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 --> 00:00:17,420 hundreds of MIT courses, visit MIT OpenCourseWare at 7 00:00:17,420 --> 00:00:21,460 ocw.mit.edu. 8 00:00:21,460 --> 00:00:25,530 PROFESSOR NELSON: Well, today we're going to continue with 9 00:00:25,530 --> 00:00:29,200 the lecture that was started last time that had the 10 00:00:29,200 --> 00:00:32,830 scintillating and descriptive title "some thermodynamic 11 00:00:32,830 --> 00:00:34,250 processes". 12 00:00:34,250 --> 00:00:37,490 And we're going to continue along those lines, and really 13 00:00:37,490 --> 00:00:39,720 there a couple of things that I want to do to follow up what 14 00:00:39,720 --> 00:00:41,000 you saw last time. 15 00:00:41,000 --> 00:00:45,150 One is that you've got introduced and led through a 16 00:00:45,150 --> 00:00:48,250 couple of elementary thermodynamic cycles. 17 00:00:48,250 --> 00:00:50,830 And I want to do a little bit more of that today. 18 00:00:50,830 --> 00:00:52,990 And one of the -- there are really two reasons. 19 00:00:52,990 --> 00:00:56,650 One is that as you saw last time doing a thermodynamic 20 00:00:56,650 --> 00:01:02,570 cycle can be extremely useful, because if you have one path 21 00:01:02,570 --> 00:01:05,800 along which some change occurs, but along which it's 22 00:01:05,800 --> 00:01:09,670 not necessarily easy to calculate all of the 23 00:01:09,670 --> 00:01:12,710 thermodynamic changes along such a path, it may be 24 00:01:12,710 --> 00:01:16,450 irreversible, lots of variables may change. 25 00:01:16,450 --> 00:01:20,850 It can be easier instead of looking at the complicated 26 00:01:20,850 --> 00:01:27,810 path to devise a cycle that involves several easier steps, 27 00:01:27,810 --> 00:01:31,520 each one of which is facile for the calculation of what 28 00:01:31,520 --> 00:01:34,730 happens to the thermodynamic properties. 29 00:01:34,730 --> 00:01:38,310 So in that case, it's much easier to go through several 30 00:01:38,310 --> 00:01:42,730 elementary step, and then the one complicated path you know 31 00:01:42,730 --> 00:01:46,140 because going around the entire cycle state functions, 32 00:01:46,140 --> 00:01:49,240 you know that they won't have any change. 33 00:01:49,240 --> 00:01:52,020 So in that sense, going through a thermodynamic cycle 34 00:01:52,020 --> 00:01:54,920 can be useful because it helps you calculate 35 00:01:54,920 --> 00:01:56,930 thermodynamic qualities. 36 00:01:56,930 --> 00:01:59,290 But the other reason to go through the thermodynamic 37 00:01:59,290 --> 00:02:02,380 cycles and really to develop great facility with them is 38 00:02:02,380 --> 00:02:06,680 because there are just an awful lot of things in nature 39 00:02:06,680 --> 00:02:10,300 and things that we build that run in cycles, where we want 40 00:02:10,300 --> 00:02:12,090 to calculate the thermodynamics, right. 41 00:02:12,090 --> 00:02:16,140 If we build an engine for a car or anything else, it 42 00:02:16,140 --> 00:02:18,720 almost always is going to have some key element that's 43 00:02:18,720 --> 00:02:20,560 operating in a cycle, otherwise it 44 00:02:20,560 --> 00:02:22,700 won't keep going, right. 45 00:02:22,700 --> 00:02:25,590 But it's not just engines that we might build. 46 00:02:25,590 --> 00:02:28,390 It's biological engines, right. 47 00:02:28,390 --> 00:02:32,680 Things that either produce or consume biological fuel to do 48 00:02:32,680 --> 00:02:34,180 different sorts of processes. 49 00:02:34,180 --> 00:02:37,290 Also, of course of key importance to understand what 50 00:02:37,290 --> 00:02:40,810 the, in that case, biological thermodynamics are. 51 00:02:40,810 --> 00:02:44,550 And again, there'll be key elements that run in cycles 52 00:02:44,550 --> 00:02:47,110 and understanding those can be extremely important in 53 00:02:47,110 --> 00:02:50,180 understanding how cellular function works. 54 00:02:50,180 --> 00:02:56,260 OK, so what I'd like to do is go through one cycle, and then 55 00:02:56,260 --> 00:02:59,540 I'll suggest another for you to work through on your own. 56 00:02:59,540 --> 00:03:02,930 So consistent with the title of the lecture, we've got a 57 00:03:02,930 --> 00:03:05,910 section from your last lecture notes entitled "some 58 00:03:05,910 --> 00:03:07,980 thermodynamic cycles". 59 00:03:07,980 --> 00:03:14,260 And what we're going to do is go through a cycle, please 60 00:03:14,260 --> 00:03:17,150 note these office hours and locations, which I think 61 00:03:17,150 --> 00:03:19,200 wasn't specified before. 62 00:03:19,200 --> 00:03:33,170 So we're going to go through a thermodynamic cycle, and 63 00:03:33,170 --> 00:03:37,630 here's what I want to calculate when we do this. 64 00:03:37,630 --> 00:03:43,610 Delta u, delta H, familiar state functions, changes in 65 00:03:43,610 --> 00:03:51,230 their values, q, w, heat and work. 66 00:03:51,230 --> 00:03:55,990 And I also want to introduce one new function. 67 00:03:55,990 --> 00:04:00,950 I won't give it a name yet, but it's going to have the 68 00:04:00,950 --> 00:04:15,500 unusual form dq over T. OK, we'll 69 00:04:15,500 --> 00:04:22,620 call it special function. 70 00:04:22,620 --> 00:04:28,350 It's so special that we'll call this quantity dS. 71 00:04:28,350 --> 00:04:34,475 OK, we'll just go through this on a whim, and this is going 72 00:04:34,475 --> 00:04:37,960 to be foreshadowing something that'll take on tremendous 73 00:04:37,960 --> 00:04:40,470 importance in subsequent lectures, but I'm going to use 74 00:04:40,470 --> 00:04:44,580 the opportunity now to just introduce the behavior of it 75 00:04:44,580 --> 00:04:45,650 in going around a cycle. 76 00:04:45,650 --> 00:04:49,780 If that's for me, tell them I'm busy please. 77 00:04:49,780 --> 00:05:01,840 OK, so let's look at the cycle we'll go through. 78 00:05:01,840 --> 00:05:04,100 We're going to look at pressure volume changes, 79 00:05:04,100 --> 00:05:05,990 similar to what you saw before. 80 00:05:05,990 --> 00:05:19,960 This time, we're going to have initial and final volumes, V1 81 00:05:19,960 --> 00:05:24,990 and V2 and different pressures where we might end up. 82 00:05:24,990 --> 00:05:31,130 And we're going to look at two ways to go to a 83 00:05:31,130 --> 00:05:35,840 final volume V2. 84 00:05:35,840 --> 00:05:40,740 One of them is going to end up at pressure p2 and the other 85 00:05:40,740 --> 00:05:43,580 is going to end up at pressure p3. 