WEBVTT 01-01 00:00:00.180 --> 00:00:04.511 Welcome to session four, solid equilibria and adsorption. 01-02 00:00:04.691 --> 00:00:10.621 This independent lesson replaces our live class while I attend the E C S meeting. 01-03 00:00:10.801 --> 00:00:21.020 Last week, you used thermodynamic consistency and liquid stability to decide whether a model and a proposed phase state were reasonable. 01-04 00:00:21.200 --> 00:00:25.445 Today, we extend that reasoning to systems containing a solid. 01-05 00:00:25.625 --> 00:00:30.084 We will distinguish bulk phase equilibrium from surface uptake. 01-06 00:00:30.264 --> 00:00:38.413 We will also separate an equilibrium composition or pressure from the amount of material in each part of a system. 01-07 00:00:38.593 --> 00:00:46.551 The lesson includes two actual Lab demonstrations, your own calculations, and independent investigations. 01-08 00:00:46.731 --> 00:00:49.845 Keep your predictions and your first answers. 01-09 00:00:50.025 --> 00:00:55.551 Correcting an explanation after checking evidence is an important part of the work. 02-01 00:00:58.081 --> 00:01:04.673 Open the six-page practice worksheet and the four-page Lab guide using the links on this slide. 02-02 00:01:04.853 --> 00:01:06.773 Keep both beside the video. 02-03 00:01:06.953 --> 00:01:13.971 The worksheet is your working record, while the guide tells you which controls to use and what each result means. 02-04 00:01:14.151 --> 00:01:21.550 The worked solutions are available for checking, but attempt each question before opening the corresponding answer. 02-05 00:01:21.730 --> 00:01:28.002 Keep your original response and add a correction, rather than replacing it without explanation. 02-06 00:01:28.182 --> 00:01:34.731 You will first work through Lab ten on bulk solid equilibria, then Lab eleven on adsorption. 02-07 00:01:34.911 --> 00:01:41.546 The demonstrations show the workflow, and the later pauses give you time to reproduce it independently. 02-08 00:01:41.726 --> 00:01:48.446 Save your calculation sheets, comparison tables and study J S O N files together. 03-01 00:01:50.976 --> 00:01:54.112 This route combines watching with active work. 03-02 00:01:54.292 --> 00:01:59.817 The fixed work budget is one hundred and twenty minutes, with a separate ten-minute break. 03-03 00:01:59.997 --> 00:02:04.136 The actual recording duration appears on the video page. 03-04 00:02:04.316 --> 00:02:11.228 Together, these make a lesson of approximately three hours, although your own completion time may differ. 03-05 00:02:11.408 --> 00:02:15.696 Start a timer so you can report your actual study time later. 03-06 00:02:15.876 --> 00:02:20.057 The table follows worksheet order, not video timestamps. 03-07 00:02:20.237 --> 00:02:27.512 Work through the process review and solubility calculation, then the finite vessel balance and Lab ten. 03-08 00:02:27.692 --> 00:02:31.745 After the break, continue with adsorption and Lab eleven. 03-09 00:02:31.925 --> 00:02:38.880 Finish with a new case that you solve before seeing the answer, then submit the exit ticket and feedback. 03-10 00:02:39.060 --> 00:02:41.705 The activities form this week's lesson. 04-01 00:02:44.235 --> 00:02:51.041 Pause here for about eight minutes and complete questions A one through A three on worksheet page one. 04-02 00:02:51.221 --> 00:02:53.532 First, classify the examples. 04-03 00:02:53.712 --> 00:02:59.301 Ask whether material moves between bulk phases or accumulates at a surface. 04-04 00:02:59.481 --> 00:03:08.527 Then identify the constraint that makes a rigid closed vessel different from the fixed-temperature, fixed-pressure examples in Week three. 04-05 00:03:08.707 --> 00:03:17.667 For the inert-gas question, write a sentence distinguishing the pressure contributed by the subliming species from the total mixture pressure. 04-06 00:03:17.847 --> 00:03:21.068 You do not need numerical values yet. 04-07 00:03:21.248 --> 00:03:26.944 The purpose is to establish which physical quantities the later equations describe. 04-08 00:03:27.124 --> 00:03:30.111 Keep your answers even if you are unsure. 04-09 00:03:30.291 --> 00:03:36.477 Use the next slide as a check only after you have made an attempt, and explain any correction. 05-01 00:03:39.007 --> 00:03:40.906 Check your classifications. 05-02 00:03:41.086 --> 00:03:47.763 Dissolution of a crystal and sublimation transfer the same substance between bulk phases. 05-03 00:03:47.943 --> 00:03:51.997 Adsorption describes uptake associated with a surface. 05-04 00:03:52.177 --> 00:03:59.409 A chemical reaction would instead change species identities and require additional reaction constraints. 05-05 00:03:59.589 --> 00:04:03.493 The relevant energy criterion depends on what is fixed. 05-06 00:04:03.673 --> 00:04:08.942 At fixed temperature and pressure, a closed system minimizes Gibbs energy. 05-07 00:04:09.122 --> 00:04:14.903 At fixed temperature and volume, the corresponding potential is Helmholtz energy. 05-08 00:04:15.083 --> 00:04:21.867 In both cases, conserved inventories and the allowed states remain constraints on the minimization. 05-09 00:04:22.047 --> 00:04:27.125 For an ideal gas mixture, total pressure is the sum of partial pressures. 05-10 00:04:27.305 --> 00:04:34.430 An inert gas can therefore raise total pressure without changing the amount of the subliming species. 05-11 00:04:34.610 --> 00:04:39.346 Keep that distinction visible when we later interpret the Lab output. 06-01 00:04:41.876 --> 00:04:45.823 We begin with a solid and a liquid at equilibrium. 06-02 00:04:46.003 --> 00:04:51.400 The chemical potential of the component must be the same in the two present phases. 06-03 00:04:51.580 --> 00:05:01.457 On the liquid side, we write its chemical potential as the pure-liquid reference plus R T times the natural logarithm of x times gamma. 06-04 00:05:01.637 --> 00:05:06.608 The reference is the subcooled pure liquid at the same temperature and pressure. 06-05 00:05:06.788 --> 00:05:15.872 That reference may be a hypothetical continuation below the melting temperature, but it defines the activity convention consistently. 06-06 00:05:16.052 --> 00:05:20.404 The pure solid has unit activity on its own standard state. 06-07 00:05:20.584 --> 00:05:26.216 Unit activity does not mean that its chemical potential or amount is zero. 06-08 00:05:26.396 --> 00:05:29.404 The equality determines a liquid activity. 06-09 00:05:29.584 --> 00:05:35.579 Determining how many moles occupy each phase still requires a material balance. 07-01 00:05:38.109 --> 00:05:43.143 The solubility equation links liquid activity to fusion properties. 07-02 00:05:43.323 --> 00:05:54.395 Read the right-hand side as negative fusion enthalpy divided by R, multiplied by the quantity one over T minus one over the melting temperature. 07-03 00:05:54.575 --> 00:05:57.455 That entire expression is dimensionless. 07-04 00:05:57.635 --> 00:06:02.627 Use temperature in kelvin and fusion enthalpy in joules per mole. 07-05 00:06:02.807 --> 00:06:07.671 With R in joules per mole per kelvin, the units cancel correctly. 07-06 00:06:07.851 --> 00:06:15.745 The expression assumes constant fusion enthalpy and neglects the heat- capacity correction between solid and liquid. 07-07 00:06:15.925 --> 00:06:22.645 It also assumes the specified pure solid, without solid solutions or chemical reactions. 07-08 00:06:22.825 --> 00:06:27.795 Below the melting temperature, the reciprocal-temperature difference is positive. 07-09 00:06:27.975 --> 00:06:32.242 The negative sign therefore gives a negative logarithm. 07-10 00:06:32.422 --> 00:06:37.777 For an ideal liquid, that predicts a saturated mole fraction below one. 08-01 00:06:40.307 --> 00:06:46.088 Pause for about six minutes and complete B one and B two on worksheet page one. 08-02 00:06:46.268 --> 00:06:50.791 The synthetic melting temperature is three hundred and fifty kelvin. 08-03 00:06:50.971 --> 00:06:58.143 The fusion enthalpy is ten thousand joules per mole, and the system temperature is three hundred kelvin. 