1 00:00:00,030 --> 00:00:02,400 The following content is provided under a creative 2 00:00:02,400 --> 00:00:03,810 commons license. 3 00:00:03,810 --> 00:00:06,860 Your support will help MIT OpenCourseWare continue to 4 00:00:06,860 --> 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 --> 00:00:16,320 hundreds of MIT courses, visit MIT OpenCourseWare at 7 00:00:16,320 --> 00:00:21,210 ocw.mit.edu. 8 00:00:21,210 --> 00:00:25,430 PROFESSOR NELSON: All right, well last time we finished our 9 00:00:25,430 --> 00:00:28,460 sort of introduction to entropy which is a difficult 10 00:00:28,460 --> 00:00:32,020 topic, but I hope we got to the point where we have some 11 00:00:32,020 --> 00:00:34,350 working understanding of physically what it's 12 00:00:34,350 --> 00:00:36,470 representing for us, and also an ability to 13 00:00:36,470 --> 00:00:37,310 just calculate it. 14 00:00:37,310 --> 00:00:39,780 So we went through at the end of the last lecture, a few 15 00:00:39,780 --> 00:00:42,890 examples where we just calculated changes in entropy 16 00:00:42,890 --> 00:00:45,470 for simple processes like heating and cooling something 17 00:00:45,470 --> 00:00:49,140 or going through a phase transition where that process, 18 00:00:49,140 --> 00:00:51,840 that sort of process is relatively simple because 19 00:00:51,840 --> 00:00:53,200 there's no temperature change. 20 00:00:53,200 --> 00:00:56,240 While the ice is melting, for example, you're putting heat 21 00:00:56,240 --> 00:00:58,530 into it, but the temperature is staying 22 00:00:58,530 --> 00:01:01,530 at zero degree Celsius. 23 00:01:01,530 --> 00:01:04,880 And we saw, how to do these calculations, we need define 24 00:01:04,880 --> 00:01:05,960 reversible paths. 25 00:01:05,960 --> 00:01:09,640 So it was extremely straightforward to calculate 26 00:01:09,640 --> 00:01:12,920 the entropy of say ice melting at zero degrees Celsius, 27 00:01:12,920 --> 00:01:16,350 because there the process is reversible, because that's the 28 00:01:16,350 --> 00:01:18,180 melting temperature. 29 00:01:18,180 --> 00:01:22,700 But if we wanted to calculate the change in entropy of ice 30 00:01:22,700 --> 00:01:27,520 melting, you know at, once it had already been cooled to ten 31 00:01:27,520 --> 00:01:31,330 degrees above the melting point, to ten degrees Celsius, 32 00:01:31,330 --> 00:01:33,420 then in order to find a reversible path, we had to 33 00:01:33,420 --> 00:01:37,650 say, OK, let's first cool it down to zero degrees Celsius. 34 00:01:37,650 --> 00:01:40,580 Then let's have it melt, and then let's warm the liquid 35 00:01:40,580 --> 00:01:43,330 back up to ten degrees Celsius so we could construct the 36 00:01:43,330 --> 00:01:46,060 sequence of reversible steps that would get from the same 37 00:01:46,060 --> 00:01:48,590 starting point to the same end point, and we could calculate 38 00:01:48,590 --> 00:01:52,770 the change in entropy through that sort of sequence. 39 00:01:52,770 --> 00:01:56,630 So now that we've got at least some experience doing 40 00:01:56,630 --> 00:01:59,470 calculations of delta S and we're just thinking a little 41 00:01:59,470 --> 00:02:03,430 bit about entropy, what I'd like to do is to try to relate 42 00:02:03,430 --> 00:02:06,230 the state variables together in a useful way. 43 00:02:06,230 --> 00:02:10,920 And and the immediate problem that I'd like to address is 44 00:02:10,920 --> 00:02:14,930 the fact that right now we have kind of a cumbersome 45 00:02:14,930 --> 00:02:18,180 expression for energy. 46 00:02:18,180 --> 00:02:29,890 So you know we have u, we look at du, right it's dq plus dw, 47 00:02:29,890 --> 00:02:34,330 and you know, I don't like those. 48 00:02:34,330 --> 00:02:38,980 They're path specific, and it would be nice to be able to do 49 00:02:38,980 --> 00:02:43,390 a calculation of changes in energy that 50 00:02:43,390 --> 00:02:44,630 didn't depend on path. 51 00:02:44,630 --> 00:02:47,650 You know it's a state function. 52 00:02:47,650 --> 00:02:51,590 So in principle it seems like it sure ought to be possible, 53 00:02:51,590 --> 00:02:54,490 and yet so far when we've actually gone through 54 00:02:54,490 --> 00:02:57,980 calculations of du, we've had to go and consider the path 55 00:02:57,980 --> 00:03:01,020 and get the heat and get the work and so forth. 56 00:03:01,020 --> 00:03:04,470 And then we found special cases, you know an ideal gas 57 00:03:04,470 --> 00:03:07,460 where the temperature, where the change is only a function 58 00:03:07,460 --> 00:03:09,860 of temperature and so forth, where we could write this as a 59 00:03:09,860 --> 00:03:12,260 function of state variables, but nothing general that 60 00:03:12,260 --> 00:03:14,270 really allows us to do the calculation under all 61 00:03:14,270 --> 00:03:17,360 circumstances. 62 00:03:17,360 --> 00:03:21,130 So let's think about how to make this better. 63 00:03:21,130 --> 00:03:24,840 So, and what I mean by that is you've seen examples like, you 64 00:03:24,840 --> 00:03:32,750 know, some special examples you saw awhile back. 65 00:03:32,750 --> 00:03:39,560 The case where du was Cv dT minus Cv and this Joule 66 00:03:39,560 --> 00:03:44,350 coefficient d v. But you know you still need to find those 67 00:03:44,350 --> 00:03:46,600 coefficients for each system. 68 00:03:46,600 --> 00:03:50,980 This isn't a general equation that tells us how energy 69 00:03:50,980 --> 00:03:54,960 changes in terms of only functions of state. 70 00:03:54,960 --> 00:03:57,600 Because of things like this and this -- what I'd really 71 00:03:57,600 --> 00:04:11,800 like is to be able to you know write du equals something. 72 00:04:11,800 --> 00:04:16,440 And that something, you know, it can have T and p and 73 00:04:16,440 --> 00:04:17,700 whatever else I need. 74 00:04:17,700 --> 00:04:23,160 It can have S, H, and of course differentials of any of 75 00:04:23,160 --> 00:04:30,550 those quantities. dT or dp or dS, dH, you name it, but all 76 00:04:30,550 --> 00:04:32,400 state variables. 77 00:04:32,400 --> 00:04:35,760 That's what I'd much rather have. 78 00:04:35,760 --> 00:04:39,260 And then, you know, if I want to, if I've got something like 79 00:04:39,260 --> 00:04:42,230 that and I want to find out how the energy changes as a 80 00:04:42,230 --> 00:04:45,970 function of volume, so I'll calculate du/dV With respect 81 00:04:45,970 --> 00:04:51,420 to some selected variable hold constant, I can do it. 82 00:04:51,420 --> 00:04:55,530 Right now, in a general sense, that's cumbersome. 