1 00:00:00,000 --> 00:00:02,750 The following content is provided under a Creative 2 00:00:02,750 --> 00:00:03,820 Commons license. 3 00:00:03,820 --> 00:00:07,065 Your support will help MIT OpenCourseWare continue to 4 00:00:07,065 --> 00:00:10,520 offer high-quality educational resources for free. 5 00:00:10,520 --> 00:00:13,390 To make a donation or view additional materials from 6 00:00:13,390 --> 00:00:17,150 hundreds of MIT courses, visit MIT OpenCourseWare at 7 00:00:17,150 --> 00:00:20,460 ocw.mit.edu. 8 00:00:20,460 --> 00:00:25,160 OK, so we ended up last time, we talked about Joule-Thomson 9 00:00:25,160 --> 00:00:28,330 expansion, which is an irreversible expansion through 10 00:00:28,330 --> 00:00:31,330 a nozzle, through a porous plug, 11 00:00:31,330 --> 00:00:33,720 constant enthalpy expansion. 12 00:00:33,720 --> 00:00:38,040 And I want to make sure everybody figured out that it 13 00:00:38,040 --> 00:00:39,740 really was an irreversible expansion. 14 00:00:39,740 --> 00:00:42,880 Anybody have any questions on that, because when we voted, 15 00:00:42,880 --> 00:00:46,830 the majority of the class thought it was reversible Yes. 16 00:00:46,830 --> 00:00:49,182 STUDENT: Just a quick question on some of the constraints, 17 00:00:49,182 --> 00:00:51,926 like isothermal, isobaric, isovolumetric expansion. 18 00:00:51,926 --> 00:00:58,080 For isothermal expansion, that means that delta u does not 19 00:00:58,080 --> 00:01:02,730 change, but delta q is equal to delta w? 20 00:01:02,730 --> 00:01:03,640 PROFESSOR BAWENDI: So the question was, for an 21 00:01:03,640 --> 00:01:08,350 isothermal expansion, delta u does not change, therefore, dq 22 00:01:08,350 --> 00:01:09,910 is equal to dw. 23 00:01:09,910 --> 00:01:13,790 The answer is that's true only for an ideal gas. 24 00:01:13,790 --> 00:01:16,360 For a real gas, that won't be true. 25 00:01:16,360 --> 00:01:19,550 Ideal gas only depends on the temperature, the energy only 26 00:01:19,550 --> 00:01:20,970 depends on the temperature. 27 00:01:20,970 --> 00:01:24,505 Therefore, if it's isothermal, the energy of the ideal gas 28 00:01:24,505 --> 00:01:25,070 doesn't change. 29 00:01:25,070 --> 00:01:29,038 For a real gas it depends on more than the temperature 30 00:01:29,038 --> 00:01:31,712 STUDENT: Are there any other constraints similar to that 31 00:01:31,712 --> 00:01:37,330 [INAUDIBLE]. 32 00:01:37,330 --> 00:01:40,055 PROFESSOR BAWENDI: So, for an ideal gas, the isothermal is 33 00:01:40,055 --> 00:01:43,370 the easy one because the energy doesn't change. 34 00:01:43,370 --> 00:01:45,450 If you have an isobaric you're going to have to calculate 35 00:01:45,450 --> 00:01:48,310 where the energy changes, and that's a calculation that's 36 00:01:48,310 --> 00:01:50,610 likely on the homework and very like on an 37 00:01:50,610 --> 00:01:51,970 exam as well, too. 38 00:01:51,970 --> 00:01:54,090 In fact, we're going to do some of that today, OK, 39 00:01:54,090 --> 00:01:55,270 calculate delta u. 40 00:01:55,270 --> 00:01:57,160 Any other questions? 41 00:01:57,160 --> 00:01:59,530 Good question. 42 00:01:59,530 --> 00:02:06,850 OK, so we ended up last time showing that for an ideal gas, 43 00:02:06,850 --> 00:02:14,620 that there was was a relationship between the heat 44 00:02:14,620 --> 00:02:17,110 capacity under constant pressure, and this is the 45 00:02:17,110 --> 00:02:20,360 molar heat capacity, that's why we have the bar on top, 46 00:02:20,360 --> 00:02:24,020 and the molar heat capacity for constant volume path. 47 00:02:24,020 --> 00:02:25,780 They're related through the gas constant. 48 00:02:25,780 --> 00:02:27,860 OK, this is only true for an ideal gas, and we went through 49 00:02:27,860 --> 00:02:34,180 that mathematically where the, with a chain rule. 50 00:02:34,180 --> 00:02:36,100 And I wanted to do it a different way which is a 51 00:02:36,100 --> 00:02:39,990 little bit convoluted, but it introduces the idea of a 52 00:02:39,990 --> 00:02:42,080 thermodynamic cycle, and it's something that we're going to 53 00:02:42,080 --> 00:02:44,430 use a lot in the class. 54 00:02:44,430 --> 00:02:47,780 Because anytime you're working with the function of state, 55 00:02:47,780 --> 00:02:50,400 and you're trying to connect two points together in a p-V 56 00:02:50,400 --> 00:02:54,020 or a T-V diagram, then it doesn't matter which way 57 00:02:54,020 --> 00:02:56,260 you're connecting those two points, as long as you're 58 00:02:56,260 --> 00:02:58,100 dealing with state functions it doesn't matter. 59 00:02:58,100 --> 00:02:59,930 It only matters if you're looking at the work or the 60 00:02:59,930 --> 00:03:01,040 heat right. 61 00:03:01,040 --> 00:03:04,900 And so sometimes, when you want to calculate what the 62 00:03:04,900 --> 00:03:09,330 energy is at the endpoint, the path of the experiment that 63 00:03:09,330 --> 00:03:11,620 you're doing may not be an easy path to calculate. 64 00:03:11,620 --> 00:03:14,170 You may have to find a different path which is an 65 00:03:14,170 --> 00:03:17,810 easier one to calculate, and it doesn't matter which path 66 00:03:17,810 --> 00:03:19,450 you take, because in the end all you care is what the 67 00:03:19,450 --> 00:03:20,980 energy is at the end. 68 00:03:20,980 --> 00:03:24,770 so we're going to use this concept of the path to go from 69 00:03:24,770 --> 00:03:27,700 the initial point to the end point. 70 00:03:27,700 --> 00:03:30,310 As long as you're dealing with state functions it doesn't 71 00:03:30,310 --> 00:03:33,630 matter to get at the same results here. 72 00:03:33,630 --> 00:03:39,100 So we're going to start with a mole of gas, at some pressure, 73 00:03:39,100 --> 00:03:44,405 some volume, temperature and some mole so V, doing it per 74 00:03:44,405 --> 00:03:46,460 mole, and we're going to do two paths here. 75 00:03:46,460 --> 00:03:48,660 We're going to take a constant volume path. 76 00:03:48,660 --> 00:03:53,550 So V is constant and we're going to change the pressure 77 00:03:53,550 --> 00:03:58,170 to some, a little change in pressure, V is constant here, 78 00:03:58,170 --> 00:04:00,920 and then temperatures is also going to change. 79 00:04:00,920 --> 00:04:05,520 So this is a path, which is either a temperature change, 80 00:04:05,520 --> 00:04:08,160 with the pressure changes at the same time, going up or 81 00:04:08,160 --> 00:04:11,170 down, constant volume. 82 00:04:11,170 --> 00:04:13,606 And since the temperature is changing, clearly Cv is going 83 00:04:13,606 --> 00:04:15,220 to have something to do with this. 84 00:04:15,220 --> 00:04:18,990 So constant volume path, where the temperature is changing. 85 00:04:18,990 --> 00:04:24,040 So somehow this is going to allow us to get Cv. 86 00:04:24,040 --> 00:04:27,030 Cv is going to be related to this path. 87 00:04:27,030 --> 00:04:32,350 And if I draw a diagram on a T-V diagram of what I'm doing 88 00:04:32,350 --> 00:04:36,130 here, so there's temperature axis. 89 00:04:36,130 --> 00:04:38,020 There's the volume axis here. 90 00:04:38,020 --> 00:04:42,580 I'm starting with some V1 here. 91 00:04:42,580 --> 00:04:47,170 And this is going to be here V plus dV. 92 00:04:47,170 --> 00:04:48,100 So I'm starting right here. 93 00:04:48,100 --> 00:04:51,100 This is my initial point. 94 00:04:51,100 --> 00:04:53,610 And the path that I'm describing then, let's assume 95 00:04:53,610 --> 00:04:56,660 that we're raising the temperature up is this path 96 00:04:56,660 --> 00:04:58,110 right here. 97 00:04:58,110 --> 00:05:01,060 OK, this is my constant volume path going 98 00:05:01,060 --> 00:05:07,780 from T to T plus dT. 99 00:05:07,780 --> 00:05:10,920 OK, we'll call this 1 here. 100 00:05:10,920 --> 00:05:14,810 All right, now I want to make a path where out of that path, 101 00:05:14,810 --> 00:05:16,450 C sub p is going to fall out. 102 00:05:16,450 --> 00:05:18,570 So that's going to have to be a constant pressure path. 103 00:05:18,570 --> 00:05:19,880 Temperature is going to change but pressure 104 00:05:19,880 --> 00:05:21,270 is going to be constant. 