WEBVTT 01-01 00:00:00.180 --> 00:00:04.724 Welcome to this review of Gibbs energy graphs and liquid stability. 01-02 00:00:04.904 --> 00:00:11.987 We will follow one calculation from the graph to the equations, then reproduce it in Lab zero four. 01-03 00:00:12.167 --> 00:00:20.272 Our central question is simple: for a fixed overall composition, which allowed state has the lowest Gibbs energy? 01-04 00:00:20.452 --> 00:00:31.137 By the end, you should be able to distinguish phase compositions from phase amounts, explain a metastable liquid, and use the Lab to check your own reasoning. 01-05 00:00:31.317 --> 00:00:33.579 Keep a calculator nearby. 01-06 00:00:33.759 --> 00:00:38.751 You will have several opportunities to pause and work before seeing the answer. 02-01 00:00:41.281 --> 00:00:46.806 Open the four-page calculation and Lab guide using the link on this slide. 02-02 00:00:46.986 --> 00:00:50.570 Also open Lab zero four in another tab. 02-03 00:00:50.750 --> 00:00:54.227 Start with sections A and B on pages one and two. 02-04 00:00:54.407 --> 00:01:01.113 Later, use section C for the screen demonstration and section D for independent practice. 02-05 00:01:01.293 --> 00:01:05.923 Your calculator must have natural logarithms, labeled L N. 02-06 00:01:06.103 --> 00:01:10.839 The suggested times on the slide describe the original reading route. 02-07 00:01:11.019 --> 00:01:16.629 The video has its own running time, and your calculation pauses add to that time. 02-08 00:01:16.809 --> 00:01:21.033 This is optional support for your original Week three worksheet. 02-09 00:01:21.213 --> 00:01:24.883 It adds no attendance or exit-ticket submission. 03-01 00:01:27.413 --> 00:01:30.388 First, separate consistency from stability. 03-02 00:01:30.568 --> 00:01:41.085 Lab zero three asks whether data are compatible with an applicable thermodynamic constraint, such as Gibbs Duhem, under the assumptions of the test. 03-03 00:01:41.265 --> 00:01:46.982 Lab zero four asks which feasible state has the lowest total Gibbs energy. 03-04 00:01:47.162 --> 00:01:49.232 These are different questions. 03-05 00:01:49.412 --> 00:01:54.212 A consistent liquid model can predict separation into two liquids. 03-06 00:01:54.392 --> 00:01:59.789 In guide A one, identify which question a common tangent helps answer. 03-07 00:01:59.969 --> 00:02:07.100 Throughout this review, temperature and pressure are fixed, and we consider only a binary liquid system. 03-08 00:02:07.280 --> 00:02:12.486 Vapor and solid phases are outside the allowed states in this example. 04-01 00:02:15.016 --> 00:02:18.194 Read the axes before interpreting the shape. 04-02 00:02:18.374 --> 00:02:26.084 The horizontal coordinate, x, is the mole fraction of component one in a possible homogeneous liquid. 04-03 00:02:26.264 --> 00:02:32.690 The vertical coordinate, g, is the molar Gibbs energy of mixing divided by R T. 04-04 00:02:32.870 --> 00:02:34.620 It is dimensionless. 04-05 00:02:34.800 --> 00:02:39.002 The symbol z describes the overall feed composition. 04-06 00:02:39.182 --> 00:02:46.666 One liquid has x equal to z, but two liquids can have different compositions whose weighted average is z. 04-07 00:02:46.846 --> 00:02:51.625 We compare states containing the same total amounts of both components. 04-08 00:02:51.805 --> 00:02:55.581 Their pure-liquid reference contributions then cancel. 04-09 00:02:55.761 --> 00:02:59.793 In guide A two, mark the feed at zero point one two. 05-01 00:03:02.323 --> 00:03:07.934 At the marked feed, the point on the curved line represents one homogeneous liquid. 05-02 00:03:08.114 --> 00:03:14.663 The point directly below it, on the common-tangent segment, represents a mixture of two liquids. 