83 00:04:55,530 --> 00:04:58,450 I've got to figure out how to do that for each particular 84 00:04:58,450 --> 00:05:01,740 case that I want to treat. 85 00:05:01,740 --> 00:05:13,520 So, let's see how we could construct such a thing. 86 00:05:13,520 --> 00:05:16,580 Let's consider just a reversible process, at 87 00:05:16,580 --> 00:05:32,120 constant pressure. 88 00:05:32,120 --> 00:05:43,400 So, OK, I've got one and I'm going to in some path wind up 89 00:05:43,400 --> 00:05:51,100 at state two and I'll write du is dq, it's reversible in this 90 00:05:51,100 --> 00:05:54,810 case, minus p dV. 91 00:05:58,600 --> 00:06:08,410 And from the second law, we know that we can write dq 92 00:06:08,410 --> 00:06:12,520 reversible as T dS. 93 00:06:12,520 --> 00:06:18,360 dq over T is dS or entropy. 94 00:06:18,360 --> 00:06:28,150 So, we can write du is T dS minus p dV. 95 00:06:35,280 --> 00:06:41,660 That's so important, we'll circle it with colored chalk. 96 00:06:41,660 --> 00:06:43,850 That's how important it is. 97 00:06:43,850 --> 00:06:48,000 You know it's a dramatic moment. 98 00:06:48,000 --> 00:06:49,640 So let's look at what we have here. 99 00:06:49,640 --> 00:06:55,930 Here's du and over on this side we have T, we have S, we 100 00:06:55,930 --> 00:07:00,770 have p and we have V. Suddenly and simply, it's only 101 00:07:00,770 --> 00:07:02,550 functions of state. 102 00:07:02,550 --> 00:07:05,940 Well that was pretty easy. 103 00:07:05,940 --> 00:07:12,680 So, what that's telling us is that we can write u this way, 104 00:07:12,680 --> 00:07:15,200 and you know, this is generally true. 105 00:07:15,200 --> 00:07:18,390 We got to this by considering a reversible 106 00:07:18,390 --> 00:07:21,670 constant pressure process. 107 00:07:21,670 --> 00:07:26,020 But we know u is a state variable right. 108 00:07:26,020 --> 00:07:29,130 So this result is going to be generally applicable. 109 00:07:29,130 --> 00:07:31,090 And it tells us a couple of things too. 110 00:07:31,090 --> 00:07:36,190 It tells us that in some sense, the natural variables 111 00:07:36,190 --> 00:07:50,980 for u are these, right, it's a function of S and V. Those are 112 00:07:50,980 --> 00:07:53,840 natural variables in the sense that then it written as 113 00:07:53,840 --> 00:08:02,430 functions of those variables, we only have state quantities 114 00:08:02,430 --> 00:08:06,000 on the right-hand side. 115 00:08:06,000 --> 00:08:09,030 Very, very valuable expression. 116 00:08:09,030 --> 00:08:12,230 And of course coming out of that then, we can take 117 00:08:12,230 --> 00:08:17,290 derivatives and at least for those particular variables, we 118 00:08:17,290 --> 00:08:26,430 can see that du/dS at constant V is minus p. 119 00:08:26,430 --> 00:08:42,300 And du/dV at constant S is T. All right, 120 00:08:42,300 --> 00:08:43,930 those fall right out. 121 00:08:43,930 --> 00:08:49,770 Now we can have a similar set of steps for H, for the 122 00:08:49,770 --> 00:09:10,090 enthalpy, so let's just look at that. 123 00:09:10,090 --> 00:09:21,150 So H, of course, it's u plus pV, so dH is just du plus 124 00:09:21,150 --> 00:09:26,220 d(pV), and now there's our expression for du. 125 00:09:26,220 --> 00:09:29,080 We're going to use it this way. 126 00:09:29,080 --> 00:09:39,380 So it's T dS minus p dV plus p dV plus V dp. 127 00:09:39,380 --> 00:09:43,370 And of course these are going to cancel. 128 00:09:43,370 --> 00:09:51,420 So we can write dH is T dS plus V dp. 129 00:09:55,210 --> 00:09:57,520 Also important enough that we'll 130 00:09:57,520 --> 00:10:00,760 highlight it a little bit. 131 00:10:00,760 --> 00:10:03,390 So again let's look at what we've got on the right-hand 132 00:10:03,390 --> 00:10:08,700 side here, T, S, V, p, only quantities that are 133 00:10:08,700 --> 00:10:13,220 functions of state. 134 00:10:13,220 --> 00:10:17,290 And of course, we can take the corresponding derivative, so 135 00:10:17,290 --> 00:10:21,180 let's also be explicit here, that means that were writing H 136 00:10:21,180 --> 00:10:28,110 as a function of the variables S and p. 137 00:10:28,110 --> 00:10:53,400 And dH/dS at constant p is T, and dH/dp at constant S is V. 138 00:10:53,400 --> 00:10:58,900 So now we've got a couple of really surprisingly simple 139 00:10:58,900 --> 00:11:03,600 expressions that we can use to describe u and H in terms of 140 00:11:03,600 --> 00:11:05,880 only state variables. 141 00:11:05,880 --> 00:11:10,080 All right? 142 00:11:10,080 --> 00:11:14,320 We also can go from these expressions, using the chain 143 00:11:14,320 --> 00:11:17,540 rule, to expressions for particularly useful 144 00:11:17,540 --> 00:11:23,260 expressions for the entropy as a function of temperature. 145 00:11:23,260 --> 00:11:32,250 So, you know, from du is T dS minus p dV. 146 00:11:32,250 --> 00:11:38,940 We can rewrite this as dS is one over T du 147 00:11:38,940 --> 00:11:44,870 plus p over T dV. 148 00:11:48,320 --> 00:11:52,430 And now we can go back, you know, if we can go back to our 149 00:11:52,430 --> 00:11:58,410 writing of u in terms of, as a function of T and V, right. 150 00:11:58,410 --> 00:12:09,320 So we can write here du as a function of T and V, Cv dT 151 00:12:09,320 --> 00:12:17,650 plus du/dV at constant T dV. 152 00:12:17,650 --> 00:12:22,440 The reason we're doing that is now we can rewrite dS is one 153 00:12:22,440 --> 00:12:28,500 over T times Cv dT, and that's the only temperature 154 00:12:28,500 --> 00:12:29,990 dependence we're going to have. 155 00:12:29,990 --> 00:12:33,760 The other part is going to be a function of volume. 156 00:12:33,760 --> 00:12:45,250 So it's, we've got p over T plus du/dV at constant T dV, 157 00:12:45,250 --> 00:12:54,170 and what this says then is that dS/dT at constant V is 158 00:12:54,170 --> 00:13:03,770 just Cv over T. Very useful, not surprising because of the 159 00:13:03,770 --> 00:13:08,280 relation between heat at constant volume and Cv, right. 