CLAUSIUS CLAPEYRON EQUATION UG Course

Transcription

1 CLAUSIUS CLAPEYRON EQUATION UG Course CLAUSIUS CLAPEYRON equation can be used to explain the relationship between equilibrium transition temperature and pressure during the phase transition of any substance. Learning Goals Interpret the effect of changes in pressure on equilibrium transition temperatures. Calculate the changes in equilibrium transition temperatures given the relevant thermodynamic parameters and values. Develop an understanding of the shape of phase boundaries as well as certain day to day phenomena using Clausius-Clapeyron equation as the basis. Author: Shrey Singh 10D UG sophomore year Dept. of ME & MS Kartikay Agarwal 10D UG sophomore year Dept. of ME & MS Mentor: Prof. M. P. Gururajan Dept. of ME & MS

2 DEFINITIONS AND KEYWORDS EQUILIBRIUM TEMPERATURE(Teq) The temperature at which phase transition takes place in a substance at certain external pressure. GIBBS FREE ENERGY (G) = H - TS ENTHALPY ( H )= U +PV ENTROPY(S)- INTERNAL ENERGY (U)- Temperature T Pressure P Volume V

3 Master Layout 1 WHY? The above diagram shows 2 containers, one is the pressure cooker and other is an open vessel.

4 Slide 5 and 6 Demo Tab 02 Tab 03 Tab 04 Tab 05 Tab 06 Tab 07 Why? Instructions/ Working area Credits

5 Step 1: Why? Description of the action Audio narration Text to be displayed Show the above picture on the screen and then the audio narration.. You must have observed that cooking the same food in an open vessel requires more time and heat than cooking it in a pressure cooker..have you ever wondered why?

6 Step 2: Lets Answer Why? Description of the action Audio narration Text to be displayed The following text appears on the screen and the text is highlighted corresponding to the audio narration After going through this learning aid, you will be able to answer WHY.. Cooking is faster in pressure cooker Espresso coffee tastes better than normal coffee Skiing is possible on hard ice even in temperatures much less than 0. And many such different day to day examples... Same as the audio narration

7 Master Layout 2 Getting Into the Equation The above figure shows a test tube or a vessel into which the experiment is carried out. The green part at the bottom is the liquid, the bubbles in green color are the vapours and the grey part covering the tube is the movable piston. Hence, label the figure accordingly.

8 Slide 9, 10 and 11 Demo Tab 02 Tab 03 Tab 04 Tab 05 Tab 06 Tab 07 Getting into the Equation Instructions/ Working area Credits

9 Step 1: Getting into the Equation Description of the action Audio narration Text to be displayed Show the above picture on the screen and show the bubbles moving in the tube upwards. Let us consider a generalised case of a weightless movable piston(weightless is considered so as to neglect gravity) on a closed tube. The tube is filled with a certain amount a substance(let s say S) with surrounding temperature T1 and pressure P1.Now if P1 is 1atm,then the substance will boil at its normal boiling point and melt at its normal melting point but as soon as P1 is varied, the boiling point or melting point will vary accordingly.

10 Step 2: Getting into the Equation Description of the action Show the same picture as on the previous slide Audio narration Text to be displayed The substance will stay in following equilibrium S (1) S (2) For any process dg = VdP SdT For equilibrium conditions dg = 0 So VdP = SdT [G is the Gibbs free energy, S is entropy, (P1,T1) and (P2,T2) are the ] Integrating both the sides Hence dp dt Also ΔS = ΔH T So P2 P1 = ΔS ΔT dp dt ΔV. dp = ΔH ΔV.T (ΔH is average of two phases) T2 = ΔS. dt T1 [CLAUSIUS-CLAPEYRON EQN.]

