1. Document that explains the chosen concept to the animator

2. Liquid-Liquid Phase Diagrams (Mixing of Two Partially Miscible Liquids) ‏ This animation is about the mixing of two partially miscible liquids with change in temperature. Authors Dhrubajyoti Samanta Mentor Subject:

3. Learning Objectives After interacting with this Learning Object, the learner will be able to: identify out the relative compositions of any two substances if they do not mix given temperature and given mole fractions. calculate the propositions in which the substances are present. identify the temperature at which the substance form a single phase.

4. Master layout or diagram Make a schematic diagram of the concept Explain to the animator about the beginning and ending of the process. Draw image big enough for explaining. In above image, identify and label different components of the process/phenomenon. (These are like characters in a film)‏ Illustrate the basic flow of action by using arrows. Use BOLD lines in the diagram, (minimum 2pts.)‏ In the slide after that, provide the definitions of ALL the labels used in the diagram 5 3 2 4 1 INSTRUCTIONS SLIDE

5. Master layout or diagram You may have multiple master layouts. –In this case, number the master layout. ( e.g. Master layout 1)‏ – Each Master layout should be followed by the stepwise description of the animation stages related to it. 5 3 2 4 1 INSTRUCTIONS SLIDE

6. Master Layout 0( for animators)‏ 5 3 2 4 1 2 T/K Mole fraction 0.20.40.60.81 260 270 280 290 300 0 (LB4) ‏ (LB3) ‏ (LB2) ‏ (LB1) ‏ (LB) ‏ (DB4) ‏ (DB3) ‏ (DB2) ‏ (DB1) ‏ (DB) ‏ 260 270 280 290 300 Liquid BLiquid A Liquid BwhiteLiquid ADark blue LB2 drops of blue to 8 drops of whiteDB2 drops of white to 8 drops of blue LB14 drops of blue to 6 drops of whiteDB14 drops of white to 6 drops of blue LB26 drops of blue to 4 drops of whiteDB26 drops of white to 4 drops of blue LB38 drops of blue to 2 drops of whiteDB38 drops of white to 2 drops of blue LB45 drops of blue to 5 drops of whiteDB45 drops of white to 5 drops of blue

7. Master Layout 1 5 3 2 4 1

8. Master Layout 2 5 3 2 4 1 T/K Mole fraction 0.20.40.60.81 260 270 280 290 300 0 (LB4) ‏ (LB3) ‏ (LB2) ‏ (LB1) ‏ (LB) ‏ (DB4) ‏ (DB3) ‏ (DB2) ‏ (DB1) ‏ (DB) ‏ 260 270 280 290 300 Liquid BLiquid A P=1 P=2 a1a2 x121

9. LIQUID – LIQUID PHASE DIAGRAMS ( Mixing of Partially Miscible Liquids)‏ Cylinder DiagramsGraph - Dhrubajyoti Samanta Please move the slider to choose the composition and temperature you want and press the Mix button to initiate the mixing: Master Layout 3

10. Master Layout 4 5 3 2 4 1 Fig AFig B

11. Theory: Content Partially miscible liquids are liquids which are miscible with each other at certain conditions of concentration and other factors like temperature. At other concentrations they form two separate solutions – one with one of the liquids in the bulk and another with the other liquid in the bulk. This is governed by the phase rule –> F = C – P +2 Where F is the number of degrees of freedom, C is the number of components and P is the number of phases. Degrees of freedom are the minimum number parameters that are required to completely define a system and to evaluate all the properties of the system. for a 2 component system at constant pressure if there are two phases then F=2-2+2=2, i.e. Two parameters are required to define it completely which are mole fraction of either of the substance and temperature. The relative composition in the two layers is given by the liver rule. a 1 l 1 =a 2 l 2 a1a1 a2a2 l1l1 l2l2

12. Explain the process 1 5 3 2 4 In this step, use an example to explain the concept. It can be an analogy, a scenario, or an action which explains this concept/process/topic Try to use examples from day-to-day life to make it more clear You have to describe what steps the animator should take to make your concept come alive as a series of moving images. Keep the examples simple to understand, and also to illustrate/animate.