105 00:05:21,270 --> 00:05:24,660 So what I can do is I can take my initial point here and 106 00:05:24,660 --> 00:05:28,720 create a new, different path, where, which is going to be a 107 00:05:28,720 --> 00:05:30,560 constant pressure path. 108 00:05:30,560 --> 00:05:33,210 So at the end of this path the pressure is constant. 109 00:05:33,210 --> 00:05:38,130 The volume is going to change though, to some new volume dV 110 00:05:38,130 --> 00:05:39,995 and the temperature is going to change also and I'm going 111 00:05:39,995 --> 00:05:42,820 to make it the same endpoint temperature as the endpoint 112 00:05:42,820 --> 00:05:44,020 temperature here. 113 00:05:44,020 --> 00:05:47,500 And out of this path, since the pressure is constant but 114 00:05:47,500 --> 00:05:50,630 the temperature is changing, I'm hoping that C sub p will 115 00:05:50,630 --> 00:05:53,260 fall out of that path of that calculation. 116 00:05:53,260 --> 00:05:55,290 All right? 117 00:05:55,290 --> 00:05:58,080 This should give us something about C sub p. 118 00:05:58,080 --> 00:06:00,470 So this will give us something about C sub v. This will give 119 00:06:00,470 --> 00:06:02,610 us something about C sub p. 120 00:06:02,610 --> 00:06:05,400 Now these two endpoints here are different. 121 00:06:05,400 --> 00:06:07,060 They have a different pressure, a different volume. 122 00:06:07,060 --> 00:06:09,770 They do have the same temperature though. 123 00:06:09,770 --> 00:06:13,790 So the connection between this endpoint here, 124 00:06:13,790 --> 00:06:15,580 and that one here. 125 00:06:15,580 --> 00:06:16,520 Same temperature. 126 00:06:16,520 --> 00:06:18,320 It's going to be an isothermal path that's 127 00:06:18,320 --> 00:06:20,270 going to connect them. 128 00:06:20,270 --> 00:06:24,760 So I'm going to connect them in an isothermal way, and as, 129 00:06:24,760 --> 00:06:27,740 since I'm dealing with an ideal gas, and as we saw from 130 00:06:27,740 --> 00:06:30,270 the first question here, isothermal always means delta 131 00:06:30,270 --> 00:06:31,270 u equals zero. 132 00:06:31,270 --> 00:06:34,090 So I'm going to write that immediately here, delta u 133 00:06:34,090 --> 00:06:37,010 equal zero, because I know the answer to that path here. 134 00:06:37,010 --> 00:06:39,040 All right, so let's go again what our paths are. 135 00:06:39,040 --> 00:06:41,290 Path number 1 I'm going straight 136 00:06:41,290 --> 00:06:42,740 up in the V-T diagram. 137 00:06:42,740 --> 00:06:48,100 Path number 2 on my diagram it's a reversible, this path 138 00:06:48,100 --> 00:06:51,900 number 2, it's a reversible constant pressure path. 139 00:06:51,900 --> 00:06:55,820 So I'm going to be keeping the pressure constant, but my 140 00:06:55,820 --> 00:07:02,450 volume is going to go up. 141 00:07:02,450 --> 00:07:03,910 My temperature is going to go up. 142 00:07:03,910 --> 00:07:11,380 Basically I'm putting heat in this system. 143 00:07:11,380 --> 00:07:14,590 That's going to be number 2 right here. 144 00:07:14,590 --> 00:07:16,400 It's going to be the same temperature as before but the 145 00:07:16,400 --> 00:07:19,370 volume is V plus dV now. 146 00:07:19,370 --> 00:07:23,830 That's 2, and then, so I'm heating up the system in this 147 00:07:23,830 --> 00:07:28,100 path here, and then to connect the 2 endpoints here, I'm 148 00:07:28,100 --> 00:07:31,440 going to connect them through a constant temperature path. 149 00:07:31,440 --> 00:07:34,300 Basically I'm going to cool down the system so that the -- 150 00:07:34,300 --> 00:07:37,060 no, I'm not going to cool down, I'm going to compresses 151 00:07:37,060 --> 00:07:42,090 the system back to the smaller volume V1 and the constant 152 00:07:42,090 --> 00:07:46,560 temperature conditions. 153 00:07:46,560 --> 00:07:48,970 So that's path number 3 here. 154 00:07:48,970 --> 00:07:50,710 And I'm only going to worry about energy 155 00:07:50,710 --> 00:07:51,850 in this case here. 156 00:07:51,850 --> 00:07:53,850 Because I know energy doesn't care about the path. 157 00:07:53,850 --> 00:07:57,400 Energy only cares about where I am on the diagram. 158 00:07:57,400 --> 00:08:00,000 And I know energy is related to Cv through Cv dT etcetera. 159 00:08:00,000 --> 00:08:04,540 So if I worry about energy I have a pretty good chance of 160 00:08:04,540 --> 00:08:09,030 extracting out these heat capacities, right, and I don't 161 00:08:09,030 --> 00:08:11,830 have to worry about exactly which path and I can really 162 00:08:11,830 --> 00:08:14,490 mix things up. 163 00:08:14,490 --> 00:08:21,180 So let's start the process. 164 00:08:21,180 --> 00:08:28,960 So the first path then, the first path, constant volume 165 00:08:28,960 --> 00:08:32,030 constant V, so I'm going to, again, let's 166 00:08:32,030 --> 00:08:33,000 just worry about energy. 167 00:08:33,000 --> 00:08:38,320 So du for this first path here, du for path 1, and write 168 00:08:38,320 --> 00:08:42,280 dq plus dw what we usually write. 169 00:08:42,280 --> 00:08:45,170 Now it's constant volume when the volume is constant. 170 00:08:45,170 --> 00:08:48,210 There's no work being done so I can immediately 171 00:08:48,210 --> 00:08:50,820 ignore this dw here. 172 00:08:50,820 --> 00:08:52,070 Constant volume dq = Cv dT. 173 00:08:55,120 --> 00:08:56,100 Now it's an ideal gas. 174 00:08:56,100 --> 00:08:59,250 So in some sense, I did too much work by writing this, 175 00:08:59,250 --> 00:09:02,500 because I knew already du = Cv dT. 176 00:09:02,500 --> 00:09:04,970 I could have written that down immediately, but I just went 177 00:09:04,970 --> 00:09:08,540 through the process of writing the first law, what's dw, what 178 00:09:08,540 --> 00:09:10,116 dq, Cv dT, etcetera. 179 00:09:10,116 --> 00:09:13,250 I'm just writing what we already know basically. 180 00:09:13,250 --> 00:09:15,670 Let's look at path number 2. 181 00:09:15,670 --> 00:09:16,880 Path number 2. 182 00:09:16,880 --> 00:09:22,000 OK, let's look at the dw, du again. du for path number 2. 183 00:09:22,000 --> 00:09:27,870 dq like the first law down, dq plus dw, then -- what's dq? 184 00:09:27,870 --> 00:09:31,770 Well it's a constant pressure path. dq is related to the 185 00:09:31,770 --> 00:09:33,995 temperature, the change in temperature. dT through the 186 00:09:33,995 --> 00:09:37,180 heat capacity, happens to be constant pressure, so this is 187 00:09:37,180 --> 00:09:38,420 exactly what we want. 188 00:09:38,420 --> 00:09:41,712 Cp, I forgot to put my little bar on top here because it's 189 00:09:41,712 --> 00:09:46,410 per mole Cp dT that's my dq here. 190 00:09:46,410 --> 00:09:47,660 The path is constant pressure. 191 00:09:47,660 --> 00:09:48,500 And then the dw. 192 00:09:48,500 --> 00:09:51,210 Well there's a change now in volume and pressure. 193 00:09:51,210 --> 00:09:53,690 So I need, well the pressure is constant, but there's a 194 00:09:53,690 --> 00:09:54,580 change in volume. 195 00:09:54,580 --> 00:09:57,690 Minus p dV is the change, is the work, right? 196 00:09:57,690 --> 00:10:07,820 So I can write that down minus p dV for that path there. 197 00:10:07,820 --> 00:10:12,320 OK, now but it's an ideal gas, so I know something about the 198 00:10:12,320 --> 00:10:15,080 relationship between dV and dT, because I've 199 00:10:15,080 --> 00:10:16,320 got dT here the dV. 200 00:10:16,320 --> 00:10:17,890 I don't want to have so many variables around. 201 00:10:17,890 --> 00:10:21,130 I've got three variables, T, p and V. I know I only need 2, 202 00:10:21,130 --> 00:10:24,950 so I can relate dV to dp through the ideal gas law. 203 00:10:24,950 --> 00:10:28,870 P dV is equal to R dT. 204 00:10:28,870 --> 00:10:34,770 pV = RT for 1 mole, so I just take dV here. dT here because 205 00:10:34,770 --> 00:10:43,190 the pressure is constant, so dV is equal to R over p dT. 206 00:10:43,190 --> 00:10:44,150 Then pressure is constant. 