11 Step 3: Getting into the Equation Descriptio n of the action Show the same picture on the screen Audio Text to be displayed For liquid-gas transition ΔV Vg Vg = nrt P So dp = ΔH dt nrt 2 [n is the number of moles, R is the universal gas constant and Vg Is the volume of gas. ] Integrating both the sides ln P2 P1 = ΔH R ( 1 T1 1 T2 ) Here ΔH is Latent Heat Of vapourization of 1 mole Now simialr case is with melting Just ΔH will be Latent Heat of fusion of 1 mole

12 Master Layout 3 Got It Finally The above is the graphical explanation of the Clausius - Clapeyron equation. The markings are as shown in the diagram

13 Slide 14,15 and 16 Demo Tab 02 Tab 03 Tab 04 Tab 05 Tab 06 Tab 07 Got IT Finally Instructions/ Working area Credits

14 Step 1: Got It Finally Description of the action Show the above graph on the screen and markings without the dotted lines. When the last line(marked In red) of the audio narration starts, then draw the dotted lines and the respective markings Audio narration This is a graph to explain how the phase boundary relates the equilibrium temperature and pressure. As the applied pressure on the substance is varied the equilibrium shifts to new parameters and leads to a change in equilibrium temperature. As in the above case pressure is changed from P to (P+ΔP) which shifts equilibrium from E1 to E2 where the temperature at E2 is (T+ΔT) Text to be displayed This is a graph to explain how the phase boundary relates the equilibrium temperature and pressure. As the applied pressure on the substance is varied the equilibrium shifts to new parameters and leads to a change in equilibrium temperature. As in the above case pressure is changed from P to (P+ΔP) which shifts equilibrium from E1 to E2 where the temperature at E2 is (T+ΔT)

15 Step 2: HOW DOES A PRESSURE COOKER WORK Description of the action Show the above graph on the screen Audio narration Since boiling point is defined as the temperature at which the external pressure equals vapour pressure, boiling point of water can be changed by changing the external pressure. When the valve is placed on the lid of pressure cooker, its mass determines the pressure inside the chamber because gas can escape through the pores only when it exceeds certain pressure. Hence the increased external pressure shifts the equilibrium point to some other position along the curve and thus the boiling point of water is increased. This reduces the consumption of heat energy and time as energy is not consumed for vapourising water at this temperature. Text to be displayed Same as the audio narration

16 Step 3: Ice skiing and espresso coffee machine Description of the action Audio narration The negative slope of the co-existence curve for phase transition in case of ice and water is the reason for ice-skiing possible even in temperatures less than 0. When at constant temperature, ice is compressed resulting in its melting, this makes it self-lubricating and hence skiing is possible. Similar is the case with a ESPRESSO COFFEE MACHINE where on application of external pressure, water doesn t boil even at 100. Water at this temperature then enhances coffee taste with the beans and hence, espresso coffee tastes better than normal coffee. Text to be displayed Same as the audio narration

17 Master Layout 4 Phase transformation in Iron The above is the graph to be displayed. Show the markings and the scale of the graph as shown

18 Slide 19 and 20 Demo Tab 02 Tab 03 Tab 04 Tab 05 Tab 06 Tab 07 Phase Transformation in Iron Instructions/ Working area Credits

19 Step 1: Phase transformation in Iron Description of the action Show the above graph on the Audio narration Above is the graph which shows the effect of pressure on the equilibrium temperature of phase transition for PURE IRON. The solid liquid phase transition in case of pure iron behaves normally and the equilibrium temperature increases gradually with increasing pressure. Text to be displayed Above is the graph which shows the effect of pressure on the equilibrium temperature of phase transition for PURE IRON. The solid liquid phase transition in case of pure iron behaves normally and the equilibrium temperature increases gradually with increasing pressure. Here, (dp/dt) =* H/ T V+ is positive since V is positive from solid to liquid transition. But the effect of pressure is different for transition of one allotrope of pure iron into other in the SOLID PHASE. There are 3 allotropes of iron, namely α, γ, δ, iron in the normal pressure range while a fourth allotrope ε-iron exists in high pressure range. Since we know, (dg/dp)=v at constant temperature, equilibrium temperature changes with change in temperature because these allotropes having different molar volumes, will have different changes in G.

20 Step 2: Phase transformation in Iron Description of the action Show the above graph on the Audio narration Text to be displayed For α- γ phase transition, increasing pressure decreases the equilibrium temperature. (dp/dt)= H/T V, Being FCC structure, closely packed γ-iron has smaller molar volume than α-iron making V negative while H is positive. This leads to a negative sign in the clausius-clapeyron equation and hence equilibrium temperature decreases with increase in pressure. Since ε-iron is stable at very high pressures as can be seen from the graph, it has the highest density amongst all the allotropes.