05-03 00:03:14.843 --> 00:03:22.587 Those liquids have the compositions at the two contacts, with amounts chosen to reproduce the same overall feed. 05-04 00:03:22.767 --> 00:03:25.156 Compare vertically at the same z. 05-05 00:03:25.336 --> 00:03:27.640 The split has lower energy here. 05-06 00:03:27.820 --> 00:03:34.668 Simply noticing that the homogeneous mixing energy is negative does not establish stability. 05-07 00:03:34.848 --> 00:03:39.328 A negative value compares mixing with the pure-component reference. 05-08 00:03:39.508 --> 00:03:48.703 Our stability question compares the homogeneous liquid with other feasible liquid states at the same overall composition. 06-01 00:03:51.233 --> 00:03:56.908 We will use the symmetric, one-parameter Margules model with A equal to three. 06-02 00:03:57.088 --> 00:04:01.738 This is a synthetic example, not a fitted real-fluid dataset. 06-03 00:04:01.918 --> 00:04:05.396 The first two terms in g describe ideal mixing. 06-04 00:04:05.576 --> 00:04:09.117 The last term describes the excess contribution. 06-05 00:04:09.297 --> 00:04:12.561 The first derivative gives the slope of the curve. 06-06 00:04:12.741 --> 00:04:16.112 The second derivative gives its local curvature. 06-07 00:04:16.292 --> 00:04:20.772 Use natural logarithms in both the energy and the slope. 06-08 00:04:20.952 --> 00:04:28.312 At a pure endpoint, evaluate the limiting value of x times natural log x, which is zero. 06-09 00:04:28.492 --> 00:04:33.782 Keep this model and A fixed while changing the feed in the later comparisons. 07-01 00:04:36.312 --> 00:04:41.048 Start with local curvature at z equal to zero point one two. 07-02 00:04:41.228 --> 00:04:49.314 Take one divided by zero point one two, add one divided by zero point eight eight, then subtract six. 07-03 00:04:49.494 --> 00:04:53.739 The result is about positive three point four six nine seven. 07-04 00:04:53.919 --> 00:05:03.860 Positive curvature means sufficiently small composition fluctuations raise the energy, so this homogeneous state is locally stable. 07-05 00:05:04.040 --> 00:05:08.819 It does not exclude a lower-energy split into more distant compositions. 07-06 00:05:08.999 --> 00:05:18.215 At the middle composition, zero point five, the curvature is negative two, so the homogeneous liquid is locally unstable. 07-07 00:05:18.395 --> 00:05:23.046 Pause for two minutes to reproduce both calculations in guide B one. 08-01 00:05:25.576 --> 00:05:28.008 Next, find the tangent contacts. 08-02 00:05:28.188 --> 00:05:32.860 The slope at each contact must equal the slope of the line joining them. 08-03 00:05:33.040 --> 00:05:40.272 For this symmetric example, the contacts are mirror images and the common tangent is horizontal. 08-04 00:05:40.452 --> 00:05:44.825 Solve the displayed slope equation for the low-composition root. 08-05 00:05:45.005 --> 00:05:51.874 It lies between zero point zero seven zero seven and zero point zero seven zero eight. 08-06 00:05:52.054 --> 00:06:01.377 Refining gives approximately zero point zero seven zero seven two, with its partner at zero point nine two nine two eight. 08-07 00:06:01.557 --> 00:06:04.181 Check the first root in guide B two. 08-08 00:06:04.361 --> 00:06:09.822 The stationary point at one half is a maximum, not a coexistence endpoint. 09-01 00:06:12.352 --> 00:06:17.558 Now calculate amounts, using a basis of one mole of total feed. 09-02 00:06:17.738 --> 00:06:20.682 The two phase fractions must add to one. 09-03 00:06:20.862 --> 00:06:24.126 Their composition-weighted sum must equal z. 09-04 00:06:24.306 --> 00:06:28.316 Solving those two balances gives the lever rule shown here. 09-05 00:06:28.496 --> 00:06:38.502 For the beta fraction, subtract the alpha composition from the feed composition, then divide by the difference between the two endpoint compositions. 09-06 00:06:38.682 --> 00:06:43.034 The result is about zero point zero five seven three nine eight. 09-07 00:06:43.214 --> 00:06:49.507 The alpha fraction is one minus that, about zero point nine four two six zero two. 09-08 00:06:49.687 --> 00:06:56.919 These are amounts of liquid phases per mole of feed, not phase compositions and not a vapor fraction. 