207 00:10:44,150 --> 00:10:46,730 So I can instead of the dV here, I can insert 208 00:10:46,730 --> 00:10:48,940 this R over p dT. 209 00:10:48,940 --> 00:11:00,970 The p's cancel out, and now I have Cp dT minus R dT. 210 00:11:00,970 --> 00:11:03,480 This is equal to Cp minus R dT. 211 00:11:07,040 --> 00:11:10,980 All right, let's complete the cycle now, the 212 00:11:10,980 --> 00:11:15,760 path number 3 here. 213 00:11:15,760 --> 00:11:18,080 Path number 3 is a constant temperature path, and I 214 00:11:18,080 --> 00:11:19,260 already wrote the answer. 215 00:11:19,260 --> 00:11:21,310 Constant temperature isothermal delta u is zero. 216 00:11:21,310 --> 00:11:23,440 There's no thinking involved. 217 00:11:23,440 --> 00:11:26,040 It's one of these things like I mentioned last time, if I 218 00:11:26,040 --> 00:11:30,350 tell you delta H in the middle of the night what you say, q 219 00:11:30,350 --> 00:11:33,170 Cp for reversible process. 220 00:11:33,170 --> 00:11:35,680 Same thing here. isothermal process, what do you say? 221 00:11:35,680 --> 00:11:37,900 Delta u is zero, right? 222 00:11:37,900 --> 00:11:39,590 It should be completely second nature. 223 00:11:39,590 --> 00:11:41,200 So delta u is zero here. 224 00:11:41,200 --> 00:11:45,040 Delta u sub 3 is, for an ideal gas. 225 00:11:45,040 --> 00:11:47,900 I gotta put that constraint in there for an ideal gas. 226 00:11:47,900 --> 00:11:49,370 Delta u is equal to zero. 227 00:11:49,370 --> 00:11:52,060 So now I'm back at the same place. 228 00:11:52,060 --> 00:11:54,090 I'm back here right. 229 00:11:54,090 --> 00:11:58,610 So path number 1 went from i, let's call this path up here. 230 00:11:58,610 --> 00:12:01,870 went to f, and this is how much energy change there was. 231 00:12:01,870 --> 00:12:05,430 The sum of path number 2 and path number 3 get me to the 232 00:12:05,430 --> 00:12:10,660 same place, so the energy change by going through this 233 00:12:10,660 --> 00:12:14,410 time path, this intermediate point here back all the way to 234 00:12:14,410 --> 00:12:18,760 final state should be the same the red path. 235 00:12:18,760 --> 00:12:24,530 So what I'm saying is that du through path 1 should be equal 236 00:12:24,530 --> 00:12:30,200 to du through path 2 plus du going through path 3. 237 00:12:30,200 --> 00:12:31,940 All right, because energy doesn't care about the path. 238 00:12:31,940 --> 00:12:34,350 It only cares about the end points. 239 00:12:34,350 --> 00:12:36,550 OK, so now let's plug what is du through path 1? 240 00:12:36,550 --> 00:12:47,440 It's Cp dT and the du through path 2 is Cp minus R dT and 241 00:12:47,440 --> 00:12:48,950 then du through path 3 is zero. 242 00:12:48,950 --> 00:12:50,810 It's the isothermal process. 243 00:12:50,810 --> 00:12:56,110 The dT is cancelled out and voila I have my answer. 244 00:12:56,110 --> 00:13:04,820 Cv is equal, oh Cv plus R is equal to Cp whichever way you 245 00:13:04,820 --> 00:13:07,280 want to do it, it's a relationship that we had up 246 00:13:07,280 --> 00:13:10,990 here that we wanted to prove. 247 00:13:10,990 --> 00:13:18,110 Now, I wanted to go to through this just to go through one 248 00:13:18,110 --> 00:13:20,570 cycle quickly because we're going to be doing these all 249 00:13:20,570 --> 00:13:23,140 the time, and the importance of the fact that the path 250 00:13:23,140 --> 00:13:26,360 doesn't matter, and you can always connect things together 251 00:13:26,360 --> 00:13:28,590 in a way, whatever you want. 252 00:13:28,590 --> 00:13:30,340 Whatever is the easiest. 253 00:13:30,340 --> 00:13:31,970 If you want to calculate a change in delta u. 254 00:13:31,970 --> 00:13:34,670 Now if you look at my derivation here, there's one 255 00:13:34,670 --> 00:13:38,780 spot where I could have just stopped and proven my point 256 00:13:38,780 --> 00:13:40,780 without going through the whole thing. 257 00:13:40,780 --> 00:13:43,750 So if you look at path number 2 here, the constant pressure 258 00:13:43,750 --> 00:13:51,240 path du is Cp minus R dT, du is Cp minus 259 00:13:51,240 --> 00:13:52,900 r dT for that path. 260 00:13:52,900 --> 00:13:56,810 What is du for an ideal gas? 261 00:13:56,810 --> 00:14:00,350 It's -- what else do we know that du is always equal to for 262 00:14:00,350 --> 00:14:02,470 an ideal gas? 263 00:14:02,470 --> 00:14:03,470 Cv dT right. 264 00:14:03,470 --> 00:14:08,140 So I could have written right here immediately equals Cv dT, 265 00:14:08,140 --> 00:14:10,010 and that was the end of my derivation. 266 00:14:10,010 --> 00:14:10,830 I could have done that. 267 00:14:10,830 --> 00:14:13,410 It would've been easier, but I wanted to go through the whole 268 00:14:13,410 --> 00:14:16,490 path right? 269 00:14:16,490 --> 00:14:23,850 OK, any questions? 270 00:14:23,850 --> 00:14:24,340 Cool. 271 00:14:24,340 --> 00:14:27,590 All right, so we're going to be doing more 272 00:14:27,590 --> 00:14:28,750 thermodynamics processes. 273 00:14:28,750 --> 00:14:32,210 We're building up to entropy and to engines, 274 00:14:32,210 --> 00:14:34,430 Carnot cycles, etcetera. 275 00:14:34,430 --> 00:14:37,850 And so we need to build our repertoire of knowledge in 276 00:14:37,850 --> 00:14:41,070 manipulating these cycles like the one I drew on the board 277 00:14:41,070 --> 00:14:44,840 that's hidden now You've already seen the simple 278 00:14:44,840 --> 00:14:47,510 process, which is the isothermal process, and you've 279 00:14:47,510 --> 00:14:52,620 seen how to functionalize, or what the function is that 280 00:14:52,620 --> 00:14:55,620 describes the isothermal expansion. 281 00:14:55,620 --> 00:15:01,520 Let's say we start from some V1 and p1 here, so high 282 00:15:01,520 --> 00:15:05,920 pressure, small volume and we end up with a high volume low 283 00:15:05,920 --> 00:15:10,780 pressure, under constant temperature condition. 284 00:15:10,780 --> 00:15:19,480 So there's some path, isothermal, so T is constant. 285 00:15:19,480 --> 00:15:25,240 And we know the functional form for this path, it's the 286 00:15:25,240 --> 00:15:28,360 ideal gas law, pV = RT, pV = nRT. 287 00:15:30,870 --> 00:15:35,965 T is constant, so the functional form here is pV is 288 00:15:35,965 --> 00:15:38,770 equal to a constant. 289 00:15:38,770 --> 00:15:45,380 All right, or p is equal to a constant divided by volume. 290 00:15:45,380 --> 00:15:47,280 If you want to write a function that describes this 291 00:15:47,280 --> 00:15:51,180 line here, it's pressure as a function of volume related to 292 00:15:51,180 --> 00:15:53,360 each other with this constant. 293 00:15:53,360 --> 00:15:54,290 So we know how to do this. 294 00:15:54,290 --> 00:15:57,210 So this allows us to calculate all sorts of things to get 295 00:15:57,210 --> 00:15:59,050 these functional forms. 296 00:15:59,050 --> 00:16:01,060 That's the isotherm here. 297 00:16:01,060 --> 00:16:02,750 We also know how to calculate the work. 298 00:16:02,750 --> 00:16:07,170 If it's a reversible process, we know the work is the area 299 00:16:07,170 --> 00:16:13,370 under the curve. w reversible. in this case here let's see 300 00:16:13,370 --> 00:16:17,570 this is it's a pressure is going down. 301 00:16:17,570 --> 00:16:22,180 It's an expansion, so we're doing work against the 302 00:16:22,180 --> 00:16:23,950 environment, or to the environment. 303 00:16:23,950 --> 00:16:26,080 So work is leaving the system. 304 00:16:26,080 --> 00:16:32,180 So minus w reversible is going to be a positive number. 305 00:16:32,180 --> 00:16:37,460 V1 to V2 and RT -- you've done this already last time or a 306 00:16:37,460 --> 00:16:45,200 couple of times ago and you end up with minus nRT log p2 307 00:16:45,200 --> 00:16:47,080 over p1, OK? 308 00:16:47,080 --> 00:16:50,890 So minus w reversible, we're doing work to the environment. 309 00:16:50,890 --> 00:16:54,260 This should be a positive number. p2 is less than p1, so 310 00:16:54,260 --> 00:16:56,940 p2 over p1 is less than one, log of something less than 1 311 00:16:56,940 --> 00:16:58,150 is negative times negative. 312 00:16:58,150 --> 00:17:01,040 So in fact this is indeed a positive number so we didn't 313 00:17:01,040 --> 00:17:03,130 make a mistake in our signs. 