Timeline for Phase diagram. When will I have a phase transition?
Current License: CC BY-SA 3.0
13 events
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| Jan 5, 2018 at 14:38 | vote | accept | StarBucK | ||
| Jan 5, 2018 at 10:57 | comment | added | StarBucK | @FGSUZ Thank you a lot for your patience. I commented under your answer below :) | |
| Jan 5, 2018 at 10:47 | comment | added | FGSUZ | Oh, yeah, true, the equilibrium points are 1 and 2, i.e. the ones along the horizontal line that cross a negative sloped part. YEs but... check the axis... That's a pressure-volume diagram! haha it's got nothing to do with the Pressure-temperature one. Your picture doesn't say anything about $T$, only about volume and pressure. | |
| Jan 5, 2018 at 10:42 | history | edited | StarBucK | CC BY-SA 3.0 |
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| Jan 5, 2018 at 10:41 | answer | added | FGSUZ | timeline score: 1 | |
| Jan 5, 2018 at 10:33 | comment | added | FGSUZ | @StarBucK Sorry I don't see any reference to what you say. Those lines separate two states. The lines are actually coexistence lines. | |
| Jan 5, 2018 at 10:33 | history | edited | Qmechanic♦ |
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| Jan 5, 2018 at 9:18 | history | edited | StarBucK | CC BY-SA 3.0 |
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| Jan 5, 2018 at 9:16 | comment | added | StarBucK | @user115350 To understand physically. Imagine I push on a piston to compress my experiment. So if I start at A, I can compress it until I reach the transition curve (vertical projection of A on the curve). Then at this moment my piston will be "stucked" and all the energy I spend on pushing on it will be transformed into heat to do the phase transition. Then when I have full gas, my piston is not stucked anymore and the energy I spend pushing on it will be indeed converted into mechanical motion (the piston will move). Is that it ? | |
| Jan 5, 2018 at 9:14 | comment | added | StarBucK | @FGSUZ I am not sure to understand because I thought that when I have two phases in equilibrium, the system is not thermodynamically stable and it will go either to full liquid or to full gas : uam.es/personal_pdi/ciencias/evelasco/master/tema_III.pdf beginning of page 64. They say that on figure 3.10 between a and b the system is unstable. Thus I can't have thermo equilibrium between gas and liquid ? | |
| Jan 5, 2018 at 5:20 | comment | added | user115350 | At the point C, as FGSUZ said, you have both liquid and vapor phase. They co-exist. Now, you might further ask how much is the liquid phase or how much is the vapor phase. The term for this is quality. It doesn't depend on where it is previously. But quality is change when you inject or retract heat from the mixture. | |
| Jan 5, 2018 at 0:25 | comment | added | FGSUZ | At the point C you will actually have BOTH in general. You'll have two coexisting states, and the quantities are such that the chemical potential or molar gibbs potential is the same. | |
| Jan 5, 2018 at 0:00 | history | asked | StarBucK | CC BY-SA 3.0 |