160 00:13:08,280 --> 00:13:15,730 And of course dS is just dq reversible over T, but this is 161 00:13:15,730 --> 00:13:19,600 telling us, in general, how the entropy changes with 162 00:13:19,600 --> 00:13:23,640 temperature at constant volume. 163 00:13:23,640 --> 00:13:28,610 We can go through the exact same procedure with the H to 164 00:13:28,610 --> 00:13:32,110 look at how entropy varies with temperature at constant 165 00:13:32,110 --> 00:13:36,480 pressure, and we'll get exactly an analogous set of 166 00:13:36,480 --> 00:13:39,650 steps that will be Cp over T, right. 167 00:13:39,650 --> 00:14:00,650 So also dS/dT at constant pressure is Cp over T. OK? 168 00:14:00,650 --> 00:14:05,200 Now I want to carry our discussion a little bit 169 00:14:05,200 --> 00:14:10,020 further and look at entropy a little more carefully and in 170 00:14:10,020 --> 00:14:15,380 particular, how it varies with temperature. 171 00:14:15,380 --> 00:14:18,480 And here's what I really want to look at. 172 00:14:18,480 --> 00:14:21,470 You know, we've talked about when we look at changes in u 173 00:14:21,470 --> 00:14:24,770 and changes in H, and we've done this under lots of 174 00:14:24,770 --> 00:14:26,360 circumstances at this point. 175 00:14:26,360 --> 00:14:29,300 And at various times I've emphasized, and I'm sure 176 00:14:29,300 --> 00:14:33,000 Professor Bawendi did too, that when we look at these 177 00:14:33,000 --> 00:14:36,350 quantities we can only define changes in them. 178 00:14:36,350 --> 00:14:41,580 There's not an absolute scales for energy or for enthalpy. 179 00:14:41,580 --> 00:14:46,640 We can set the zero in a particular problem, 180 00:14:46,640 --> 00:14:49,350 arbitrarily. 181 00:14:49,350 --> 00:14:55,050 And so, for example, when we talked about thermochemistry, 182 00:14:55,050 --> 00:15:00,490 we defined heats of formation, and then we said, well, the 183 00:15:00,490 --> 00:15:05,310 heat of formation of an element in its natural state 184 00:15:05,310 --> 00:15:08,390 at room temperature and pressure we'll call zero. 185 00:15:08,390 --> 00:15:09,790 We called it zero. 186 00:15:09,790 --> 00:15:13,950 If we wanted to put some number on it, and put energy 187 00:15:13,950 --> 00:15:15,390 in it, we could have done that. 188 00:15:15,390 --> 00:15:18,300 We defined the zero. 189 00:15:18,300 --> 00:15:22,230 So far, that's, well not just so far, that's always the way 190 00:15:22,230 --> 00:15:25,970 it will be for quantities like energy and enthalpy. 191 00:15:25,970 --> 00:15:30,130 Entropy is different. 192 00:15:30,130 --> 00:15:33,520 So let's just see how that works. 193 00:15:33,520 --> 00:15:52,230 So, let's consider the entropy, we'll consider as a 194 00:15:52,230 --> 00:16:00,780 function of temperature and pressure. 195 00:16:00,780 --> 00:16:03,360 First let's just see how it varies with pressure. 196 00:16:03,360 --> 00:16:05,220 We're going to see -- what we'll do is consider its 197 00:16:05,220 --> 00:16:08,110 variation to both pressure and temperature, and the objective 198 00:16:08,110 --> 00:16:10,830 is to say all right, if I've got some substance at any 199 00:16:10,830 --> 00:16:15,020 arbitrary temperature and pressure, can I define and 200 00:16:15,020 --> 00:16:19,020 calculate an absolute number for the entropy? 201 00:16:19,020 --> 00:16:22,190 Not just a change in entropy, unlike the cases with delta u 202 00:16:22,190 --> 00:16:26,470 and delta H, but an absolute number that says in absolute 203 00:16:26,470 --> 00:16:29,480 terms the entropy of this substance at room temperature 204 00:16:29,480 --> 00:16:31,340 and pressure or whatever temperature and pressure is 205 00:16:31,340 --> 00:16:33,140 this amount. 206 00:16:33,140 --> 00:16:39,160 Something that I can do by choice of a zero for energy or 207 00:16:39,160 --> 00:16:41,680 for u or H, but here, I want to look 208 00:16:41,680 --> 00:16:43,920 for an absolute answer. 209 00:16:43,920 --> 00:16:46,510 All right, so let's start by looking at the pressure 210 00:16:46,510 --> 00:16:48,620 dependence. 211 00:16:48,620 --> 00:16:59,450 So we're going to start with du is T dS minus p dV, so dS 212 00:16:59,450 --> 00:17:07,130 is du plus p dV over T. 213 00:17:07,130 --> 00:17:14,230 Now let's look, T being constant. 214 00:17:14,230 --> 00:17:16,910 OK, and now let's specify a little bit. 215 00:17:16,910 --> 00:17:20,610 I want to make it something as tractable as possible. 216 00:17:20,610 --> 00:17:27,760 Let's go to an ideal gas. 217 00:17:27,760 --> 00:17:31,660 So then at constant temperature, that says du is 218 00:17:31,660 --> 00:17:34,650 equal to zero. 219 00:17:34,650 --> 00:17:41,680 So dS at constant temperature is just p over T dV. 220 00:17:45,450 --> 00:17:57,750 And in the case of an ideal gas, that's nR dV over V. 221 00:17:57,750 --> 00:18:08,110 And at constant temperature, that means that d(nRT), which 222 00:18:08,110 --> 00:18:19,470 is the same as d(pV) is equal to zero, but this 223 00:18:19,470 --> 00:18:23,790 is p dV plus V dp. 224 00:18:27,190 --> 00:18:34,210 So this says that dV over V, that I've got there, is the 225 00:18:34,210 --> 00:18:39,550 same thing as negative dp over p. 226 00:18:39,550 --> 00:18:59,900 Right, so I can write that dS as constant temperature is 227 00:18:59,900 --> 00:19:06,980 minus nR dp over p. 228 00:19:06,980 --> 00:19:07,710 So that's great. 229 00:19:07,710 --> 00:19:10,720 That says now if I know the entropy at some particular 230 00:19:10,720 --> 00:19:13,350 pressure, I can calculate how it changes as a function of 231 00:19:13,350 --> 00:19:20,930 pressure, right. 232 00:19:20,930 --> 00:19:33,910 If I know S at some standard pressure that we can define, 233 00:19:33,910 --> 00:19:42,050 then S at some arbitrary pressure, is just S of p 234 00:19:42,050 --> 00:19:49,420 naught and T minus the integral from p naught to p of 235 00:19:49,420 --> 00:19:53,220 nR dp over p. 236 00:19:53,220 --> 00:20:01,940 All right, which is to say it's S of p naught T minus nR 237 00:20:01,940 --> 00:20:06,800 log of p over p naught. 