08-04 00:06:58.323 --> 00:07:01.651 Assume an ideal liquid, so gamma is one. 08-05 00:07:01.831 --> 00:07:10.322 First calculate the exponent in the solubility equation, then take the exponential to obtain the saturated mole fraction. 08-06 00:07:10.502 --> 00:07:15.430 Write the intermediate value so you can check the sign and units later. 08-07 00:07:15.610 --> 00:07:21.327 For B two, decide whether the resulting number gives moles of remaining solid. 08-08 00:07:21.507 --> 00:07:27.139 Explain what additional inventory or composition information you would need. 08-09 00:07:27.319 --> 00:07:32.823 Make that distinction before continuing to the numerical check on the next slide. 09-01 00:07:35.353 --> 00:07:41.135 The exponent is approximately negative zero point five seven two seven three. 09-02 00:07:41.315 --> 00:07:48.803 Its exponential gives a saturated liquid mole fraction of about zero point five six four. 09-03 00:07:48.983 --> 00:07:52.289 This is a composition within the liquid phase. 09-04 00:07:52.469 --> 00:08:00.789 It does not tell us the number of moles dissolved, the total amount of liquid, or the remaining solid inventory. 09-05 00:08:00.969 --> 00:08:07.348 For a nonideal liquid, the same thermodynamic relation constrains x times gamma. 09-06 00:08:07.528 --> 00:08:16.872 Since gamma can depend on composition, finding the composition may require solving an implicit equation within the admitted phase model. 09-07 00:08:17.052 --> 00:08:21.233 We still need material balances to determine phase amounts. 09-08 00:08:21.413 --> 00:08:27.984 A eutectic introduces additional solid-phase conditions and a second solubility branch. 09-09 00:08:28.164 --> 00:08:35.595 That extension is available in the optional appendix, but it is not required for today's investigation. 10-01 00:08:38.125 --> 00:08:42.499 Now consider a pure bulk solid in equilibrium with its vapor. 10-02 00:08:42.679 --> 00:08:47.607 Equality of chemical potential applies while both phases are present. 10-03 00:08:47.787 --> 00:08:57.984 With the ideal-gas and solid-pressure approximations stated here, this gives a vapor partial pressure equal to the sublimation pressure at that temperature. 10-04 00:08:58.164 --> 00:09:00.980 The phrase while solid remains matters. 10-05 00:09:01.160 --> 00:09:08.776 A small initial solid inventory may completely enter the gas phase before the vapor reaches that pressure. 10-06 00:09:08.956 --> 00:09:17.468 The final state can then have a lower partial pressure because there is no bulk solid left to enforce the two-phase equality. 10-07 00:09:17.648 --> 00:09:20.784 Also note which pressure appears in the condition. 10-08 00:09:20.964 --> 00:09:24.761 It is the partial pressure of the subliming species. 10-09 00:09:24.941 --> 00:09:32.233 An inert gas contributes to total pressure, but it is not part of that species' sublimation inventory. 11-01 00:09:34.763 --> 00:09:40.950 Start the vessel calculation by finding the gas capacity at sublimation equilibrium. 11-02 00:09:41.130 --> 00:09:47.423 Multiply sublimation pressure by accessible gas volume, then divide by R T. 11-03 00:09:47.603 --> 00:09:52.617 Compare that capacity with the total amount of the subliming species. 11-04 00:09:52.797 --> 00:09:58.301 Gas moles equal the smaller of the available inventory and that capacity. 11-05 00:09:58.481 --> 00:10:01.766 Solid moles are the remaining inventory. 11-06 00:10:01.946 --> 00:10:13.039 Use gas moles in the ideal-gas equation to obtain the species partial pressure, and add the inert contribution only when calculating total pressure. 11-07 00:10:13.219 --> 00:10:21.945 With kilopascals and liters, R has the numerical value eight point three one four four six in compatible units. 11-08 00:10:22.125 --> 00:10:34.007 The gas chemical potential minus the pure-solid reference, divided by R T, is the natural logarithm of partial pressure divided by sublimation pressure. 11-09 00:10:34.187 --> 00:10:36.769 Zero applies with present solid. 11-10 00:10:36.949 --> 00:10:41.002 A negative value is allowed after solid exhaustion. 12-01 00:10:43.532 --> 00:10:50.913 Follow the worked case at three hundred kelvin, one liter and a sublimation pressure of ten kilopascals. 12-02 00:10:51.093 --> 00:10:56.256 The capacity is ten times one, divided by R times three hundred. 12-03 00:10:56.436 --> 00:11:02.175 This gives approximately zero point zero zero four zero zero nine moles. 12-04 00:11:02.355 --> 00:11:08.200 An inventory of zero point zero one zero moles exceeds the capacity. 12-05 00:11:08.380 --> 00:11:18.343 About zero point zero zero five nine nine one moles therefore remain solid, and the vapor partial pressure is ten kilopascals. 12-06 00:11:18.523 --> 00:11:24.112 An inventory of zero point zero zero two moles is below capacity. 12-07 00:11:24.292 --> 00:11:30.308 All of it enters the gas phase, giving about four point nine eight nine kilopascals. 12-08 00:11:30.488 --> 00:11:40.003 Adding zero point zero zero two moles of inert gas adds the same pressure contribution at this fixed temperature and volume. 13-01 00:11:42.533 --> 00:11:48.997 Now pause for about ten minutes and complete C one through C three on worksheet page two. 13-02 00:11:49.177 --> 00:11:56.835 This is a different vessel: two liters at three hundred kelvin, with a sublimation pressure of eight kilopascals. 13-03 00:11:57.015 --> 00:12:04.781 Compare total inventories of zero point zero one zero and zero point zero zero four moles. 13-04 00:12:04.961 --> 00:12:11.809 Calculate the capacity first, then gas amount, remaining solid and species partial pressure. 13-05 00:12:11.989 --> 00:12:19.327 Add zero point zero zero one mole of inert gas to each case and calculate total pressure. 13-06 00:12:19.507 --> 00:12:26.526 Finally, calculate the natural logarithm of partial pressure divided by sublimation pressure. 13-07 00:12:26.706 --> 00:12:35.410 Explain the difference between the equality condition for a present solid and a valid state at the exhausted-solid boundary. 13-08 00:12:35.590 --> 00:12:39.707 Show your substitutions and units before opening the check. 14-01 00:12:42.237 --> 00:12:48.659 The gas capacity is approximately zero point zero zero six four one five moles. 14-02 00:12:48.839 --> 00:13:00.871 Therefore, the larger inventory retains about zero point zero zero three five eight five moles of solid and has a species partial pressure of eight kilopascals. 14-03 00:13:01.051 --> 00:13:08.155 The smaller inventory completely vaporizes and gives about four point nine eight nine kilopascals. 14-04 00:13:08.335 --> 00:13:13.647 The inert contributes about one point two four seven kilopascals in each case. 14-05 00:13:13.827 --> 00:13:18.819 Add that contribution to obtain the total pressures shown in the table. 14-06 00:13:18.999 --> 00:13:30.579 The logarithmic chemical-potential difference is zero with solid present, and about negative zero point four seven two three for the exhausted-solid case. 14-07 00:13:30.759 --> 00:13:33.575 A negative value is appropriate there. 14-08 00:13:33.755 --> 00:13:39.301 Check that gas plus solid equals the volatile-species inventory in each row. 15-01 00:13:41.831 --> 00:13:46.610 Pause for about five minutes and complete D one on worksheet page three. 15-02 00:13:46.790 --> 00:13:53.250 Read the first two pages of the Lab guide, but make your predictions before running the simulation. 15-03 00:13:53.430 --> 00:13:58.017 We will use three cases at three hundred kelvin and one liter. 15-04 00:13:58.197 --> 00:14:03.893 Set the sublimation reference pressure to ten kilopascals at three hundred kelvin. 15-05 00:14:04.073 --> 00:14:11.091 Case ten A has zero point zero one zero moles of volatile species and no inert. 15-06 00:14:11.271 --> 00:14:17.138 Case ten B reduces that inventory to zero point zero zero two moles. 