10-01 00:06:59.449 --> 00:07:01.966 Check the result before interpreting it. 10-02 00:07:02.146 --> 00:07:05.778 First, the alpha and beta amounts add to one mole. 10-03 00:07:05.958 --> 00:07:13.019 Next, multiply each phase amount by its component-one mole fraction and add the contributions. 10-04 00:07:13.199 --> 00:07:18.148 The result is approximately zero point one two moles of component one. 10-05 00:07:18.328 --> 00:07:22.872 Component two then accounts for zero point eight eight moles. 10-06 00:07:23.052 --> 00:07:26.743 Small differences in the final digits come from rounding. 10-07 00:07:26.923 --> 00:07:32.214 A successful balance establishes feasibility, not minimum Gibbs energy. 10-08 00:07:32.394 --> 00:07:40.052 If either proposed phase fraction is negative or greater than one, those endpoints cannot represent this feed. 10-09 00:07:40.232 --> 00:07:43.475 Record the component balance in guide B three. 11-01 00:07:46.005 --> 00:07:49.610 Calculate the homogeneous energy term by term. 11-02 00:07:49.790 --> 00:07:54.228 Multiply zero point one two by its natural logarithm. 11-03 00:07:54.408 --> 00:07:59.613 This contributes about negative zero point two five four four three two. 11-04 00:07:59.793 --> 00:08:08.049 The corresponding term for zero point eight eight contributes about negative zero point one one two four nine three. 11-05 00:08:08.229 --> 00:08:17.232 The excess term, three times zero point one two times zero point eight eight, is positive zero point three one six eight. 11-06 00:08:17.412 --> 00:08:24.046 Adding all three gives approximately negative zero point zero five zero one two five. 11-07 00:08:24.226 --> 00:08:31.949 This dimensionless value is total mixing Gibbs energy divided by the number of moles times R T. 11-08 00:08:32.129 --> 00:08:34.902 It is not yet an energy in joules. 12-01 00:08:37.432 --> 00:08:46.350 Both tangent contacts have the same dimensionless energy, approximately negative zero point zero five eight three four one. 12-02 00:08:46.530 --> 00:08:53.356 Weighting these equal values by phase fractions that add to one gives that same value for the split. 12-03 00:08:53.536 --> 00:08:57.547 Subtract the split energy from the homogeneous energy. 12-04 00:08:57.727 --> 00:09:03.060 The positive decrease is about zero point zero zero eight two one six. 12-05 00:09:03.240 --> 00:09:11.184 The homogeneous feed is therefore metastable: locally stable, but a feasible split lowers the total energy. 12-06 00:09:11.364 --> 00:09:21.966 For an actual total energy decrease, multiply the displayed dimensionless decrease by the total number of moles, R, and the absolute temperature. 12-07 00:09:22.146 --> 00:09:26.733 Pause for five minutes to complete and check guide B three and B four. 13-01 00:09:29.263 --> 00:09:33.380 This close view shows the same calculation geometrically. 13-02 00:09:33.560 --> 00:09:42.264 Locate the upper point on the homogeneous curve and the lower point on the equilibrium segment, both at the feed composition. 13-03 00:09:42.444 --> 00:09:46.839 Their vertical separation is the dimensionless energy decrease. 13-04 00:09:47.019 --> 00:09:51.798 It is not the horizontal distance between the two phase compositions. 13-05 00:09:51.978 --> 00:09:59.898 The endpoint spacing enters the lever rule, while this vertical gap measures the energy advantage of the feasible split. 13-06 00:10:00.078 --> 00:10:06.649 On guide B four, label the upper point homogeneous and the lower point equilibrium split. 13-07 00:10:06.829 --> 00:10:14.765 Keep the direction of subtraction clear: homogeneous minus split is positive when separation lowers the energy. 14-01 00:10:17.295 --> 00:10:21.818 A candidate common tangent must support the whole allowed curve. 14-02 00:10:21.998 --> 00:10:26.264 Subtract the line from g at every possible composition. 