238 00:20:06,800 --> 00:20:18,400 Right, now normally we'll define p naught as 239 00:20:18,400 --> 00:20:23,930 equal to one bar. 240 00:20:23,930 --> 00:20:28,750 And often you'll see this simply written as nR log p. 241 00:20:28,750 --> 00:20:31,060 I don't particularly like to do that because of course, 242 00:20:31,060 --> 00:20:33,630 then, formally speaking were looking at something that's 243 00:20:33,630 --> 00:20:36,760 written that has units inside as the argument of a log. 244 00:20:36,760 --> 00:20:39,660 Of course it's understood when you see that, and you're 245 00:20:39,660 --> 00:20:42,780 likely to see it in various places, it's understood when 246 00:20:42,780 --> 00:20:46,830 you see that the quantity p is always supposed to be divided 247 00:20:46,830 --> 00:20:53,050 by one bar, and the units then are taken care of. 248 00:20:53,050 --> 00:21:01,600 For one mole, we can write the molar quantities S of p and T, 249 00:21:01,600 --> 00:21:16,400 is S, S naught of T minus R log p over p naught. 250 00:21:16,400 --> 00:21:24,330 All right, so that's our pressure dependence. 251 00:21:24,330 --> 00:21:25,820 What about that? 252 00:21:25,820 --> 00:21:29,630 We still don't really have a formulation for calculating 253 00:21:29,630 --> 00:21:32,620 this, or you know, defining it or whatever we're going to do 254 00:21:32,620 --> 00:21:57,000 to allow us to know it. 255 00:21:57,000 --> 00:22:02,670 Well, let's just consider the entropy as a function of 256 00:22:02,670 --> 00:22:06,770 temperature, starting all the way down at zero degrees 257 00:22:06,770 --> 00:22:09,600 Kelvin, and going up to whatever temperature we want 258 00:22:09,600 --> 00:22:23,460 to consider. 259 00:22:23,460 --> 00:22:28,140 Now, we certainly do know how to calculate delta S for all 260 00:22:28,140 --> 00:22:30,750 that because we've seen how to calculate delta S if you just 261 00:22:30,750 --> 00:22:34,760 heat something up, and we've seen how to calculate delta S 262 00:22:34,760 --> 00:22:37,640 when something under goes a phase transition, right. 263 00:22:37,640 --> 00:22:39,840 Presumably, if we're starting at zero Kelvin, we're starting 264 00:22:39,840 --> 00:22:42,780 in a solid state. 265 00:22:42,780 --> 00:22:46,490 As we heat it up, depending on the material, it may melt at 266 00:22:46,490 --> 00:22:47,090 some temperature. 267 00:22:47,090 --> 00:22:49,510 If we keep keep heating it up, it'll boil at some 268 00:22:49,510 --> 00:22:54,110 temperature, but we know how to treat all of that, right. 269 00:22:54,110 --> 00:22:56,670 So let's just consider something that undergoes that 270 00:22:56,670 --> 00:22:58,130 set of changes. 271 00:22:58,130 --> 00:23:05,180 So, we've got some substance A, solid, zero 272 00:23:05,180 --> 00:23:13,480 degrees Kelvin, one bar. 273 00:23:13,480 --> 00:23:16,200 Here's process one. 274 00:23:16,200 --> 00:23:20,340 It goes to A, it's a solid at the melting 275 00:23:20,340 --> 00:23:29,500 temperature and one bar. 276 00:23:29,500 --> 00:23:34,570 Process two is it turns into a liquid at the melting 277 00:23:34,570 --> 00:23:41,430 temperature and one bar. 278 00:23:41,430 --> 00:23:50,750 Process three is we heat it up some more, up to the boiling 279 00:23:50,750 --> 00:23:56,190 temperature at one bar. 280 00:23:56,190 --> 00:24:03,010 Process four is it evaporates, so now it's a gas at the 281 00:24:03,010 --> 00:24:07,620 boiling temperature and one bar. 282 00:24:07,620 --> 00:24:13,350 Finally, we heat it up some more, so now it's a gas at 283 00:24:13,350 --> 00:24:19,730 temperature T and one bar. 284 00:24:19,730 --> 00:24:21,510 And if we wanted to, we can go further. 285 00:24:21,510 --> 00:24:26,200 We can make it a gas at temperature and whatever 286 00:24:26,200 --> 00:24:26,900 pressure we want. 287 00:24:26,900 --> 00:24:36,420 That part we already know how to take care of, right. 288 00:24:36,420 --> 00:24:41,410 Well, let's look at what happens to S, all right. 289 00:24:41,410 --> 00:24:46,110 S, a molar enthalpy at T and p, where we're going to 290 00:24:46,110 --> 00:24:56,000 finally end up, is, well it's s zero at zero Kelvin and one 291 00:24:56,000 --> 00:25:02,470 bar or one bar is implied by the superscripts here. 292 00:25:02,470 --> 00:25:07,080 And then we have delta S for step one, and delta S for step 293 00:25:07,080 --> 00:25:09,730 two and so on. 294 00:25:09,730 --> 00:25:13,470 So all right, let's, we can label this six so we to all 295 00:25:13,470 --> 00:25:37,890 the way to delta S for step six. 296 00:25:37,890 --> 00:25:40,400 Well, so we can do that. 297 00:25:40,400 --> 00:25:51,300 It's S of the material at T and p is S naught at zero 298 00:25:51,300 --> 00:25:55,620 Kelvin, plus, here's for process one. 299 00:25:55,620 --> 00:26:01,310 We heat it up from zero Kelvin up to the melting point. 300 00:26:01,310 --> 00:26:08,690 Cp of the solid more heat capacity, divided by T dT. 301 00:26:08,690 --> 00:26:15,050 We can calculate delta S for heating something up, right. 302 00:26:15,050 --> 00:26:21,230 Plus, now we've got the heat of fusion to melt the stuff, 303 00:26:21,230 --> 00:26:28,310 so it's just delta H naught of fusion, divided by Tm right. 304 00:26:28,310 --> 00:26:29,250 We saw that last time. 305 00:26:29,250 --> 00:26:32,890 In other words, remember, we're just looking at q 306 00:26:32,890 --> 00:26:38,160 reversible over T to get delta S, and it's just given by the 307 00:26:38,160 --> 00:26:40,640 heat of fusion. 308 00:26:40,640 --> 00:26:44,910 All right, then let's go from the melting point to the 309 00:26:44,910 --> 00:26:45,830 boiling point. 310 00:26:45,830 --> 00:26:50,200 So it's Cp now it's the heat capacity, the molar heat 311 00:26:50,200 --> 00:26:54,920 capacity of the liquid, divided by T dT. 312 00:26:54,920 --> 00:26:59,540 We're heating up the liquid. 313 00:26:59,540 --> 00:27:01,550 And then there's vaporization. 314 00:27:01,550 --> 00:27:11,200 Delta H of vaporization over T at the boiling point. 315 00:27:11,200 --> 00:27:14,050 Then we can go from the boiling point to our final 316 00:27:14,050 --> 00:27:19,880 temperature T. Now it's the molar heat capacity of the gas 317 00:27:19,880 --> 00:27:29,440 over T dT minus R log p over p naught. 318 00:27:29,440 --> 00:27:34,200 OK, so that's everything, and these are all things that we 319 00:27:34,200 --> 00:27:37,160 know how to do. 320 00:27:37,160 --> 00:27:38,790 Just about. 321 00:27:38,790 --> 00:27:42,300 OK, this one we're going to have to think about, but all 322 00:27:42,300 --> 00:27:44,530 the changes we know how to calculate, right. 