15-07 00:14:17.318 --> 00:14:25.915 Case ten C keeps the smaller volatile inventory and adds zero point zero zero two moles of inert gas. 15-08 00:14:26.095 --> 00:14:30.938 Predict phase presence, partial pressure and total pressure for each case. 15-09 00:14:31.118 --> 00:14:35.683 Identify the one input changed between successive cases. 16-01 00:14:38.213 --> 00:14:41.307 We are now in the real Lab ten interface. 16-02 00:14:41.487 --> 00:14:46.713 Keep worksheet page three and Lab guide pages one and two beside the browser. 16-03 00:14:46.893 --> 00:14:53.315 In the Experiment menu, choose gas solid equilibrium, sublimation in a rigid vessel. 16-04 00:14:53.495 --> 00:14:57.505 The lab initially supplies its own synthetic defaults. 16-05 00:14:57.685 --> 00:15:02.699 Those initial results are not our worksheet case, so do not copy them. 16-06 00:15:02.879 --> 00:15:06.825 Selecting an experiment may also reset the controls. 16-07 00:15:07.005 --> 00:15:11.677 Choose the experiment first, then enter every value in the guide. 16-08 00:15:11.857 --> 00:15:18.535 Our first record will be case ten A, with enough volatile material for some solid to remain. 16-09 00:15:18.715 --> 00:15:24.496 Enter a temperature of three hundred kelvin and a free gas volume of one liter. 16-10 00:15:24.676 --> 00:15:32.740 Set the total volatile inventory to zero point zero one zero mole and the inert gas inventory to zero. 16-11 00:15:32.920 --> 00:15:36.141 These are total amounts, not mole fractions. 16-12 00:15:36.321 --> 00:15:40.652 The free gas volume is the space available to the gas. 16-13 00:15:40.832 --> 00:15:44.715 In this teaching model we neglect the solid volume. 16-14 00:15:44.895 --> 00:15:50.953 Keep the declared temperature interval from two hundred fifty to three hundred fifty kelvin. 16-15 00:15:51.133 --> 00:16:00.328 The source field identifies these numbers as synthetic teaching definitions rather than measured properties of a real chemical. 16-16 00:16:00.508 --> 00:16:04.305 The reference temperature is also three hundred kelvin. 16-17 00:16:04.485 --> 00:16:13.168 Enter a reference sublimation pressure of ten kilopascals and a sublimation enthalpy of fifty thousand joules per mole. 16-18 00:16:13.348 --> 00:16:19.321 At the reference temperature the correlation therefore returns ten kilopascals directly. 16-19 00:16:19.501 --> 00:16:24.088 The enthalpy controls the temperature dependence away from that point. 16-20 00:16:24.268 --> 00:16:32.417 For real properties, the declared interval must exclude melting and solid transitions and match the source correlation. 16-21 00:16:32.597 --> 00:16:38.101 Notice that changing inputs clears the previous results and disables export. 16-22 00:16:38.281 --> 00:16:43.231 Press Calculate equilibrium only after checking the values and units. 16-23 00:16:43.411 --> 00:16:46.867 The result for case ten A is solid plus gas. 16-24 00:16:47.047 --> 00:16:56.028 The volatile partial pressure and the total pressure both equal ten kilopascals because this case contains no inert gas. 16-25 00:16:56.208 --> 00:16:58.384 Read the two amounts separately. 16-26 00:16:58.564 --> 00:17:09.935 There are about zero point zero zero four zero zero nine mole of volatile gas and zero point zero zero five nine nine one mole of solid. 16-27 00:17:10.115 --> 00:17:14.253 Their sum is the entered zero point zero one zero mole. 16-28 00:17:14.433 --> 00:17:19.532 The chemical potential difference is zero within numerical precision. 16-29 00:17:19.712 --> 00:17:23.893 That equality is appropriate because both phases are present. 16-30 00:17:24.073 --> 00:17:34.036 A small balance residual is a useful calculation check, but it does not validate the physical assumptions for an unknown real material. 16-31 00:17:34.216 --> 00:17:37.715 Change the View menu to Gas and solid amounts. 16-32 00:17:37.895 --> 00:17:42.055 Read the horizontal axis before interpreting the curves. 16-33 00:17:42.235 --> 00:17:46.459 It is temperature, not time and not added inventory. 16-34 00:17:46.639 --> 00:17:54.511 Each point is a separate equilibrium calculation at that temperature with the same total volatile inventory. 16-35 00:17:54.691 --> 00:17:59.704 As temperature rises, more of the volatile material can enter the gas. 16-36 00:17:59.884 --> 00:18:05.964 Eventually the solid amount reaches zero and all the volatile material is gaseous. 16-37 00:18:06.144 --> 00:18:11.797 The gas and solid amounts sum to the same inventory throughout the admitted range. 16-38 00:18:11.977 --> 00:18:16.095 This plot does not tell us how quickly sublimation occurs. 16-39 00:18:16.275 --> 00:18:20.755 Our selected operating point remains three hundred kelvin. 16-40 00:18:20.935 --> 00:18:29.148 Now change only the total volatile inventory to zero point zero zero two mole and calculate again. 16-41 00:18:29.328 --> 00:18:31.248 This is case ten B. 16-42 00:18:31.428 --> 00:18:35.503 The heading changes to unsaturated gas, no solid. 16-43 00:18:35.683 --> 00:18:44.387 All the volatile material is now gas, and the volatile partial pressure is about four point nine eight nine kilopascals. 16-44 00:18:44.567 --> 00:18:51.863 The sublimation pressure remains ten kilopascals because the temperature and correlation have not changed. 16-45 00:18:52.043 --> 00:18:56.459 There simply is not enough material to saturate this vessel. 16-46 00:18:56.639 --> 00:19:02.783 Do not force the gas pressure to ten kilopascals or report a negative solid amount. 16-47 00:19:02.963 --> 00:19:10.408 The negative gas minus solid chemical potential difference is compatible with the solid being absent. 16-48 00:19:10.588 --> 00:19:21.468 For case ten C, keep the volatile inventory at zero point zero zero two mole and add zero point zero zero two mole of inert gas. 16-49 00:19:21.648 --> 00:19:26.149 Calculate again without changing temperature or free volume. 16-50 00:19:26.329 --> 00:19:38.357 The volatile partial pressure remains about four point nine eight nine kilopascals, but the total gas pressure becomes about nine point nine seven seven kilopascals. 16-51 00:19:38.537 --> 00:19:41.716 The volatile gas mole fraction is one half. 16-52 00:19:41.896 --> 00:19:43.923 No solid has reappeared. 16-53 00:19:44.103 --> 00:19:55.260 The inert gas contributes its own ideal partial pressure without changing the volatile inventory or its saturation capacity under these assumptions. 16-54 00:19:55.440 --> 00:20:00.197 Always identify whether a reported pressure is partial or total. 16-55 00:20:00.377 --> 00:20:05.455 In the pressure view, distinguish the three curves using their legend labels. 16-56 00:20:05.635 --> 00:20:18.712 The sublimation pressure is a reference property at each temperature, the actual volatile partial pressure obeys the finite inventory constraint, and the total pressure includes the inert gas. 16-57 00:20:18.892 --> 00:20:27.255 At low temperatures, enough volatile material can remain solid for the partial pressure to follow the sublimation curve. 16-58 00:20:27.435 --> 00:20:33.259 After the solid is exhausted, the partial pressure falls below that reference curve. 16-59 00:20:33.439 --> 00:20:38.281 These are equilibrium states across temperature, not a time record. 16-60 00:20:38.461 --> 00:20:43.731 Use the numerical table for precise values at three hundred kelvin. 16-61 00:20:43.911 --> 00:20:47.196 Keep a separate record for each calculated case. 16-62 00:20:47.376 --> 00:20:54.480 In Your interpretation, write the phase label, the volatile balance, and the reason for any pressure change. 16-63 00:20:54.660 --> 00:21:02.111 Expand Study workspace, reopen and compare cases, and enter a clear label for the next captured case. 16-64 00:21:02.291 --> 00:21:07.219 Use Export study JSON after calculating that case. 16-65 00:21:07.399 --> 00:21:14.333 The export downloads the inputs and results and also retains the case in this temporary comparison. 16-66 00:21:14.513 --> 00:21:21.126 Here the three labels distinguish solid remaining, solid exhausted, and inert gas added. 