01:18,250 then this is zero. 18 00:01:18,250 --> 00:01:21,830 Now go all the way around a cycle, of course, this is just 19 00:01:21,830 --> 00:01:26,160 delta S for the cycle. 20 00:01:26,160 --> 00:01:27,890 And that's consistent with the idea that 21 00:01:27,890 --> 00:01:29,070 entropy is a state functions. 22 00:01:29,070 --> 00:01:31,430 So if we go around cycle the cycle, we have the same ending 23 00:01:31,430 --> 00:01:34,936 point, the starting point, same state, then the change in 24 00:01:34,936 --> 00:01:39,420 the state function has to be zero. 25 00:01:39,420 --> 00:01:48,140 But we saw that for an irreversible path around a 26 00:01:48,140 --> 00:01:55,540 cycle, then we have that, this gives 27 00:01:55,540 --> 00:01:57,920 something less than zero. 28 00:01:57,920 --> 00:02:01,610 And so that was expressed in Clausius inequality that 29 00:02:01,610 --> 00:02:08,140 includes both these cases. 30 00:02:08,140 --> 00:02:10,830 And now I just want to go through a number of 31 00:02:10,830 --> 00:02:12,440 calculations of entropy. 32 00:02:12,440 --> 00:02:16,480 How entropy changes for a number processes, both to just 33 00:02:16,480 --> 00:02:19,830 learn more about it in general terms and also just to see how 34 00:02:19,830 --> 00:02:23,060 it's calculated and what sort of changes in undergoes for 35 00:02:23,060 --> 00:02:25,770 different sorts of changes in state. 36 00:02:25,770 --> 00:02:31,560 So the first very general example that I want to show is 37 00:02:31,560 --> 00:02:33,270 the following. 38 00:02:33,270 --> 00:02:46,180 Let's say we've got an isolated system, and it 39 00:02:46,180 --> 00:02:57,780 undergoes some irreversible change. 40 00:02:57,780 --> 00:03:09,050 So we'll start at one, and we'll go irreversibly to some 41 00:03:09,050 --> 00:03:19,070 new state, two. 42 00:03:19,070 --> 00:03:29,090 OK, now we can always return the system back to the initial 43 00:03:29,090 --> 00:03:36,760 state reversibly. 44 00:03:36,760 --> 00:03:40,040 If we do that, then we can't keep the system isolated. 45 00:03:40,040 --> 00:03:45,760 So it won't be isolated anymore. 46 00:03:45,760 --> 00:03:48,260 In other words, in order to arrange things so that you 47 00:03:48,260 --> 00:03:51,460 could have a return along a reversible 48 00:03:51,460 --> 00:03:53,130 path, it can't be isolated. 49 00:03:53,130 --> 00:03:56,190 In some sense you know that, because when it was isolated 50 00:03:56,190 --> 00:03:59,520 it spontaneously irreversibly went from one to two. 51 00:03:59,520 --> 00:04:03,260 And example would be ice melting at some temperature 52 00:04:03,260 --> 00:04:06,500 that might be higher than the freezing point. 53 00:04:06,500 --> 00:04:09,310 So you've got solid ice, but the temperature is above zero 54 00:04:09,310 --> 00:04:12,180 degrees Celsius, so irreversibly it'll melt. 55 00:04:12,180 --> 00:04:14,710 Systems isolated, it'll melt. 56 00:04:14,710 --> 00:04:17,790 Then you could put the system in contact with a cold 57 00:04:17,790 --> 00:04:23,930 reservoir reversibly re-freeze the water to form ice again. 58 00:04:23,930 --> 00:04:26,230 But of course to do that you couldn't keep it isolated. 59 00:04:26,230 --> 00:04:30,270 You need to have it in contact with a low temperature bath. 60 00:04:30,270 --> 00:04:34,190 But again, this is just completely 61 00:04:34,190 --> 00:04:35,100 general at this point. 62 00:04:35,100 --> 00:04:38,380 It's some irreversible process returned then along a 63 00:04:38,380 --> 00:04:40,520 reversible path. 64 00:04:40,520 --> 00:04:43,210 So let's just look at what happens. 65 00:04:43,210 --> 00:04:51,360 So here's our path B. Along path A, well 66 00:04:51,360 --> 00:04:56,580 it's an isolated system. 67 00:04:56,580 --> 00:05:01,480 So the heat exchanged is zero. 68 00:05:01,480 --> 00:05:10,230 And then, for the cycle based on the Clausius inequality 69 00:05:10,230 --> 00:05:21,480 expressed here, the integral around the cycle of dq over T 70 00:05:21,480 --> 00:05:24,610 in general, it's less than or equal to zero. 71 00:05:24,610 --> 00:05:27,890 If there was an irreversible process involved, then in 72 00:05:27,890 --> 00:05:31,810 fact, it'll be less than zero. 73 00:05:31,810 --> 00:05:34,940 I'll write it here in the more general case, but we know what 74 00:05:34,940 --> 00:05:40,110 it's really going to be with the irreversible step. 75 00:05:40,110 --> 00:05:47,380 Well, so we can write this as the integral of dq 76 00:05:47,380 --> 00:05:58,920 irreversible over T, going from state one to state two. 77 00:05:58,920 --> 00:06:04,720 Oh, let me rewrite this down here. 78 00:06:04,720 --> 00:06:12,450 So around the cycle, dq over T, it's integral from state 79 00:06:12,450 --> 00:06:20,390 one to two of dq irreversible over T, plus the integral from 80 00:06:20,390 --> 00:06:28,120 two back to one of dq reversible over T. And of 81 00:06:28,120 --> 00:06:32,970 course, we know that that's less than or equal to zero. 82 00:06:32,970 --> 00:06:36,740 But this, we know is zero, right, because this 83 00:06:36,740 --> 00:06:38,990 irreversible step is taken in isolation. 84 00:06:38,990 --> 00:06:40,070 The system is isolated. 85 00:06:40,070 --> 00:06:42,550 We've already seen that there's no heat crossing the 86 00:06:42,550 --> 00:06:45,420 boundary of the system. 87 00:06:45,420 --> 00:06:48,110 So we just have this. 88 00:06:48,110 --> 00:06:57,060 So of course this dq reversible over T from two to 89 00:06:57,060 --> 00:07:00,930 one, that's, what now we've a reversible path, so that's 90 00:07:00,930 --> 00:07:04,830 just the entropy at the delta S going from two to one. 91 00:07:04,830 --> 00:07:11,660 It's S1 minus S2 it's delta S, we can write it as delta S 92 00:07:11,660 --> 00:07:15,450 backwards, right, we're going back along 93 00:07:15,450 --> 00:07:17,460 that reversible path. 94 00:07:17,460 --> 00:07:23,580 Which is of course negative delta S forward. 95 00:07:23,580 --> 00:07:29,880 Because around the whole cycle delta s has to be zero. 96 00:07:29,880 --> 00:07:43,580 So what that says is delta S forward is greater than or 97 00:07:43,580 --> 00:07:50,120 equal to zero. 98 00:07:50,120 --> 00:07:52,260 That's interesting. 