314 00:17:03,130 --> 00:17:05,460 You always want to check your signs at the and because it's 315 00:17:05,460 --> 00:17:10,190 so easy to get the sign wrong, and I was hoping that it was 316 00:17:10,190 --> 00:17:11,750 right here because I wasn't sure, but it 317 00:17:11,750 --> 00:17:13,520 turned out to be right. 318 00:17:13,520 --> 00:17:16,120 Always check your sign at the end. 319 00:17:16,120 --> 00:17:17,580 All right so we know how to do this. 320 00:17:17,580 --> 00:17:19,440 Now we are going to do the same thing with a different 321 00:17:19,440 --> 00:17:19,940 constraint. 322 00:17:19,940 --> 00:17:23,290 Our constraint is going to be an adiabatic expansion or 323 00:17:23,290 --> 00:17:23,950 compression. 324 00:17:23,950 --> 00:17:26,960 Adiabatic meaning there's no heat involved, and we're going 325 00:17:26,960 --> 00:17:30,390 to see how that differs from the isothermal expansion and 326 00:17:30,390 --> 00:17:31,000 compression. 327 00:17:31,000 --> 00:17:37,610 OK, so this is the experiment then. 328 00:17:37,610 --> 00:17:40,740 We're going to take a piston here which 329 00:17:40,740 --> 00:17:42,350 is going to be insulated. 330 00:17:42,350 --> 00:17:45,240 It's going to have some volume, temperature to begin 331 00:17:45,240 --> 00:17:50,570 with, and then we're going to do something to it. 332 00:17:50,570 --> 00:17:54,130 We're going to change the pressure above, right now 333 00:17:54,130 --> 00:17:59,270 there's a p external, which is equal to p on the inside. 334 00:17:59,270 --> 00:18:02,790 OK, we're going to do this reversibly, which means we're 335 00:18:02,790 --> 00:18:06,270 going to slowly change the external pressure very, very 336 00:18:06,270 --> 00:18:10,040 slightly at a time, so that at every point we're basically in 337 00:18:10,040 --> 00:18:13,430 equilibrium, until the pressure reaches a new smaller 338 00:18:13,430 --> 00:18:15,850 pressure p2. 339 00:18:15,850 --> 00:18:21,640 So that's going to results in an expansion where the new 340 00:18:21,640 --> 00:18:25,240 volume new temperature new pressure and an external 341 00:18:25,240 --> 00:18:28,730 pressure, which is p2 which is a smaller pressure. 342 00:18:28,730 --> 00:18:31,730 It's really important that this, that you remember that 343 00:18:31,730 --> 00:18:33,470 this is the reversible case. 344 00:18:33,470 --> 00:18:36,730 We're going to also look at the irreversible case briefly. 345 00:18:36,730 --> 00:18:42,590 So on the p-V diagram, then, there's a V1 here a V2 here, a 346 00:18:42,590 --> 00:18:44,210 p1 here a p2 here. 347 00:18:44,210 --> 00:18:49,830 This V2 is going to be different than this one here. 348 00:18:49,830 --> 00:18:52,250 A priori we don't know what it is. 349 00:18:52,250 --> 00:18:53,680 We don't know if it's bigger or smaller. 350 00:18:53,680 --> 00:18:55,810 We kind of feel like it's going to be different because 351 00:18:55,810 --> 00:18:56,840 it's a different constraint. 352 00:18:56,840 --> 00:18:58,200 This is isothermal. 353 00:18:58,200 --> 00:19:00,300 This is adiabatic, there's no heat. 354 00:19:00,300 --> 00:19:02,340 So there's going to be a line that's going to connect the 355 00:19:02,340 --> 00:19:08,540 initial point to the final point, and that line 356 00:19:08,540 --> 00:19:12,050 mathematically is not going to be the same as this one here. 357 00:19:12,050 --> 00:19:14,620 And our job is to find out what is the mathematical 358 00:19:14,620 --> 00:19:19,420 description of this path, this line in p-V's case that 359 00:19:19,420 --> 00:19:21,200 connects these two point. 360 00:19:21,200 --> 00:19:27,250 And again, I want to stress this is a reversible path. 361 00:19:27,250 --> 00:19:32,730 If it was non-reversible, I would be allowed to put an 362 00:19:32,730 --> 00:19:35,450 initial point and a final point, but I wouldn't be 363 00:19:35,450 --> 00:19:38,060 allowed to put a path between them like this, connecting 364 00:19:38,060 --> 00:19:39,570 them together. 365 00:19:39,570 --> 00:19:42,230 Because if it's irreversible, it's very likely that I don't 366 00:19:42,230 --> 00:19:45,510 know what the pressure inside the system is doing while this 367 00:19:45,510 --> 00:19:46,030 is happening. 368 00:19:46,030 --> 00:19:47,200 It could be very chaotic. 369 00:19:47,200 --> 00:19:49,860 The pressure could not be, might not even be constant 370 00:19:49,860 --> 00:19:50,900 throughout the system. 371 00:19:50,900 --> 00:19:53,110 It could have eddies and all sorts of things. 372 00:19:53,110 --> 00:19:56,340 So for an irreversible process, I wouldn't really be 373 00:19:56,340 --> 00:19:57,830 allowed to put a path there. 374 00:19:57,830 --> 00:20:00,670 I can do that because it's reversible, and I can get a 375 00:20:00,670 --> 00:20:03,130 functional form out. 376 00:20:03,130 --> 00:20:09,110 What we're going to find in this process is that the 377 00:20:09,110 --> 00:20:12,070 functional form from this is that p is related to the 378 00:20:12,070 --> 00:20:16,380 volume through a constant, but instead of being V to the 379 00:20:16,380 --> 00:20:19,820 first power as it was for the isothermal, it's now a V to 380 00:20:19,820 --> 00:20:22,480 some constant gamma. 381 00:20:22,480 --> 00:20:31,620 What we're going to find is that gamma is greater than 1. 382 00:20:31,620 --> 00:20:39,440 And if gamma is greater than 1, and you're changing the 383 00:20:39,440 --> 00:20:41,990 pressure to the same final point, what we're going to 384 00:20:41,990 --> 00:20:44,450 find then is the volume that you go to is going to be a 385 00:20:44,450 --> 00:20:47,030 smaller volume than for the isothermal. 386 00:20:47,030 --> 00:20:49,350 We're going to go through these points, so don't worry 387 00:20:49,350 --> 00:20:53,130 about getting at all straight right now. 388 00:20:53,130 --> 00:20:54,120 All right, so let's do this. 389 00:20:54,120 --> 00:20:55,370 Let's try to show this. 390 00:20:55,370 --> 00:20:57,900 Let's try to show that in fact this is the right functional 391 00:20:57,900 --> 00:21:02,990 form, that pV to the gamma is equal to constant is the right 392 00:21:02,990 --> 00:21:05,880 form that describes the path in an adiabatic expansion or 393 00:21:05,880 --> 00:21:08,270 compression. 394 00:21:08,270 --> 00:21:09,970 So let's do the experiment. 395 00:21:09,970 --> 00:21:11,940 Let's write everything that we know down. 396 00:21:11,940 --> 00:21:15,630 This is where, the way that we should start every problem set 397 00:21:15,630 --> 00:21:17,110 or ever problem. 398 00:21:17,110 --> 00:21:23,760 It's writing everything we know and what it means. 399 00:21:23,760 --> 00:21:27,250 OK, so adiabatic. 400 00:21:27,250 --> 00:21:31,050 All the words mean things mathematically. 401 00:21:31,050 --> 00:21:32,850 Adiabatic means dq equals zero. 402 00:21:32,850 --> 00:21:38,280 Let's write that down. dq equals zero. 403 00:21:38,280 --> 00:21:38,940 Reversible. 404 00:21:38,940 --> 00:21:44,550 Reversible means that if I looked at dw it's minus p 405 00:21:44,550 --> 00:21:46,310 reversible. 406 00:21:46,310 --> 00:21:50,940 It's minus p external dV, that's what it is, but in the 407 00:21:50,940 --> 00:21:56,690 reversible process p external is equal to p. 408 00:21:56,690 --> 00:21:59,110 That's the only time you can set p external equal to p is 409 00:21:59,110 --> 00:22:01,580 if it's reversible. 410 00:22:01,580 --> 00:22:02,710 So this is the external pressure. 411 00:22:02,710 --> 00:22:05,130 This is the internal pressure of the system. 412 00:22:05,130 --> 00:22:09,830 You can only do this because it's reversible. 413 00:22:09,830 --> 00:22:10,950 What else do I know? 414 00:22:10,950 --> 00:22:15,790 It's an ideal gas. 415 00:22:15,790 --> 00:22:24,660 So I can write du is Cv dT and pV is equal to RT. 416 00:22:27,350 --> 00:22:30,540 OK. 417 00:22:30,540 --> 00:22:32,170 So what else can I write? 