13. Analogy / Scenario / Action 1 5 3 2 4

14. Stepwise description of process The goal of the document is to provide instructions to an animator who is not a expert. You have to describe what steps the animator should take to make your concept come alive as a moving visualization. Use one slide per step. This will ensure clarity of the explanation. Add a image of the step in the box, and the details in the table below the box. You can use any images for reference, but mention about it's copyright status The animator will have to re-draw / re-create the drawings Add more slides as per the requirement of the animation 1 5 3 2 4

15. Step 1: 1 5 2 4 Temperature Constant I nstruction for the animator Boundary limits (if any) ‏ T ext to be displayed in the working area (DT) ‏ Show the above liquid and graph. Once the learner selects a value on the slider pour the white liquid into the blue liquid beaker. (Master Layout 2) ‏ Once the learner click on mix show the point on the graph and shake the solution. 0 – 1  The temperature being fixed at 280 K select the value on the slider and click on mix to view the change in the liquids. T/K Mole fraction 0.20.40.60.81 260 270 280 290 300 0 (LB4) ‏ (LB3) ‏ (LB2) ‏ (LB1) ‏ (LB) ‏ (DB4) ‏ (DB3) ‏ (DB2) ‏ (DB1) ‏ (DB) ‏ 260 270 280 290 300 Liquid BLiquid A Liquid B (0 blue in a scale of 0-255) ‏ Liquid A (255 blue in a scale of 0-255) ‏ 1/00/1 0.8/ 0.2 0.6/ 0.4 0.4/ 0.6 0.2/ 0.8 3

16. Step 2: I nstruction for the animator Boundary limits (if any) ‏ T ext to be displayed in the working area (DT) ‏ If learner selects any point before the blue dot on the black line then the mixture has one colour which is lighter than (LB2). (fig A above) ‏ If learner selects any point between the blue dot and red dot on the black line then the mixture has two colours which are (LB2) for liquid B and (DB2) for liquid A. (fig B above) ‏ If learner selects any point after the red dot on the black line then the mixture has one colour which is darker than (DB2). (fig A above) ‏ 0 – 1  The temperature being fixed at 280 K select the value on the slider and click on mix to view the change in the liquids. 1 5 3 2 4 Temperature Constant Fig AFig B T/K Mole fraction 0.20.40.60.81 260 270 280 290 300 0 (LB4) ‏ (LB3) ‏ (LB2) ‏ (LB1) ‏ (LB) ‏ (DB4) ‏ (DB3) ‏ (DB2) ‏ (DB1) ‏ (DB) ‏ 260 270 280 290 300 Liquid BLiquid A Liquid B (0 blue in a scale of 0-255) ‏ Liquid A (255 blue in a scale of 0-255) ‏ 1/00/1 0.8/ 0.2 0.6/ 0.4 0.4/ 0.6 0.2/ 0.8

17. Step 3: 1 5 3 2 4 Mixture constant I nstruction for the animator Boundary limits (if any) ‏ T ext to be displayed in the working area (DT) ‏ Show the above liquid and graph. Once the learner selects a value on the slider pour the white liquid into the blue liquid beaker. Once the learner click on mix show the point on the graph and shake the solution. Temperature – 260 K- 300 K  The mixture being fixed at ratio 40 : 60 (liquid B :liquid A) select the temperature and click on mix to view the change in the liquids. (LB) ‏ T/K Mole fraction 0.20.40.60.81 260 270 280 290 300 0 (DB4) ‏ (DB3) ‏ (DB2) ‏ (DB1) ‏ (DB) ‏ 260 270 280 290 300 Liquid A (LB4) ‏ (LB3) ‏ (LB2) ‏ (LB1) ‏ Liquid B Temperature 260K270K280K290K300K