99 00:07:52,260 --> 00:07:55,160 What that's saying is -- remember, we didn't specify 100 00:07:55,160 --> 00:07:55,760 the process. 101 00:07:55,760 --> 00:08:00,600 It's just for any completely general irreversible process. 102 00:08:00,600 --> 00:08:06,320 For any such process, for any spontaneous process, this is 103 00:08:06,320 --> 00:08:08,070 going to be the case. 104 00:08:08,070 --> 00:08:14,360 So in other words, for an isolated system, delta S tells 105 00:08:14,360 --> 00:08:18,710 us the direction of spontaneous change. 106 00:08:18,710 --> 00:08:29,180 In particular, for the isolated system, delta S is 107 00:08:29,180 --> 00:08:40,590 greater than zero for something that's irreversible. 108 00:08:40,590 --> 00:08:51,330 Delta S equals zero for something that's reversible, 109 00:08:51,330 --> 00:09:02,200 and delta S is never less than zero. 110 00:09:02,200 --> 00:09:05,180 In a sense, this is a direct consequence of the Clausius 111 00:09:05,180 --> 00:09:11,720 inequality. 112 00:09:11,720 --> 00:09:17,200 That's a pretty useful result. 113 00:09:17,200 --> 00:09:21,450 So remember when we started this whole discussion, I tried 114 00:09:21,450 --> 00:09:25,200 to emphasize that the first law of thermodynamics told us 115 00:09:25,200 --> 00:09:27,020 about conservation of energy. 116 00:09:27,020 --> 00:09:30,280 But it didn't tell us which way something would go 117 00:09:30,280 --> 00:09:34,620 spontaneously if you just left the system to its own devices 118 00:09:34,620 --> 00:09:36,430 under certain conditions. 119 00:09:36,430 --> 00:09:40,440 This is telling us that. 120 00:09:40,440 --> 00:09:43,450 We can tell which way the system will evolve in which 121 00:09:43,450 --> 00:09:46,680 direction it'll go. 122 00:09:46,680 --> 00:09:50,530 This is for an isolated system and we'll generalize this in 123 00:09:50,530 --> 00:10:00,570 subsequent lectures. 124 00:10:00,570 --> 00:10:06,090 Now, let's look at what happens if the 125 00:10:06,090 --> 00:10:08,920 system isn't isolated. 126 00:10:08,920 --> 00:10:25,710 So now let's go from one to two not isolated. 127 00:10:25,710 --> 00:10:31,550 So of course, delta S I'm going to specify for the 128 00:10:31,550 --> 00:10:36,420 system for reasons that'll soon be obvious is S2 for the 129 00:10:36,420 --> 00:10:39,610 system minus S1 for the system. 130 00:10:39,610 --> 00:10:42,360 Of course it doesn't depend on the path, even though to 131 00:10:42,360 --> 00:10:52,610 calculate it, we need to find a reversible path. 132 00:10:52,610 --> 00:10:55,610 Just to add emphasis to what we just did here, of course, 133 00:10:55,610 --> 00:10:57,590 it was the same thing. 134 00:10:57,590 --> 00:11:02,440 In order to calculate delta S along that irreversible path 135 00:11:02,440 --> 00:11:06,380 for that irreversible process we needed to construct some 136 00:11:06,380 --> 00:11:09,270 reversible path. 137 00:11:09,270 --> 00:11:12,670 And what we did is imagined the reversible path going 138 00:11:12,670 --> 00:11:15,310 backwards and then said, OK, fine then delta S for the 139 00:11:15,310 --> 00:11:18,180 irreversible process must be the opposite of delta S that 140 00:11:18,180 --> 00:11:19,120 we calculated along the 141 00:11:19,120 --> 00:11:21,850 reversible bath going backwards. 142 00:11:21,850 --> 00:11:24,820 The point is that to calculate delta S in either direction, 143 00:11:24,820 --> 00:11:27,690 we needed to construct some reversible path. 144 00:11:27,690 --> 00:11:31,650 All right, and of course it'll be the same here. 145 00:11:31,650 --> 00:11:35,960 Now, in this case since the system isn't isolated, that 146 00:11:35,960 --> 00:11:38,830 means heat's going to flow across the boundary. 147 00:11:38,830 --> 00:11:43,580 So that means heat is going between, it's being exchanged 148 00:11:43,580 --> 00:11:45,110 between the system and the surroundings. 149 00:11:45,110 --> 00:11:47,150 So what's happening to the surroundings? 150 00:11:47,150 --> 00:11:50,430 Generally, we've worry a lot about the system, but not much 151 00:11:50,430 --> 00:11:54,560 in recent discussions about the surroundings. 152 00:11:54,560 --> 00:12:01,520 But delta S surroundings must be something. 153 00:12:01,520 --> 00:12:04,360 We could calculate it. 154 00:12:04,360 --> 00:12:07,500 And that's depends on the path. 155 00:12:07,500 --> 00:12:10,450 Not because it isn't the state function in the surroundings, 156 00:12:10,450 --> 00:12:15,300 of course it is, but because depending on the path, the 157 00:12:15,300 --> 00:12:16,650 final state of the 158 00:12:16,650 --> 00:12:22,060 surroundings will be different. 159 00:12:22,060 --> 00:12:25,550 After all, depending on how much heat got exchanged, the 160 00:12:25,550 --> 00:12:30,970 surroundings got more or less of that heat. 161 00:12:30,970 --> 00:12:33,780 Work might have been done on the surroundings. 162 00:12:33,780 --> 00:12:37,000 So the surroundings are going to respond, are going to 163 00:12:37,000 --> 00:12:40,040 change also, when the system changes in this way and it's 164 00:12:40,040 --> 00:12:41,990 not isolated. 165 00:12:41,990 --> 00:12:44,640 And how the surroundings change will depend on what 166 00:12:44,640 --> 00:12:45,820 path was taken. 167 00:12:45,820 --> 00:12:47,950 And you've seen that before in calculating things like 168 00:12:47,950 --> 00:12:51,450 pressure, volume, work along different paths. 169 00:12:51,450 --> 00:12:54,580 Where you calculated delta u for the system and saw that it 170 00:12:54,580 --> 00:12:59,230 was the same for different paths but q and w were not the 171 00:12:59,230 --> 00:13:01,330 same for the system, and therefore, of course, not for 172 00:13:01,330 --> 00:13:07,710 the surroundings either. 173 00:13:07,710 --> 00:13:21,450 So let's just consider an irreversible path. 174 00:13:21,450 --> 00:13:23,300 And in order to think about what happens to the 175 00:13:23,300 --> 00:13:29,890 surroundings, let's take the entire universe which consists 176 00:13:29,890 --> 00:13:36,550 of the system and the surroundings -- that is a big 177 00:13:36,550 --> 00:13:40,400 isolated system, because it's everything. 