323 00:27:44,530 --> 00:27:56,000 So if we plot this, S, and let's just do this as a 324 00:27:56,000 --> 00:27:58,000 function of temperature. 325 00:27:58,000 --> 00:28:02,360 I don't have pressure in here explicitly. 326 00:28:02,360 --> 00:28:08,770 Well, it's going to change as I warm up the solid, soon 327 00:28:08,770 --> 00:28:14,750 we're really starting at zero Kelvin. 328 00:28:14,750 --> 00:28:17,690 This stuff is all positive, right, so the change in 329 00:28:17,690 --> 00:28:19,810 entropy is going to be positive. 330 00:28:19,810 --> 00:28:23,070 Entropy is going to increase as this happens, and then 331 00:28:23,070 --> 00:28:26,450 there's a change right at some fixed temperature as the 332 00:28:26,450 --> 00:28:30,210 material melts. 333 00:28:30,210 --> 00:28:33,570 So here is step one. 334 00:28:33,570 --> 00:28:37,220 Here is step two, right, this must be the melting 335 00:28:37,220 --> 00:28:40,170 temperature. 336 00:28:40,170 --> 00:28:45,130 And then there's another heating step. 337 00:28:45,130 --> 00:28:48,590 Well, strictly speaking, I'm going to run out of space here 338 00:28:48,590 --> 00:28:52,430 if I'm not careful, so I'm going to be a little more 339 00:28:52,430 --> 00:28:55,470 careful here. 340 00:28:55,470 --> 00:28:59,930 One, two, three. 341 00:28:59,930 --> 00:29:01,690 I'm heating it up a little more. 342 00:29:01,690 --> 00:29:03,650 Entropy is still increasing, right. 343 00:29:03,650 --> 00:29:05,520 So I've done this. 344 00:29:05,520 --> 00:29:06,670 I've done this. 345 00:29:06,670 --> 00:29:09,870 Now I've heated up the liquid. 346 00:29:09,870 --> 00:29:13,760 Now, I'm going to boil the liquid, so it's going to have 347 00:29:13,760 --> 00:29:17,100 some change in entropy. 348 00:29:17,100 --> 00:29:21,770 This must be my boiling point, and now there's some further 349 00:29:21,770 --> 00:29:25,550 change in the gas, and that gets me to whatever my final 350 00:29:25,550 --> 00:29:32,040 temperature is, right, that I'm going to reach. 351 00:29:32,040 --> 00:29:36,700 Four and five, great. 352 00:29:36,700 --> 00:29:42,020 So there is monotonic increase in the entropy. 353 00:29:42,020 --> 00:29:52,380 OK, so we're there, except for this value. 354 00:29:52,380 --> 00:29:59,760 That one stinking little number -- 355 00:29:59,760 --> 00:30:03,530 S naught at zero Kelvin. 356 00:30:03,530 --> 00:30:07,320 That's the only thing we don't know so far. 357 00:30:07,320 --> 00:30:13,470 So, for this we need some additional input. 358 00:30:13,470 --> 00:30:17,610 We got some input of the sort that we need 359 00:30:17,610 --> 00:30:18,820 in 1905 from Nernst. 360 00:30:18,820 --> 00:30:26,810 Nernst deduced that as you go down from zero Kelvin for any 361 00:30:26,810 --> 00:30:31,130 process, the change in entropy gets smaller and smaller. 362 00:30:31,130 --> 00:30:35,690 It approaches zero. 363 00:30:35,690 --> 00:30:38,490 Now, that actually was certainly an important 364 00:30:38,490 --> 00:30:43,360 advance, but it was superseded by such an important advance 365 00:30:43,360 --> 00:30:47,280 that I'm not even going to reward it by placing it on the 366 00:30:47,280 --> 00:30:49,380 blackboard. 367 00:30:49,380 --> 00:30:52,430 Forget highlight, color, forget it. 368 00:30:52,430 --> 00:30:59,550 Because Planck, six years later, in 1911, deduced a 369 00:30:59,550 --> 00:31:04,480 stronger statement which is extremely useful, 370 00:31:04,480 --> 00:31:06,200 and it's the following. 371 00:31:06,200 --> 00:31:11,120 What he showed is that as temperature approaches zero 372 00:31:11,120 --> 00:31:15,730 Kelvin, for a pure substance in it's 373 00:31:15,730 --> 00:31:20,620 crystalline state, S is zero. 374 00:31:20,620 --> 00:31:22,990 A much stronger statement, right. 375 00:31:22,990 --> 00:31:27,350 A stronger statement than the idea that changes in S get 376 00:31:27,350 --> 00:31:30,180 very small as you approach zero Kelvin. 377 00:31:30,180 --> 00:31:34,100 No, he's saying we can make a statement about that absolute 378 00:31:34,100 --> 00:31:39,240 number S goes to zero as temperature goes to zero. 379 00:31:39,240 --> 00:31:43,380 For a pure substance in it's crystalline state. 380 00:31:43,380 --> 00:31:48,310 So that is monumentally important. 381 00:31:48,310 --> 00:32:14,680 So as T goes to zero Kelvin, S goes to zero, for every pure 382 00:32:14,680 --> 00:32:21,600 substance in its, and I'll sort of interject here, 383 00:32:21,600 --> 00:32:30,640 perfect crystalline state. 384 00:32:30,640 --> 00:32:32,690 That's really an amazing result. 385 00:32:32,690 --> 00:32:36,420 So what it's saying is I'm down at zero Kelvin. 386 00:32:36,420 --> 00:32:40,310 Minimally, I've somehow cooled it as much as I possibly 387 00:32:40,310 --> 00:32:44,930 could, and I've got my perfect crystal lattice. 388 00:32:44,930 --> 00:32:47,380 It could be an atomic crystal like this, or, you know, it 389 00:32:47,380 --> 00:32:50,560 could be molecules. 390 00:32:50,560 --> 00:32:54,060 But they're all exactly where they belong in their locations 391 00:32:54,060 --> 00:32:57,520 in the crystal, and the absolute entropy is something 392 00:32:57,520 --> 00:33:02,960 I can define and it's zero. 393 00:33:02,960 --> 00:33:07,030 So S equals zero. 394 00:33:07,030 --> 00:33:19,160 Perfect, pure crystal, all right. 395 00:33:19,160 --> 00:33:26,860 OK, well this came out of a microscopic description of 396 00:33:26,860 --> 00:33:32,180 entropy that I briefly alluded to last lecture, and again 397 00:33:32,180 --> 00:33:36,620 we'll go into in more detail in a few lectures hence. 398 00:33:36,620 --> 00:33:42,400 But the result that I mentioned, the general result, 399 00:33:42,400 --> 00:33:48,120 was that S was R over Na Avogadro's number, times the 400 00:33:48,120 --> 00:34:07,800 log of this omega number of microscopic states available 401 00:34:07,800 --> 00:34:10,210 to the system that I'm considering. 402 00:34:10,210 --> 00:34:14,000 Now normally for a macroscopic system, I've got just an 403 00:34:14,000 --> 00:34:19,530 astronomical number of microscopic states. 404 00:34:19,530 --> 00:34:22,680 You know, that could mean in a liquid, different little 405 00:34:22,680 --> 00:34:25,930 configurations of the molecules around each other. 