14-03 00:10:26.444 --> 00:10:28.599 No value may be negative. 14-04 00:10:28.779 --> 00:10:37.766 Equal chemical potentials identify a candidate pair, and the material balance establishes whether its amounts are feasible. 14-05 00:10:37.946 --> 00:10:43.002 Global support completes the minimum-energy argument within this liquid model. 14-06 00:10:43.182 --> 00:10:51.114 Checking only the two contacts is insufficient, because another part of the curve could lie below the candidate line. 14-07 00:10:51.294 --> 00:10:54.985 A coarse plotted grid could also miss a narrow dip. 14-08 00:10:55.165 --> 00:11:02.418 Here, we can examine all stationary points and both endpoints to verify the support condition directly. 15-01 00:11:04.948 --> 00:11:10.815 For this specific model, the stationary compositions are the two contacts and one half. 15-02 00:11:10.995 --> 00:11:14.686 Include the pure endpoints as boundary candidates. 15-03 00:11:14.866 --> 00:11:32.572 Subtracting the horizontal common-tangent value gives approximately positive zero point zero five eight three four one at each pure endpoint, zero at each contact, and positive zero point one one five one nine four at one half. 15-04 00:11:32.752 --> 00:11:34.331 None is negative. 15-05 00:11:34.511 --> 00:11:41.039 Because these cover the stationary points and boundaries, they establish support across this curve. 15-06 00:11:41.219 --> 00:11:45.123 This is stronger than checking a few arbitrary graph points. 15-07 00:11:45.303 --> 00:11:51.383 Pause for two minutes to reproduce the middle-composition value and complete guide B five. 16-01 00:11:53.913 --> 00:11:56.132 Now change the reference line. 16-02 00:11:56.312 --> 00:12:04.440 Tangent-plane distance, or T P D, uses the tangent to the homogeneous curve at the feed composition. 16-03 00:12:04.620 --> 00:12:11.084 At a trial composition w, subtract that feed-tangent value from g of w. 16-04 00:12:11.264 --> 00:12:16.640 A negative result shows that the homogeneous feed is not globally stable. 16-05 00:12:16.820 --> 00:12:22.964 This feed tangent is different from the coexistence common tangent we just verified. 16-06 00:12:23.144 --> 00:12:28.989 Its minimizing trial composition is not automatically a binodal endpoint. 16-07 00:12:29.169 --> 00:12:42.954 In guide C one, keep these two tests distinct: support of the coexistence line verifies the equilibrium construction, while feed T P D tests the original homogeneous liquid. 17-01 00:12:45.484 --> 00:12:50.241 Open Lab zero four and go to the liquid stability experiment. 17-02 00:12:50.421 --> 00:12:59.189 Under Explore a state, choose the metastable example with A equal to three and feed composition zero point one two. 17-03 00:12:59.369 --> 00:13:04.492 Confirm Margules, one parameter, and check both numerical inputs. 17-04 00:13:04.672 --> 00:13:06.614 Click Find equilibrium. 17-05 00:13:06.794 --> 00:13:10.399 Select the Gibbs energy and common tangent view. 17-06 00:13:10.579 --> 00:13:12.926 The graph should match our worked case. 17-07 00:13:13.106 --> 00:13:18.610 Open Verify the equilibrium to inspect the energies and component balance. 17-08 00:13:18.790 --> 00:13:24.059 Compare the two endpoint compositions and the phase amounts with guide B. 17-09 00:13:24.239 --> 00:13:32.410 Opening the page alone does not guarantee these settings, so selecting and checking the preset is an essential first step. 18-01 00:13:34.940 --> 00:13:38.630 Compare the displayed results with the calculation table. 18-02 00:13:38.810 --> 00:13:48.607 The alpha liquid accounts for about ninety-four point two six percent of the feed, and the beta liquid about five point seven four percent. 18-03 00:13:48.787 --> 00:13:54.867 The equilibrium energy is lower by about zero point zero zero eight two one six. 18-04 00:13:55.047 --> 00:13:57.692 Now switch View to Local curvature. 18-05 00:13:57.872 --> 00:14:00.347 The feed has positive curvature. 