178 00:13:40,400 --> 00:13:45,950 So it's not somehow in contact with another additional body 179 00:13:45,950 --> 00:13:48,000 or universe. 180 00:13:48,000 --> 00:13:49,720 We can treat the entire universe 181 00:13:49,720 --> 00:13:54,100 as an isolated system. 182 00:13:54,100 --> 00:14:05,690 So, we're going to think about the entire universe. 183 00:14:05,690 --> 00:14:13,890 We can think about it as an isolated system. 184 00:14:13,890 --> 00:14:16,050 Great. 185 00:14:16,050 --> 00:14:23,400 And we know that delta S for an isolated system, for a 186 00:14:23,400 --> 00:14:28,230 spontaneous process, is greater than zero. 187 00:14:28,230 --> 00:14:30,130 That's right there. 188 00:14:30,130 --> 00:14:38,050 So delta S for the universe which is delta S for the 189 00:14:38,050 --> 00:14:44,950 system plus delta S for the surroundings, and I'll specify 190 00:14:44,950 --> 00:14:49,130 since we're considering an irreversible process, specify 191 00:14:49,130 --> 00:14:52,150 that, it's greater than zero. 192 00:14:52,150 --> 00:14:55,390 The whole thing happened spontaneously, and the whole 193 00:14:55,390 --> 00:14:58,790 universe is an isolated system. 194 00:14:58,790 --> 00:15:03,190 So if we wanted to, we can write that delta S 195 00:15:03,190 --> 00:15:07,320 irreversible for the surroundings must be bigger 196 00:15:07,320 --> 00:15:15,340 than minus delta S for the system in order that there sum 197 00:15:15,340 --> 00:15:19,590 be positive. 198 00:15:19,590 --> 00:15:21,640 All right? 199 00:15:21,640 --> 00:15:33,150 Now, if we consider a reversible process, then of 200 00:15:33,150 --> 00:15:37,160 course we already know what happens for an isolated system 201 00:15:37,160 --> 00:15:43,450 in a reversible process, delta S is zero. 202 00:15:43,450 --> 00:15:47,820 So delta S in this case of the whole universe, which is delta 203 00:15:47,820 --> 00:15:53,860 S for the system plus delta S for the surroundings in the 204 00:15:53,860 --> 00:16:02,210 reversible case must be equal to zero. 205 00:16:02,210 --> 00:16:08,050 In other words, delta S of the surroundings in the reversible 206 00:16:08,050 --> 00:16:15,350 case is exactly the opposite of delta S of the system. 207 00:16:15,350 --> 00:16:18,230 Delta S of the system is the same in either case because 208 00:16:18,230 --> 00:16:21,380 reversible or irreversible, we're specifying the system, 209 00:16:21,380 --> 00:16:24,940 goes between the same states, but what happens to the 210 00:16:24,940 --> 00:16:30,790 surroundings is different in the two cases. 211 00:16:30,790 --> 00:16:36,590 So one way of recasting this statement or the Clausius 212 00:16:36,590 --> 00:16:40,080 inequality is the entropy of the whole 213 00:16:40,080 --> 00:16:44,720 universe never decreases. 214 00:16:44,720 --> 00:16:47,370 Of course, in some sense, that's obvious from this 215 00:16:47,370 --> 00:16:49,800 statement, since we're considering the universe an 216 00:16:49,800 --> 00:16:51,670 isolated system. 217 00:16:51,670 --> 00:16:56,900 So the entropy in any process can, it can, the change in 218 00:16:56,900 --> 00:17:04,950 entropy can either be zero or positive and never negative. 219 00:17:10,220 --> 00:17:27,700 So entropy off the whole universe never decreases. 220 00:17:27,700 --> 00:17:30,610 That's actually a really quite profound conclusion. 221 00:17:30,610 --> 00:17:36,580 All sorts of philosophical and other consequences of it that 222 00:17:36,580 --> 00:17:39,670 are sometimes considered in things like evolutionary 223 00:17:39,670 --> 00:17:41,090 biology and other things. 224 00:17:41,090 --> 00:17:43,135 And how things can change altogether an 225 00:17:43,135 --> 00:17:48,710 entire system of things. 226 00:17:48,710 --> 00:17:55,140 Now, I'd like to just go through a few calculations of 227 00:17:55,140 --> 00:17:59,140 delta S for common processes. 228 00:17:59,140 --> 00:18:01,970 We've seen a few general features of entropy. 229 00:18:01,970 --> 00:18:05,980 Let's calculate it for a few things. 230 00:18:05,980 --> 00:18:15,640 So calculations of delta S. And we 231 00:18:15,640 --> 00:18:17,700 know the general procedure. 232 00:18:17,700 --> 00:18:20,240 Whatever the process is, we're going to need to find 233 00:18:20,240 --> 00:18:24,830 reversible paths, and along a reversible path then, we'll be 234 00:18:24,830 --> 00:18:28,480 able to calculate it. 235 00:18:28,480 --> 00:18:39,230 One, let's just take two pieces of material at 236 00:18:39,230 --> 00:18:43,590 different temperatures. 237 00:18:43,590 --> 00:18:56,350 The entire system is going to be isolated. 238 00:18:56,350 --> 00:19:01,010 And then let's connect them. 239 00:19:01,010 --> 00:19:02,830 And then let's connect them. 240 00:19:02,830 --> 00:19:03,480 Put some material between them. 241 00:19:03,480 --> 00:19:05,500 So now heat can flow from one to the other. 242 00:19:05,500 --> 00:19:07,410 And you know what's going to happen which is if they start 243 00:19:07,410 --> 00:19:09,340 out at unequal temperatures, and you wait a little while, 244 00:19:09,340 --> 00:19:11,020 eventually they're going to wind up at the same 245 00:19:11,020 --> 00:19:12,810 temperature. 246 00:19:12,810 --> 00:19:14,850 In other words, you know the direction 247 00:19:14,850 --> 00:19:17,950 of spontaneous change. 248 00:19:17,950 --> 00:19:23,950 So let's just see what happens to delta S. 249 00:19:23,950 --> 00:19:33,550 So isolated system, so there's no work being done. 250 00:19:33,550 --> 00:19:35,250 There's no heat that's being exchanged. 251 00:19:35,250 --> 00:19:42,060 So delta u is zero. 252 00:19:42,060 --> 00:19:44,810 What's dS? 253 00:19:44,810 --> 00:19:49,360 Well it's just the sum of the changes in entropy for the two 254 00:19:49,360 --> 00:19:53,450 sub-systems the two separate pieces. 255 00:19:53,450 --> 00:20:03,880 So it's dq1 over T1 plus dq2 over T2, I think there's a 256 00:20:03,880 --> 00:20:11,720 type in your notes. it's plus not minus of course. 257 00:20:11,720 --> 00:20:16,410 But in this case, dq1 is just the opposite of dq2. 258 00:20:16,410 --> 00:20:20,280 Because whatever heat is flowing into T1, into this 259 00:20:20,280 --> 00:20:22,770 block, is coming out of this block, or vice versa, 260 00:20:22,770 --> 00:20:32,490 depending on which one is hotter. 