18. Step 4: 1 5 3 2 4 Mixture Constant I nstruction for the animator Boundary limits (if any) ‏ T ext to be displayed in the working area (DT) ‏ If learner selects any point between the blue dot and red dot on the black line then the mixture has two colours at each temperature At 260K it will be LB and DB, At 270K it will be LB1 and DB1, At 280K it will be LB2 and DB2, At 290K it will be LB3 and DB3, At 300K it will be LB4 or DB4(which is the same colour) ‏ If learner selects any point after the red dot on the black line then the mixture has one colour which is lighter than (DB4/LB4). Temperature – 260 K- 300 K  The mixture being fixed at ratio 40 : 60 (liquid B :liquid A) select the temperature and click on mix to view the change in the liquids. (LB) ‏ T/K Mole fraction 0.20.40.60.81 260 270 280 290 300 0 (DB4) ‏ (DB3) ‏ (DB2) ‏ (DB1) ‏ (DB) ‏ 260 270 280 290 300 Liquid A (LB4) ‏ (LB3) ‏ (LB2) ‏ (LB1) ‏ Liquid B

19. 1 5 3 2 4 Interactivity option 1: Step No: 1 Interactivity type (IO 1/IO 2) ‏ Instruction to learners Boundary limits & options Instruction to animators Results and output SlidersChoose the temperature and the ratio mix and click on the mix button 0.1 – 1 for slider 1 Temperatu re slider 260K – 300K Once the learner chooses the value and clicks on mix then show the point on the graph. If the point fall inside the curve the liquids do not mix only change colours. If the point lies outside the curve the liquids mix completely to form one solution. At temperature 260K if the slider 1 is at 0.1/0.9 then plot on 0.1 in the graph, if the slider 1 is at 0.2/0.8 then plot on 0.2 in the graph and so on. – The colour of the liquids (fig B Master layout 3) at this temperature for points inside the curve are: liquid B (LB) liquid A (DB). – The colour of the liquid (fig A- Master layout 3) at this temperature for points outside (before blue dot) the curve is: entire liquid little lighter than LB (more drops of white) ‏ – The colour of the liquid (fig A- Master layout 3) at this temperature for points outside (after red dot) the curve is: entire liquid little darker than DB (more drops of blue) ‏ Liquid B (0 blue in a scale of 0-255) ‏ Liquid A (255 blue in a scale of 0-255) ‏ 1/00/1 0.8/ 0.2 0.6/ 0.4 0.4/ 0.6 0.2/ 0.8 Temperature 260K270K280K290K300K 2 T/K Mole fraction 0.20.40.60.81 260 270 280 290 300 0 (LB4) ‏ (LB3) ‏ (LB2) ‏ (LB1) ‏ (LB) ‏ (DB4) ‏ (DB3) ‏ (DB2) ‏ (DB1) ‏ (DB) ‏ 260 270 280 290 300 Liquid BLiquid A

20. 1 5 3 2 4 Interactivity option 1: Step No: 2 Interactivity type (IO 1/IO 2) ‏ Instruction to learners Boundary limits & options Instruction to animators Results and output SlidersChoose the temperature and the ratio mix and click on the mix button 0.1 – 1 for slider 1 Temperatu re slider 260K – 300K Once the learner chooses the value and clicks on mix then show the point on the graph. If the point fall inside the curve the liquids do not mix only change colours. If the point lies outside the curve the liquids mix completely to form one solution. At temperature 270K if the slider 1 is at 0.1/0.9 then plot on 0.1 in the graph, if the slider 1 is at 0.2/0.8 then plot on 0.2 in the graph and so on. – The colour of the liquids (fig B Master layout 3) at this temperature for points inside the curve are: liquid B (LB1) liquid A (DB1). – The colour of the liquid (fig A- Master layout 3) at this temperature for points outside (before blue dot) the curve is: entire liquid little lighter than LB1 (more drops of white) ‏ – The colour of the liquid (fig A- Master layout 3) at this temperature for points outside (after red dot) the curve is: entire liquid little darker than DB1 (more drops of blue) ‏ Liquid B (0 blue in a scale of 0-255) ‏ Liquid A (255 blue in a scale of 0-255) ‏ 1/00/1 0.8/ 0.2 0.6/ 0.4 0.4/ 0.6 0.2/ 0.8 Temperature 260K270K280K290K300K 2 T/K Mole fraction 0.20.40.60.81 260 270 280 290 300 0 (LB4) ‏ (LB3) ‏ (LB2) ‏ (LB1) ‏ (LB) ‏ (DB4) ‏ (DB3) ‏ (DB2) ‏ (DB1) ‏ (DB) ‏ 260 270 280 290 300 Liquid BLiquid A