16-67 00:21:21.306 --> 00:21:28.026 Saving current inputs only is a different action and does not contain a calculated result. 16-68 00:21:28.206 --> 00:21:33.390 Select Download comparison to preserve all three cases in one file. 16-69 00:21:33.570 --> 00:21:39.394 The comparison uses recorded field names, and some values use internal units. 16-70 00:21:39.574 --> 00:21:50.753 For example, the gas solid calculation stores pressure in pascals and volume in cubic meters, whereas the main controls display kilopascals and liters. 16-71 00:21:50.933 --> 00:21:54.410 Ten thousand pascals is ten kilopascals. 16-72 00:21:54.590 --> 00:21:59.070 Compare quantities only after checking their units and physical meaning. 16-73 00:21:59.250 --> 00:22:04.002 The comparison is recorded evidence, not a fresh calculation. 16-74 00:22:04.182 --> 00:22:10.091 Keep the downloaded file with your worksheet so that you can reproduce the result later. 16-75 00:22:10.271 --> 00:22:15.626 We will now reopen the saved file from the beginning of the restore workflow. 16-76 00:22:15.806 --> 00:22:23.635 Because our three synthetic cases have already been downloaded, I clear only this temporary comparison from memory. 16-77 00:22:23.815 --> 00:22:26.354 The downloaded file is unchanged. 16-78 00:22:26.534 --> 00:22:34.576 Under Load studies or a comparison JSON, choose the solids comparison file that was just saved. 16-79 00:22:34.756 --> 00:22:41.284 You can instead select an individual study JSON if you only need one case. 16-80 00:22:41.464 --> 00:22:44.878 Loading a file should produce case labels to restore. 16-81 00:22:45.058 --> 00:22:50.882 It does not mean the main calculator has already recalculated the imported inputs. 16-82 00:22:51.062 --> 00:22:55.520 The three case labels have returned after loading the comparison. 16-83 00:22:55.700 --> 00:22:59.711 Select Restore beside case ten A, solid remains. 16-84 00:22:59.891 --> 00:23:11.240 Check that the controls return to one liter, zero point zero one zero mole of volatile material, zero inert gas, and the correct reference pressure. 16-85 00:23:11.420 --> 00:23:14.983 The stored interpretation should return with the inputs. 16-86 00:23:15.163 --> 00:23:20.880 Keep the distinction between opening a recorded result and solving the current state. 16-87 00:23:21.060 --> 00:23:27.887 In particular, switching the experiment during restoration can briefly display default results. 16-88 00:23:28.067 --> 00:23:33.315 Always finish restoration by pressing Calculate equilibrium again. 16-89 00:23:33.495 --> 00:23:46.120 The recalculated case again reports solid plus gas, a volatile partial pressure of ten kilopascals, and the same gas and solid amounts as our original record. 16-90 00:23:46.300 --> 00:23:49.543 That is the reproducibility check we wanted. 16-91 00:23:49.723 --> 00:23:59.764 In worksheet section D, retain the three rows, your independent material balance, and one explanation of partial versus total pressure. 16-92 00:23:59.944 --> 00:24:04.296 Use the next activity stop to repeat this workflow yourself. 16-93 00:24:04.476 --> 00:24:16.146 Your evidence should show that you can identify which phase is present, apply the finite inventory constraint, and reopen the inputs without relying on a screenshot alone. 17-01 00:24:18.676 --> 00:24:22.942 Use this list to decide what evidence belongs in your record. 17-02 00:24:23.122 --> 00:24:28.772 Start with the active mode and inputs, because they define the problem being solved. 17-03 00:24:28.952 --> 00:24:36.248 Next, record the gas amount and remaining solid, which establish whether both bulk phases are present. 17-04 00:24:36.428 --> 00:24:40.204 Keep species partial pressure separate from total pressure. 17-05 00:24:40.384 --> 00:24:48.490 Then check the volatile inventory independently: gas moles plus solid moles must equal the volatile total. 17-06 00:24:48.670 --> 00:24:51.358 The inert has its own conserved amount. 17-07 00:24:51.538 --> 00:24:54.717 The Lab plots sweep temperature, not time. 17-08 00:24:54.897 --> 00:25:01.340 The volatile partial pressure follows the sublimation-pressure curve wherever solid remains. 17-09 00:25:01.520 --> 00:25:09.882 A crossing or boundary on that plot describes a change across the stated conditions, rather than a measured time history. 17-10 00:25:10.062 --> 00:25:14.862 Save the study file so that the chosen conditions can be reproduced. 18-01 00:25:17.392 --> 00:25:23.494 Pause for about twenty-two minutes and complete D two through D four on worksheet page three. 18-02 00:25:23.674 --> 00:25:28.644 Open Lab ten using the link, and follow Lab guide pages one and two. 18-03 00:25:28.824 --> 00:25:31.960 Run all three specified cases yourself. 18-04 00:25:32.140 --> 00:25:38.028 Record phase presence, gas and solid amounts, partial pressure and total pressure. 18-05 00:25:38.208 --> 00:25:45.803 Use your calculator to check the capacity and the inert contribution, rather than copying the output alone. 18-06 00:25:45.983 --> 00:25:52.106 Inspect the temperature plots and explain where the finite inventory becomes important. 18-07 00:25:52.286 --> 00:25:55.848 Export your studies and reopen one saved file. 18-08 00:25:56.028 --> 00:26:00.995 Restore its controls, verify their values, and calculate again. 18-09 00:26:01.175 --> 00:26:04.418 Confirm that the result agrees with the saved case. 18-10 00:26:04.598 --> 00:26:08.523 Keep the comparison table and the calculation evidence. 18-11 00:26:08.703 --> 00:26:13.994 If a result differs from your prediction, explain the difference before moving on. 19-01 00:26:16.524 --> 00:26:19.020 Check the three cases against the table. 19-02 00:26:19.200 --> 00:26:30.186 Case ten A retains about zero point zero zero five nine nine one moles of solid, so its volatile partial pressure is ten kilopascals. 19-03 00:26:30.366 --> 00:26:37.193 Case ten B has no solid left and gives about four point nine eight nine kilopascals. 19-04 00:26:37.373 --> 00:26:45.160 Case ten C has the same volatile inventory and the same volatile partial pressure as case ten B. 19-05 00:26:45.340 --> 00:26:52.529 The added inert gas increases total pressure to about nine point nine seven seven kilopascals. 19-06 00:26:52.709 --> 00:26:59.813 This is a direct example of why total pressure cannot replace partial pressure in the sublimation condition. 19-07 00:26:59.993 --> 00:27:04.793 The values describe equilibrium endpoints under the stated model. 19-08 00:27:04.973 --> 00:27:13.250 They do not predict a sublimation rate, a transport resistance or the time needed for the vessel to reach equilibrium. 20-01 00:27:15.780 --> 00:27:17.850 Take a ten-minute break here. 20-02 00:27:18.030 --> 00:27:22.531 Save your Lab ten files and your worksheet before leaving the screen. 20-03 00:27:22.711 --> 00:27:26.359 There is no calculation to complete during this break. 20-04 00:27:26.539 --> 00:27:31.936 When you return, open worksheet page four and Lab guide pages three and four. 20-05 00:27:32.116 --> 00:27:35.956 We will move from bulk sublimation to surface adsorption. 20-06 00:27:36.136 --> 00:27:41.427 That change requires a new variable: uptake per unit mass of adsorbent. 20-07 00:27:41.607 --> 00:27:45.319 The material-balance question will remain familiar. 20-08 00:27:45.499 --> 00:27:56.464 We will first evaluate an equilibrium relation at a specified pressure, then determine the pressure that can exist in a closed vessel with a fixed inventory. 20-09 00:27:56.644 --> 00:28:01.380 Keep the distinction between a state quantity and an amount in mind. 20-10 00:28:01.560 --> 00:28:05.422 Resume when you are ready to begin the adsorption section. 21-01 00:28:07.952 --> 00:28:12.112 Adsorption describes material associated with a surface. 21-02 00:28:12.292 --> 00:28:19.289 An adsorption isotherm gives uptake as a function of gas pressure at a specified temperature. 21-03 00:28:19.469 --> 00:28:25.592 In this lesson, q is absolute uptake measured in moles per kilogram of adsorbent. 