261 00:20:32,490 --> 00:20:39,070 So this is just equal to dq1 times one over T1 minus one 262 00:20:39,070 --> 00:20:45,700 over T2, which is T2 minus T1 over T1 times T2. 263 00:20:49,460 --> 00:20:53,910 So that's dS, and then we could integrate it to get the 264 00:20:53,910 --> 00:20:56,380 delta S, but this is sufficient for what I'd like 265 00:20:56,380 --> 00:20:59,850 to illustrate, which is just the sign of things. 266 00:20:59,850 --> 00:21:09,140 We know dS must be greater than zero for the spontaneous 267 00:21:09,140 --> 00:21:14,920 change that's going to occur. 268 00:21:14,920 --> 00:21:18,860 So let's just make sure that common sense prevails here. 269 00:21:18,860 --> 00:21:23,810 That says if T2 is greater than T1, right T2 is hotter. 270 00:21:23,810 --> 00:21:27,110 Then dq must be positive, so this is positive, this is 271 00:21:27,110 --> 00:21:30,240 positive, dS we know has to be positive. 272 00:21:30,240 --> 00:21:35,440 So, it says T2 is hotter. 273 00:21:35,440 --> 00:21:40,790 This being positive means heat is flowing in from this one to 274 00:21:40,790 --> 00:21:45,230 this one, from the hotter to the colder body. 275 00:21:45,230 --> 00:21:47,210 So it looks right. 276 00:21:47,210 --> 00:21:51,780 If T2 is lower than T1, if this one is colder, then in 277 00:21:51,780 --> 00:21:55,120 order for dS to be positive, which it has to be, this 278 00:21:55,120 --> 00:21:57,210 better be negative. 279 00:21:57,210 --> 00:22:01,160 In other words, again heat is going to flow from the hotter 280 00:22:01,160 --> 00:22:05,810 toward the colder, into the colder body. 281 00:22:05,810 --> 00:22:11,890 So, as promised, the condition that dS must be greater than 282 00:22:11,890 --> 00:22:17,160 zero, guides us and tells us the direction that spontaneous 283 00:22:17,160 --> 00:22:21,050 change is going to occur. 284 00:22:21,050 --> 00:22:24,170 We can tell just from this condition which way things 285 00:22:24,170 --> 00:22:44,530 have to evolve. 286 00:22:44,530 --> 00:22:48,120 All right, let's try another pretty simple process. 287 00:22:48,120 --> 00:22:59,690 Let's just take a gas in some volume V and over here is 288 00:22:59,690 --> 00:23:05,400 going to be vacuum of equal volume, and we're going to 289 00:23:05,400 --> 00:23:07,630 remove the barrier. 290 00:23:07,630 --> 00:23:10,260 You know what's going to happen spontaneously, right? 291 00:23:10,260 --> 00:23:14,490 The gas is going to fill the available volume once it 292 00:23:14,490 --> 00:23:16,540 becomes available. 293 00:23:16,540 --> 00:23:23,870 So this, we'll make this a Joule expansion, and we'll 294 00:23:23,870 --> 00:23:29,230 make it an ideal gas. 295 00:23:29,230 --> 00:23:37,590 So the process is one mole of gas at our initial volume and 296 00:23:37,590 --> 00:23:46,200 some temperature, and we'll make this adiabatic, and this 297 00:23:46,200 --> 00:23:53,360 goes to one mole of gas at a new volume, 2V and at the same 298 00:23:53,360 --> 00:23:55,750 temperature. 299 00:23:55,750 --> 00:24:01,350 All right, so there's no work done. 300 00:24:01,350 --> 00:24:02,220 It's adiabatic. 301 00:24:02,220 --> 00:24:03,440 There's no heat exchanged. 302 00:24:03,440 --> 00:24:08,330 Delta u is zero. 303 00:24:08,330 --> 00:24:10,870 Everything's zero. 304 00:24:10,870 --> 00:24:13,700 Almost everything is zero. 305 00:24:13,700 --> 00:24:16,360 What isn't zero? 306 00:24:16,360 --> 00:24:16,680 STUDENT: Entropy. 307 00:24:16,680 --> 00:24:17,300 PROFESSOR NELSON: Entropy. 308 00:24:17,300 --> 00:24:19,860 It's going to tell us which way this whole thing goes. 309 00:24:19,860 --> 00:24:22,730 So this is an irreversible process. 310 00:24:22,730 --> 00:24:26,020 In order to calculate delta S we need to construct a 311 00:24:26,020 --> 00:24:30,790 reversible path along which this can go. 312 00:24:30,790 --> 00:24:34,830 So here's a way to do it. 313 00:24:34,830 --> 00:24:38,480 We could compress it back, isothermally and reversibly. 314 00:24:38,480 --> 00:24:41,190 Just like the example I gave, the general example, where I 315 00:24:41,190 --> 00:24:42,750 mentioned the ice melting. 316 00:24:42,750 --> 00:24:44,140 In that case, it won't be adiabatic. 317 00:24:44,140 --> 00:24:48,240 We'd have to put it in contact with a heat 318 00:24:48,240 --> 00:24:51,840 source of some sort. 319 00:24:51,840 --> 00:25:00,420 So reversible process. 320 00:25:00,420 --> 00:25:21,870 We could compress it back to volume V, isothermally, and 321 00:25:21,870 --> 00:25:22,880 reversibly. 322 00:25:22,880 --> 00:25:25,860 Let me say, there's no -- we wouldn't have to consider this 323 00:25:25,860 --> 00:25:26,880 process in reverse. 324 00:25:26,880 --> 00:25:29,400 Of course, we could also consider the forward process 325 00:25:29,400 --> 00:25:34,480 in the presence of some heat source with isn't isolated and 326 00:25:34,480 --> 00:25:37,670 do this, but let's just consider the compression. 327 00:25:37,670 --> 00:25:46,450 So, OK, now we'll have the -- well, once 328 00:25:46,450 --> 00:25:48,930 again, delta u is zero. 329 00:25:48,930 --> 00:25:50,090 It's an ideal gas. 330 00:25:50,090 --> 00:25:52,530 There's no temperature change. 331 00:25:52,530 --> 00:25:55,150 But this time q and w aren't going to be zero. 332 00:25:55,150 --> 00:25:57,270 They're going to be some finite numbers that are 333 00:25:57,270 --> 00:25:59,310 opposites of each other. 334 00:25:59,310 --> 00:26:05,750 Because we're no longer isolated. 335 00:26:05,750 --> 00:26:19,360 So let's just write out delta S to go backward is dq 336 00:26:19,360 --> 00:26:28,210 reversible over T, and that's minus dw 337 00:26:28,210 --> 00:26:32,550 over T, not zero anymore. 338 00:26:32,550 --> 00:26:39,420 And where we're going is from 2V to V, we're compressing, 339 00:26:39,420 --> 00:26:41,670 and it's an ideal gas. 340 00:26:41,670 --> 00:26:59,620 So this is the same as p dV, but it's p dV over T which is 341 00:26:59,620 --> 00:27:02,960 the same thing as going from 2V to V. Now we're going to 342 00:27:02,960 --> 00:27:07,110 replace this with RT over V, and the T's will cancel. 