86 00:05:43,580 --> 00:05:50,010 One of them is going to be a reversible isothermal path. 87 00:05:50,010 --> 00:05:56,060 This is going to be our path we'll label A, reversible 88 00:05:56,060 --> 00:06:05,950 isothermal which is to say constant T, right? 89 00:06:05,950 --> 00:06:10,860 And we're going to start at temperature T1, so since it's 90 00:06:10,860 --> 00:06:12,940 constant temperature, this is going to end up 91 00:06:12,940 --> 00:06:15,490 at temperature T1. 92 00:06:15,490 --> 00:06:18,480 And then we're also going to consider a reversible 93 00:06:18,480 --> 00:06:25,760 adiabatic path. 94 00:06:25,760 --> 00:06:36,820 This'll be our path B, reversible adiabatic. 95 00:06:36,820 --> 00:06:40,410 And then we'll close this circle, this cycle. 96 00:06:40,410 --> 00:06:42,080 This is going to end up at a different 97 00:06:42,080 --> 00:06:43,140 temperature by the way. 98 00:06:43,140 --> 00:06:46,140 You saw this last time in a slightly different way. 99 00:06:46,140 --> 00:06:49,600 Last time what you saw is we compared isothermal and 100 00:06:49,600 --> 00:06:53,720 adiabatic paths that ended up at the same final pressure, 101 00:06:53,720 --> 00:06:55,650 and what you saw is that therefore, they ended up in 102 00:06:55,650 --> 00:06:57,460 different final volumes. 103 00:06:57,460 --> 00:06:59,590 So this is just a little bit different from that. 104 00:06:59,590 --> 00:07:03,560 So this is going to end up at a different temperature, we'll 105 00:07:03,560 --> 00:07:05,380 call it T2. 106 00:07:05,380 --> 00:07:08,740 And then we're going to close the cycle. 107 00:07:08,740 --> 00:07:12,010 And so this is a constant volume path then. 108 00:07:12,010 --> 00:07:19,030 This is path C, constant volume. 109 00:07:19,030 --> 00:07:21,280 OK? 110 00:07:21,280 --> 00:07:26,540 So now let's go around the cycle and just compare notes 111 00:07:26,540 --> 00:07:28,400 on what happens to the thermodynamic 112 00:07:28,400 --> 00:07:30,660 quantities as we do that. 113 00:07:30,660 --> 00:07:39,960 So here's path A, it's isothermal. 114 00:07:39,960 --> 00:07:43,610 And I didn't specify, but let's make sure to do so, 115 00:07:43,610 --> 00:07:52,740 we've got a, it's going to be an ideal gas. 116 00:07:52,740 --> 00:07:59,790 OK, so A is constant temperature. 117 00:07:59,790 --> 00:08:07,890 What does that mean for delta u in path A? 118 00:08:07,890 --> 00:08:10,010 Zero right? 119 00:08:10,010 --> 00:08:14,870 Ideal gas, no temperature change, right has to be zero 120 00:08:14,870 --> 00:08:19,920 because da is Cv dT for an ideal gas. dT is zero, there's 121 00:08:19,920 --> 00:08:22,800 no temperature change. 122 00:08:22,800 --> 00:08:31,540 OK, how about delta H, zero Cp dT for an ideal gas. 123 00:08:31,540 --> 00:08:33,530 dT is zero. 124 00:08:33,530 --> 00:08:37,370 All right, now, it's reversible. 125 00:08:37,370 --> 00:08:41,570 So that means we can immediately write down an 126 00:08:41,570 --> 00:08:45,170 expression for the work. 127 00:08:45,170 --> 00:08:53,960 Since it's reversible, it's negative p dV right? 128 00:08:53,960 --> 00:09:00,810 And since it's an ideal gas then we can replace p, right, 129 00:09:00,810 --> 00:09:05,010 with RT over V, right? 130 00:09:05,010 --> 00:09:07,500 We want to do that because we have too many variables here. 131 00:09:07,500 --> 00:09:12,130 We've already got dV we'll get rid of p as an additional 132 00:09:12,130 --> 00:09:18,590 variable and replace it with V which is already in here. 133 00:09:18,590 --> 00:09:27,520 So it's minus R T1 dV over V, right? 134 00:09:27,520 --> 00:09:33,060 I can put in T1 because we're a constant temperature. so now 135 00:09:33,060 --> 00:09:38,260 we can just integrate straight away and find that the work in 136 00:09:38,260 --> 00:09:44,160 path A it's just the integral of that quantity minus 137 00:09:44,160 --> 00:09:54,430 integral of R T1 going from V1 to V2 dV over V. so it's minus 138 00:09:54,430 --> 00:10:02,550 R T1 log V2 over V1. 139 00:10:02,550 --> 00:10:06,070 All right, so let's just do a reality check here. 140 00:10:06,070 --> 00:10:08,550 The way I've got this drawn, the volume is 141 00:10:08,550 --> 00:10:10,880 going up in the process. 142 00:10:10,880 --> 00:10:13,210 It's an expansion. 143 00:10:13,210 --> 00:10:17,660 So is this system doing work on the surroundings, or are 144 00:10:17,660 --> 00:10:25,730 the surroundings doing work on the system? 145 00:10:25,730 --> 00:10:30,230 Somebody say it real loud. 146 00:10:30,230 --> 00:10:32,210 Yes, the system is working on the surroundings, right? 147 00:10:32,210 --> 00:10:34,810 It's pushing out against them. 148 00:10:34,810 --> 00:10:37,890 Work is defined as the work that the surroundings do on 149 00:10:37,890 --> 00:10:38,760 the system. 150 00:10:38,760 --> 00:10:40,660 So this is the negative number, right? 151 00:10:40,660 --> 00:10:42,250 V2 is bigger than V1. 152 00:10:42,250 --> 00:10:43,060 This is positive. 153 00:10:43,060 --> 00:10:46,650 This whole thing is negative. 154 00:10:46,650 --> 00:10:56,220 All right, q has to be just the opposite of this because 155 00:10:56,220 --> 00:10:59,870 we've already figured out that there's no change in u. 156 00:10:59,870 --> 00:11:03,470 This is just q plus w. 157 00:11:03,470 --> 00:11:12,520 There's w, q has to be R T1 log of V2 over V1. 158 00:11:12,520 --> 00:11:20,960 OK, so that means heat is being imparted to the system, 159 00:11:20,960 --> 00:11:26,170 right, from the surroundings in a manner that compensates 160 00:11:26,170 --> 00:11:28,780 exactly the amount of work done. 161 00:11:28,780 --> 00:11:33,350 OK, All right, so these are the thermodynamic quantities 162 00:11:33,350 --> 00:11:36,010 that you're familiar with already. 163 00:11:36,010 --> 00:11:43,420 Let's just quickly look at our special function. 164 00:11:43,420 --> 00:11:45,320 It's not going to be hard to calculate because it's a 165 00:11:45,320 --> 00:11:47,550 constant temperature path, so that T in the 166 00:11:47,550 --> 00:11:50,280 bottom is just T1, right. 167 00:11:50,280 --> 00:11:55,060 You've already looked at q, right. 168 00:11:55,060 --> 00:12:20,830 So dq A over T1 it's just R log V2 over V1 right. 