21-04 00:28:25.772 --> 00:28:29.185 The pressure is the partial pressure of the adsorbate. 21-05 00:28:29.365 --> 00:28:36.746 To turn uptake into an amount in moles, multiply q by the adsorbent mass expressed in kilograms. 21-06 00:28:36.926 --> 00:28:46.630 For example, a value of one mole per kilogram on ten grams of adsorbent corresponds to zero point zero one mole adsorbed. 21-07 00:28:46.810 --> 00:28:50.650 That distinction will matter when we change adsorbent mass. 21-08 00:28:50.830 --> 00:28:57.294 A lower uptake per kilogram does not automatically mean fewer total adsorbed moles. 21-09 00:28:57.474 --> 00:29:02.808 The total amount depends on both uptake and the amount of adsorbent present. 22-01 00:29:05.338 --> 00:29:14.383 The Langmuir equation writes uptake as the saturation uptake multiplied by b P, divided by one plus b P. 22-02 00:29:14.563 --> 00:29:18.616 Saturation uptake has units of moles per kilogram. 22-03 00:29:18.796 --> 00:29:24.940 The parameter b has reciprocal-pressure units, so the product b P is dimensionless. 22-04 00:29:25.120 --> 00:29:31.158 The ratio of uptake to saturation uptake is the fractional coverage in this model. 22-05 00:29:31.338 --> 00:29:35.476 It rises from zero toward one as pressure increases. 22-06 00:29:35.656 --> 00:29:39.411 At b P equal to one, the coverage is one half. 22-07 00:29:39.591 --> 00:29:47.378 The idealized interpretation involves a finite set of equivalent sites without lateral interactions. 22-08 00:29:47.558 --> 00:29:56.855 These assumptions help us understand the equation, but fitting the equation successfully does not establish a microscopic mechanism. 22-09 00:29:57.035 --> 00:30:04.224 Later, we will use it as a specified synthetic model and check its consequences through a vessel balance. 23-01 00:30:06.754 --> 00:30:11.064 At sufficiently low pressure, b P is much smaller than one. 23-02 00:30:11.244 --> 00:30:20.417 The denominator of the Langmuir equation is then approximately one, so uptake becomes saturation uptake times b times pressure. 23-03 00:30:20.597 --> 00:30:27.189 This gives the Henry adsorption coefficient as saturation uptake multiplied by b. 23-04 00:30:27.369 --> 00:30:33.854 With our chosen units, that coefficient is measured in moles per kilogram per kilopascal. 23-05 00:30:34.034 --> 00:30:40.648 A matched Henry line and Langmuir curve therefore share the same slope as pressure approaches zero. 23-06 00:30:40.828 --> 00:30:44.284 They do not necessarily agree at finite pressure. 23-07 00:30:44.464 --> 00:30:55.066 The word Henry also appears in gas-solubility equations, sometimes written as partial pressure equals a constant times liquid mole fraction. 23-08 00:30:55.246 --> 00:30:58.489 That convention defines a different constant. 23-09 00:30:58.669 --> 00:31:05.816 Always state the equation, phases and concentration basis rather than relying on the name alone. 24-01 00:31:08.346 --> 00:31:16.367 Use the synthetic saturation uptake of two moles per kilogram and b equal to zero point one per kilopascal. 24-02 00:31:16.547 --> 00:31:23.822 Their product gives a Henry adsorption coefficient of zero point two moles per kilogram per kilopascal. 24-03 00:31:24.002 --> 00:31:27.884 At one kilopascal, b P is zero point one. 24-04 00:31:28.064 --> 00:31:36.658 The Langmuir uptake is about zero point one eight one eight moles per kilogram, while the Henry value is zero point two. 24-05 00:31:36.838 --> 00:31:40.230 At ten kilopascals, b P equals one. 24-06 00:31:40.410 --> 00:31:47.087 Langmuir uptake is then one mole per kilogram, exactly half of its saturation uptake. 24-07 00:31:47.267 --> 00:31:50.616 The Henry line predicts two moles per kilogram. 24-08 00:31:50.796 --> 00:31:55.682 This growing difference follows from the denominator in the Langmuir equation. 24-09 00:31:55.862 --> 00:32:03.136 Ten kilopascals is already well beyond a convincing small-b-P approximation for these parameters. 25-01 00:32:05.666 --> 00:32:11.469 Pause for about ten minutes and complete E one through E three on worksheet page four. 25-02 00:32:11.649 --> 00:32:17.857 Keep the same synthetic parameters and compare pressures of five and forty kilopascals. 25-03 00:32:18.037 --> 00:32:24.778 Calculate both Langmuir and matched Henry uptake, then calculate Langmuir fractional coverage. 25-04 00:32:24.958 --> 00:32:28.670 Write the dimensionless product b P at each pressure. 25-05 00:32:28.850 --> 00:32:35.634 Use that product to judge the low-pressure assumption rather than deciding from the pressure number alone. 25-06 00:32:35.814 --> 00:32:40.578 Finally, describe what each model predicts as pressure becomes large. 25-07 00:32:40.758 --> 00:32:51.680 Ask whether the equation contains a saturation limit and whether a mathematical value outside that limit is physically credible for the model's intended domain. 25-08 00:32:51.860 --> 00:32:54.228 Show the substitutions and units. 25-09 00:32:54.408 --> 00:33:00.318 Attempt the calculation before continuing to the comparison table on the next slide. 26-01 00:33:02.848 --> 00:33:07.093 At five kilopascals, b P is zero point five. 26-02 00:33:07.273 --> 00:33:12.884 Langmuir uptake is two thirds of a mole per kilogram, with coverage one third. 26-03 00:33:13.064 --> 00:33:16.818 The matched Henry line predicts one mole per kilogram. 26-04 00:33:16.998 --> 00:33:20.100 At forty kilopascals, b P is four. 26-05 00:33:20.280 --> 00:33:26.638 Langmuir uptake is one point six moles per kilogram and coverage is zero point eight. 26-06 00:33:26.818 --> 00:33:30.252 The Henry value is eight moles per kilogram. 26-07 00:33:30.432 --> 00:33:39.307 That large value comes from extending a linear low-pressure expression into a range where its assumption is not satisfied. 26-08 00:33:39.487 --> 00:33:47.679 Langmuir uptake approaches the specified saturation value of two moles per kilogram as pressure increases. 26-09 00:33:47.859 --> 00:33:50.568 The Henry equation has no plateau. 26-10 00:33:50.748 --> 00:34:00.604 The comparison illustrates model domains, rather than evidence that a real material actually follows either curve over this entire range. 27-01 00:34:03.134 --> 00:34:10.580 For a closed adsorption vessel, total adsorbate inventory equals gas moles plus adsorbed moles. 27-02 00:34:10.760 --> 00:34:16.434 The gas term is pressure times accessible gas volume divided by R T. 27-03 00:34:16.614 --> 00:34:22.033 The adsorbed term is adsorbent mass multiplied by absolute uptake. 27-04 00:34:22.213 --> 00:34:31.578 Use the accessible gas volume in the ideal-gas term and convert adsorbent mass to kilograms before multiplying by uptake. 27-05 00:34:31.758 --> 00:34:35.492 The unknown pressure is the adsorbate partial pressure. 27-06 00:34:35.672 --> 00:34:43.565 At fixed temperature, volume, mass and inventory, the isotherm and this balance determine pressure together. 27-07 00:34:43.745 --> 00:34:52.705 For positive Langmuir parameters, both amount terms increase with nonnegative pressure, giving a single nonnegative solution. 27-08 00:34:52.885 --> 00:34:55.466 The balance uses absolute uptake. 27-09 00:34:55.646 --> 00:35:03.113 An excess-adsorption measurement needs a density and volume convention before it can replace that quantity. 28-01 00:35:05.643 --> 00:35:11.254 This worked example starts with a known pressure, rather than a known total inventory. 28-02 00:35:11.434 --> 00:35:22.783 At ten kilopascals, with saturation uptake two and b equal to zero point one per kilopascal, the Langmuir uptake is one mole per kilogram. 28-03 00:35:22.963 --> 00:35:30.238 Ten grams is zero point zero one kilograms, so adsorbed moles equal zero point zero one. 28-04 00:35:30.418 --> 00:35:37.607 The gas amount in one liter at three hundred kelvin is about four point zero zero nine millimoles. 28-05 00:35:37.787 --> 00:35:46.064 Adding them gives a required total inventory of about zero point zero one four zero zero nine moles. 28-06 00:35:46.244 --> 00:35:54.116 The next task reduces total inventory to zero point zero one mole while keeping the other parameters fixed. 