406 00:34:25,930 --> 00:34:27,940 They're all different states. 407 00:34:27,940 --> 00:34:33,660 Huge amounts of possible states, and the gas even more. 408 00:34:33,660 --> 00:34:36,060 But if I go to zero Kelvin, and I've got a perfect 409 00:34:36,060 --> 00:34:40,080 crystal, every atom, every molecule is 410 00:34:40,080 --> 00:34:42,680 exactly in its place. 411 00:34:42,680 --> 00:34:46,900 How many possible different states is that? 412 00:34:46,900 --> 00:34:48,600 It's one. 413 00:34:48,600 --> 00:34:51,230 There aren't any more possible states. 414 00:34:51,230 --> 00:34:55,880 I've localized every identical lateral molecule in its 415 00:34:55,880 --> 00:34:59,850 particular place, and it's done. 416 00:34:59,850 --> 00:35:03,490 And you know, if I start worrying about the various 417 00:35:03,490 --> 00:35:06,780 things that would matter under ordinary conditions, right, 418 00:35:06,780 --> 00:35:09,080 you know, maybe at higher temperature, I'd have some 419 00:35:09,080 --> 00:35:13,100 molecules in excited vibration levels or 420 00:35:13,100 --> 00:35:14,740 maybe electronic levels. 421 00:35:14,740 --> 00:35:17,970 Maybe if it's hydrogen, maybe everything isn't 422 00:35:17,970 --> 00:35:18,860 in the ground state. 423 00:35:18,860 --> 00:35:21,770 It's not all in the 1s orbital but in higher levels. 424 00:35:21,770 --> 00:35:24,790 Then there's be lots of states available, right, even of only 425 00:35:24,790 --> 00:35:28,350 one atom in the whole crystal is excited, well there's one 426 00:35:28,350 --> 00:35:30,690 state that would have it be this atom. 427 00:35:30,690 --> 00:35:33,020 A different one would have it be this atom, and so forth. 428 00:35:33,020 --> 00:35:36,960 Already, there'd be an enormous number of states, but 429 00:35:36,960 --> 00:35:41,320 at zero Kelvin, there's no thermal lexitation of any of 430 00:35:41,320 --> 00:35:42,960 that stuff. 431 00:35:42,960 --> 00:35:46,230 Things are in the lowest states, and they're in their 432 00:35:46,230 --> 00:35:48,450 proper positions. 433 00:35:48,450 --> 00:35:52,100 There's only one state for the whole system, so that's why 434 00:35:52,100 --> 00:35:56,030 the entropy is zero in a perfect crystal. 435 00:35:56,030 --> 00:36:01,610 At zero degrees Kelvin. 436 00:36:01,610 --> 00:36:05,310 Now, there are things that may appear to violate that. 437 00:36:05,310 --> 00:36:07,480 Now of course you can make measurements of entropy, 438 00:36:07,480 --> 00:36:14,800 right, so it can be verified that this is the case. 439 00:36:14,800 --> 00:36:18,440 There are some things that would appear to 440 00:36:18,440 --> 00:36:21,750 violate that result. 441 00:36:21,750 --> 00:36:25,670 You know, you can make measurements of entropies, and 442 00:36:25,670 --> 00:36:29,860 for example let's take carbon monoxide crystal, CO. 443 00:36:29,860 --> 00:36:32,200 So let's say this is a crystal lattice, 444 00:36:32,200 --> 00:36:33,800 it's diatomic molecules. 445 00:36:33,800 --> 00:36:38,500 It's carbon oxygen carbon oxygen carbon oxygen. 446 00:36:38,500 --> 00:36:42,370 All there in perfect place in the crystal lattice. 447 00:36:42,370 --> 00:36:45,710 Well it turns out when you form the crystal every now and 448 00:36:45,710 --> 00:36:51,880 then -- you know, let's put it in color. 449 00:36:51,880 --> 00:36:56,690 Let's put it in the color that we'll use to signify something 450 00:36:56,690 --> 00:37:01,250 in some way evil. 451 00:37:01,250 --> 00:37:03,140 No insult to people who like that color. 452 00:37:03,140 --> 00:37:05,100 I kind of like it in fact. 453 00:37:05,100 --> 00:37:12,020 OK, so you know, you're making the crystal, cooling it, 454 00:37:12,020 --> 00:37:14,500 started out maybe in the gas phase, start cooling it, 455 00:37:14,500 --> 00:37:15,140 starts crystallizing. 456 00:37:15,140 --> 00:37:22,870 Gets colder and colder, but you know, carbon monoxide is 457 00:37:22,870 --> 00:37:25,820 pretty easy for those things to flip sides. 458 00:37:25,820 --> 00:37:28,030 And even in the crystalline state, even though it's a 459 00:37:28,030 --> 00:37:30,990 crystal, so the molecules center of mass are all where 460 00:37:30,990 --> 00:37:35,150 they belong, still at ordinary temperatures they will be able 461 00:37:35,150 --> 00:37:36,890 to rotate a bit. 462 00:37:36,890 --> 00:37:40,280 So even in the crystalline state, when it's originally 463 00:37:40,280 --> 00:37:43,380 formed, not at zero Kelvin, there's thermal energy around. 464 00:37:43,380 --> 00:37:45,360 These things can to get knocked around, and the 465 00:37:45,360 --> 00:37:47,020 orientations can change. 466 00:37:47,020 --> 00:37:50,350 Now you start cooling it, and you know, by and large they'll 467 00:37:50,350 --> 00:37:54,010 all go into the proper orientation. 468 00:37:54,010 --> 00:37:57,500 Right, that's the lowest energy state, but you know, 469 00:37:57,500 --> 00:37:59,640 there are all sorts of kinetic things involved. 470 00:37:59,640 --> 00:38:01,710 There it takes time for the flipping, depending on how 471 00:38:01,710 --> 00:38:05,310 long, how slowly it was cooled, and so forth. 472 00:38:05,310 --> 00:38:08,390 May never happen, and then it's cooled, and then anything 473 00:38:08,390 --> 00:38:11,230 that's left in the other orientation 474 00:38:11,230 --> 00:38:12,370 is frozen in there. 475 00:38:12,370 --> 00:38:13,680 There's no thermal energy anymore. 476 00:38:13,680 --> 00:38:16,470 It can't find a way to reorient. 477 00:38:16,470 --> 00:38:18,210 That's it. 478 00:38:18,210 --> 00:38:21,010 All right, let's say we're down to zero Kelvin, and out 479 00:38:21,010 --> 00:38:24,040 of the whole crystal, we've got a mole of molecules. 480 00:38:24,040 --> 00:38:27,560 One of them is in the wrong orientation. 481 00:38:27,560 --> 00:38:30,080 Now how many states do we have like that 482 00:38:30,080 --> 00:38:32,130 that would be possible? 483 00:38:32,130 --> 00:38:34,020 We'd have a mole of states, right. 484 00:38:34,020 --> 00:38:34,690 It could be this one. 485 00:38:34,690 --> 00:38:36,470 It could be that one, right. 486 00:38:36,470 --> 00:38:38,480 Or in general, of course really there's a whole 487 00:38:38,480 --> 00:38:41,160 distribution of them, and they could be anywhere, and pairs 488 00:38:41,160 --> 00:38:43,790 of them could be, and it doesn't take long to get to 489 00:38:43,790 --> 00:38:47,340 really large numbers. 