343 00:27:07,110 --> 00:27:16,770 So it's R dV over V going from 2V to one V. So it's just R 344 00:27:16,770 --> 00:27:23,440 times the log of 1/2. 345 00:27:23,440 --> 00:27:27,190 So now we've just calculated the value of delta S by 346 00:27:27,190 --> 00:27:30,080 constructing this reversible path. 347 00:27:30,080 --> 00:27:33,260 Now this is delta S backward. 348 00:27:33,260 --> 00:27:38,070 To do the compression, so of course delta S forward is just 349 00:27:38,070 --> 00:27:46,930 the opposite of that. 350 00:27:46,930 --> 00:27:58,460 So, delta S forward is R log 2. 351 00:27:58,460 --> 00:28:12,920 Reassuringly, that's a positive number. 352 00:28:12,920 --> 00:28:16,830 So this process that we considered originally, this 353 00:28:16,830 --> 00:28:20,370 irreversibly process in this isolated system, remember, for 354 00:28:20,370 --> 00:28:23,720 the irreversible expansion we considered the system 355 00:28:23,720 --> 00:28:26,880 isolated, Happened spontaneously, and of course 356 00:28:26,880 --> 00:28:30,430 you know that's the case. 357 00:28:30,430 --> 00:28:37,570 Now, I'll just mention, and later on we'll go into this in 358 00:28:37,570 --> 00:28:40,310 considerably more detail, but I'll just mention at this 359 00:28:40,310 --> 00:28:45,130 stage through a little bit of sort of foreshadowing that 360 00:28:45,130 --> 00:28:49,040 everything we've discussed so far about entropy is with a 361 00:28:49,040 --> 00:28:50,720 macroscopic picture. 362 00:28:50,720 --> 00:28:54,010 We started with heat engines and Carnot cycles and 363 00:28:54,010 --> 00:28:56,690 macroscopic processes, reversible and irreversible 364 00:28:56,690 --> 00:28:58,790 processes and so on. 365 00:28:58,790 --> 00:29:02,620 But there's a, and that's certainly the way it was all 366 00:29:02,620 --> 00:29:04,080 formulated originally. 367 00:29:04,080 --> 00:29:06,960 Part of the power of thermodynamics is that it 368 00:29:06,960 --> 00:29:12,190 doesn't depend on a specific microscopic model of matter. 369 00:29:12,190 --> 00:29:14,220 That said though, we have a pretty good 370 00:29:14,220 --> 00:29:16,670 microscopic model of matter. 371 00:29:16,670 --> 00:29:20,170 We know about atoms and molecules, and in fact there 372 00:29:20,170 --> 00:29:25,890 is an entirely microscopic formulation of entropy that 373 00:29:25,890 --> 00:29:31,310 has to do with disorder and the number of available 374 00:29:31,310 --> 00:29:36,920 states, microscopic states available to a system. 375 00:29:36,920 --> 00:29:42,050 So I'll just state what we'll see in much more 376 00:29:42,050 --> 00:29:44,010 detail later on. 377 00:29:44,010 --> 00:29:56,940 Which is in microscopic terms, it's going to turn out that 378 00:29:56,940 --> 00:30:04,270 the entropy of a system can be given by R over Avogadro's 379 00:30:04,270 --> 00:30:21,160 number times the log of the number of distinct microscopic 380 00:30:21,160 --> 00:30:24,840 states that are available to a system. 381 00:30:24,840 --> 00:30:28,030 Right so in this isolated system, when we double that 382 00:30:28,030 --> 00:30:34,240 volume, each molecule suddenly has twice as many states 383 00:30:34,240 --> 00:30:36,430 available to it. 384 00:30:36,430 --> 00:30:38,140 You could formulate that in a lot of ways. 385 00:30:38,140 --> 00:30:42,350 You can sort of divide up the volume into tiny little 386 00:30:42,350 --> 00:30:46,240 molecule-sized cubes and realize well, now, there are 387 00:30:46,240 --> 00:30:49,730 twice as many of them. 388 00:30:49,730 --> 00:30:52,950 Now if I have one molecule, the number of states doubles. 389 00:30:52,950 --> 00:30:56,080 If I have n molecules, the number of states goes 390 00:30:56,080 --> 00:30:57,540 up by 2 to the n. 391 00:30:57,540 --> 00:30:59,690 Because for each one of the molecules, I can select any 392 00:30:59,690 --> 00:31:01,150 one of those states, and then the next one I 393 00:31:01,150 --> 00:31:04,110 can select any one. 394 00:31:04,110 --> 00:31:05,570 So you have this enormous increase in 395 00:31:05,570 --> 00:31:08,370 the number of states. 396 00:31:08,370 --> 00:31:19,740 So delta S in this case looks like R over NA log of 2 to the 397 00:31:19,740 --> 00:31:22,110 power NA, it's Avogadro's number, 398 00:31:22,110 --> 00:31:25,300 for a mole of molecules. 399 00:31:25,300 --> 00:31:30,690 This can come out and cancel this. 400 00:31:30,690 --> 00:31:34,560 There's our R log 2 result. 401 00:31:34,560 --> 00:31:37,210 But the point I'm making here is that could actually be 402 00:31:37,210 --> 00:31:42,800 derived from entirely microscopic consideration. 403 00:31:42,800 --> 00:31:44,580 We haven't gone through that yet. 404 00:31:44,580 --> 00:31:48,040 We've done entirely macroscopic thermodynamics so 405 00:31:48,040 --> 00:31:51,480 far, but we'll do a little bit of treatment of statistical 406 00:31:51,480 --> 00:31:56,000 mechanics, is what is called the whole microscopic 407 00:31:56,000 --> 00:31:59,620 formulation of entropy in thermodynamics. 408 00:31:59,620 --> 00:32:03,680 And that's a really important fundamental result. 409 00:32:03,680 --> 00:32:08,540 And we'll get to it in much more detail in a while. 410 00:32:08,540 --> 00:32:12,580 But for now, I just want to continue calculating entropy 411 00:32:12,580 --> 00:32:31,680 changes for a few more kinds of processes. 412 00:32:31,680 --> 00:32:37,940 So, how about if we have two different substances and we 413 00:32:37,940 --> 00:32:50,280 mix them, entropy of mixing. 414 00:32:50,280 --> 00:33:00,530 So, we start with some number of moles of substance A in 415 00:33:00,530 --> 00:33:05,480 some volume VA, and some other number of moles of substance B 416 00:33:05,480 --> 00:33:09,420 in volume VB, separated. 417 00:33:09,420 --> 00:33:11,800 And then we remove the barrier, of course we know 418 00:33:11,800 --> 00:33:14,580 they're going to mix. 419 00:33:14,580 --> 00:33:19,940 So we'll have some total number of moles, nA plus nB, 420 00:33:19,940 --> 00:33:23,650 and they'll be filling some total volume, which is the sum 421 00:33:23,650 --> 00:33:26,940 of the two original volumes. 422 00:33:26,940 --> 00:33:28,750 Certainly this is what we expect to happen 423 00:33:28,750 --> 00:33:30,760 spontaneously. 