418 00:22:32,170 --> 00:22:42,220 Well from the first law, du is equal to dq plus dw, and I 419 00:22:42,220 --> 00:22:44,600 wrote down everything I knew at the beginning here. 420 00:22:44,600 --> 00:22:47,900 I wrote dq equals zero I wrote what dw was. 421 00:22:47,900 --> 00:22:51,320 So du is just minus p dV. 422 00:22:51,320 --> 00:22:53,540 Now from the ideal gas, I have another 423 00:22:53,540 --> 00:22:55,260 expression for du, Cv dT. 424 00:22:55,260 --> 00:22:57,360 So now I have two expressions for du. 425 00:22:57,360 --> 00:23:00,440 That's a good thing, because I can them equal to each other. 426 00:23:00,440 --> 00:23:08,250 So if you get these two guys together you get Cv dT is 427 00:23:08,250 --> 00:23:11,170 minus p dV. 428 00:23:11,170 --> 00:23:13,540 So we're connecting temperature and changes in 429 00:23:13,540 --> 00:23:21,780 temperature and changes in volume together. 430 00:23:21,780 --> 00:23:24,380 OK, now we don't really want this p here. 431 00:23:24,380 --> 00:23:26,750 This is going to be a function, I only need two 432 00:23:26,750 --> 00:23:28,670 variables, and here I've got three variables down. 433 00:23:28,670 --> 00:23:31,240 So I know p is going to be a function of T and V, and I 434 00:23:31,240 --> 00:23:33,740 know how to relate p to T and V because I know 435 00:23:33,740 --> 00:23:35,040 the ideal gas law. 436 00:23:35,040 --> 00:23:44,230 So instead of p, here I'm going to put nRT over V. or RT 437 00:23:44,230 --> 00:23:47,980 over V bar. 438 00:23:47,980 --> 00:23:56,640 So now, this expression becomes Cv dT over T is equal 439 00:23:56,640 --> 00:24:05,210 to minus R dV over V. And I also did a little bit of 440 00:24:05,210 --> 00:24:05,980 rearrangement. 441 00:24:05,980 --> 00:24:06,748 Yes? 442 00:24:06,748 --> 00:24:22,100 STUDENT: [INAUDIBLE] 443 00:24:22,100 --> 00:24:25,920 PROFESSOR BAWENDI: Well, let's see. du is, for an ideal gas, 444 00:24:25,920 --> 00:24:30,240 it's always equal to Cv dT. 445 00:24:30,240 --> 00:24:33,580 Constant volume process tells you that if you have a 446 00:24:33,580 --> 00:24:37,440 constant volume, then dq is Cv dT. 447 00:24:37,440 --> 00:24:38,970 This is what the constant volume path tells you, that 448 00:24:38,970 --> 00:24:40,770 you can equate Cv dT with dq. 449 00:24:40,770 --> 00:24:45,620 The ideal gas tells you that this is always true here. 450 00:24:45,620 --> 00:24:49,920 It's an ideal gas in adiabatic expansion dq is equal to zero. 451 00:24:49,920 --> 00:24:54,500 Adiabatic expansion means you can't use this. dq is zero. 452 00:24:54,500 --> 00:24:55,690 The that is adiabatic. 453 00:24:55,690 --> 00:24:57,890 It's not constant volume. 454 00:24:57,890 --> 00:25:01,680 You're allowed Cv comes out here for this adiabatic 455 00:25:01,680 --> 00:25:04,090 expansion, which is not a constant volume only because 456 00:25:04,090 --> 00:25:06,510 this is always true for an ideal gas. 457 00:25:06,510 --> 00:25:09,740 We didn't use a constraint that V is constant. 458 00:25:09,740 --> 00:25:16,640 Just use this as an ideal gas here. 459 00:25:16,640 --> 00:25:20,620 OK, so we've got dT here dV here. 460 00:25:20,620 --> 00:25:22,250 We want to get rid of derivatives. 461 00:25:22,250 --> 00:25:23,850 We want to integrate. 462 00:25:23,850 --> 00:25:27,620 So let's take the integral of both sides, going from the 463 00:25:27,620 --> 00:25:29,220 initial point to the final point. 464 00:25:29,220 --> 00:25:35,750 We've got Cv integral from T1 to T2, dT over T is equal to 465 00:25:35,750 --> 00:25:44,050 minus R from V1 to V2 dV over V. OK, we do take that 466 00:25:44,050 --> 00:25:51,600 integral, and that leads us to Cv log of T2 over T1 is equal 467 00:25:51,600 --> 00:25:58,150 to minus R log of V2 over V1, and now we know how to 468 00:25:58,150 --> 00:26:01,950 rearrange this, because the Cv times a log of T2 over T1 is 469 00:26:01,950 --> 00:26:04,360 the same thing as the log of T2 over T1 to 470 00:26:04,360 --> 00:26:06,820 the Cv power, right. 471 00:26:06,820 --> 00:26:12,480 So by rearranging these expression of the logs here, 472 00:26:12,480 --> 00:26:17,660 this is the same thing is writing log T2 over T1 to the 473 00:26:17,660 --> 00:26:25,110 Cv power is equal to log of V1 over V2 to the R power. 474 00:26:25,110 --> 00:26:27,910 And I've reversed the fraction here because of the 475 00:26:27,910 --> 00:26:33,780 negative sign here. 476 00:26:33,780 --> 00:26:43,080 OK, get rid of the logs, and you get that T2 over T1 is 477 00:26:43,080 --> 00:26:52,390 equal to V1 over V2 to the R over Cv by taking this Cv and 478 00:26:52,390 --> 00:26:56,520 bringing it on the other side here. 479 00:26:56,520 --> 00:27:01,780 OK, so now we've got the initial and final points, a 480 00:27:01,780 --> 00:27:04,530 relationship between the temperature and volume for the 481 00:27:04,530 --> 00:27:08,320 initial and final points. 482 00:27:08,320 --> 00:27:09,650 Eventually that's not what we want. 483 00:27:09,650 --> 00:27:12,830 Eventually we want the same relationship in the pressure 484 00:27:12,830 --> 00:27:13,430 and volume. 485 00:27:13,430 --> 00:27:18,600 We want a relationship in p-V space, not in T-V space. 486 00:27:18,600 --> 00:27:20,870 So we're going to have to do something about that. 487 00:27:20,870 --> 00:27:23,670 But first, it turns out that now we have this R over Cv. 488 00:27:23,670 --> 00:27:27,260 and remember, we have this relationship between R Cv and 489 00:27:27,260 --> 00:27:30,680 Cp here, which is going to be interesting. 490 00:27:30,680 --> 00:27:36,630 So let's get rid of R in this expression here. 491 00:27:36,630 --> 00:27:41,280 So R over Cv, R is Cp minus Cv. 492 00:27:41,280 --> 00:27:44,950 We proved that first thing this morning, divided by Cv so 493 00:27:44,950 --> 00:27:52,700 this is Cp over Cv minus 1 and because I know the answer of 494 00:27:52,700 --> 00:27:56,510 this whole thing here, I'm going to call this thing here, 495 00:27:56,510 --> 00:27:59,440 Cp over Cv. 496 00:27:59,440 --> 00:28:01,280 I'm going to call it gamma. 497 00:28:01,280 --> 00:28:02,530 Because I know the answer, right. 498 00:28:02,530 --> 00:28:02,980 I know the answer. 499 00:28:02,980 --> 00:28:06,070 I'm going to call this gamma. 500 00:28:06,070 --> 00:28:13,520 By definition I'm going to define gamma by Cp over Cv by 501 00:28:13,520 --> 00:28:15,180 definition. 502 00:28:15,180 --> 00:28:19,730 OK, so that means that this is really instead of R over Cv. 503 00:28:19,730 --> 00:28:22,410 it's really gamma minus one. 504 00:28:22,410 --> 00:28:29,300 So now I've got this gamma placed in there, gamma minus 505 00:28:29,300 --> 00:28:30,960 1, so I've ben feeling a little bit better. 506 00:28:30,960 --> 00:28:33,350 I've gotten V to the gamma power almost to the gamma 507 00:28:33,350 --> 00:28:34,640 power, it's minus 1 here. 508 00:28:34,640 --> 00:28:39,120 But it's almost right and I'm looking for something like 509 00:28:39,120 --> 00:28:42,550 this here, V to the gamma power pV to the gamma power is 510 00:28:42,550 --> 00:28:45,880 a constant to describe that line. 511 00:28:45,880 --> 00:28:47,580 So I just got to get rid of the temperature now. 512 00:28:47,580 --> 00:28:49,450 Somehow I've gotta get rid of the temperature and make it 513 00:28:49,450 --> 00:28:58,260 pressure instead. 514 00:28:58,260 --> 00:29:04,500 OK, so let's -- let me get rid of this here, let's use the 515 00:29:04,500 --> 00:29:08,230 ideal gas law to get rid of the temperature. 516 00:29:08,230 --> 00:29:17,800 So we have T is pV over R. So now if I look at my V1 over V2 517 00:29:17,800 --> 00:29:24,360 to the gamma minus 1, that's T2 over T1. 518 00:29:24,360 --> 00:29:27,750 I'll replace the ideal gas law for those two T's. 519 00:29:27,750 --> 00:29:34,770 That p2 V2 over R, and then I have p1 V1 over R, the R's 520 00:29:34,770 --> 00:29:35,620 cancel out. 521 00:29:35,620 --> 00:29:39,390 Well let's see, there's a V1 over V2 to the gamma minus 1. 522 00:29:39,390 --> 00:29:42,790 There's a V1 over V2 here or V2 over V1. 