169 00:12:20,830 --> 00:12:23,470 So there's our special function for 170 00:12:23,470 --> 00:12:25,980 this particular path. 171 00:12:25,980 --> 00:12:41,770 OK, so that's path A. Now let's compare to path B. So if 172 00:12:41,770 --> 00:12:49,260 path B is an adibat, what's zero? q, it's adiabatic, that 173 00:12:49,260 --> 00:12:52,470 means there's no heat exchanged between the system 174 00:12:52,470 --> 00:12:54,040 and the surroundings. 175 00:12:54,040 --> 00:13:08,480 So what happens, q B is zero. 176 00:13:08,480 --> 00:13:12,600 There is a change in temperature, right? 177 00:13:12,600 --> 00:13:16,650 We've got, if we say it's a mole of gas, it's going from 178 00:13:16,650 --> 00:13:32,450 p1, V1, and T1 to one mole of our gas at p3, V2, and T2. 179 00:13:32,450 --> 00:13:33,560 So unlike before now 180 00:13:33,560 --> 00:13:35,920 temperature is going to change. 181 00:13:35,920 --> 00:13:43,680 So that means that delta u isn't zero this time. 182 00:13:43,680 --> 00:13:46,140 du, it's an ideal gas. 183 00:13:46,140 --> 00:13:52,110 So this is Cv dT and of course we can just integrate this 184 00:13:52,110 --> 00:13:53,370 straight away. 185 00:13:53,370 --> 00:14:09,740 So delta u B is Cv times T2 minus T1, right. and similarly 186 00:14:09,740 --> 00:14:15,730 dH is Cp dT right. 187 00:14:15,730 --> 00:14:29,940 So delta H in pathway B is just Cp T2 minus T1 okay? 188 00:14:29,940 --> 00:14:34,040 All right, what we've already got that q is zero. 189 00:14:34,040 --> 00:14:35,860 Delta u is q plus w. 190 00:14:35,860 --> 00:14:37,340 Delta u isn't zero. 191 00:14:37,340 --> 00:14:49,780 So this must be equal to work, right? 192 00:14:49,780 --> 00:14:52,680 And now we can look at our special function, but since 193 00:14:52,680 --> 00:14:58,840 this is zero right dq B is zero, along path B, so our 194 00:14:58,840 --> 00:15:09,594 special function, dS, so we're going to go, dq B over T is 195 00:15:09,594 --> 00:15:12,330 equal to zero. 196 00:15:12,330 --> 00:15:21,450 OK? 197 00:15:21,450 --> 00:15:27,400 OK, finally let's look at our third path, this constant 198 00:15:27,400 --> 00:15:33,540 volume path, that's going to connect T1 and T2, right? 199 00:15:33,540 --> 00:15:41,450 Constant volume, what's zero? 200 00:15:41,450 --> 00:15:41,800 STUDENT: [INAUDIBLE] 201 00:15:41,800 --> 00:15:44,460 PROFESSOR NELSON: A little more noise here. 202 00:15:44,460 --> 00:15:46,160 What's zero? 203 00:15:46,160 --> 00:15:47,270 STUDENT: Work. 204 00:15:47,270 --> 00:15:50,170 PROFESSOR NELSON: Work, great. 205 00:15:50,170 --> 00:16:02,260 So, path C constant volume means our work in past C is 206 00:16:02,260 --> 00:16:05,420 zero, right? 207 00:16:05,420 --> 00:16:11,680 Now, our temperature is going to change from T2 to T1, and 208 00:16:11,680 --> 00:16:16,170 we just saw what happens to u and H when that happens going 209 00:16:16,170 --> 00:16:17,840 from T1 to T2, and it's no different. 210 00:16:17,840 --> 00:16:20,410 It's an ideal gas going along these path. 211 00:16:20,410 --> 00:16:25,150 So we can immediately write delta u C is Cv 212 00:16:25,150 --> 00:16:28,190 times T1 minus T2. 213 00:16:28,190 --> 00:16:40,220 Delta Hc C is Cp times T1 minus T2, right? 214 00:16:40,220 --> 00:16:42,580 Work is zero. 215 00:16:42,580 --> 00:16:47,390 Delta u is w plus q, work plus heat. 216 00:16:47,390 --> 00:16:49,540 This is zero, this isn't. 217 00:16:49,540 --> 00:16:55,550 This must be heat q B, right. 218 00:16:55,550 --> 00:16:58,980 So again, there are our familiar thermodynamic 219 00:16:58,980 --> 00:17:04,300 quantities, but now let's also look at our special function. 220 00:17:04,300 --> 00:17:08,080 It's not going to be zero this time because we have non zero 221 00:17:08,080 --> 00:17:10,090 heat exchange between the system and the 222 00:17:10,090 --> 00:17:12,990 environment, right. 223 00:17:12,990 --> 00:17:25,980 So are integral of dq C over T, right, which is our heat, 224 00:17:25,980 --> 00:17:28,700 but in this case we can do this calculation easily 225 00:17:28,700 --> 00:17:36,960 because since work is zero, we can equate dq with du, right. 226 00:17:36,960 --> 00:17:44,240 So we can write this as our integral from T1 to T2, du 227 00:17:44,240 --> 00:18:06,330 over T. du is just Cv dT, right. 228 00:18:06,330 --> 00:18:15,480 So this is just Cv log T2 over T1. 229 00:18:15,480 --> 00:18:18,510 OK? 230 00:18:18,510 --> 00:18:26,750 So now we've completed the cycle. 231 00:18:26,750 --> 00:18:28,280 So let's just compare. 232 00:18:28,280 --> 00:18:34,510 Let's compare what happened in path A to what happened in 233 00:18:34,510 --> 00:18:43,030 paths B and C. Yes? 234 00:18:43,030 --> 00:18:43,740 STUDENT: [INAUDIBLE] 235 00:18:43,740 --> 00:18:46,250 PROFESSOR NELSON: Ah sorry, yes of course. 236 00:18:46,250 --> 00:18:56,360 Thank you. 237 00:18:56,360 --> 00:19:04,630 Yes, oh yes, so. 238 00:19:04,630 --> 00:19:05,940 STUDENT: [INAUDIBLE] 239 00:19:05,940 --> 00:19:07,600 PROFESSOR NELSON: And it's right because, thank you 240 00:19:07,600 --> 00:19:11,930 again, it's because we're going from T1 to T2. 241 00:19:11,930 --> 00:19:17,270 So I'm getting confused by the oh sorry, we're 242 00:19:17,270 --> 00:19:20,880 going from T2 to T1. 243 00:19:20,880 --> 00:19:25,810 So it's in reverse of the order of the subscripts and I 244 00:19:25,810 --> 00:19:27,850 let this confuse me. 245 00:19:27,850 --> 00:19:31,870 There it is. 246 00:19:31,870 --> 00:19:37,510 There it is. 247 00:19:37,510 --> 00:19:39,200 There. 248 00:19:39,200 --> 00:19:44,670 Thank you, any other details which should be pointed out? 249 00:19:44,670 --> 00:19:46,150 All right, then let's proceed. 250 00:19:46,150 --> 00:19:48,670 So what we're going to do is our comparison of what 251 00:19:48,670 --> 00:19:54,790 happened on pathway A to what happened in combined pathways 252 00:19:54,790 --> 00:19:59,780 B and C, right? 253 00:19:59,780 --> 00:20:07,700 So here are results for pathway A, right, for delta u 254 00:20:07,700 --> 00:20:09,850 zero delta H zero. 255 00:20:09,850 --> 00:20:18,210 I didn't actually explicitly write it or did I? 256 00:20:18,210 --> 00:20:20,440 Let's just write it again. 257 00:20:20,440 --> 00:20:27,133 Here's our work in path A, and here's our heat exchange in 258 00:20:27,133 --> 00:20:40,820 path A. Now let's look at the sum of B and C. So B plus C, 259 00:20:40,820 --> 00:20:48,210 here is delta uB Cv times T2 minus T1. 