490 00:38:47,340 --> 00:38:49,390 So the entropy won't be zero. 491 00:38:49,390 --> 00:38:51,310 Entropy of a perfect crystal would mean 492 00:38:51,310 --> 00:38:54,460 it's perfectly ordered. 493 00:38:54,460 --> 00:38:56,970 So that's zero. 494 00:38:56,970 --> 00:38:58,980 But things like that not withstanding, and of course 495 00:38:58,980 --> 00:39:00,520 it's the same if you have a mixed crystal. 496 00:39:00,520 --> 00:39:03,920 In a sense I've described a mixed crystal where the 497 00:39:03,920 --> 00:39:06,590 mixture is a mixture of carbon monoxide pointing this way, 498 00:39:06,590 --> 00:39:08,680 and carbon monoxide pointing this way. 499 00:39:08,680 --> 00:39:12,580 But a real mixed crystal with two different constituents, 500 00:39:12,580 --> 00:39:14,970 well of course, then you have all the possible 501 00:39:14,970 --> 00:39:17,790 configurations where, you know, they could be here and 502 00:39:17,790 --> 00:39:19,800 here and here, and then you could move one of them around 503 00:39:19,800 --> 00:39:21,480 and so forth, there are zillions 504 00:39:21,480 --> 00:39:25,880 and zillions of states. 505 00:39:25,880 --> 00:39:30,920 But for a pure crystal, in perfect form, you really have 506 00:39:30,920 --> 00:39:32,895 only one configuration, and your entropy 507 00:39:32,895 --> 00:39:34,600 is therefore zero. 508 00:39:34,600 --> 00:39:44,490 OK, so now, we can go back and we can do this. 509 00:39:44,490 --> 00:39:47,100 And at least in principle, even for things that don't 510 00:39:47,100 --> 00:39:49,720 form perfect crystals, we could calculate the change in 511 00:39:49,720 --> 00:39:54,190 entropy going from the perfect ordered crystal to something 512 00:39:54,190 --> 00:39:56,930 else with some degree of disorder and keep going and 513 00:39:56,930 --> 00:40:00,130 change the temperature and do all these things. 514 00:40:00,130 --> 00:40:03,970 So the real point is that this is extremely powerful because 515 00:40:03,970 --> 00:40:07,400 given this, we really can calculate absolute numbers for 516 00:40:07,400 --> 00:40:10,640 the entropies of substances, at ordinary, not just at zero 517 00:40:10,640 --> 00:40:13,280 Kelvin, but using this which, you know, this is a really 518 00:40:13,280 --> 00:40:15,390 very straightforward procedure. 519 00:40:15,390 --> 00:40:18,790 And in fact these things are really quite easy to measure. 520 00:40:18,790 --> 00:40:21,530 You know, you do calorimetry, you can measure those delta H 521 00:40:21,530 --> 00:40:25,240 of fusions, right, delta H of vaporization. 522 00:40:25,240 --> 00:40:27,400 You can measure the heat capacities, the things in the 523 00:40:27,400 --> 00:40:28,160 calorimeter. 524 00:40:28,160 --> 00:40:30,370 You'll see how much heat is needed to raise the 525 00:40:30,370 --> 00:40:31,230 temperature a degree. 526 00:40:31,230 --> 00:40:33,405 That give you your heat capacity for the gas or the 527 00:40:33,405 --> 00:40:34,630 liquid or the solid. 528 00:40:34,630 --> 00:40:37,930 So in fact, it's extremely practical to make all those 529 00:40:37,930 --> 00:40:40,680 measurements, and you can easily find those values of 530 00:40:40,680 --> 00:40:44,710 the heat capacities and the delta H's tabulated for a huge 531 00:40:44,710 --> 00:40:47,190 number of substances. 532 00:40:47,190 --> 00:40:50,620 So this is, in fact, the practical procedure then of 533 00:40:50,620 --> 00:40:54,140 protocol for calculating absolute entropies of all 534 00:40:54,140 --> 00:41:00,960 sorts of materials at ordinary temperatures and pressures. 535 00:41:00,960 --> 00:41:03,690 Very important. 536 00:41:03,690 --> 00:41:12,000 Now, one of those corollaries to this law is that in fact 537 00:41:12,000 --> 00:41:16,010 it's impossible to reduce the temperature of any substance, 538 00:41:16,010 --> 00:41:20,680 any system, all the way to absolute, exact, no 539 00:41:20,680 --> 00:41:23,000 approximations, zero Kelvin. 540 00:41:23,000 --> 00:41:29,000 Because you can't quite get down to zero Kelvin. 541 00:41:29,000 --> 00:41:31,600 And there are various ways that you can see that this 542 00:41:31,600 --> 00:41:43,280 must be the case. 543 00:41:43,280 --> 00:41:53,250 But here's one way to think about it. 544 00:41:53,250 --> 00:42:08,300 So, let's just write that first. 545 00:42:08,300 --> 00:42:16,420 All right, can't get quite down to zero Kelvin. 546 00:42:16,420 --> 00:42:29,130 All right, let's consider a mole of an ideal gas. 547 00:42:29,130 --> 00:42:41,160 So p is RT over V. And let's start at T1 and V1, and now 548 00:42:41,160 --> 00:42:45,900 let's bring it down to some lower temperature, T2 in some 549 00:42:45,900 --> 00:42:48,640 spontaneous process. 550 00:42:48,640 --> 00:42:50,490 We'll make it adiabatic so it's like an 551 00:42:50,490 --> 00:42:53,450 irreversible expansion. 552 00:42:53,450 --> 00:42:57,440 I just want to calculate what delta S would be there in 553 00:42:57,440 --> 00:42:59,310 terms of T and V. 554 00:42:59,310 --> 00:43:21,400 So, well, du is T dS minus p dV. 555 00:43:25,630 --> 00:43:33,700 dS is du over T plus p over T dV. 556 00:43:33,700 --> 00:43:42,370 But we can also write du is Cv dT in this case, right? 557 00:43:42,370 --> 00:43:55,080 So that says that p over T, that's R over V, and we can 558 00:43:55,080 --> 00:44:31,106 write, dS is Cv dT over T plus R dV over V. It's Cv, I'll 559 00:44:31,106 --> 00:44:49,300 write it as d(log T) plus R d(log V), and so delta S, Cv 560 00:44:49,300 --> 00:44:56,180 log of T2 minus log of T1, if T2 is my final temperature, 561 00:44:56,180 --> 00:45:07,350 plus R log of V2 minus log of V1, okay. 562 00:45:07,350 --> 00:45:16,940 Or Cv log of T2 over T1 plus R log of V2 over V1. 563 00:45:16,940 --> 00:45:35,290 Well, what does it mean when T2 is zero? 564 00:45:35,290 --> 00:45:41,510 Well, I don't know what it means. 565 00:45:41,510 --> 00:45:51,880 This turns into negative infinity. 566 00:45:51,880 --> 00:45:57,010 So we're going to write it again as our either evil or at 567 00:45:57,010 --> 00:46:00,030 least unattainable color. 568 00:46:00,030 --> 00:46:02,590 What's going to happen? 