1. Driving force for solidification This Learning object ‏ will introduce the concept of undercooling and driving force for phase transformations Subject: MEMS Mentor: Prof. Gururajan Authors:

2. Learning Objectives After interacting with this Learning Object, the learner will be able to: calculate the change in entropy of the system calculate the driving force of the system during phase transformation

3. Definitions: 1) Gibbs free energy: The part of the energy of a system that is available as work during chemical change is known as free energy of the system. The Gibbs free energy of a system is defined by the equation G = H – TS where H is the enthalpy, T the absolute temperature and S the entropy of the system. 2) Undercooling Cooling a material below the temperature of an equilibrium phase change fast enough to not allow the occurrence of the transformation. 3) Driving force: The gain in free energy of a system during transformation from one phase to another. It is given by the formula ∆G ∆H f – T 4) Melting point of ice: The temperature at which free energy of water (liquid) is equal to the free energy of ice (solid) ‏

4. Definitions: 5) Boiling point of water The temperature at which free energy of water vapour (gaseous state) is equal to the free energy of water (liquid) 6) Transformation temperature/ Temperature of transformation: The temperature at which free energy of state is equal to free energy of state of transformations. 7) Specific heat of a substance: It is the quantity of heat (in joules) required to raise the temperature of 1 gram of pure substance by one degree Kelvin. 8) Latent heat of fusion/vaporisation: The quantity of heat absorbed or released when a substance changes its physical phase at constant temperature. 9) Entropy of fusion: It is the change in entropy at the time of melting of a material It is defined by the equation

5. Derivation: The free energies of the liquid and solid at temperature T are given by G L = H L – TS L G S = H S – TS S Therefore, at a temperature T, ∆G = ∆H – T ∆S ………………. (1) ‏ Where ∆H = H L – H S and ∆S = S L – S S At the equilibrium melting temperature T m the free energies of the solid and liquid are equal i.e. ∆G = 0 Therefore, ∆G = ∆H – T m ∆S = 0

6. Derivation: Therefore at T m ∆S = ……………. (2) ‏ This is known as “Entropy of fusion”. For small undercoolings (∆T) the difference in the specific heats of the liquid and solid (C L p - C S p ) can be ignored. Therefore, ∆H and ∆S are approximately independent of temperature. Combining equations (1) and (2), we get ∆G ΔH f – T ( for small ΔT) ‏

7. Diagram for reference: For the animator – Draw this image As temperature increases, enthalpy increases with slope C p i.e specific heat at constant pressure 2) As temperature increases, Gibbs free energy decreases at a rate given by (-S) ‏

8. Diagram for reference: For the animator–Draw this image and the let the text come below the image Variations of the free energies of solid and liquid phases with temperature at constant pressure solid liquid solidliquid TmTm The Gibbs free energy of the liquid decreases more rapidly with increasing temperature than that of the solid For temperatures upto T m, the solid phase has the lowest free energy and is therefore the stable equilibrium phase For temperatures above T m, the liquid phase has the lowest free energy and is therefore the stable equilibrium phase At both the phases have the same value of G and both solid and liquid can exist in equilibrium.

9. Master Layout Difference in free energy between water (liquid) and ice (solid) close to melting point The curvature of the G w and G i lines have been ignored w i Title of the graph Footnote T m - Ice point

10. 1 5 3 2 4 Step No: 1 Instructions for the animatorAudio narration/ text to be displayed Display the image of water in glass first then the thermometer Can we keep water below its freezing point without allowing it to turn into ice? The red line of the thermometer goes down from 30 0 C to - 30 0 C Click on the NEXT button to find out the answer. NEXT Images copyrighted

11. 1 5 3 2 4 Step No: 2 Instructions for the animatorAudio narration/ text to be displayed Display the animation of green box. If we keep water in the refrigerator eventually it will turn into ice. Display master layout as aboveIf we plot the graph of Gibbs free energy for cooling of water as a function of temperature it will look like this Master layout do not show - G i line, title of the graph and footnote Show water in vessel being kept in the refrigerator for cooling. Vessel is taken out and show ice in the vessel.

12. 1 5 3 2 4 Step No: 3 Master layout http://en.wikipedia.org/wiki/File:M elting_icecubes.gif Instructions for the animatorAudio narration/ text to be displayed Display the animation of green box. Now if we keep ice at room temperature to melt eventually it will turn into water. Display master layout as aboveIf we plot the graph of Gibbs free energy for melting of ice as a function of temperature it will look like this

13. 1 5 3 2 4 Step No: 4 Master layout Instructions for the animator Audio narration/ text to be displayed The master layout remains on screen as in previous slide Observe the graph of Gibbs free energy v/s temperature as shown above. the two lines intersect at a point called T m below T m ice has lower free energy hence it exists in stable equilibrium above T m water has lower free energy hence it exists in stable equilibrium at both water and ice have the same free energies and co-exist in equilibrium.