260 00:20:48,210 --> 00:20:57,380 Delta uC Cv times T1 minus T2, the sum is zero right. 261 00:20:57,380 --> 00:21:01,840 That's the some of delta u in these two paths, and if we 262 00:21:01,840 --> 00:21:05,390 look at delta u in just path A, it's zero. 263 00:21:05,390 --> 00:21:07,660 And it's going to be the same without delta H. The only 264 00:21:07,660 --> 00:21:12,920 difference is it'll be Cp instead of Cv, but there it is 265 00:21:12,920 --> 00:21:18,110 for pathway B. There it is for a pathway C. So the state 266 00:21:18,110 --> 00:21:22,250 functions that we're familiar with are doing what we expect 267 00:21:22,250 --> 00:21:23,470 they ought to be doing, right? 268 00:21:23,470 --> 00:21:27,830 If you go around in a cycle, starting and ending at the 269 00:21:27,830 --> 00:21:29,730 same place, the state functions 270 00:21:29,730 --> 00:21:30,660 have to stay the same. 271 00:21:30,660 --> 00:21:32,890 They only depend on the state the system is in. 272 00:21:32,890 --> 00:21:36,320 They don't depend on the path, and that's what 273 00:21:36,320 --> 00:21:38,150 we've shown, right. 274 00:21:38,150 --> 00:21:41,320 Similarly if we go from this point to this point through 275 00:21:41,320 --> 00:21:44,290 path A or if we do it through the combination of these two 276 00:21:44,290 --> 00:21:48,790 paths, and the change in u and H in any state function, those 277 00:21:48,790 --> 00:21:49,660 have to be the same. 278 00:21:49,660 --> 00:21:52,990 It turns out they're zero in both cases, and that's what 279 00:21:52,990 --> 00:21:56,260 we've seen, right? 280 00:21:56,260 --> 00:22:01,360 Now let's compare what happens to work and heat. 281 00:22:01,360 --> 00:22:04,160 So here are expressions for work and heat. 282 00:22:04,160 --> 00:22:12,615 They're opposites for the pathway A. Here's pathway B 283 00:22:12,615 --> 00:22:17,960 and C. In C the work is zero. 284 00:22:17,960 --> 00:22:24,920 In B it isn't, it's Cv times T2 minus T1, right. 285 00:22:24,920 --> 00:22:31,910 The work isn't equal to the work in pathway A, right, 286 00:22:31,910 --> 00:22:35,050 because you know work is not a state function it depends on 287 00:22:35,050 --> 00:22:39,890 the path right, and there are different amounts of work done 288 00:22:39,890 --> 00:22:44,420 on the system, or done by the system on the surroundings in 289 00:22:44,420 --> 00:22:46,190 these two different processes. 290 00:22:46,190 --> 00:22:47,150 They're both expansions. 291 00:22:47,150 --> 00:22:50,510 They'll both have net work done on the surroundings, but 292 00:22:50,510 --> 00:22:53,080 not the same amount, right. 293 00:22:53,080 --> 00:22:54,830 And of course it's going to be the same with heat. 294 00:22:54,830 --> 00:22:58,360 We've already seen that delta u is zero. 295 00:22:58,360 --> 00:23:00,920 So we know that in each case the heat is going to be the 296 00:23:00,920 --> 00:23:03,050 opposite of the work, but the work isn't the same in these 297 00:23:03,050 --> 00:23:08,950 two different ways of getting from here to here, right. 298 00:23:08,950 --> 00:23:10,640 So let's just see it explicitly. 299 00:23:10,640 --> 00:23:12,340 Here's our qA. 300 00:23:12,340 --> 00:23:18,460 Here's heat exchanged in pathway A and in pathway B 301 00:23:18,460 --> 00:23:23,810 heat is zero, and in pathway C, here is qC 302 00:23:23,810 --> 00:23:27,110 it's Cv T1 minus T2. 303 00:23:27,110 --> 00:23:29,980 So again, for both heat and work we don't 304 00:23:29,980 --> 00:23:32,570 get the same result. 305 00:23:32,570 --> 00:23:36,630 Now let's look at our special function, right. 306 00:23:36,630 --> 00:23:42,670 So here's path A. We found that it's R log V2 over V1. 307 00:23:46,850 --> 00:23:49,770 Pathway B is zero. 308 00:23:49,770 --> 00:23:53,620 Pathway C it's Cv log T1 over T2. 309 00:23:53,620 --> 00:24:08,930 All right, let's just look at that a little more carefully. 310 00:24:08,930 --> 00:24:12,500 R log V2 over V1. 311 00:24:20,080 --> 00:24:31,940 B plus C, Cv log of T1 over T2, right. 312 00:24:31,940 --> 00:24:37,340 Just some weird combination of functions, right? 313 00:24:37,340 --> 00:24:42,550 But, of course, it's not quite like that. 314 00:24:42,550 --> 00:24:44,410 Why? 315 00:24:44,410 --> 00:24:50,520 Because this path is an adiabatic path. 316 00:24:50,520 --> 00:24:54,530 And you already saw last time there was this relationship 317 00:24:54,530 --> 00:24:59,360 between the temperature and volume changes along an 318 00:24:59,360 --> 00:25:01,000 adiabatic path. 319 00:25:01,000 --> 00:25:07,620 So let me just write that down. 320 00:25:07,620 --> 00:25:16,830 Adiabatic reversible path. 321 00:25:16,830 --> 00:25:27,110 What you saw is that T1 over T2 was V2 over V1 to the 322 00:25:27,110 --> 00:25:32,310 power R over Cv. 323 00:25:32,310 --> 00:25:36,340 Isn't that something? 324 00:25:36,340 --> 00:25:39,780 So what does that mean? 325 00:25:39,780 --> 00:25:45,590 We take this Cv and put in the exponent here, right. 326 00:25:45,590 --> 00:25:50,290 And we put this R up in the exponent here. 327 00:25:50,290 --> 00:25:51,740 What are we going to discover? 328 00:25:51,740 --> 00:25:58,790 Those two things are equal right, T1 over T2 to the power 329 00:25:58,790 --> 00:26:10,380 Cv is V2 over V1 to the power R, so dS over path A equals dS 330 00:26:10,380 --> 00:26:17,570 over paths B plus C, okay? 331 00:26:17,570 --> 00:26:23,120 So what this suggests is that whatever this funny special 332 00:26:23,120 --> 00:26:28,700 function is, at least for this one cycle that we've, tried 333 00:26:28,700 --> 00:26:33,440 it's behaving like a state function, right. 334 00:26:33,440 --> 00:26:38,400 It seems like its change is independent of path. 335 00:26:38,400 --> 00:26:40,970 Start here, end up here, do it through two different 336 00:26:40,970 --> 00:26:45,480 pathways, end up in the same place, right. 337 00:26:45,480 --> 00:26:50,120 OK, now that's all the foreshadowing I'm going to 338 00:26:50,120 --> 00:26:51,220 give it for right now. 339 00:26:51,220 --> 00:26:53,500 We will certainly come back to this very 340 00:26:53,500 --> 00:26:56,720 special function shortly. 