28-07 00:35:54.296 --> 00:35:58.456 Do not carry the ten-kilopascal pressure across unchanged. 28-08 00:35:58.636 --> 00:36:01.047 It must satisfy the new balance. 29-01 00:36:03.577 --> 00:36:09.593 Pause for about five minutes and complete F one through F three on worksheet page four. 29-02 00:36:09.773 --> 00:36:14.701 Read Lab guide pages three and four before watching the demonstration. 29-03 00:36:14.881 --> 00:36:25.270 First predict the Langmuir pressure when the inventory is zero point zero one mole, less than the amount needed for ten kilopascals in the worked example. 29-04 00:36:25.450 --> 00:36:28.885 Next compare Langmuir with the matched Henry model. 29-05 00:36:29.065 --> 00:36:34.334 At the same positive pressure, the Henry expression predicts greater uptake. 29-06 00:36:34.514 --> 00:36:42.130 Use the closed balance to decide which model should therefore require a lower pressure for the same inventory. 29-07 00:36:42.310 --> 00:36:48.006 Finally, return to Langmuir and double adsorbent mass from ten to twenty grams. 29-08 00:36:48.186 --> 00:36:52.880 Keep temperature, accessible volume and total inventory fixed. 29-09 00:36:53.060 --> 00:36:56.238 Predict the pressure direction and explain why. 30-01 00:36:58.768 --> 00:37:02.288 We are now in the real Lab eleven interface. 30-02 00:37:02.468 --> 00:37:07.268 Keep worksheet page five and Lab guide pages three and four open. 30-03 00:37:07.448 --> 00:37:11.182 Begin in Explore isotherms, the upper calculator. 30-04 00:37:11.362 --> 00:37:16.268 The data fitting section lower on the page is optional for this lesson. 30-05 00:37:16.448 --> 00:37:22.123 Our main task is to compare two adsorption models and close a vessel balance. 30-06 00:37:22.303 --> 00:37:30.576 The chart displays four model curves, but the model selected in the menu is the one used to solve the closed vessel. 30-07 00:37:30.756 --> 00:37:36.303 The other curves are not additional phases and are not combined into that balance. 30-08 00:37:36.483 --> 00:37:44.654 Select Langmuir and check a capacity of two moles per kilogram and an affinity of zero point one per kilopascal. 30-09 00:37:44.834 --> 00:37:46.924 Use three hundred kelvin. 30-10 00:37:47.104 --> 00:37:53.184 The pressure units matter because the product of affinity and pressure must be dimensionless. 30-11 00:37:53.364 --> 00:38:02.452 Entering a pressure in pascals while retaining an affinity per kilopascal would change that product by a factor of one thousand. 30-12 00:38:02.632 --> 00:38:07.710 Identify the parameters as synthetic definitions at this temperature. 30-13 00:38:07.890 --> 00:38:14.524 Changing the temperature field does not generate a new temperature dependent parameter set. 30-14 00:38:14.704 --> 00:38:24.731 For our first comparison, set the minimum pressure to zero and the maximum to ten kilopascals so that the lower pressure curves are easy to see. 30-15 00:38:24.911 --> 00:38:31.119 The worksheet uses zero to one hundred kilopascals for your independent investigation. 30-16 00:38:31.299 --> 00:38:38.168 All three vessel roots are below ten kilopascals, so either admitted window gives the same roots. 30-17 00:38:38.348 --> 00:38:40.652 We will widen the range later. 30-18 00:38:40.832 --> 00:38:45.440 Expand Closed vessel, fixed temperature and free gas volume. 30-19 00:38:45.620 --> 00:38:57.468 Enter one liter of free gas volume, ten grams of dry adsorbent, zero point zero one zero mole of total adsorbate, and zero inert gas. 30-20 00:38:57.648 --> 00:39:02.405 This total inventory is the one used in the Lab investigation. 30-21 00:39:02.585 --> 00:39:12.740 It is different from the earlier worked example in which a pressure of ten kilopascals was given and the required total inventory was calculated. 30-22 00:39:12.920 --> 00:39:15.010 Here the pressure is unknown. 30-23 00:39:15.190 --> 00:39:21.313 The solver must divide the specified total amount between gas and adsorbed material. 30-24 00:39:21.493 --> 00:39:25.440 Press Compare and solve vessel after checking these inputs. 30-25 00:39:25.620 --> 00:39:34.281 For case eleven A, the equilibrium adsorbate partial pressure is about six point zero eight zero kilopascals. 30-26 00:39:34.461 --> 00:39:39.453 The uptake is about zero point seven five six two moles per kilogram. 30-27 00:39:39.633 --> 00:39:48.684 To obtain the total adsorbed amount, multiply that uptake by zero point zero one zero kilogram, not by ten. 30-28 00:39:48.864 --> 00:39:54.368 The result is about zero point zero zero seven five six two mole adsorbed. 30-29 00:39:54.548 --> 00:39:59.668 The gas contains about zero point zero zero two four three eight mole. 30-30 00:39:59.848 --> 00:40:04.712 Together these recover the total zero point zero one zero mole. 30-31 00:40:04.892 --> 00:40:10.265 Record the selected model, pressure, uptake, both amounts, and residual. 30-32 00:40:10.445 --> 00:40:17.293 Before changing the model, save the calculated Langmuir case with its interpretation and label. 30-33 00:40:17.473 --> 00:40:20.289 Then select Henry in the model menu. 30-34 00:40:20.469 --> 00:40:26.400 Enter a Henry coefficient of zero point two mole per kilogram per kilopascal. 30-35 00:40:26.580 --> 00:40:35.241 This is the product of the Langmuir capacity and affinity, so the two isotherms have the same limiting slope near zero pressure. 30-36 00:40:35.421 --> 00:40:41.453 Keep ten grams of adsorbent, the same volume, and the same total inventory. 30-37 00:40:41.633 --> 00:40:44.193 Press Compare and solve vessel again. 30-38 00:40:44.373 --> 00:40:50.965 Matching the low pressure slope does not make the two models identical over the full pressure range. 30-39 00:40:51.145 --> 00:40:58.292 The Henry calculation gives about four point one six five kilopascals for case eleven B. 30-40 00:40:58.472 --> 00:41:09.117 It predicts about zero point zero zero eight three three zero mole adsorbed and zero point zero zero one six seven zero mole in the gas. 30-41 00:41:09.297 --> 00:41:16.397 Its material balance closes, but that does not establish that the approximation is physically adequate. 30-42 00:41:16.577 --> 00:41:24.998 At this pressure, the Langmuir product of affinity and pressure is about zero point four two, which is not very small. 30-43 00:41:25.178 --> 00:41:33.840 The linear model stores more material on the adsorbent at a given pressure, so it reaches the fixed inventory at a lower pressure. 30-44 00:41:34.020 --> 00:41:36.281 Save this second case separately. 30-45 00:41:36.461 --> 00:41:45.293 For case eleven C, return the selected model to Langmuir and increase the dry adsorbent mass from ten to twenty grams. 30-46 00:41:45.473 --> 00:41:52.769 Keep the capacity, affinity, temperature, gas volume, and total adsorbate inventory unchanged. 30-47 00:41:52.949 --> 00:42:03.616 Compare this case with the original Langmuir case, eleven A, so that adsorbent mass is the only physical variable being changed in that comparison. 30-48 00:42:03.796 --> 00:42:07.998 Notice that editing a field invalidates the old result. 30-49 00:42:08.178 --> 00:42:12.018 Recalculate before recording or exporting anything. 30-50 00:42:12.198 --> 00:42:20.348 We are adding adsorbent surface capacity to a vessel with the same amount of adsorbate, not adding more adsorbate. 30-51 00:42:20.528 --> 00:42:28.037 With twenty grams of adsorbent, the Langmuir pressure falls to about two point eight four five kilopascals. 30-52 00:42:28.217 --> 00:42:34.873 The total adsorbed amount increases to about zero point zero zero eight eight five nine mole. 30-53 00:42:35.053 --> 00:42:42.626 Yet the uptake per kilogram decreases to about zero point four four three zero mole per kilogram. 30-54 00:42:42.806 --> 00:42:45.174 These changes are consistent. 30-55 00:42:45.354 --> 00:42:51.452 There is more adsorbent, but the lower gas pressure leaves each kilogram less heavily loaded. 30-56 00:42:51.632 --> 00:42:58.352 Multiply uptake by zero point zero two zero kilogram to recover the adsorbed amount. 