424 00:33:30,760 --> 00:33:41,370 So are process is nA moles of A, gas, initial volume VA and 425 00:33:41,370 --> 00:33:48,380 T, plus nB of substance B, I think this is miswritten as A 426 00:33:48,380 --> 00:33:49,980 there in your notes. 427 00:33:49,980 --> 00:34:01,780 Gas VB and T goes to nA moles of A plus nB moles of B. Gas, 428 00:34:01,780 --> 00:34:11,490 total volume and T. And we'll have constant 429 00:34:11,490 --> 00:34:15,310 temperature and pressure. 430 00:34:15,310 --> 00:34:19,350 Well, there was no p v worked done. 431 00:34:19,350 --> 00:34:22,370 The temperature didn't change and it's ideal gases, so delta 432 00:34:22,370 --> 00:34:28,340 u is zero, but what about delta S? 433 00:34:28,340 --> 00:34:30,880 Well again, to calculate it, we need to find 434 00:34:30,880 --> 00:34:31,890 a reversible path. 435 00:34:31,890 --> 00:34:34,200 This is irreversible if we just remove the barrier and 436 00:34:34,200 --> 00:34:37,080 let it all go. 437 00:34:37,080 --> 00:34:40,620 But we could at least imagine a reversible path, and we 438 00:34:40,620 --> 00:34:44,730 could actually construct this for the right substances. 439 00:34:44,730 --> 00:34:53,560 We could imagine that we can find some sort of piston here. 440 00:34:53,560 --> 00:35:00,920 And we're going to prepare to push it inward, with a 441 00:35:00,920 --> 00:35:07,880 membrane that's permeable only to b. 442 00:35:07,880 --> 00:35:14,410 And over here we could set up the exact same thing, but this 443 00:35:14,410 --> 00:35:23,580 one's permeable only to a. 444 00:35:23,580 --> 00:35:27,840 And then we could just push them to recover this state. 445 00:35:27,840 --> 00:35:29,470 And we can do it reversibly. 446 00:35:29,470 --> 00:35:33,480 So we could construct a reversible process that does 447 00:35:33,480 --> 00:35:36,900 the reverse of this. 448 00:35:36,900 --> 00:35:40,810 And in this case, it'll be isothermal still, constant 449 00:35:40,810 --> 00:35:45,950 pressure, reversible compression of the two gases. 450 00:35:45,950 --> 00:36:22,580 So, reversible compression and demixing. 451 00:36:22,580 --> 00:36:23,980 All right, so what happens? 452 00:36:23,980 --> 00:36:28,800 Well, of course, our delta S of demixing is going to be 453 00:36:28,800 --> 00:36:34,900 minus delta S of mixing. 454 00:36:34,900 --> 00:36:40,800 Delta u of demixing is still zero. 455 00:36:40,800 --> 00:36:47,370 Temperature didn't change, ideal gases, and there's some 456 00:36:47,370 --> 00:36:52,000 reversible work and heat. 457 00:36:52,000 --> 00:37:04,730 So dw in this reversible case, is just minus pA dVA minus pB 458 00:37:04,730 --> 00:37:12,940 dVB, the infinitesimal amount of work done as these things 459 00:37:12,940 --> 00:37:15,380 are gradually moved toward each other. 460 00:37:15,380 --> 00:37:27,000 So delta S for demixing, of course it's dq reversible over 461 00:37:27,000 --> 00:37:30,960 T, but just like we did right there, of course, in this case 462 00:37:30,960 --> 00:37:33,070 that's just negative dw. 463 00:37:33,070 --> 00:37:40,390 So it's integral from V to VA, right, were compressing. 464 00:37:40,390 --> 00:37:43,320 So for substance a, we're going to go back to there. 465 00:37:43,320 --> 00:37:45,620 It was occupying the whole volume, and then it's going to 466 00:37:45,620 --> 00:37:48,340 end up in only part of the volume, VA. 467 00:37:48,340 --> 00:38:06,840 pA dVA over T. Same thing for B, going to volume B, pB dVB 468 00:38:06,840 --> 00:38:11,980 over T. And now we can substitute for pressure, 469 00:38:11,980 --> 00:38:22,360 right? pV is nRT, so it's nA times R log of VA over V plus 470 00:38:22,360 --> 00:38:33,580 nB of R log of VB over V. 471 00:38:33,580 --> 00:38:37,720 Now, so this is a suitable answer, but I'm going to make 472 00:38:37,720 --> 00:38:42,020 if a little easier by putting in terms of mole fractions. 473 00:38:42,020 --> 00:39:05,700 So, mole fractions, of course, XA is nA over n, and XB is nB 474 00:39:05,700 --> 00:39:13,750 over n and XA is also equal to VA over V and XB is equal to 475 00:39:13,750 --> 00:39:17,140 VB over V for ideal gases, right? 476 00:39:17,140 --> 00:39:18,940 The molar volumes are the same. 477 00:39:18,940 --> 00:39:21,600 We're at the same temperature and pressure. 478 00:39:21,600 --> 00:39:32,140 So that immediately gives us the delta X of demixing is nR 479 00:39:32,140 --> 00:39:42,660 times XA log XA plus XB log XB. 480 00:39:42,660 --> 00:39:47,030 And of course delta S of mixing is the opposite of 481 00:39:47,030 --> 00:39:58,930 this, minus nR XA log XA plus XB log XB. 482 00:39:58,930 --> 00:40:01,520 Now XA and XB are mole fractions. 483 00:40:01,520 --> 00:40:05,880 They are between zero and one, So their log 484 00:40:05,880 --> 00:40:08,590 rhythms are both negative. 485 00:40:08,590 --> 00:40:10,160 There's a negative sign. 486 00:40:10,160 --> 00:40:13,660 So delta S of mixing is positive. 487 00:40:13,660 --> 00:40:15,070 That's reassuring. 488 00:40:15,070 --> 00:40:18,530 That tells us as we expect that the mixing should happen 489 00:40:18,530 --> 00:40:28,130 spontaneously and irreversibly when we remove the barrier. 490 00:40:28,130 --> 00:40:31,130 Any questions so far about how we're going about these 491 00:40:31,130 --> 00:40:34,640 calculations of delta S? 492 00:40:34,640 --> 00:40:52,740 OK, let's just do a couple more examples. 493 00:40:52,740 --> 00:40:54,900 Here's a really straightforward one. 494 00:40:54,900 --> 00:40:58,270 What if we just heat stuff up or cool it down. 495 00:40:58,270 --> 00:41:02,430 It happens all the time, but it's pretty important, right? 496 00:41:02,430 --> 00:41:12,320 So let's just heat or cool at constant volume. 497 00:41:12,320 --> 00:41:17,120 So, we're going to go from substance A at T1 and some 498 00:41:17,120 --> 00:41:22,280 volume going to substance A at T2 at some volume. 499 00:41:22,280 --> 00:41:25,940 We can do this. 500 00:41:25,940 --> 00:41:35,520 Delta S is integral of dq reversible over T. Going from 501 00:41:35,520 --> 00:41:43,400 T1 to T2, but we know how to do this, right? 