341 00:26:56,720 --> 00:26:59,520 Before I move on, I'm just going to put on the board 342 00:26:59,520 --> 00:27:04,250 another cycle, and I'm going to urge you to work through 343 00:27:04,250 --> 00:27:05,080 that on your own. 344 00:27:05,080 --> 00:27:10,230 It's worked through already, let's see, I believe it's 345 00:27:10,230 --> 00:27:17,720 worked through in the notes, yes. 346 00:27:17,720 --> 00:27:21,210 So if you need the help of the notes it's in there, but I 347 00:27:21,210 --> 00:27:26,020 would urge you to work through it on your own, and it's the 348 00:27:26,020 --> 00:27:49,190 following: let's start with the same path A, all right? 349 00:27:49,190 --> 00:28:00,480 So we've got our reversible adiabatic path right. 350 00:28:00,480 --> 00:28:06,190 And now, try working through what happens if we close the 351 00:28:06,190 --> 00:28:09,530 cycle in a different way, right, like this. 352 00:28:09,530 --> 00:28:18,310 So here's a path that's constant pressure, and here is 353 00:28:18,310 --> 00:28:24,140 a path that's constant volume, similar to the constant volume 354 00:28:24,140 --> 00:28:28,280 path that we did before, but not between the same 355 00:28:28,280 --> 00:28:30,440 pressures, right. 356 00:28:30,440 --> 00:28:37,460 So we can label this one D and this one E, and I urge you to 357 00:28:37,460 --> 00:28:41,500 just try going through that and verifying for yourself 358 00:28:41,500 --> 00:28:45,760 that again the state functions will behave the way state 359 00:28:45,760 --> 00:28:50,760 functions should, and you can see whether this special 360 00:28:50,760 --> 00:28:54,880 function once again behaves like a state function, right. 361 00:28:54,880 --> 00:28:57,270 And of course you should expect to see that the work 362 00:28:57,270 --> 00:28:59,540 and the heat again won't behave like state functions 363 00:28:59,540 --> 00:29:01,850 which they are, then you'll see you have different results 364 00:29:01,850 --> 00:29:07,420 for those, depending on the path, OK? 365 00:29:07,420 --> 00:29:12,140 Any questions about going through these cycles and using 366 00:29:12,140 --> 00:29:16,770 expressions like this and so forth? 367 00:29:16,770 --> 00:29:20,310 Expressions like this you know turn up in various places, 368 00:29:20,310 --> 00:29:22,980 like in the equations sheet that appears at the end of 369 00:29:22,980 --> 00:29:24,650 exams, right? 370 00:29:24,650 --> 00:29:27,960 Which means that you may not need to be 371 00:29:27,960 --> 00:29:30,410 committing it to memory. 372 00:29:30,410 --> 00:29:33,080 On the other hand it means that you should, it's the sort 373 00:29:33,080 --> 00:29:36,490 of thing that you should be familiar with so that the need 374 00:29:36,490 --> 00:29:40,610 to use it is something that you'll conjure when you're 375 00:29:40,610 --> 00:29:43,260 working on a problem and there's a reversible adiabat, 376 00:29:43,260 --> 00:29:46,870 and you have temperatures and volumes that change that you 377 00:29:46,870 --> 00:29:53,170 might like to relate to each other. 378 00:29:53,170 --> 00:29:56,750 Cycles that are suggested that you go through and aren't gone 379 00:29:56,750 --> 00:29:59,350 through in class also sometimes turn up on exams 380 00:29:59,350 --> 00:30:00,770 too, by the way. 381 00:30:00,770 --> 00:30:06,270 So that said, let's move on to the next topic, which is 382 00:30:06,270 --> 00:30:08,180 thermochemistry. 383 00:30:08,180 --> 00:30:14,170 So, so far what you've seen in most of the examples in the 384 00:30:14,170 --> 00:30:19,080 class are essentially mechanical kinds of changes, 385 00:30:19,080 --> 00:30:22,240 pressure volume sorts of changes, where the system is 386 00:30:22,240 --> 00:30:24,960 doing essentially a kind of mechanical work, pressure 387 00:30:24,960 --> 00:30:27,580 volume work on the surroundings or vice versa and 388 00:30:27,580 --> 00:30:31,110 heat is being exchanged, and you're calculating the basic 389 00:30:31,110 --> 00:30:33,385 thermodynamics just like we went through, of 390 00:30:33,385 --> 00:30:36,250 processes of that sort. 391 00:30:36,250 --> 00:30:40,230 Now I'd like to start introducing something that's 392 00:30:40,230 --> 00:30:43,760 really central to chemical thermodynamics, mainly 393 00:30:43,760 --> 00:30:45,420 chemistry, right. 394 00:30:45,420 --> 00:30:48,550 Let's start looking at chemical reactions, right, and 395 00:30:48,550 --> 00:30:53,870 understanding what the changes in thermodynamic properties 396 00:30:53,870 --> 00:30:57,170 are that occur when chemical reactions occur, and the 397 00:30:57,170 --> 00:31:01,490 immediate application is to calculate the change in 398 00:31:01,490 --> 00:31:04,480 enthalpy, which, as you've seen it, at constant pressure 399 00:31:04,480 --> 00:31:08,200 which is the condition under which an awful lot of 400 00:31:08,200 --> 00:31:11,420 chemistry is done, that's just the heat. 401 00:31:11,420 --> 00:31:13,755 Which means we're going to calculate heats 402 00:31:13,755 --> 00:31:16,310 of reaction, right. 403 00:31:16,310 --> 00:31:17,520 You're running some reaction. 404 00:31:17,520 --> 00:31:19,260 It's in the atmosphere. 405 00:31:19,260 --> 00:31:21,140 It's at constant pressure. 406 00:31:21,140 --> 00:31:23,220 You know, you mix acid and base together and the thing 407 00:31:23,220 --> 00:31:26,400 heats up like crazy, right. 408 00:31:26,400 --> 00:31:28,600 Or other things might react spontaneously. 409 00:31:28,600 --> 00:31:31,700 You can feel something cooling off right. 410 00:31:31,700 --> 00:31:34,155 I mean simple examples of these happen when you, you 411 00:31:34,155 --> 00:31:36,350 know, if you buy cold or hot packs. 412 00:31:36,350 --> 00:31:38,940 You break the seal between them and feel the thing get 413 00:31:38,940 --> 00:31:42,780 cold, for example, right. 414 00:31:42,780 --> 00:31:43,770 What's happening there? 415 00:31:43,770 --> 00:31:46,730 Well, the selection of reactants has been done 416 00:31:46,730 --> 00:31:50,680 judiciously to provide either heat or to provide something 417 00:31:50,680 --> 00:31:51,890 that's cool. 418 00:31:51,890 --> 00:31:56,530 And all of that is coming out of the heat of reaction, 419 00:31:56,530 --> 00:31:59,450 whether it's positive or negative. 420 00:31:59,450 --> 00:32:02,130 Of course the biggest practical application is 421 00:32:02,130 --> 00:32:04,580 burning of fuel. 422 00:32:04,580 --> 00:32:08,570 You might use the heat to, and convert it into pressure 423 00:32:08,570 --> 00:32:10,640 volume work, right? 