30-57 00:42:58.532 --> 00:43:06.916 Add the gas amount, about zero point zero zero one one four one mole, and check the same total inventory. 30-58 00:43:07.096 --> 00:43:16.632 Save the third case, then widen the maximum chart pressure to one hundred kilopascals and calculate again to inspect the model shapes. 30-59 00:43:16.812 --> 00:43:21.270 The Henry line continues upward without a finite capacity. 30-60 00:43:21.450 --> 00:43:27.210 The Langmuir curve bends toward the specified capacity of two moles per kilogram. 30-61 00:43:27.390 --> 00:43:34.238 This helps explain the worksheet comparison at forty kilopascals, where the two predictions differ strongly. 30-62 00:43:34.418 --> 00:43:39.432 A wider plotting range does not supply experimental validation. 30-63 00:43:39.612 --> 00:43:45.585 The Freundlich and BET curves use their own synthetic parameters and domains. 30-64 00:43:45.765 --> 00:43:53.484 They are visible for context, but fitting them and interpreting isosteric heat are optional extensions today. 30-65 00:43:53.664 --> 00:44:03.349 The Study workspace now contains three named model studies: Langmuir with ten grams, Henry with ten grams, and Langmuir with twenty grams. 30-66 00:44:03.529 --> 00:44:10.270 For each one, use Your interpretation to record the balance and the reason for the pressure change. 30-67 00:44:10.450 --> 00:44:17.640 Export model study downloads a calculated study and retains it in this temporary comparison. 30-68 00:44:17.820 --> 00:44:23.324 Do not use Export data study from the separate fitting section for this activity. 30-69 00:44:23.504 --> 00:44:34.917 The saved cases keep the original zero to ten kilopascal calculation window even though we subsequently widened the current chart to inspect the isotherm shapes. 30-70 00:44:35.097 --> 00:44:40.025 Select Download comparison to save these three model studies together. 30-71 00:44:40.205 --> 00:44:42.829 Read the recorded fields carefully. 30-72 00:44:43.009 --> 00:44:53.718 In this lab the equilibrium pressure is recorded in kilopascals, the uptake in moles per kilogram, and the gas and adsorbed amounts in moles. 30-73 00:44:53.898 --> 00:44:59.530 The vessel input mass is stored in kilograms and the volume in cubic meters. 30-74 00:44:59.710 --> 00:45:05.812 This differs from some display units and from the pressure convention in the gas solid export. 30-75 00:45:05.992 --> 00:45:12.093 Do not compare raw field values across different labs without checking their definitions. 30-76 00:45:12.273 --> 00:45:15.388 A blank field is not automatically zero. 30-77 00:45:15.568 --> 00:45:22.288 To check that the study is reusable, clear the temporary comparison only after saving the file. 30-78 00:45:22.468 --> 00:45:30.190 Under Load studies or a comparison JSON, choose the downloaded adsorption comparison file. 30-79 00:45:30.370 --> 00:45:35.938 The file is read by the browser and the saved cases become available to restore. 30-80 00:45:36.118 --> 00:45:39.873 Keep the actual file on your computer with the worksheet. 30-81 00:45:40.053 --> 00:45:45.024 An in memory comparison alone may disappear when you leave the page. 30-82 00:45:45.204 --> 00:45:56.842 Saving a screenshot is useful for discussion, but it cannot replace the inputs, units, model selection, and interpretation needed to reproduce the calculation. 30-83 00:45:57.022 --> 00:46:02.643 After import, select Restore beside case eleven A, Langmuir ten grams. 30-84 00:46:02.823 --> 00:46:11.685 Verify that the selected model, capacity, affinity, temperature, inventory, and adsorbent mass have returned correctly. 30-85 00:46:11.865 --> 00:46:17.049 The plot range should also return to the saved zero to ten kilopascals. 30-86 00:46:17.229 --> 00:46:21.367 Restoring inputs is not the final calculation step. 30-87 00:46:21.547 --> 00:46:25.643 Press Compare and solve vessel to calculate the current state. 30-88 00:46:25.823 --> 00:46:34.869 If you edit any parameter after restoration, calculate again before treating the displayed result as evidence for that new case. 30-89 00:46:35.049 --> 00:46:46.846 The restored and recalculated Langmuir case again gives approximately six point zero eight zero kilopascals and the same gas and adsorbed amounts as before. 30-90 00:46:47.026 --> 00:46:53.106 This completes the full workflow from choosing a model to saving and reopening the result. 30-91 00:46:53.286 --> 00:47:00.027 In worksheet section G, reproduce the three cases yourself and verify each balance by hand. 30-92 00:47:00.207 --> 00:47:07.845 Explain why more adsorbent can increase the total adsorbed amount while decreasing uptake per kilogram. 30-93 00:47:08.025 --> 00:47:12.611 State the absolute uptake basis and the ideal gas assumption. 30-94 00:47:12.791 --> 00:47:17.975 The next activity stop gives you time to create and check your own evidence. 31-01 00:47:20.505 --> 00:47:25.092 A useful comparison needs more than a screenshot of a curve. 31-02 00:47:25.272 --> 00:47:31.949 Record the selected model, units and temperature because they define the meaning of each parameter. 31-03 00:47:32.129 --> 00:47:37.271 Then record pressure and uptake separately from the gas and adsorbed amounts. 31-04 00:47:37.451 --> 00:47:42.443 The uptake is measured per kilogram, while the vessel balance uses moles. 31-05 00:47:42.623 --> 00:47:45.076 Check the mass conversion explicitly. 31-06 00:47:45.256 --> 00:47:54.024 The sum of gas moles and adsorbed moles should reproduce the chosen inventory, within rounding and numerical precision. 31-07 00:47:54.204 --> 00:48:01.031 Keep a saved study and verify that reopening and recalculating it reproduces the result. 31-08 00:48:01.211 --> 00:48:07.355 Remember that two models can share a low-pressure slope without agreeing over the full range. 31-09 00:48:07.535 --> 00:48:19.285 Also, a correct numerical inventory balance verifies the calculation, but it does not prove that the chosen isotherm accurately represents a real material. 32-01 00:48:21.815 --> 00:48:28.535 Pause for about twenty-four minutes and complete G one through G four on worksheet page five. 32-02 00:48:28.715 --> 00:48:33.238 Open Lab eleven and follow Lab guide pages three and four. 32-03 00:48:33.418 --> 00:48:39.370 Run three cases with the same temperature, accessible volume and total inventory. 32-04 00:48:39.550 --> 00:48:43.945 Case eleven A uses Langmuir with ten grams of adsorbent. 32-05 00:48:44.125 --> 00:48:48.882 Case eleven B uses the matched Henry model with the same mass. 32-06 00:48:49.062 --> 00:48:53.521 Case eleven C returns to Langmuir and uses twenty grams. 32-07 00:48:53.701 --> 00:49:01.146 Record pressure, uptake, gas moles and adsorbed moles, then check each balance with your calculator. 32-08 00:49:01.326 --> 00:49:05.486 Compare the findings with your predictions and inspect the graph. 32-09 00:49:05.666 --> 00:49:11.021 Save your studies, reopen one, verify its controls and calculate again. 32-10 00:49:11.201 --> 00:49:18.859 Explain the difference between uptake per kilogram and total adsorbed amount when interpreting the mass change. 33-01 00:49:21.389 --> 00:49:27.448 The Langmuir case with ten grams gives about six point zero eight zero kilopascals. 33-02 00:49:27.628 --> 00:49:35.564 With the same inventory and mass, the matched Henry model gives about four point one six five kilopascals. 33-03 00:49:35.744 --> 00:49:43.573 Henry predicts more uptake at a common positive pressure, so it reaches this fixed inventory at a lower pressure. 33-04 00:49:43.753 --> 00:49:52.415 Returning to Langmuir and increasing mass to twenty grams lowers pressure to about two point eight four five kilopascals. 33-05 00:49:52.595 --> 00:49:59.503 Uptake per kilogram decreases, but total adsorbed moles increase because there is more adsorbent. 