502 00:41:43,400 --> 00:41:44,890 It's the heat that we need. 503 00:41:44,890 --> 00:41:47,710 It's just given by the constant volume heat capacity 504 00:41:47,710 --> 00:41:49,320 times the change in temperature. 505 00:41:49,320 --> 00:42:04,230 So this is just T1 to T2, Cv dT over T. And that is just Cv 506 00:42:04,230 --> 00:42:16,990 times the log of T2 over T1 if Cv is temperature independent. 507 00:42:16,990 --> 00:42:18,720 That's not always the case. 508 00:42:18,720 --> 00:42:21,770 Usually if it's not too big an excursion of temperature then 509 00:42:21,770 --> 00:42:25,950 it's a reasonable approximation. 510 00:42:25,950 --> 00:42:33,080 So that's straightforward. 511 00:42:33,080 --> 00:42:43,560 By the way, notice that delta S is greater than zero if T2 512 00:42:43,560 --> 00:42:44,780 is greater than T1. 513 00:42:44,780 --> 00:42:45,080 1 514 00:42:45,080 --> 00:42:50,060 In other words, if we heated it up, delta S is positive. 515 00:42:50,060 --> 00:42:55,040 Notice also delta S is less than zero if T2 516 00:42:55,040 --> 00:42:57,880 is less than T1. 517 00:42:57,880 --> 00:43:01,860 Delta S is negative if we cooled it. 518 00:43:01,860 --> 00:43:06,520 Now, under certain conditions we've seen that delta S can't 519 00:43:06,520 --> 00:43:07,950 be negative. 520 00:43:07,950 --> 00:43:09,930 Why can it be negative here? 521 00:43:09,930 --> 00:43:14,530 STUDENT: [UNINTELLIGIBLE]. 522 00:43:14,530 --> 00:43:17,250 PROFESSOR NELSON: Yes, I heard it. 523 00:43:17,250 --> 00:43:19,810 It's because the system isn't isolated, right. 524 00:43:19,810 --> 00:43:23,290 It was for the isolated system that delta S is always greater 525 00:43:23,290 --> 00:43:24,590 than or equal to zero. 526 00:43:24,590 --> 00:43:27,020 And you know, for the whole universe entropy never 527 00:43:27,020 --> 00:43:30,210 decreases and so on. 528 00:43:30,210 --> 00:43:32,980 In this case, though, you know we're taking a system and 529 00:43:32,980 --> 00:43:35,510 we're putting it in contact with a cold bath 530 00:43:35,510 --> 00:43:36,380 and cooling it down. 531 00:43:36,380 --> 00:43:38,150 It's certainly not isolated. 532 00:43:38,150 --> 00:43:39,830 Heat is being exchanged. 533 00:43:39,830 --> 00:43:43,880 It's going from the system to the surroundings, to the bath. 534 00:43:43,880 --> 00:43:44,570 So it's fine. 535 00:43:44,570 --> 00:43:48,080 Delta S can be negative. 536 00:43:48,080 --> 00:43:53,110 What would happen if we calculated delta S of a new 537 00:43:53,110 --> 00:43:55,860 entire system consisting of the original system 538 00:43:55,860 --> 00:43:58,810 plus the heat bath? 539 00:43:58,810 --> 00:44:00,120 What would delta S be then? 540 00:44:00,120 --> 00:44:07,790 We already kind of did that with those two blocks, T1 and 541 00:44:07,790 --> 00:44:11,030 T2 in an isolated systems. 542 00:44:11,030 --> 00:44:13,570 What do you think is going to happen? 543 00:44:13,570 --> 00:44:17,690 In other words, if we call our system the stuff that is put 544 00:44:17,690 --> 00:44:21,010 in contact with the heat bath, well OK, then we've seen delta 545 00:44:21,010 --> 00:44:24,570 S can be plus, can be positive or negative. 546 00:44:24,570 --> 00:44:28,200 Now let's call our system the original system plus the heat 547 00:44:28,200 --> 00:44:32,160 bath, and we'll put them all in some isolating box, that's 548 00:44:32,160 --> 00:44:34,080 thermally isolating and so on. 549 00:44:34,080 --> 00:44:37,020 And now we'll put our original system, which is now sort of a 550 00:44:37,020 --> 00:44:40,590 sub, part of the system, and it's now going to be in 551 00:44:40,590 --> 00:44:45,450 contact with the cold bath. 552 00:44:45,450 --> 00:44:48,440 The bath that it's in contact with is colder than it is. 553 00:44:48,440 --> 00:44:50,920 That's why its delta S is less than zero. 554 00:44:50,920 --> 00:44:53,950 But what's delta S to be for the entire new system 555 00:44:53,950 --> 00:44:58,350 including the cold bath? 556 00:44:58,350 --> 00:44:59,740 STUDENT: Wouldn't it greater than zero? 557 00:44:59,740 --> 00:45:01,540 PROFESSOR NELSON: Yes, it's going to be positive, because 558 00:45:01,540 --> 00:45:04,000 it's an isolated system and something happened 559 00:45:04,000 --> 00:45:09,040 spontaneously in it, namely the original system got cooler 560 00:45:09,040 --> 00:45:11,250 and presumably the heat bath got at least 561 00:45:11,250 --> 00:45:12,900 a little bit warmer. 562 00:45:12,900 --> 00:45:15,370 And it happened spontaneously, which immediately means delta 563 00:45:15,370 --> 00:45:20,660 S had to be positive for that entire assembly, of you know 564 00:45:20,660 --> 00:45:25,160 the original system plus the bath. 565 00:45:25,160 --> 00:45:32,580 Here's another important sort of process. 566 00:45:32,580 --> 00:45:44,110 Reversible phase change, we'll melt something or freeze it. 567 00:45:44,110 --> 00:45:50,190 How about water -- liquid 100 degrees Celsius 568 00:45:50,190 --> 00:45:54,270 one bar goes to water. 569 00:45:54,270 --> 00:46:01,500 Gas 100 degrees Celsius one bar. 570 00:46:01,500 --> 00:46:05,510 Well in that case, you know what the heat is. 571 00:46:05,510 --> 00:46:11,750 It's just the delta H of vaporization, right, at 572 00:46:11,750 --> 00:46:15,060 constant pressure. 573 00:46:15,060 --> 00:46:17,550 So there's nothing really much to calculate here. 574 00:46:17,550 --> 00:46:27,480 Delta S of vaporization for H2O at 100 degrees Celsius is 575 00:46:27,480 --> 00:46:36,080 just q of vaporization over the boiling temperature, which 576 00:46:36,080 --> 00:46:43,820 is delta H of vaporization over the boiling temperature. 577 00:46:43,820 --> 00:46:45,800 So that was easy. 578 00:46:45,800 --> 00:46:49,320 One last thing. 579 00:46:49,320 --> 00:46:54,130 What if we want to calculate the entropy of melting or some 580 00:46:54,130 --> 00:46:56,870 phase transition not at equilibrium? 581 00:46:56,870 --> 00:46:59,920 In this case it's at equilibrium because it's water 582 00:46:59,920 --> 00:47:04,190 and water vapor at 100 degrees Celsius and one 583 00:47:04,190 --> 00:47:05,870 atmosphere or one bar. 584 00:47:05,870 --> 00:47:08,710 So it's at its boiling point. 585 00:47:08,710 --> 00:47:14,690 But what if it isn't at its boiling point? 