424 00:32:10,640 --> 00:32:12,060 So if it's an internal combustion 425 00:32:12,060 --> 00:32:13,870 engine, you'll do a reaction. 426 00:32:13,870 --> 00:32:15,820 There will be the heat of reaction. 427 00:32:15,820 --> 00:32:18,640 There'll be a volume change, depending on the conditions 428 00:32:18,640 --> 00:32:22,990 under which is occurs, and you can extract work through the 429 00:32:22,990 --> 00:32:24,500 reaction and so forth. 430 00:32:24,500 --> 00:32:27,410 So let's just to see how you run through some of those 431 00:32:27,410 --> 00:32:28,350 calculations. 432 00:32:28,350 --> 00:32:34,120 Now let me just ask who here took 5.111? 433 00:32:34,120 --> 00:32:36,910 And who here took 5.112? 434 00:32:36,910 --> 00:32:43,950 OK, now you've seen some thermal chemistry 435 00:32:43,950 --> 00:32:46,040 in 5.111 and 5.112. 436 00:32:46,040 --> 00:32:48,590 So I I'm going to do some review of that, but I'm also 437 00:32:48,590 --> 00:32:51,270 going to call on the thermochemistry that I'm 438 00:32:51,270 --> 00:32:53,700 pretty sure you're familiar with. 439 00:32:53,700 --> 00:32:58,000 So I won't go painstakingly over every element of the 440 00:32:58,000 --> 00:32:58,830 notes here. 441 00:32:58,830 --> 00:33:02,350 Subsequently we'll go on into additional topics that you 442 00:33:02,350 --> 00:33:24,520 haven't seen, and I'll treat them in full detail. 443 00:33:24,520 --> 00:33:33,690 OK, so let's do some thermochemistry. 444 00:33:36,590 --> 00:33:39,910 What we want to do is we'd like to be able to predict for 445 00:33:39,910 --> 00:33:43,720 any reaction what's the heat of reaction. 446 00:33:43,720 --> 00:33:59,140 And so what we're going to calculator is delta H. So for 447 00:33:59,140 --> 00:34:03,920 any kind of reaction we should be able to do that, right? 448 00:34:03,920 --> 00:34:09,080 And we're going to treat constant pressure situations 449 00:34:09,080 --> 00:34:17,090 certainly at first and most of the time, right? 450 00:34:17,090 --> 00:34:32,030 Constant pressure, right, reversible delta H is going to 451 00:34:32,030 --> 00:34:37,840 give us our heat of reaction, OK. 452 00:34:37,840 --> 00:34:44,600 And then if we can also determine delta u, then we 453 00:34:44,600 --> 00:34:50,412 know this, we know delta u is q plus w, then we can 454 00:34:50,412 --> 00:35:07,710 determine work as well, right? 455 00:35:07,710 --> 00:35:14,270 And typically, we'll be treating at least some cases 456 00:35:14,270 --> 00:35:16,670 where we're dealing with ideal gases in which case we can 457 00:35:16,670 --> 00:35:20,160 easily get delta u. 458 00:35:20,160 --> 00:35:22,080 And then we'll be able to in a very 459 00:35:22,080 --> 00:35:24,960 straightforward way get w. 460 00:35:24,960 --> 00:35:30,030 So it's a really powerful and simple formalism, once we set 461 00:35:30,030 --> 00:35:34,050 up what is needed the go through it, right? 462 00:35:34,050 --> 00:35:56,400 So let's just consider any reaction. 463 00:35:56,400 --> 00:35:59,300 The one that I've got written down for you is it's the 464 00:35:59,300 --> 00:36:02,070 reverse of the rusting of iron, right? 465 00:36:02,070 --> 00:36:07,390 So iron left to own devices in the atmosphere in the presence 466 00:36:07,390 --> 00:36:09,210 of a little bit of water, and the atmosphere will start 467 00:36:09,210 --> 00:36:11,970 combining with the water form iron oxide. 468 00:36:11,970 --> 00:36:13,810 There's an equilibrium between the two. 469 00:36:13,810 --> 00:36:19,830 So we'll have an iron oxide species. 470 00:36:19,830 --> 00:36:27,740 It's a solid. 471 00:36:27,740 --> 00:36:38,110 It's combining with hydrogen which is a gas to give iron, 472 00:36:38,110 --> 00:36:49,180 also a solid and water, right? 473 00:36:49,180 --> 00:36:52,110 So the way I've written this we're imagining the iron, the 474 00:36:52,110 --> 00:36:57,020 solid iron immersed in the water as a liquid, could be 475 00:36:57,020 --> 00:37:00,510 calculated otherwise as well, but in this case were imaging 476 00:37:00,510 --> 00:37:03,810 the water as liquid and going in this direction then it's 477 00:37:03,810 --> 00:37:12,390 forming iron oxide and evolving hydrogen gas, okay? 478 00:37:12,390 --> 00:37:21,690 And our heat of reaction or enthalpy of reaction is 479 00:37:21,690 --> 00:37:26,190 defined as the enthalpy at constant pressure. 480 00:37:26,190 --> 00:37:31,410 Isothermal conditions, temperature won't change and 481 00:37:31,410 --> 00:37:38,480 reversible work. 482 00:37:38,480 --> 00:37:49,850 For Isothermal reaction constant 483 00:37:49,850 --> 00:38:00,230 pressure reversible, right? 484 00:38:00,230 --> 00:38:02,390 Of course, they're all sorts of conditions under which a 485 00:38:02,390 --> 00:38:04,640 reaction could be wrong in the lab or 486 00:38:04,640 --> 00:38:08,330 outdoors or however, right. 487 00:38:08,330 --> 00:38:09,350 But this is the way we're going to 488 00:38:09,350 --> 00:38:11,650 define delta H of reaction. 489 00:38:11,650 --> 00:38:13,830 We want to have that definition clear because in 490 00:38:13,830 --> 00:38:16,750 fact we're going to, we might want tabulate heats of 491 00:38:16,750 --> 00:38:19,360 reaction, right, and of course want to know what the 492 00:38:19,360 --> 00:38:23,980 conditions are for the tabulated values apply. 493 00:38:23,980 --> 00:38:26,050 And we're going to want to calculate them from other 494 00:38:26,050 --> 00:38:28,300 quantities, and again, we're going to need to know each 495 00:38:28,300 --> 00:38:30,790 case what are the relevant conditions? 496 00:38:30,790 --> 00:38:37,090 OK, so what happens? 497 00:38:37,090 --> 00:38:40,910 Well, we should be able to get this. 498 00:38:40,910 --> 00:38:44,100 We should be able to calculate delta H. 499 00:38:44,100 --> 00:38:46,770 It's a state function. 500 00:38:46,770 --> 00:38:55,410 If we know the enthalpy of the products minus the enthalpy of 501 00:38:55,410 --> 00:39:00,520 the reactants, right. 502 00:39:00,520 --> 00:39:05,400 It's a state function. 503 00:39:05,400 --> 00:39:10,280 And we can we can do this in principle except for one 504 00:39:10,280 --> 00:39:14,450 important detail, which is that enthalpy, just like 505 00:39:14,450 --> 00:39:19,120 energy u isn't measured on an absolute scale but on a 506 00:39:19,120 --> 00:39:21,550 relative scale, right? 