523 00:29:42,790 --> 00:29:45,470 That's going to get rid of this minus 1 here. 524 00:29:45,470 --> 00:29:49,180 Then I'm going to get, be left with a p2 over p1. 525 00:29:49,180 --> 00:29:54,650 So this guy here gets rid of this, and I have my answer as 526 00:29:54,650 --> 00:30:01,860 V1 over V2 to the gamma power is equal to p2 over p1. 527 00:30:01,860 --> 00:30:08,630 Or I can re-write this as p1, V1 to the gamma is equal to p2 528 00:30:08,630 --> 00:30:11,310 V2 to the gamma. 529 00:30:11,310 --> 00:30:18,750 That means that p1 and -- where I am, the point number 1 530 00:30:18,750 --> 00:30:22,150 and point number 2 were completely arbitrary. 531 00:30:22,150 --> 00:30:24,390 They happen to be on the path. 532 00:30:24,390 --> 00:30:30,060 So any point I pick on that path will be equal to p1, V1 533 00:30:30,060 --> 00:30:30,570 to the gamma. 534 00:30:30,570 --> 00:30:34,400 So if I take p times V to the gamma, anywhere on the path, 535 00:30:34,400 --> 00:30:36,350 it's going to be equal to the same relation 536 00:30:36,350 --> 00:30:38,140 from my first point. 537 00:30:38,140 --> 00:30:40,590 This is going to be true for any point on the path. 538 00:30:40,590 --> 00:30:44,000 As long as I'm on the path, pV to the gamma will be whatever 539 00:30:44,000 --> 00:30:47,110 it was for the first point, which is going to be some sort 540 00:30:47,110 --> 00:30:52,130 of constant. 541 00:30:52,130 --> 00:30:53,540 All right, so I've proven my point. 542 00:30:53,540 --> 00:30:55,750 I've proven what I was trying to do, which was to show that 543 00:30:55,750 --> 00:31:03,230 on this adiabat, everywhere on the adiabat, if you take a 544 00:31:03,230 --> 00:31:06,500 functional form that relates p and V together, it's going to 545 00:31:06,500 --> 00:31:09,540 have this relationship with gamma as related to the heat 546 00:31:09,540 --> 00:31:10,570 capacities. 547 00:31:10,570 --> 00:31:18,020 Gamma is Cp over Cv. 548 00:31:18,020 --> 00:31:26,660 OK, now there's more we can say because we know that's Cp 549 00:31:26,660 --> 00:31:28,533 is related to Cv through this R here, and R 550 00:31:28,533 --> 00:31:30,310 is a positive number. 551 00:31:30,310 --> 00:31:34,610 So Cp is always bigger than Cv. 552 00:31:34,610 --> 00:31:42,810 In fact, if it's an ideal gas Cv and Cp are well defined 553 00:31:42,810 --> 00:31:44,530 numbers, it turns out. 554 00:31:44,530 --> 00:31:56,970 If you have a monotomic ideal gas, OK, so an atomic version, 555 00:31:56,970 --> 00:32:02,330 not a molecular ideal gas, then it turns out that Cv is 556 00:32:02,330 --> 00:32:09,380 equal to 3/2 R, and therefore, Cp which is just R plus this 557 00:32:09,380 --> 00:32:13,430 is five halves R, which means that gamma for a mono atomic 558 00:32:13,430 --> 00:32:17,470 ideal gas is 5/3. 559 00:32:17,470 --> 00:32:22,630 So if you have a gas of argon atoms, you know what gamma is. 560 00:32:22,630 --> 00:32:26,490 This is something that you're going to prove in statistical 561 00:32:26,490 --> 00:32:28,870 mechanics, and so we're not going to worry about where 562 00:32:28,870 --> 00:32:29,840 this comes from. 563 00:32:29,840 --> 00:32:33,200 We're just going to take it for granted. 564 00:32:33,200 --> 00:32:37,190 All right, so gamma is for ideal gas, is bigger than one. 565 00:32:37,190 --> 00:32:38,820 In fact we have a number for it if it's an 566 00:32:38,820 --> 00:32:40,790 atomic ideal gas. 567 00:32:40,790 --> 00:32:47,030 So what it means then is if I look at if this relationship 568 00:32:47,030 --> 00:32:51,780 here, gamma is bigger than 1, so gamma something bigger 569 00:32:51,780 --> 00:32:52,460 than 1 minus 1. 570 00:32:52,460 --> 00:32:56,440 This is, so this is positive here. 571 00:32:56,440 --> 00:32:57,820 It's positive. 572 00:32:57,820 --> 00:33:04,480 So if I have an expansion where V2 is greater than V1, 573 00:33:04,480 --> 00:33:15,510 so V2 is greater than V1, so this is a number which is less 574 00:33:15,510 --> 00:33:24,860 than one, then I expect then T2 is going to be less and T1, 575 00:33:24,860 --> 00:33:29,630 T2 is less than T1. 576 00:33:29,630 --> 00:33:37,800 OK, I have this number here to a power, which is positive, a 577 00:33:37,800 --> 00:33:39,160 positive power. 578 00:33:39,160 --> 00:33:40,760 And its number is less than one. 579 00:33:40,760 --> 00:33:41,970 This is going to give me something which 580 00:33:41,970 --> 00:33:43,060 is less than one. 581 00:33:43,060 --> 00:33:49,090 So T2 is less than T1, and I have a cooling of a gas. 582 00:33:49,090 --> 00:33:54,490 So whenever you have an adiabatic expansion, the 583 00:33:54,490 --> 00:33:56,980 temperature cools, it gets colder. 584 00:33:56,980 --> 00:34:06,740 It gets colder, and if I look at my graph here, then if the 585 00:34:06,740 --> 00:34:11,925 volume of the expansion, if the temperature of the 586 00:34:11,925 --> 00:34:18,120 adiabatic expansion was colder, that means that -- 587 00:34:18,120 --> 00:34:23,540 this is the wrong graph here. 588 00:34:23,540 --> 00:34:27,710 If I want to compare it with the isothermal expansion, 589 00:34:27,710 --> 00:34:32,320 which is sitting here. 590 00:34:32,320 --> 00:34:41,650 All right, so gamma, the gas is cooling so V2 is going to 591 00:34:41,650 --> 00:34:44,080 be less than it what would be if the 592 00:34:44,080 --> 00:34:46,050 temperature kept constant. 593 00:34:46,050 --> 00:34:48,720 So finally, on the same graph, I put down what an isothermal 594 00:34:48,720 --> 00:34:53,650 expansion would be, if my final pressure is the same 595 00:34:53,650 --> 00:35:02,570 pressure, my volume for an isothermal expansion will be 596 00:35:02,570 --> 00:35:08,530 bigger than for an adiabatic expansion. 597 00:35:08,530 --> 00:35:10,950 So when I expand this gas adiabatically and it cools 598 00:35:10,950 --> 00:35:16,650 down, why do you think it might cool down? 599 00:35:16,650 --> 00:35:19,270 It cools down because I'm doing work through the 600 00:35:19,270 --> 00:35:23,360 environment, but the energy has to go somewhere, right? 601 00:35:23,360 --> 00:35:26,440 There's an interplay between the energy inside the gas 602 00:35:26,440 --> 00:35:29,480 which is the heat energy which is allowing me to do all that 603 00:35:29,480 --> 00:35:34,100 work to be outside, and so I'm using up some of the energy 604 00:35:34,100 --> 00:35:37,420 that's inside the gas to do the work on the outside. 605 00:35:37,420 --> 00:35:42,980 And here, in an isothermal expansion, The bath is putting 606 00:35:42,980 --> 00:35:49,380 back the energy that the gas is expanding or using to 607 00:35:49,380 --> 00:35:53,620 expand, and so the energy is flowing back into the gas 608 00:35:53,620 --> 00:35:59,310 through the environment in the isothermal expansion. 609 00:35:59,310 --> 00:36:02,800 In the opposite case, if you have a compression, then it's 610 00:36:02,800 --> 00:36:03,750 the opposite of expansion. 611 00:36:03,750 --> 00:36:08,300 Compression you're expected to heat up, right? 612 00:36:08,300 --> 00:36:12,880 So in the compression, you're going to have V2 613 00:36:12,880 --> 00:36:13,720 is less than V1. 614 00:36:13,720 --> 00:36:19,420 So that T2 is going to be greater than T1. 615 00:36:19,420 --> 00:36:23,990 That's going to be your heating up of the gas. 616 00:36:23,990 --> 00:36:25,860 And that really is the bicycle pump. 617 00:36:25,860 --> 00:36:29,410 That example I gave last time with the bicycle pump was not 618 00:36:29,410 --> 00:36:30,330 quite the right example. 619 00:36:30,330 --> 00:36:32,900 But this time this really is the bicycle pump. 620 00:36:32,900 --> 00:36:34,350 That you're pushing really hard on it. 621 00:36:34,350 --> 00:36:37,650 So you start with the bicycle pump at one 622 00:36:37,650 --> 00:36:39,060 bar, let's say, right? 623 00:36:39,060 --> 00:36:40,760 And you compress the air in the bicycle pump. 624 00:36:40,760 --> 00:36:48,060 You do it so quickly that the heat flow between the inside 625 00:36:48,060 --> 00:36:50,700 of the bicycle pump and the outside is too slow compared 626 00:36:50,700 --> 00:36:53,620 to the speed at which you compress. 