33-06 00:49:59.683 --> 00:50:03.139 Check the gas and adsorbed amounts in each row. 33-07 00:50:03.319 --> 00:50:06.369 Their sum is zero point zero one mole. 33-08 00:50:06.549 --> 00:50:16.135 The Henry result is useful for model sensitivity, but its finite b-P value requires caution about the low-pressure approximation. 34-01 00:50:18.665 --> 00:50:25.104 The Lab includes additional models and analysis options, but they need their own assumptions. 34-02 00:50:25.284 --> 00:50:32.196 Freundlich is an empirical expression over a stated pressure range and has no saturation plateau. 34-03 00:50:32.376 --> 00:50:39.693 BET uses relative pressure and requires a justified fitting range and reference pressure. 34-04 00:50:39.873 --> 00:50:46.871 Isosteric heat requires comparing pressures at the same absolute loading across temperatures. 34-05 00:50:47.051 --> 00:50:51.787 Comparing arbitrary points on separate isotherms is not enough. 34-06 00:50:51.967 --> 00:50:58.751 Experimental fitting also needs measured data, uncertainty, units and a clear objective. 34-07 00:50:58.931 --> 00:51:05.885 Today's examples use synthetic parameters to examine equilibrium and inventory reasoning. 34-08 00:51:06.065 --> 00:51:11.484 They do not establish an adsorption mechanism or validate extrapolation. 34-09 00:51:11.664 --> 00:51:22.672 You can explore the extensions later, but no new fit, BET calculation or isosteric-heat calculation is required to complete this lesson. 34-10 00:51:22.852 --> 00:51:26.564 Choose one relevant limitation for your reflection. 35-01 00:51:29.094 --> 00:51:33.617 The final calculation transfers the method to new parameters. 35-02 00:51:33.797 --> 00:51:39.962 Keep the worked solutions closed and read H one through H four on worksheet page six. 35-03 00:51:40.142 --> 00:51:48.803 For the gas-solid case, use three hundred and ten kelvin, two liters and a sublimation pressure of five kilopascals. 35-04 00:51:48.983 --> 00:51:56.621 Compare volatile inventories of zero point zero zero two and zero point zero zero eight moles. 35-05 00:51:56.801 --> 00:52:01.515 Add zero point zero zero one mole of inert gas to each. 35-06 00:52:01.695 --> 00:52:12.746 For adsorption, use a different synthetic material with saturation uptake three moles per kilogram and b equal to zero point zero five per kilopascal. 35-07 00:52:12.926 --> 00:52:22.526 At twenty kilopascals and three hundred and ten kelvin, use two liters of accessible gas volume and five grams of adsorbent. 35-08 00:52:22.706 --> 00:52:26.269 Calculate the inventory required by that state. 36-01 00:52:28.799 --> 00:52:33.257 Pause for about twenty minutes and complete H one through H four. 36-02 00:52:33.437 --> 00:52:40.345 Use the equations from the earlier slides, but make your own calculation before opening the answers. 36-03 00:52:40.525 --> 00:52:45.240 For gas-solid equilibrium, determine the gas capacity first. 36-04 00:52:45.420 --> 00:52:52.567 Then find gas and solid amounts, volatile partial pressure and total pressure with the inert included. 36-05 00:52:52.747 --> 00:52:55.563 Check the species inventory in each case. 36-06 00:52:55.743 --> 00:53:00.180 For adsorption, evaluate the isotherm at the specified pressure. 36-07 00:53:00.360 --> 00:53:07.805 Convert five grams to kilograms, then calculate gas moles, adsorbed moles and total inventory. 36-08 00:53:07.985 --> 00:53:13.959 Predict what happens to pressure if adsorbent mass doubles while that inventory stays fixed. 36-09 00:53:14.139 --> 00:53:16.976 An exact new pressure is not required. 36-10 00:53:17.156 --> 00:53:25.071 Finally, state one shared assumption and one difference between bulk gas-solid equilibrium and surface adsorption. 37-01 00:53:27.601 --> 00:53:34.193 The gas capacity is approximately zero point zero zero three eight eight zero moles. 37-02 00:53:34.373 --> 00:53:39.023 The smaller volatile inventory therefore completely vaporizes. 37-03 00:53:39.203 --> 00:53:51.150 The larger inventory leaves about zero point zero zero four one two zero moles of solid and maintains a volatile partial pressure of five kilopascals. 37-04 00:53:51.330 --> 00:53:56.407 The inert contribution is about one point two eight nine kilopascals. 37-05 00:53:56.587 --> 00:53:59.382 Add it to obtain the total pressures shown. 37-06 00:53:59.562 --> 00:54:05.941 For adsorption, uptake at twenty kilopascals is one point five moles per kilogram. 37-07 00:54:06.121 --> 00:54:15.536 Gas moles are about zero point zero one five five one nine, and adsorbed moles are zero point zero zero seven five. 37-08 00:54:15.716 --> 00:54:20.473 Their sum is about zero point zero two three zero one nine moles. 37-09 00:54:20.653 --> 00:54:24.963 Doubling adsorbent mass at that fixed inventory lowers pressure. 38-01 00:54:27.493 --> 00:54:32.613 Use these four questions whenever you approach a new equilibrium problem. 38-02 00:54:32.793 --> 00:54:35.704 First, identify the physical process. 38-03 00:54:35.884 --> 00:54:42.135 Bulk phase transfer, surface uptake and chemical reaction require different descriptions. 38-04 00:54:42.315 --> 00:54:44.357 Second, state the constraints. 38-05 00:54:44.537 --> 00:54:51.065 Temperature and pressure are not interchangeable with temperature and accessible volume. 38-06 00:54:51.245 --> 00:55:01.018 Third, identify the model and its conventions, including standard states, gas ideality, uptake basis, units and pressure range. 38-07 00:55:01.198 --> 00:55:08.643 Finally, use material balances to determine amounts and check whether the proposed phases are actually present. 38-08 00:55:08.823 --> 00:55:17.250 A solubility composition, a sublimation pressure or an isotherm loading supplies only part of the description. 38-09 00:55:17.430 --> 00:55:20.182 None replaces an inventory balance. 38-10 00:55:20.362 --> 00:55:29.429 In your reflection, report a numerical result with its units, one supporting check and the assumptions needed for the interpretation. 38-11 00:55:29.609 --> 00:55:33.790 That combination makes the result useful and reproducible. 39-01 00:55:36.320 --> 00:55:45.045 Pause for about ten minutes to complete section I on worksheet page six and submit the Session four exit ticket and feedback. 39-02 00:55:45.225 --> 00:55:48.233 Use the link or scan the Q R code. 39-03 00:55:48.413 --> 00:55:52.765 Explain what happens to gas pressure when bulk solid runs out. 39-04 00:55:52.945 --> 00:56:01.585 In the evidence answer, include one Lab ten comparison and one Lab eleven numerical balance, with settings and units. 39-05 00:56:01.765 --> 00:56:11.216 Report your actual study time, including the activities, and tell me about any access problems or explanations that need improvement. 39-06 00:56:11.396 --> 00:56:15.812 The form does not ask for your name, student I D or email. 39-07 00:56:15.992 --> 00:56:20.024 Submit once and wait for the message Response received. 39-08 00:56:20.204 --> 00:56:25.751 The next slide is a receipt and saved-work check, not another submission request. 40-01 00:56:28.281 --> 00:56:31.161 Check that the form shows Response received. 40-02 00:56:31.341 --> 00:56:36.354 If you already have that confirmation, do not submit a second response. 40-03 00:56:36.534 --> 00:56:46.801 Save your completed worksheet, including the predictions, calculations and corrections, together with the Lab ten and Lab eleven comparison tables. 40-04 00:56:46.981 --> 00:56:56.560 Keep the study J S O N exports and a note of one result that you reproduced by reopening and recalculating a saved case. 40-05 00:56:56.740 --> 00:56:59.577 Record your actual study time as well. 40-06 00:56:59.757 --> 00:57:05.091 The recording, worksheet and Lab investigations together form this week's lesson. 40-07 00:57:05.271 --> 00:57:09.367 There is no additional compulsory after-class assignment. 40-08 00:57:09.547 --> 00:57:14.091 The appendix remains available if you want to examine an extension. 40-09 00:57:14.271 --> 00:57:26.808 In the next session, we will continue using the same habits: identify the process and constraints, calculate carefully, and check the interpretation against conserved quantities.