18-06 00:14:00.527 --> 00:14:04.346 Next select Tangent-plane distance, feed stability. 18-07 00:14:04.526 --> 00:14:09.731 Its negative minimum rejects global stability of this homogeneous feed. 18-08 00:14:09.911 --> 00:14:14.924 These observations are compatible, and together explain metastability. 18-09 00:14:15.104 --> 00:14:18.838 Allow for display rounding when checking residuals. 18-10 00:14:19.018 --> 00:14:25.439 Pause for four minutes to reproduce both views and record your comparisons in guide C one. 19-01 00:14:27.969 --> 00:14:31.596 Predict two new cases before revealing the results. 19-02 00:14:31.776 --> 00:14:33.653 Keep A equal to three. 19-03 00:14:33.833 --> 00:14:44.218 For z equal to zero point zero two, decide whether the feed lies between the binodal endpoints and whether the lever-rule amounts are feasible. 19-04 00:14:44.398 --> 00:14:51.161 For z equal to one half, predict the curvature sign and the phase amounts suggested by symmetry. 19-05 00:14:51.341 --> 00:14:54.114 Write both predictions in guide C two. 19-06 00:14:54.294 --> 00:15:01.420 Then select the corresponding Stable and Unstable presets and click Find equilibrium for each. 19-07 00:15:01.600 --> 00:15:03.392 Pause for five minutes. 19-08 00:15:03.572 --> 00:15:10.896 Record your prediction, the Lab result, and a reason for any correction before continuing to the comparison. 20-01 00:15:13.426 --> 00:15:19.208 Select the Stable preset at zero point zero two and run Find equilibrium. 20-02 00:15:19.388 --> 00:15:22.844 The result is one liquid at the feed composition. 20-03 00:15:23.024 --> 00:15:30.426 Using the two binodal endpoints would require a negative beta amount, so that split is infeasible. 20-04 00:15:30.606 --> 00:15:36.772 Now select the Unstable preset at zero point five and run Find equilibrium. 20-05 00:15:36.952 --> 00:15:43.932 The homogeneous curvature is negative two, and equilibrium has equal amounts of the two liquids. 20-06 00:15:44.112 --> 00:15:47.014 The endpoint compositions stay unchanged. 20-07 00:15:47.194 --> 00:15:49.583 Return to the comparison table. 20-08 00:15:49.763 --> 00:15:55.672 Stable, metastable, and unstable classify the initial homogeneous feed. 20-09 00:15:55.852 --> 00:16:02.423 They must not be confused with the stable equilibrium obtained after allowing separation. 21-01 00:16:04.953 --> 00:16:07.620 Locate the same cases on this curve. 21-02 00:16:07.800 --> 00:16:12.109 The binodal contacts bound the two-liquid equilibrium interval. 21-03 00:16:12.289 --> 00:16:21.974 The spinodals, near zero point two one one three and zero point seven eight eight seven, mark where local curvature changes sign. 21-04 00:16:22.154 --> 00:16:28.000 Between a binodal and its neighboring spinodal, the homogeneous liquid is metastable. 21-05 00:16:28.180 --> 00:16:31.764 Inside the spinodals it is locally unstable. 21-06 00:16:31.944 --> 00:16:36.445 Outside the binodal interval it is stable in this example. 21-07 00:16:36.625 --> 00:16:43.650 Place zero point zero two, zero point one two, and zero point five in these regions. 21-08 00:16:43.830 --> 00:16:52.555 The equilibrium common-tangent segment applies throughout the binodal interval, with phase amounts determined by the feed. 22-01 00:16:55.085 --> 00:16:58.477 Work the new case in guide D one on page four. 22-02 00:16:58.657 --> 00:17:03.329 Keep A equal to three and set the feed to zero point three five. 22-03 00:17:03.509 --> 00:17:06.005 First calculate local curvature. 22-04 00:17:06.185 --> 00:17:13.502 Then use the same endpoint compositions to calculate both phase fractions and check the component balance. 22-05 00:17:13.682 --> 00:17:18.333 Calculate the homogeneous energy and its decrease on splitting. 22-06 00:17:18.513 --> 00:17:25.660 Only after your hand calculation, enter the feed in Lab zero four and run Find equilibrium. 