21 Slide 22 Demo Working Interactions Tab 03 Tab 04 Tab 05 Tab 06 Tab 07 Clausius Clapeyron Activity Area Instructions/ Working area Credits

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23 DESCRIPTION Make the 3 boxes on top of screen, 2 in green and one in grey as tabs to be selected by the user. Once they select 1 tab, accordingly the graph has to be selected from the 2 graphs given in another file. Then user will drag the piston up and down and as the piston moves,the pressure will change on x axis by changing the level of the red colour and the corresponding equilibrium temperature rises on the y axis by changing the level of the colour.the scale will be chosen according to the given graph. you have to accordingly fill the entries of p vs t Graph and the pressure will increase on dragging down and decrease on dragging up. Also on dragging down the liquid in the figure shown increases but very gradually and molecules decreases and vice versa. Also for each reading two perpendicular dash lines will be made one to x and one to y which will cut axis on the level of red colour and write the values in a box near the intersection of these perpendicular lines indicating pressure and equilibrium temperature. If the person wishes to perform the same experiment for different material, then he can once again select the other tab. AUDIO NARRATION Select a material from the given list to analise its pressure temperature relationship. now, move the piston either up or down to decrease or increase the pressure inside the vessel accordingly. This will enable you to view the equilibrium temperature inside the given box on variation of pressure. There is also a graph tab on the screen which on clicking will directly display the graph of the selected material. If you wish to do the same for other material, then you can once again select a different material from the given list. TEXT

24 Also, there will be one graph tab which on clicking will directly display the graph of the selected material. In that graph, also mark separately the normal melting/boiling point, which is the temperature at 1 atmosphere pressure and can be easily seen from the graph

25 Title LETS TRY THIS OUT! DESCRIPTION The questionnaire series from. next slide begins with the title written above,then appears this the question in red colour and next is this image. Now the heavy load attached, is pulled in the direction shown with a hand and the fine wire smoothly cuts the ice block. This cutting has to be shown with dark blue or white color. As the wire keeps on moving downwards, the ice block refreezes till the place where this wire just passed by. AUDIO NARR ATION TEXT TO BE DISPLAYED Same as on next slide

26 Questionnaire LETS TRY THIS OUT! 1. Which phenomenon is shown in the animation below? This is known as REGELATION. When ice melts on application of high pressure and then it again refreezes when pressure is relieved, the concept is known as regelation. An example is cutting a block of ice with a fine wire which is wound around it and pulled with high pressure. This enables the wire to pass through the block of ice and as soon as the wire passes through, block of ice refreezes.

27 2. Calculate the change in enthalpy ( H) for Methyl Chloride given that it s transition temperature from liquid to gaseous phase at 1 atm. Pressure is -25⁰C and at 5 atm. Pressure is 23 ⁰C. ANS J 3. For the given conditions, calculate the transition temperature of fluorobenzene from liquid to gaseous phase at 0.8 atm if at.1 atm. pressure,the temperature is 25⁰C and ( H) for this is J. ANS 76 ⁰C.

28 SUMMARY CLAUSIUS CLAPEYRON equation can be used to explain the relationship between equilibrium transition temperature and pressure during the phase transition of any substance. There are different graphs for different elements having unique properties. Iron has a unique property of its existence in 4 allotropic forms in different physical conditions, helium has a graph which has a negative slope and then becomes positive after a minima. Unlike most materials, water has a negative slope in solid- liquid phase transition and so on. After going through this learning aid, we are able to answer why pressure cooker cooks faster, espresso coffee tastes better, ice skiing is possible in temperatures less than 0 degree celsius and many such queries.

29 Slide 28 Demo Working Interactions Summary Tab 04 Tab 05 Tab 06 Tab 07 Summary Instructions/ Working area Credits

30 References Phase transformations in metals and alloys, D. A. Porter and K. E. Easterling, 2 nd edition, Chapman And Hall.