569 00:46:02,590 --> 00:46:06,850 I mean you could say, well, we can counteract it by having 570 00:46:06,850 --> 00:46:08,890 this go to plus infinity, right. 571 00:46:08,890 --> 00:46:11,930 Make the volume infinite. 572 00:46:11,930 --> 00:46:13,945 In other words, have the expansion be in through an 573 00:46:13,945 --> 00:46:14,970 infinite volume. 574 00:46:14,970 --> 00:46:18,180 Of course that's impossible. 575 00:46:18,180 --> 00:46:20,620 That would be the only way to counteract the 576 00:46:20,620 --> 00:46:24,280 divergence of this term. 577 00:46:24,280 --> 00:46:29,370 In practice, you really cannot get to absolute zero, but it 578 00:46:29,370 --> 00:46:32,660 is possible to get extremely close. 579 00:46:32,660 --> 00:46:34,110 That's doable experimentally. 580 00:46:34,110 --> 00:46:38,040 It's possible to get down to some micro or nano Kelvin 581 00:46:38,040 --> 00:46:41,560 temperatures, right. 582 00:46:41,560 --> 00:46:49,420 In fact, our MIT physicist, Wolfgang Ketterle, by bringing 583 00:46:49,420 --> 00:46:53,370 atoms and molecules down to extremely low temperatures, 584 00:46:53,370 --> 00:46:57,290 was able to see them all reach the very lowest possible 585 00:46:57,290 --> 00:46:59,100 quantum state available to them. 586 00:46:59,100 --> 00:47:02,500 All sorts of unusual properties emerge in that kind 587 00:47:02,500 --> 00:47:06,510 of state, where the atoms and molecules behave coherently. 588 00:47:06,510 --> 00:47:10,290 You can make matter waves, right, and see interferences 589 00:47:10,290 --> 00:47:12,740 among them because of the fact that they're all in this 590 00:47:12,740 --> 00:47:14,730 lowest quantum state. 591 00:47:14,730 --> 00:47:17,780 So it is possible to get extremely low temperatures, 592 00:47:17,780 --> 00:47:23,130 but never absolute zero. 593 00:47:23,130 --> 00:47:34,240 Here's another way to think about it. 594 00:47:34,240 --> 00:47:37,320 You could consider what happens to the absolute 595 00:47:37,320 --> 00:47:56,790 entropy, starting at T equals zero, right. 596 00:47:56,790 --> 00:48:00,220 So we already saw that, you know, that we can go, the 597 00:48:00,220 --> 00:48:04,310 first step will go from zero to some temperature of Cp over 598 00:48:04,310 --> 00:48:11,880 T for the solid dT if we start at zero Kelvin and warm up. 599 00:48:11,880 --> 00:48:16,710 Now, already we've got a problem. 600 00:48:16,710 --> 00:48:23,390 If this initial temperature really is zero, what happens? 601 00:48:23,390 --> 00:48:26,530 This diverge, right. 602 00:48:26,530 --> 00:48:31,820 Well, in fact, what that suggests is as you approach 603 00:48:31,820 --> 00:48:40,380 zero Kelvin, the heat capacity also approaches zero. 604 00:48:40,380 --> 00:48:54,340 So, does this go to infinity as T approaches zero Kelvin? 605 00:48:54,340 --> 00:49:05,450 Well, not if Cp of the solid approaches zero as T 606 00:49:05,450 --> 00:49:09,320 approaches zero Kelvin. 607 00:49:09,320 --> 00:49:15,910 And in fact, we can measure Cp, heat capacities at very 608 00:49:15,910 --> 00:49:20,050 low temperatures, and what we find is, they do. 609 00:49:20,050 --> 00:49:27,860 They do go to zero as you approach zero Kelvin. 610 00:49:27,860 --> 00:49:35,530 So in fact, Cp of T approaches zero as T 611 00:49:35,530 --> 00:49:38,700 approaches zero Kelvin. 612 00:49:38,700 --> 00:49:41,320 Good! 613 00:49:41,320 --> 00:49:49,020 But one thing that this says is, remember, that dT is dq at 614 00:49:49,020 --> 00:49:51,700 the heat in divided by Cp. 615 00:49:55,990 --> 00:49:58,200 And this is getting really, really small. 616 00:49:58,200 --> 00:49:59,720 It's going to zero as temperature 617 00:49:59,720 --> 00:50:01,370 goes to zero, right. 618 00:50:01,370 --> 00:50:03,440 Which means that this is enormous. 619 00:50:03,440 --> 00:50:09,770 What it says is even the tiniest amount the heat input 620 00:50:09,770 --> 00:50:15,130 leads to a significant change increase in the temperature. 621 00:50:15,130 --> 00:50:17,540 So, this is another way of understanding why it just 622 00:50:17,540 --> 00:50:21,380 becomes impossible to lower the temperature of a system to 623 00:50:21,380 --> 00:50:25,700 absolute zero, because any kind of contact, and I mean 624 00:50:25,700 --> 00:50:27,470 any kind, like let's say you've got the 625 00:50:27,470 --> 00:50:29,000 system in some cryostat. 626 00:50:29,000 --> 00:50:31,790 Of course the walls of the cryostat aren't at zero 627 00:50:31,790 --> 00:50:33,650 Kelvin, but somewhere in here it's at zero Kelvin. 628 00:50:33,650 --> 00:50:36,300 You can say, okay I won't won't make it in 629 00:50:36,300 --> 00:50:37,870 contact with the walls. 630 00:50:37,870 --> 00:50:44,540 You can try, but actually light, photons that emit from 631 00:50:44,540 --> 00:50:47,450 the walls go into the sample. 632 00:50:47,450 --> 00:50:50,800 You wouldn't think that heats something up very much, and it 633 00:50:50,800 --> 00:50:53,160 doesn't heat it up very much, but we're not 634 00:50:53,160 --> 00:50:55,170 talking about very much. 635 00:50:55,170 --> 00:50:57,400 It does heat it up by enough that you can't get the 636 00:50:57,400 --> 00:50:58,850 absolute zero. 637 00:50:58,850 --> 00:51:02,660 In other words, somehow it will be in contact with stuff 638 00:51:02,660 --> 00:51:06,880 around that is not at absolute zero Kelvin. 639 00:51:06,880 --> 00:51:10,910 And even that kind of radiative contact is enough, 640 00:51:10,910 --> 00:51:16,730 in fact, to make it not reach absolute zero. 641 00:51:16,730 --> 00:51:19,610 So the point is, one way or another, you can easily see 642 00:51:19,610 --> 00:51:23,090 that it becomes impossible to keep pulling heat out of 643 00:51:23,090 --> 00:51:26,770 something and keep it down at the temperature that's right 644 00:51:26,770 --> 00:51:30,310 there at zero, but again it's possible to get very close. 645 00:51:30,310 --> 00:51:33,620 All right, what we're going to be able to do next time is 646 00:51:33,620 --> 00:51:37,180 take what we've seen so far, and develop the conditions for 647 00:51:37,180 --> 00:51:38,200 reaching equilibrium. 648 00:51:38,200 --> 00:51:41,640 So in a general sense, we'll be able to tell which way the 649 00:51:41,640 --> 00:51:45,480 processes go left unto themselves to move toward 650 00:51:45,480 --> 00:51:46,730 equilibrium.