22-07 00:17:25.840 --> 00:17:30.789 The endpoints stay fixed because the model and conditions are unchanged. 22-08 00:17:30.969 --> 00:17:36.409 Pause for eight minutes to attempt all five steps before watching the worked answer. 23-01 00:17:38.939 --> 00:17:42.438 Check your calculation for zero point three five. 23-02 00:17:42.618 --> 00:17:50.554 The curvature is about negative one point six zero four four, so the homogeneous feed is locally unstable. 23-03 00:17:50.734 --> 00:17:57.454 The beta fraction is about zero point three two five two eight nine, and alpha is the remainder. 23-04 00:17:57.634 --> 00:18:01.922 Their weighted composition returns zero point three five. 23-05 00:18:02.102 --> 00:18:14.539 The homogeneous energy is approximately positive zero point zero three five zero five three, and the decrease is zero point zero nine three three nine five. 23-06 00:18:14.719 --> 00:18:20.948 Enter zero point three five in Feed mole fraction and click Find equilibrium. 23-07 00:18:21.128 --> 00:18:25.314 Compare the same endpoints, new amounts, and energy decrease. 23-08 00:18:25.494 --> 00:18:28.460 Correct guide D one in a different color. 24-01 00:18:30.990 --> 00:18:34.083 Write a short explanation in Your explanation. 24-02 00:18:34.263 --> 00:18:43.394 State the model, A, feed composition, classification, and phase amounts, then include one balance or energy comparison. 24-03 00:18:43.574 --> 00:18:48.395 Click Export study JSON and keep the downloaded file. 24-04 00:18:48.575 --> 00:18:53.183 To reopen it, expand Study workspace, reopen and compare cases. 24-05 00:18:53.363 --> 00:18:59.123 Choose Load studies or a comparison JSON and select your saved file. 24-06 00:18:59.303 --> 00:19:01.500 Click Restore beside the case. 24-07 00:19:01.680 --> 00:19:06.480 Check the restored inputs, then click Find equilibrium to recompute. 24-08 00:19:06.660 --> 00:19:11.832 If something differs, check the model, A, feed, selected view, and rounding. 24-09 00:19:12.012 --> 00:19:15.809 You do not need to edit the JSON manually. 24-10 00:19:15.989 --> 00:19:20.256 Keep this file with your worksheet as evidence you can explain. 25-01 00:19:22.786 --> 00:19:26.732 Finish with guide D two, without using the Lab output. 25-02 00:19:26.912 --> 00:19:32.672 Explain why positive curvature alone does not establish global stability. 25-03 00:19:32.852 --> 00:19:38.442 State what stays fixed and what changes when the feed moves inside this tie line. 25-04 00:19:38.622 --> 00:19:44.680 Explain why equal chemical potentials are only part of the equilibrium test. 25-05 00:19:44.860 --> 00:19:56.231 Finally, explain why the binodal split cannot represent a feed of zero point zero two, even though extending the tangent line places it below the curve there. 25-06 00:19:56.411 --> 00:19:59.291 Use one short sentence for each answer. 25-07 00:19:59.471 --> 00:20:03.076 Pause for three minutes before checking the final slide. 25-08 00:20:03.256 --> 00:20:08.952 Keep these corrections with your original worksheet; no additional form is required. 26-01 00:20:11.482 --> 00:20:14.170 Compare your reasoning with these answers. 26-02 00:20:14.350 --> 00:20:22.521 Positive curvature establishes local stability, while a distant feasible split may still lower the total energy. 26-03 00:20:22.701 --> 00:20:28.909 Inside a fixed tie line, the phase compositions stay fixed and the amounts change. 26-04 00:20:29.089 --> 00:20:36.257 Equality of chemical potentials gives a candidate, which still needs feasible amounts and global support. 26-05 00:20:36.437 --> 00:20:45.589 The feed at zero point zero two lies outside the two endpoints, so their mixture would require a negative phase amount. 26-06 00:20:45.769 --> 00:20:52.148 If you can connect these statements to your calculation and the Lab views, you have the key reasoning. 26-07 00:20:52.328 --> 00:20:56.317 Return to the Week three study route whenever you are ready.