507 00:39:21,550 --> 00:39:24,310 You know, if you want to measure the potential energy 508 00:39:24,310 --> 00:39:26,820 of something in a gravitational field, you have 509 00:39:26,820 --> 00:39:30,090 to define the zero somewhere, right, because it's arbitrary. 510 00:39:30,090 --> 00:39:32,120 You can set it anywhere you want. 511 00:39:32,120 --> 00:39:33,030 It's the same with enthalpy. 512 00:39:33,030 --> 00:39:39,890 Enthalpy is just u plus p V. So there is an arbitrary set 513 00:39:39,890 --> 00:39:43,690 point that needs to be defined, right? 514 00:39:43,690 --> 00:39:47,250 Because what you actually measure in the lab are changes 515 00:39:47,250 --> 00:39:51,420 in enthalpy, just like what you measure when you look at 516 00:39:51,420 --> 00:39:54,350 energy change of some sort, you measure the change in 517 00:39:54,350 --> 00:39:55,960 energy, right. 518 00:39:55,960 --> 00:39:59,010 The absolute number that you assign to it is something 519 00:39:59,010 --> 00:39:59,880 that's arbitrary. 520 00:39:59,880 --> 00:40:03,260 You have to set what the zero is. 521 00:40:03,260 --> 00:40:07,450 And so there's a well-understood convention for 522 00:40:07,450 --> 00:40:09,010 what the zero is. 523 00:40:09,010 --> 00:40:17,290 What we define as zero is the enthalpy of every element in 524 00:40:17,290 --> 00:40:19,310 its natural state at room 525 00:40:19,310 --> 00:40:21,560 temperature and ambient pressure. 526 00:40:21,560 --> 00:40:26,780 In other words, we choose a convention for the zero of 527 00:40:26,780 --> 00:40:30,330 entropy, so that then we can write entropies of products 528 00:40:30,330 --> 00:40:35,660 and reactants always referring to the same standard state. 529 00:40:35,660 --> 00:40:41,920 And then we calculate changes, the convention is understood 530 00:40:41,920 --> 00:40:45,330 with respect to what is the zero, right. 531 00:40:45,330 --> 00:40:48,800 And so our tabulated values, they'll all work. 532 00:40:48,800 --> 00:40:52,000 They'll all refer to the same standard state. 533 00:40:52,000 --> 00:40:55,230 And we'll be able to use the formalism that we set up. 534 00:40:55,230 --> 00:41:36,030 So our reference point for H, it's 298.15 Kelvin, one bar 535 00:41:36,030 --> 00:41:44,730 and in that standard state our molar enthalpy is defined as 536 00:41:44,730 --> 00:41:59,260 zero for every element in its stable form. 537 00:41:59,260 --> 00:42:05,410 OK, very important. 538 00:42:05,410 --> 00:42:14,450 OK, now, given that reference point, all we need to do to 539 00:42:14,450 --> 00:42:18,870 get the value of enthalpy that we're going to use for each 540 00:42:18,870 --> 00:42:25,280 reactant and each product is calculate how much does the 541 00:42:25,280 --> 00:42:31,730 enthalpy change to form it from the elements in their 542 00:42:31,730 --> 00:42:33,370 stable form at room temperature 543 00:42:33,370 --> 00:42:37,540 and pressure, right. 544 00:42:37,540 --> 00:42:42,160 So we're going to replace the H or we're going to put in for 545 00:42:42,160 --> 00:42:46,050 the H what we'll call our heat of formation or delta H of 546 00:42:46,050 --> 00:42:49,260 formation starting from the elements in their stable 547 00:42:49,260 --> 00:42:51,580 states at room temperature and pressure, for 548 00:42:51,580 --> 00:42:53,500 each of these things. 549 00:42:53,500 --> 00:42:57,440 And we can do that because now we're going to have instead of 550 00:42:57,440 --> 00:43:00,650 just sort of H, it's going to be delta H of formation for 551 00:43:00,650 --> 00:43:03,570 each of these things, is going to appear in our calculation 552 00:43:03,570 --> 00:43:06,040 of H, but that's OK. 553 00:43:06,040 --> 00:43:11,310 Delta H of formation means the enthalpy of this compound 554 00:43:11,310 --> 00:43:16,720 minus the enthalpy of its constituent elements in their 555 00:43:16,720 --> 00:43:20,440 most stable state at room temperature and pressure. 556 00:43:20,440 --> 00:43:24,160 But we've defined the enthalpy of those elements in their 557 00:43:24,160 --> 00:43:26,930 stable state at room temperature and pressure as 558 00:43:26,930 --> 00:43:29,800 zero, right? 559 00:43:29,800 --> 00:43:33,930 So we're just subtracting, in effect, zero, right, from the 560 00:43:33,930 --> 00:43:37,160 enthalpy of the product, but of course it's important have 561 00:43:37,160 --> 00:43:40,070 that established because the heat of formation is something 562 00:43:40,070 --> 00:43:42,120 you could measure, right? 563 00:43:42,120 --> 00:43:46,050 You could run the reaction, take solid iron, gaseous 564 00:43:46,050 --> 00:43:51,880 oxygen, form iron oxide, measure the heat of formation 565 00:43:51,880 --> 00:43:54,680 of it, tabulate it. 566 00:43:54,680 --> 00:43:55,420 We know it. 567 00:43:55,420 --> 00:43:58,440 We know it forever, right. 568 00:43:58,440 --> 00:44:02,080 And for any of the other compounds. 569 00:44:02,080 --> 00:44:05,960 So in other words, by defining that reference state, we can 570 00:44:05,960 --> 00:44:11,390 then figure out or measure heats of formation of a vast 571 00:44:11,390 --> 00:44:13,370 number of compounds. 572 00:44:13,370 --> 00:44:14,450 We can tabulate them. 573 00:44:14,450 --> 00:44:17,250 We can know them, and then when we have reactions that 574 00:44:17,250 --> 00:44:20,920 inter-convert different compounds, we can calculate 575 00:44:20,920 --> 00:44:23,256 the heat of reaction is just the difference between the 576 00:44:23,256 --> 00:44:26,260 heat of formation of the reactants, and the heat of 577 00:44:26,260 --> 00:44:30,720 formation of the products, right. 578 00:44:30,720 --> 00:44:37,300 So let's just write that out. 579 00:44:37,300 --> 00:44:48,080 So first of all, for example, right, molar enthalpy of 580 00:44:48,080 --> 00:44:57,950 hydrogen gas at 298.15 K is zero. 581 00:44:57,950 --> 00:45:03,310 Hydrogen gas it's in its most stable state, right at room 582 00:45:03,310 --> 00:45:04,470 temperature and pressure. 583 00:45:04,470 --> 00:45:10,370 That little zero, that little superscript 584 00:45:10,370 --> 00:45:17,290 means one bar always. 585 00:45:17,290 --> 00:45:17,810 OK. 586 00:45:17,810 --> 00:45:33,750 A molar enthalpy of or whatever, iron, as a solid at 587 00:45:33,750 --> 00:45:38,030 298.15 Kelvin is zero. 588 00:45:38,030 --> 00:45:42,660 Iron as an element is a solid. 