21. 1 5 3 2 4 Interactivity option 1: Step No: 3 Interactivity type (IO 1/IO 2) ‏ Instruction to learners Boundary limits & options Instruction to animators Results and output SlidersChoose the temperature and the ratio mix and click on the mix button 0.1 – 1 for slider 1 Temperatu re slider 260K – 300K Once the learner chooses the value and clicks on mix then show the point on the graph. If the point fall inside the curve the liquids do not mix only change colours. If the point lies outside the curve the liquids mix completely to form one solution. At temperature 280K if the slider 1 is at 0.1/0.9 then plot on 0.1 in the graph, if the slider 1 is at 0.2/0.8 then plot on 0.2 in the graph and so on. – The colour of the liquids (fig B Master layout 3) at this temperature for points inside the curve are: liquid B (LB2) liquid A (DB2). – The colour of the liquid (fig A- Master layout 3) at this temperature for points outside (before blue dot) the curve is: entire liquid little lighter than LB2 (more drops of white) ‏ – The colour of the liquid (fig A- Master layout 3) at this temperature for points outside (after red dot) the curve is: entire liquid little darker than DB2 (more drops of blue) ‏ Liquid B (0 blue in a scale of 0-255) ‏ Liquid A (255 blue in a scale of 0-255) ‏ 1/00/1 0.8/ 0.2 0.6/ 0.4 0.4/ 0.6 0.2/ 0.8 Temperature 260K270K280K290K300K 2 T/K Mole fraction 0.20.40.60.81 260 270 280 290 300 0 (LB4) ‏ (LB3) ‏ (LB2) ‏ (LB1) ‏ (LB) ‏ (DB4) ‏ (DB3) ‏ (DB2) ‏ (DB1) ‏ (DB) ‏ 260 270 280 290 300 Liquid BLiquid A

22. 1 5 3 4 Interactivity option 1: Step No: 4 Interactivity type (IO 1/IO 2) ‏ Instruction to learners Boundary limits & options Instruction to animators Results and output SlidersChoose the temperature and the ratio mix and click on the mix button 0.1 – 1 for slider 1 Temperatu re slider 260K – 300K Once the learner chooses the value and clicks on mix then show the point on the graph. If the point fall inside the curve the liquids do not mix only change colours. If the point lies outside the curve the liquids mix completely to form one solution. At temperature 290K if the slider 1 is at 0.1/0.9 then plot on 0.1 in the graph, if the slider 1 is at 0.2/0.8 then plot on 0.2 in the graph and so on. – The colour of the liquids (fig B Master layout 3) at this temperature for points inside the curve are: liquid B (LB3) liquid A (DB3). – The colour of the liquid (fig A- Master layout 3) at this temperature for points outside (before blue dot) the curve is: entire liquid little lighter than LB3 (more drops of white) ‏ – The colour of the liquid (fig A- Master layout 3) at this temperature for points outside (after red dot) the curve is: entire liquid little darker than DB3 (more drops of blue) ‏ Liquid B (0 blue in a scale of 0-255) ‏ Liquid A (255 blue in a scale of 0-255) ‏ 1/00/1 0.8/ 0.2 0.6/ 0.4 0.4/ 0.6 0.2/ 0.8 Temperature 260K270K280K290K300K 2 T/K Mole fraction 0.20.40.60.81 260 270 280 290 300 0 (LB4) ‏ (LB3) ‏ (LB2) ‏ (LB1) ‏ (LB) ‏ (DB4) ‏ (DB3) ‏ (DB2) ‏ (DB1) ‏ (DB) ‏ 260 270 280 290 300 Liquid BLiquid A