627 00:36:53,620 --> 00:36:55,770 But you don't compress it so quickly that you're not in 628 00:36:55,770 --> 00:37:01,100 there reversible process so you compress that if you were 629 00:37:01,100 --> 00:37:03,590 to feel the temperature of the air and you can feel it 630 00:37:03,590 --> 00:37:07,010 through the nozzle gives you an idea of the temperature of 631 00:37:07,010 --> 00:37:08,080 the air inside your bicycle pump. 632 00:37:08,080 --> 00:37:09,360 You'll feel that it's warmer. 633 00:37:09,360 --> 00:37:13,290 You've just done an adiabatic compression of the ideal gas, 634 00:37:13,290 --> 00:37:15,300 you can pretend there is an ideal gas. 635 00:37:15,300 --> 00:37:19,040 And that gives rise to the heating that you 636 00:37:19,040 --> 00:37:22,160 can actually measure. 637 00:37:22,160 --> 00:37:36,930 OK, any questions on this part here. 638 00:37:36,930 --> 00:37:41,470 All right, so we've just gone through the reversible 639 00:37:41,470 --> 00:37:46,140 adiabatic, and I'm not going to do this in detail, but I 640 00:37:46,140 --> 00:37:50,200 want to go through the irreversible adiabatic fairly 641 00:37:50,200 --> 00:37:58,000 quickly, and see where the difference is here. 642 00:37:58,000 --> 00:38:01,960 OK, so now we're going to do the same kind of experiment, 643 00:38:01,960 --> 00:38:02,270 but irreversibly. 644 00:38:02,270 --> 00:38:09,270 An irreversible adiabatic. 645 00:38:09,270 --> 00:38:17,870 OK, so the experiment is going to be to take my cylinder now, 646 00:38:17,870 --> 00:38:21,480 and the external pressure and the internal pressure are not 647 00:38:21,480 --> 00:38:22,680 going to be in equilibrium with each 648 00:38:22,680 --> 00:38:24,180 other during the process. 649 00:38:24,180 --> 00:38:25,890 I'm going to start with my cylinder here, I'm going to 650 00:38:25,890 --> 00:38:29,460 put little pegs here so it doesn't fly up. 651 00:38:29,460 --> 00:38:30,970 There's going to be some external pressure. 652 00:38:30,970 --> 00:38:32,140 I'm going to set that equal to p2. 653 00:38:32,140 --> 00:38:36,410 There is going to be an internal pressure where p1 is 654 00:38:36,410 --> 00:38:40,360 less than p2 and there's V1 and T1 here. 655 00:38:40,360 --> 00:38:44,170 The whole thing is going to be insulated. 656 00:38:44,170 --> 00:38:48,170 Then I'm going to release these little pegs here and my 657 00:38:48,170 --> 00:38:50,730 piston is going to shoot up now because p2 658 00:38:50,730 --> 00:38:52,300 is less than p1. 659 00:38:52,300 --> 00:38:57,790 So p2 is less than p1, the external pressures is less 660 00:38:57,790 --> 00:38:58,660 than the internal pressure. 661 00:38:58,660 --> 00:39:01,220 So it's going to shoot up until the internal pressure 662 00:39:01,220 --> 00:39:03,190 and the external pressure are in equilibrium. 663 00:39:03,190 --> 00:39:13,300 So they're both p2. external is p2 and I have p2, V2, T2, 664 00:39:13,300 --> 00:39:17,470 on the other side. 665 00:39:17,470 --> 00:39:28,640 OK, so in my diagram now, I have p1, V1 for my gas. p2, V2 666 00:39:28,640 --> 00:39:32,560 for my gas, so I know that I'm starting here and I know that 667 00:39:32,560 --> 00:39:35,930 I'm ending here, but I can't connect this path here. 668 00:39:35,930 --> 00:39:38,470 I don't know how to do that. 669 00:39:38,470 --> 00:39:41,950 It's an irreversible expansion. 670 00:39:41,950 --> 00:39:43,160 I don't know what the pressure is doing in 671 00:39:43,160 --> 00:39:45,030 there, doing that expansion. 672 00:39:45,030 --> 00:39:46,100 It could be quite chaotic. 673 00:39:46,100 --> 00:39:51,900 It could be non-spacially constant. 674 00:39:51,900 --> 00:39:56,710 OK, so but you still know a number of things. 675 00:39:56,710 --> 00:40:00,450 We're now going to be able to write you know dw is minus p 676 00:40:00,450 --> 00:40:01,960 dV, but we're still going to be able to 677 00:40:01,960 --> 00:40:03,520 write a bunch of things. 678 00:40:03,520 --> 00:40:05,890 We're still going to be able to write that it's an adiabat 679 00:40:05,890 --> 00:40:11,210 so that dq equals zero. 680 00:40:11,210 --> 00:40:16,530 We're still going to be able to write dw instead of pV 681 00:40:16,530 --> 00:40:19,980 where p is the internal pressure. dw now is actually 682 00:40:19,980 --> 00:40:20,590 much easier. 683 00:40:20,590 --> 00:40:22,400 It's minus p external. 684 00:40:22,400 --> 00:40:24,650 Well p external is actually p2, so that makes it actually 685 00:40:24,650 --> 00:40:29,070 easier. dV we're still going to write that 686 00:40:29,070 --> 00:40:29,900 it's an ideal gas. 687 00:40:29,900 --> 00:40:34,940 So du is still going to be equal to Cv dT, and we're 688 00:40:34,940 --> 00:40:36,590 still going to be able to use the first law, all these 689 00:40:36,590 --> 00:40:40,500 things don't matter where the path is. 690 00:40:40,500 --> 00:40:44,240 Still know that du is dq plus dw. 691 00:40:44,240 --> 00:40:48,390 dq is zero. dw is minus p2 dV. 692 00:40:48,390 --> 00:40:51,140 So this is what we know, just like what we wrote there. 693 00:40:51,140 --> 00:40:52,690 We write what we know. 694 00:40:52,690 --> 00:40:54,610 I'm not going to turn the crank here. 695 00:40:54,610 --> 00:40:56,340 I'm going to give you the answer. 696 00:40:56,340 --> 00:40:59,320 OK, you use the ideal gas law, etc., then you get a 697 00:40:59,320 --> 00:41:05,960 relationship that connects the pressure and the temperature, 698 00:41:05,960 --> 00:41:07,720 like here we got a relationship that connected 699 00:41:07,720 --> 00:41:10,160 the temperatures and the volumes together. 700 00:41:10,160 --> 00:41:15,630 You can get a relationship that connects the temperature 701 00:41:15,630 --> 00:41:27,780 so this is T2 times Cv plus R is equal to T1 on Cv plus p2 702 00:41:27,780 --> 00:41:31,810 over p1 times R. 703 00:41:31,810 --> 00:41:34,620 This is going to be the connect, what connects the 704 00:41:34,620 --> 00:41:36,580 pressures and the temperature. 705 00:41:36,580 --> 00:41:43,570 And p2 is less than p1, so this number right here 706 00:41:43,570 --> 00:41:45,470 is less than 1. 707 00:41:45,470 --> 00:41:50,240 So and Cv plus R, Cv plus something less than R, so T1 708 00:41:50,240 --> 00:41:52,370 better be bigger than T2. 709 00:41:52,370 --> 00:41:57,690 OK, so T2 is less than T1. 710 00:41:57,690 --> 00:41:59,770 Same qualitative result as before. 711 00:41:59,770 --> 00:42:01,270 You have an expansion. 712 00:42:01,270 --> 00:42:03,060 It's and adiabatic expansion. 713 00:42:03,060 --> 00:42:05,750 You get a cooling of the gas. 714 00:42:05,750 --> 00:42:07,560 OK? 715 00:42:07,560 --> 00:42:13,290 Now here's a question for you guys, which we're going to 716 00:42:13,290 --> 00:42:18,710 vote on, so you better start thinking about it. 717 00:42:18,710 --> 00:42:26,050 Is the temperature T2 in this process smaller or larger than 718 00:42:26,050 --> 00:42:31,250 if I were to do the process reversibly with the same 719 00:42:31,250 --> 00:42:35,120 endpoint pressure. 720 00:42:35,120 --> 00:42:38,100 So now, instead of having p external equals to p2 here, I 721 00:42:38,100 --> 00:42:42,970 put p1 and I let this whole thing go reversibly. 722 00:42:42,970 --> 00:42:49,760 And I compare T2 irreversible to T2 reversible. 723 00:42:49,760 --> 00:42:51,130 They're both less than T1. 724 00:42:51,130 --> 00:42:52,400 They're supposed, there's a cooling 725 00:42:52,400 --> 00:42:54,220 happening in both cases. 726 00:42:54,220 --> 00:42:57,410 One is going to be colder than the other, maybe. 727 00:42:57,410 --> 00:42:58,330 Maybe they're going to be the same. 728 00:42:58,330 --> 00:42:58,750 I don't know. 729 00:42:58,750 --> 00:43:01,500 I'm going to let you try to figure that out qualitatively 730 00:43:01,500 --> 00:43:03,790 without doing any math. 731 00:43:03,790 --> 00:43:05,230 See if you can figure it out. 732 00:43:05,230 --> 00:43:15,540 OK, so which is colder? 