589 00:45:42,660 --> 00:45:45,220 That's it's most stable state at room temperature and 590 00:45:45,220 --> 00:45:52,580 pressure, right, and so on. 591 00:45:52,580 --> 00:45:55,290 And then we can figure out heats of formation. 592 00:45:55,290 --> 00:45:59,440 Now in this particular reaction, I've got hydrogen 593 00:45:59,440 --> 00:46:03,290 gas, iron solid. 594 00:46:03,290 --> 00:46:07,380 Those already are elements in their most stable forms at 595 00:46:07,380 --> 00:46:09,200 room temperature and pressure. 596 00:46:09,200 --> 00:46:13,880 But this is a compound, right, it has some non-zero heat of 597 00:46:13,880 --> 00:46:16,400 formation from the elements. 598 00:46:16,400 --> 00:46:19,940 So is water, right? 599 00:46:19,940 --> 00:46:23,470 So I need to find out the heats of formation of the iron 600 00:46:23,470 --> 00:46:26,500 oxide and the water. 601 00:46:26,500 --> 00:46:29,470 And if I do that then I can find out the change in 602 00:46:29,470 --> 00:46:31,120 enthalpy of this reaction. 603 00:46:31,120 --> 00:46:36,440 It's just going to be the heat of formation of these three 604 00:46:36,440 --> 00:46:38,990 moles of water, minus the heat of 605 00:46:38,990 --> 00:46:42,180 formation of the iron oxide. 606 00:46:42,180 --> 00:46:48,340 OK. 607 00:46:48,340 --> 00:46:50,270 So how am I going to do that? 608 00:46:50,270 --> 00:46:54,920 Well, I need to write the reaction that forms that 609 00:46:54,920 --> 00:46:57,220 compound from it's elements, right? 610 00:46:57,220 --> 00:47:00,930 And I want to tabulate that for an enormous number of 611 00:47:00,930 --> 00:47:02,890 materials, right? 612 00:47:02,890 --> 00:47:10,160 So for example, if I want to look at HBr, there's a simple 613 00:47:10,160 --> 00:47:18,640 case, right, hydrogen bromine. 614 00:47:18,640 --> 00:47:21,480 What's my heat of formation? 615 00:47:21,480 --> 00:47:22,900 Delta Hf. 616 00:47:22,900 --> 00:47:28,420 It's going to have our little zero, right? 617 00:47:28,420 --> 00:47:29,610 How do I calculate it? 618 00:47:29,610 --> 00:47:33,040 Well, I need to write the reaction that forms this from 619 00:47:33,040 --> 00:47:36,140 its constituent elements. 620 00:47:36,140 --> 00:47:36,960 What is it? 621 00:47:36,960 --> 00:47:48,010 Well, 1/2 H2 as a gas, temperature, at one bar, plus 622 00:47:48,010 --> 00:47:56,390 1/2 bromine, no actually bromine is a liquid at room 623 00:47:56,390 --> 00:48:00,430 temperature and one atmosphere. 624 00:48:00,430 --> 00:48:07,440 They form HBr which is a gas at room temperature and one 625 00:48:07,440 --> 00:48:11,050 bar, right. 626 00:48:11,050 --> 00:48:17,370 I can measure that. 627 00:48:17,370 --> 00:48:25,220 And the heat of reaction for this, delta H of reaction is 628 00:48:25,220 --> 00:48:36,520 equal to delta H of formation, of HBr as a gas at our 629 00:48:36,520 --> 00:48:42,500 temperature and one bar, okay? 630 00:48:42,500 --> 00:48:46,900 So that's how we determine our heats of formation. 631 00:48:46,900 --> 00:48:53,670 We measure them for all these compounds. 632 00:48:53,670 --> 00:48:58,570 And then we go back to reactions like this, and we 633 00:48:58,570 --> 00:49:12,330 can just very simply determine the heat of reaction because 634 00:49:12,330 --> 00:49:26,350 all we're doing is the following cycle. 635 00:49:26,350 --> 00:49:37,180 We go from reactants to products and there's some heat 636 00:49:37,180 --> 00:49:42,960 of reaction, and here's the cycle that we have. 637 00:49:42,960 --> 00:49:48,790 We go to the elements in their standard 638 00:49:48,790 --> 00:49:57,460 states, in both cases. 639 00:49:57,460 --> 00:50:10,000 So we're really just doing our delta H is the negative sum of 640 00:50:10,000 --> 00:50:16,260 delta H of formation for the reactants, all right? 641 00:50:16,260 --> 00:50:23,330 Because here what we're doing is we're going to take apart 642 00:50:23,330 --> 00:50:27,210 our reactants and form the elements from them, right? 643 00:50:27,210 --> 00:50:29,880 So written this way, for example, I'm going to pull 644 00:50:29,880 --> 00:50:32,460 these things apart, and I'm going to have solid iron, and 645 00:50:32,460 --> 00:50:36,270 I'm going to have gaseous oxygen, and I know the 646 00:50:36,270 --> 00:50:38,780 enthalpy of that reaction. 647 00:50:38,780 --> 00:50:42,005 That's just given by the heat of formation, the enthalpy of 648 00:50:42,005 --> 00:50:43,060 formation of iron oxide. 649 00:50:43,060 --> 00:50:50,680 Here, now I'm going to take those same elements, and I'm 650 00:50:50,680 --> 00:50:51,610 going to make the products, right. 651 00:50:51,610 --> 00:50:54,180 I know it's going to work because I already had this 652 00:50:54,180 --> 00:50:56,370 reaction, so all the, you know, I'm going to conserve 653 00:50:56,370 --> 00:50:59,380 atoms in the right way. 654 00:50:59,380 --> 00:51:03,780 So here, I'm going to have delta H, is just the sum for 655 00:51:03,780 --> 00:51:10,500 all the products of delta heat of formation, right? 656 00:51:10,500 --> 00:51:17,500 Here I'm going to put together all the products. 657 00:51:17,500 --> 00:51:19,830 So this is a positive heat of formation. 658 00:51:19,830 --> 00:51:22,560 This is the negative heat of formation, right? 659 00:51:22,560 --> 00:51:24,440 In other words, I've got reactants, 660 00:51:24,440 --> 00:51:25,560 and I've got products. 661 00:51:25,560 --> 00:51:27,510 What's delta H of reaction? 662 00:51:27,510 --> 00:51:32,040 It's delta H of formation of the products minus delta H of 663 00:51:32,040 --> 00:51:35,770 formation of the reactants. 664 00:51:35,770 --> 00:51:42,490 That's the important message, delta H of formation of the 665 00:51:42,490 --> 00:51:54,340 products, minus delta H of formation of the reactants. 666 00:51:54,340 --> 00:52:00,070 Any questions? 667 00:52:00,070 --> 00:52:03,600 All right, tomorrow I'll go through the remaining details. 668 00:52:03,600 --> 00:52:08,110 The truth is after this, it's all arithmetic, right? 669 00:52:08,110 --> 00:52:09,750 We're just going to go through it and 670 00:52:09,750 --> 00:52:11,660 execute it a little bit. 671 00:52:11,660 --> 00:52:14,200 And then we'll move on and talk about how we'd actually 672 00:52:14,200 --> 00:52:16,840 make some of these measurements of delta H and 673 00:52:16,840 --> 00:52:19,130 compare them to what we calculate. 674 00:52:19,130 --> 00:52:20,810 See you tomorrow.