23. 1 5 3 2 4 Interactivity option 1: Step No: 5 Interactivit y type (IO 1/IO 2) ‏ Instruction to learners Boundary limits & options Instruction to animators Results and output SlidersChoose the temperature and the ratio mix and click on the mix button 0.1 – 1 for slider 1 Temperatu re slider 260K – 300K Once the learner chooses the value and clicks on mix then show the point on the graph. If the point fall inside the curve the liquids do not mix only change colours. If the point lies outside the curve the liquids mix completely to form one solution. After the animation is the label (CST) should appear along with text in blue. At temperature 300K if the slider 1 is at 0.1/0.9 then plot on 0.1 in the graph, if the slider 1 is at 0.2/0.8 then plot on 0.2 in the graph and so on. – The colour of the liquid (fig A- Master layout 3) at this temperature for a point on the curve is LB4/DB4 (which are the same colour) ‏ – The colour of the liquid (fig A- Master layout 3) at this temperature for points outside (before blue dot) the curve is: entire liquid little lighter than LB4 (more drops of white) ‏ – The colour of the liquid (fig A- Master layout 3) at this temperature for points outside (after red dot) the curve is: entire liquid little darker than DB4 (more drops of blue. “The tip of the phase diagram is called the upper critical solution temperature (CST), there exist only one composition for any given mixture at which this upper CST occurs. This is the maximum temperature at which phase separation can occur. This point does not depend on any parameters and can be attained if and only if the specific concentrations are mixed at specific temperature and pressure.” A lower CST may also be observed under few cases. Liquid B (0 blue in a scale of 0-255) ‏ Liquid A (255 blue in a scale of 0-255) ‏ 1/00/1 0.8/ 0.2 0.6/ 0.4 0.4/ 0.6 0.2/ 0.8 Temperature 260K270K280K290K300K Mole fraction T/K 0.20.40.60.81 260 270 280 290 300 0 (LB4) ‏ (LB3) ‏ (LB2) ‏ (LB1) ‏ (LB) ‏ (DB4) ‏ (DB3) ‏ (DB2) ‏ (DB1) ‏ (DB) ‏ 260 270 280 290 300 Liquid BLiquid A CST

24. T1: Lets look at a real example I nstruction for the animator Audio narrations (if any) ‏ Text to be displayed T1 appears. The diagram appears with labels and the curve only. The text to be displayed appears.  A mixture of 50 g of hexane (0.59 mol C 6 H 14 ) and 50g of nitrobenzene (0.41 mol C 6 H 14 No 2 ) was prepared at 290K. What are the compositions of the phases? What proportions do they occur? To what temperature must the sample be heated in order to obtain a single phase? Step No: 6

25. T1: Lets look at a real example I nstruction for the animator Audio narrations (if any) ‏ Text to be displayed  What do we need to find?  What are the compositions of the phases?  What proportions do they occur?  To what temperature must the sample be heated in order to obtain a single phase?  Let us begin with the first question Step No: 7

26. T1: Lets look at a real example I nstruction for the animator Audio narrations (if any) ‏ Text to be displayed Vertical dotted line (blue) at 0.41 appears. The lever rule gets solved step by step highlighting the  What are the compositions of the phases?  Method: The composition of phases in equilibrium are given by the points where the tie- line representing the temperature intersects the phase boundaries.  Answer: We denote hexane as H and nitrobenzene by N. The point X N = 0.41, T= 290K occurs in two phase region of phase diagram.  The horizontal tie line cuts the phase boundary at X N = 0.35 and X N = 0.83, so those are the compositions of the two phases. Step No: 8

27. T1: Lets look at a real example I nstruction for the animator Audio narrations (if any) ‏ Text to be displayed Show lever rule a 1 l 1 = a 2 l 2 The sum (in text to be displayed) gets solved step by step highlighting the black dot and red dot and then the red dot and blue dot.  What proportions do they occur?  Method: The proportions of the phases are given by the Lever rule.  Answer: According to the lever rule the ratio of amounts of each phase is equal to the ratio of the distance l 1 and l 2.  Therefore: a 1 = l 2 = 0.83 -0.41 = 0.42 = 7 a 2 l 1 0.41 – 0.35 0.06  That is there is about 7 times more hexane-rich phase than nitrobenzene – rich phase. Step No: 9

28. T1: Lets look at a real example I nstruction for the animator Audio narrations (if any) ‏ Text to be displayed The blue line extends upwards from the red dot and green dot and green line is shown.  To what temperature must the sample be heated in order to obtain a single phase?  Method: The temperature at which the components are completely miscible is found by following the isopleth upward and noting the temperature at which it enters the one phase region of the phase diagrams.  Answer: Heating the sample to 292K takes the components into the single-phase region.  Because the phase diagram has been constricted experimentally, these conclusions are not based on any assumptions about ideality. They would be modified if the system were subjected to a different pressure. Step No: 10