586 00:47:14,690 --> 00:47:24,570 So what if instead, let's take H2O, liquid, at minus ten 587 00:47:24,570 --> 00:47:29,580 degrees Celsius and one bar so it's cold. 588 00:47:29,580 --> 00:47:31,300 It's going to freeze. 589 00:47:31,300 --> 00:47:35,620 So it's going to turn into solid water at minus 10 590 00:47:35,620 --> 00:47:38,540 degrees Celsius and one bar, we're going to have it be 591 00:47:38,540 --> 00:47:40,930 isothermal. 592 00:47:40,930 --> 00:47:44,320 Now you know that's going to happen spontaneously, and it's 593 00:47:44,320 --> 00:47:52,730 irreversible. 594 00:47:52,730 --> 00:47:56,320 Irreversible, which means we can't just straight away 595 00:47:56,320 --> 00:47:59,520 calculate delta S along this path because it's 596 00:47:59,520 --> 00:48:00,760 irreversible. 597 00:48:00,760 --> 00:48:04,100 But what we can do, just like we've done before for making 598 00:48:04,100 --> 00:48:07,610 cycles, we can make some other set of events that are all 599 00:48:07,610 --> 00:48:08,240 reversible. 600 00:48:08,240 --> 00:48:12,410 We can construct a reversible pass that will get there. 601 00:48:12,410 --> 00:48:13,560 How do we do it? 602 00:48:13,560 --> 00:48:17,020 Well you know for the phase change, for the freezing to be 603 00:48:17,020 --> 00:48:20,890 reversible, it has to happen at zero degrees Celsius right. 604 00:48:20,890 --> 00:48:23,030 You know that's where you have reversible, 605 00:48:23,030 --> 00:48:25,230 freezing and melting. 606 00:48:25,230 --> 00:48:28,750 Reversible equilibrium between liquid and solid water. 607 00:48:28,750 --> 00:48:33,160 So surely that's got to get involved here. 608 00:48:33,160 --> 00:48:39,520 H2O liquid zero degrees Celsius, one bar, in 609 00:48:39,520 --> 00:48:44,550 equilibrium with H2O solid at zero degrees Celsius, one bar. 610 00:48:44,550 --> 00:48:48,590 Whatever reversible path we construct that's going to go 611 00:48:48,590 --> 00:48:52,640 from liquid to solid water, somewhere along the set of 612 00:48:52,640 --> 00:48:55,570 steps, that better be one of them. 613 00:48:55,570 --> 00:48:58,500 So now what we need to do is go from liquid at minus 10 614 00:48:58,500 --> 00:49:02,850 degrees Celsius and one bar to liquid at zero degrees Celsius 615 00:49:02,850 --> 00:49:05,520 and one bar. 616 00:49:05,520 --> 00:49:08,140 That's called heating. 617 00:49:08,140 --> 00:49:09,770 We have to heat it. 618 00:49:09,770 --> 00:49:15,240 And here we have to cool it. 619 00:49:15,240 --> 00:49:18,170 And we've already seen all these three processes, right. 620 00:49:18,170 --> 00:49:23,890 So for the phase change, it's reversible now. 621 00:49:23,890 --> 00:49:26,790 This is just delta heat of fusion. 622 00:49:26,790 --> 00:49:30,130 Great. 623 00:49:30,130 --> 00:49:33,490 Because this is now reversible. 624 00:49:33,490 --> 00:49:35,010 This is going to be reversible. 625 00:49:35,010 --> 00:49:37,400 This is going to be reversible. 626 00:49:37,400 --> 00:49:38,320 Great. 627 00:49:38,320 --> 00:49:43,630 So dq reversible, it's going to be the heat capacity at 628 00:49:43,630 --> 00:49:46,350 constant pressure, the example we did before it was constant 629 00:49:46,350 --> 00:49:51,380 volume for the liquid dT. 630 00:49:51,380 --> 00:49:59,650 Here, dq reversible is going to be Cp solid, dT. 631 00:49:59,650 --> 00:50:01,180 Those won't be the same right? 632 00:50:01,180 --> 00:50:03,600 It's one thing to say the heat capacity of something doesn't 633 00:50:03,600 --> 00:50:07,170 change over some small excursion in temperature, but 634 00:50:07,170 --> 00:50:09,520 it's another thing when the thing actually changes phase. 635 00:50:09,520 --> 00:50:11,350 The heat capacity of the solid and the liquid 636 00:50:11,350 --> 00:50:14,690 won't be the same. 637 00:50:14,690 --> 00:50:18,020 So this we can just finish up in a jiffy, because we're just 638 00:50:18,020 --> 00:50:23,310 going to add the three things. 639 00:50:23,310 --> 00:50:30,720 So delta S is delta S of heating, minus, I think 640 00:50:30,720 --> 00:50:33,960 there's a plus that should be a minus there, minus delta S 641 00:50:33,960 --> 00:50:36,150 of fusion because it's going in the direction of freezing 642 00:50:36,150 --> 00:50:38,070 the way we've written it. 643 00:50:38,070 --> 00:50:42,730 Plus delta S of cooling. 644 00:50:42,730 --> 00:50:52,770 So integral from T1 to the melting point of Cp of the 645 00:50:52,770 --> 00:51:01,930 liquid dT over T, minus delta H of fusion over the melting 646 00:51:01,930 --> 00:51:03,720 temperature. 647 00:51:03,720 --> 00:51:07,750 Plus the integral going from the heat of the temperature of 648 00:51:07,750 --> 00:51:15,100 melting to T1 of Cp of the solid dT over T. 649 00:51:15,100 --> 00:51:18,770 That's it, all right? 650 00:51:18,770 --> 00:51:24,870 So delta S is minus delta H of fusion over T. That's what it 651 00:51:24,870 --> 00:51:27,300 would be for just the reversible phase change 652 00:51:27,300 --> 00:51:30,190 happening at zero degrees Celsius. 653 00:51:30,190 --> 00:51:34,227 And then there's this additional part which is, we 654 00:51:34,227 --> 00:51:39,370 can write it from T1 to melting point of Cp of the 655 00:51:39,370 --> 00:51:47,140 liquid, minus Cp of the solid, dT over T. 656 00:51:47,140 --> 00:51:50,320 Now usually we can assume that these will be temperature 657 00:51:50,320 --> 00:51:55,080 independent in their own phases. 658 00:51:55,080 --> 00:51:59,950 So we can usually write this as minus delta H of fusion 659 00:51:59,950 --> 00:52:10,080 over the melting temperature plus Cp of the liquid minus Cp 660 00:52:10,080 --> 00:52:21,970 of the solid log of T of fusion over T1. 661 00:52:21,970 --> 00:52:24,580 Any questions? 662 00:52:24,580 --> 00:52:25,900 What did we do? 663 00:52:25,900 --> 00:52:28,785 We constructed a reversible path going from here to here 664 00:52:28,785 --> 00:52:31,380 and calculated delta S that way, and of course that has to 665 00:52:31,380 --> 00:52:36,110 be the same as delta S in the one irreversible step. 666 00:52:36,110 --> 00:52:40,360 All right, more on entropy and its consequences for how to 667 00:52:40,360 --> 00:52:43,930 figure out what happened spontaneously next time.