733 00:43:15,540 --> 00:43:18,020 I'm going to let you think about it for about thirty 734 00:43:18,020 --> 00:43:35,300 seconds or so. 735 00:43:35,300 --> 00:43:37,000 You can talk to each other if you don't, 736 00:43:37,000 --> 00:43:57,000 can't figure it out. 737 00:43:57,000 --> 00:43:58,750 All right, let's take a vote. 738 00:43:58,750 --> 00:44:01,750 How many people say that T2 irreversible is colder than T2 739 00:44:01,750 --> 00:44:04,240 reversible? 740 00:44:04,240 --> 00:44:14,320 One, two, don't be shy, three four, anybody else, five. 741 00:44:14,320 --> 00:44:15,370 OK, I've got five votes here. 742 00:44:15,370 --> 00:44:15,710 Professor Nelson? 743 00:44:15,710 --> 00:44:19,070 PROFESSOR NELSON: Oh, I'm abstaining. 744 00:44:19,070 --> 00:44:22,190 PROFESSOR BAWENDI: You're abstaining, OK, good choice. 745 00:44:22,190 --> 00:44:27,000 OK, what about T1 cooler than T2? 746 00:44:27,000 --> 00:44:29,590 OK, I've got a lot of votes here. 747 00:44:29,590 --> 00:44:32,300 So the majority of the people, so this is 748 00:44:32,300 --> 00:44:34,160 much bigger than five. 749 00:44:34,160 --> 00:44:35,830 OK. 750 00:44:35,830 --> 00:44:37,900 So remember last time the majority was wrong, right? 751 00:44:37,900 --> 00:44:39,860 The majority was wrong last time. 752 00:44:39,860 --> 00:44:41,580 That doesn't mean that the majority is wrong here. 753 00:44:41,580 --> 00:44:43,000 It could be right. 754 00:44:43,000 --> 00:44:45,660 Let me give you another piece of information here that you 755 00:44:45,660 --> 00:44:46,880 already know. 756 00:44:46,880 --> 00:44:54,380 Minus w irreversible, this is the work which is done to the 757 00:44:54,380 --> 00:44:59,300 environment by the system, minus w irreversible is always 758 00:44:59,300 --> 00:45:01,900 smaller than minus w reversible. 759 00:45:01,900 --> 00:45:04,550 You learned that a little while ago. 760 00:45:04,550 --> 00:45:08,630 OK, so this is a little hint. 761 00:45:08,630 --> 00:45:09,620 Let's vote again. 762 00:45:09,620 --> 00:45:12,520 Let's think about it for ten seconds why this could be 763 00:45:12,520 --> 00:45:19,410 interesting and relevant, and I want to ask if anybody 764 00:45:19,410 --> 00:45:22,740 changed their mind. 765 00:45:22,740 --> 00:45:24,530 Anybody change their mind, change their vote? 766 00:45:24,530 --> 00:45:27,540 Yes You want to change your vote? 767 00:45:27,540 --> 00:45:29,022 To which one? 768 00:45:29,022 --> 00:45:33,224 STUDENT: From T2 reversible is greater than T2 irreversible, 769 00:45:33,224 --> 00:45:36,280 saying that T2 reversible is [UNINTELLIGIBLE]. 770 00:45:36,280 --> 00:45:37,440 PROFESSOR BAWENDI: So the question then, 771 00:45:37,440 --> 00:45:40,482 which one is colder? 772 00:45:40,482 --> 00:45:42,447 STUDENT: T2 reversible should be colder. 773 00:45:42,447 --> 00:45:44,750 PROFESSOR BAWENDI: T2 irreversible should be colder? 774 00:45:44,750 --> 00:45:47,710 STUDENT: Yes. 775 00:45:47,710 --> 00:45:49,600 PROFESSOR BAWENDI: All right. 776 00:45:49,600 --> 00:45:50,710 OK. 777 00:45:50,710 --> 00:45:52,690 I'm going to keep you there then, the majority. 778 00:45:52,690 --> 00:45:52,900 Yes? 779 00:45:52,900 --> 00:45:54,810 STUDENT: I'm switching to [INAUDIBLE]. 780 00:45:54,810 --> 00:45:56,720 PROFESSOR BAWENDI: You're switching this way here? 781 00:45:56,720 --> 00:46:00,460 So now we have six here. 782 00:46:00,460 --> 00:46:04,840 Anybody else want to switch? 783 00:46:04,840 --> 00:46:08,700 All right, let's take the example of, the extreme 784 00:46:08,700 --> 00:46:12,350 example, let's go to the extreme example where p 785 00:46:12,350 --> 00:46:17,250 external is really small. 786 00:46:17,250 --> 00:46:22,180 And I have this adiabatic expansion where p external is 787 00:46:22,180 --> 00:46:26,030 really small. it's kind of like the Joule 788 00:46:26,030 --> 00:46:30,850 expansion, an ideal gas. 789 00:46:30,850 --> 00:46:32,240 What happened to the temperature in a Joule 790 00:46:32,240 --> 00:46:33,310 expansion in ideal gas? 791 00:46:33,310 --> 00:46:37,340 Anybody remember? 792 00:46:37,340 --> 00:46:39,860 Anybody remember? 793 00:46:39,860 --> 00:46:44,800 Joule expansion, this is where we proved that delta u was 794 00:46:44,800 --> 00:46:46,130 only dependent on temperature. 795 00:46:46,130 --> 00:46:51,780 Bob Field taught that lecture. 796 00:46:51,780 --> 00:46:52,740 Nobody remembers? 797 00:46:52,740 --> 00:46:56,430 Well, all right, delta u is equal to zero for that. 798 00:46:56,430 --> 00:46:59,960 What happened to the temperature in that expansion? 799 00:46:59,960 --> 00:47:04,080 It's an adiabatic expansion. 800 00:47:04,080 --> 00:47:04,980 Well, we'll revisit that. 801 00:47:04,980 --> 00:47:07,520 Let me ask you another question here. 802 00:47:07,520 --> 00:47:11,920 So w, the work is less for the irreversible process than the 803 00:47:11,920 --> 00:47:13,190 reversible process. 804 00:47:13,190 --> 00:47:14,960 There's less work done to the outside. 805 00:47:14,960 --> 00:47:19,080 So there's less energy expanded by the system. 806 00:47:19,080 --> 00:47:22,070 The energy expanded by the system is smaller for the 807 00:47:22,070 --> 00:47:23,240 irreversible process. 808 00:47:23,240 --> 00:47:25,150 It's not doing as much work. 809 00:47:25,150 --> 00:47:27,410 It's not pressing against as much pressure. 810 00:47:27,410 --> 00:47:31,550 Right? p external equals p2 is less than p 811 00:47:31,550 --> 00:47:34,300 external equals p internal. 812 00:47:34,300 --> 00:47:36,080 So it doesn't have to do as much work. 813 00:47:36,080 --> 00:47:39,400 It doesn't have to expend as much energy, therefore, the 814 00:47:39,400 --> 00:47:40,680 majority in this case is right. 815 00:47:40,680 --> 00:47:43,710 The temperature is not going to cool as much for the 816 00:47:43,710 --> 00:47:44,640 irreversible process. 817 00:47:44,640 --> 00:47:46,710 The reversible process is going to be colder because it 818 00:47:46,710 --> 00:47:48,000 has to expend more energy. 819 00:47:48,000 --> 00:47:50,840 It's going to have a higher pressure. 820 00:47:50,840 --> 00:47:53,460 The pressure is going to decrease along the way, but 821 00:47:53,460 --> 00:47:56,090 it's going to have to do more work because 822 00:47:56,090 --> 00:47:57,260 of this right here. 823 00:47:57,260 --> 00:47:58,850 So the majority in this case is right. 824 00:47:58,850 --> 00:48:01,680 Now the Joule expansion, the temperature doesn't change, T 825 00:48:01,680 --> 00:48:08,440 equals zero. eta sub J, the Joule coefficient for an ideal 826 00:48:08,440 --> 00:48:17,380 gas, is equal to zero and eta sub J was dT/dV 827 00:48:17,380 --> 00:48:21,210 under constant energy. 828 00:48:21,210 --> 00:48:23,340 All right, any questions? 829 00:48:23,340 --> 00:48:23,730 Yes? 830 00:48:23,730 --> 00:48:29,980 STUDENT: [INAUDIBLE]. 831 00:48:29,980 --> 00:48:32,660 PROFESSOR BAWENDI: The only -- a priori, that's true. 832 00:48:32,660 --> 00:48:36,800 But it's hard to imagine an irreversible compression the 833 00:48:36,800 --> 00:48:44,540 way we can just imagine an irreversible expansion. 834 00:48:44,540 --> 00:48:47,350 You have to do it awfully fast to get your irreversible 835 00:48:47,350 --> 00:48:47,800 compression. 836 00:48:47,800 --> 00:48:49,790 I don't know how to do it really. 837 00:48:49,790 --> 00:48:54,720 I think it's, but it's more complicated, but you're 838 00:48:54,720 --> 00:48:57,230 basically right, basically right. 839 00:48:57,230 --> 00:48:59,880 That's why we always talk about irreversible expansion, 840 00:48:59,880 --> 00:49:01,210 because it's easy to write down an 841 00:49:01,210 --> 00:49:02,330 irreversible expansion. 842 00:49:02,330 --> 00:49:04,680 A compression is a little bit more complicated. 843 00:49:04,680 --> 00:49:08,540 Any other questions? 844 00:49:08,540 --> 00:49:10,680 OK, great. 845 00:49:10,680 --> 00:49:15,450 Monday Professor Nelson who is sitting here will be taking 846 00:49:15,450 --> 00:49:18,160 over for a few weeks and then you'll see 847 00:49:18,160 --> 00:49:20,170 me again after that.