29. Animation design Please see the design template provided in the next slide. This is a sample template, and you are free to change as per your design requirements. Try and recreate the sections/subsections as shown in the template. 1 5 2 4 3

30. Want to know more… (Further Reading) ‏ Definitions Formula with derivation (if any) ‏ Graphs/Diagram (for reference) ‏ Animation Area Test your understanding (questionnaire) ‏ Lets Learn! Concepts Assumptions (if any) ‏ Lets Sum up (summary) ‏ Instructions/ Working area Radio buttons (if any)/Drop down (if any) ‏ Interactivity options Sliders(IO1) ‏ / Input Boxes(IO2) ‏ /Drop down(IO3) ‏ (if any) ‏ Play/pauseRestart Output result of interactivity (if any) ‏ What will you learn Credits

31. Interactivity and Boundary limits In this section, you will add the ‘Interactivity’ options to the animation. Use the template in the next slide to give the details. Insert the image of the step/s (explained earlier in the Section 3) in the box, and provide the details in the table below. The details of Interactivity could be: Types: Drop down, Slider bar, Data inputs etc. Options: Select one, Multiple selections etc Boundary Limits: Values of the parameters, which won’t show results after a particular point Results: Explain the effect of the interaction in this column Add more slides if necessary 1 2 5 3 4

32. 1 5 3 2 4 For interactivity see slide No 15 to 22 Instructions for the animator Instruction to the learner Results and Output Boundary limits Interactivity type (IO1/IO2..) ‏ Interactivity option 1: Step No:

33. INSTRUCTIONS SLIDE Self- Assessment Questionnaire for Learners Please provide a set of questions that a user can answer based on the LO. They can be of the following types: –These questions should be 5 in number and can be of objective type (like MCQ, Match the columns, Yes or No, Sequencing, Odd One Out). –The questions can also be open-ended. The user would be asked to think about the question. The author is requested to provide hints if possible, but a full answer is not necessary. –One can include questions, for which the user will need to interact with the LO (with certain parameters) in order to answer it.

34. INSTRUCTIONS SLIDE Please make sure that the questions can be answered by interacting with the LO. It is better to avoid questions based purely on recall. Questionnaire for users to test their understanding

35. Questionnaire 1. What is F in the phase rule Answers: a) no of direction in which molecule may move b)number of parameters required to describe a system completelyc) number of phases d) ‏ none of the above 2. The composition of the to phases at two phases after mixing at a fixed temperature remains same irrespective of amount of components if they mixtures are not too dilute. Answers: a)trueb)false c)cant say d) ‏ depends on viscosity of substance 3. By looking at the phase diagram what can u say about the miscibility of two substance with increase in temperature? Answers: a) Miscibility increases b) Miscibility decreases c) Remains the same d) ‏ Cant say 4. Do you think phase rule F=C-P+2 valid for solid liquid equilibrium as well? Answers: a)Nob)Cant sayc)Yesd) ‏ It is univeral 5. On basis of phase rule, prove that there can be only one triple point of water (the condition at which water,ice and vapour are in equilibrium with each other. Answers: a)c=1,P=3=> F= 0, hence proved b) c=1, P=2, => F=1 hence proved c) cant not be provedd) ‏ phase rule is not valid for more than two phases. 1 5 2 4 3

36. Links for further reading Books:Physical Chemistry,third indian edition G.W.Castellen, Narosa publications, New Delhi. Physical chemistry, 6 edition, P W Atkins, Oxford university press, london. Principles of physical chemistry, third indian edition, Marron and Prutton, Narosa publications, New Delhi.

37. INSTRUCTIONS SLIDE Please provide points to remember to understand the concept/ key terms of the animation The summary will help the user in the quick review of the concept. Summary

38. We have visualised how the composition of each phase changes as two partially miscible liquids mix with each other. If phase separation occurs that at a fixed temperature the composition of the phases do not change certain change in composition. (Note that the volume of the two phases does change. But composition of the two phases remain the same defined by the isotherm in the phase diagram) ‏ CST is fixed for a particular mixture. And is attained only at a particular composition, temperature and pressure.