14. Master Layout 1 Difference in free energy between water (liquid) and ice (solid) close to melting point The curvature of the G w and G i lines have been ignored w i T3T3 T2T2 T1T1 T2T2 T m - Ice point

15. Master Layout 1 Difference in free energy between water (liquid) and ice (solid) close to melting point The curvature of the G w and G i lines have been ignored w i T1T1 ∆T ∆G T m - Ice point

16. 1 5 3 2 4 Step No: 5 Master layout 1 Instructions for the animator Audio narration/ text to be displayed Display master layout as explained above If water is cooled at temperature T 1 below T m, the driving force ∆G is small and the barrier for water to form ice is large.

17. Master Layout 2 Difference in free energy between water (liquid) and ice (solid) close to melting point The curvature of the G w and G i lines have been ignored w i T1T1 ∆T ∆G T2T2 T m - Ice point

18. 1 5 3 2 4 Step No: 5 Master layout 2 Instructions for the animator Audio narration/ text to be displayed Display master layout as explained above If water is cooled at temperature T 2 below T m, the driving force ∆G is higher than that at T 1 It means that the gain of free energy by the system during the transformation of water to ice is more.

19. Master Layout 3 Difference in free energy between water (liquid) and ice (solid) close to melting point The curvature of the G w and G i lines have been ignored w i T1T1 ∆T ∆G T2T2 T m - Ice point

20. 1 5 3 2 4 Step No: 5 Master layout 3 Instructions for the animator Audio narration/ text to be displayed Display master layout as explained above If water is cooled at temperature T 3 below T m, the driving force ∆G is much higher than that at T 1 and T 2 It means that the gain of free energy by the system during the transformation of water to ice is very large i.e the driving force ∆G is very large.

21. 1 5 3 2 4 Step No: 5 Master layout 3 Instructions for the animator Audio narration/ text to be displayed Display master layout as explained above You must have observed that higher the undercooling ∆T, the driving force is large. In this case, undercooling of water is how much below T m can we keep water without allowing it to turn into ice The driving force is given by ∆G L - T

22. 1 5 3 2 4 Step No: 5 diagram on slide 7 Instructions for the animator Audio narration/ text to be displayed Display diagram as explained above In general, the graph of Gibbs free energy as a function of temperature for any material (in solid and liquid state) looks like this. If the driving force is large the system will move to its stable state quickly. If the driving force is small then the system will take a longer time to move to its stable state. Thus water can be kept in its liquid state below 0 0 C.

23. Want to know more… (Further Reading) ‏ Graphs/Diagram (for reference) ‏ Animation Area Test your understanding (questionnaire) ‏ Lets Learn! 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 Definitions Derivation

24. 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

25. Questionnaire 1)NPTEL link will be provided by Prof. Gururajan 2) Questions: (Raghavan, V., Material Science and Engineering, Prentice Hall of India, V th edition) ‏ a) The free energy change during melting of ice at 0 o C is equal to i) enthalpy of melting – entropy of melting ii) 0 iii) 273 iv) can’t say without more data b) Calculate the entropy increase when one mole of ice melts into water at 0 o C (Hint: Use formula for ∆S) ‏ Answers: ∆S = 22.04 J mol -1 K -1 c) Calculate the entropy change and the free energy change during the melting of gold at its melting point. The enthalpy of fusion for gold is 12.6 kJmol -1 Answers: ∆S = 9.43 J mol -1 K -1, ∆G = 0 1 5 2 4 3

26. Links for further reading Books: 1) Raghavan, V., Materials Science and Engineering, a first course, Second edition, Prentice Hall of India, New Delhi 2) Jena, A. K., Chaturvdei, M.C., (1992), Phase transformation in materials, Englewoods Cliffs: Prentice Hall 3) Porter, D., Easterling, K., Phase Transformation in metals and alloys, 3 rd edition Weblinks: NPTEL link will be provided by Prof. Gururajan

27. Summary Driving force is the gain in free energy of the system during transformation from one state to another. Undercooling is cooling the material below its melting point (transformation temperature) without allowing it to convert into the stable state. The larger the undercooling the associated driving force is also large. The rate of transformation depends upon the driving force. Driving force is given by ∆G ∆H f - T