Melting Points as Function of P

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SUMMARY

The discussion focuses on determining the melting points of materials, specifically Teflon, in a vacuum environment. It highlights that in an ideal vacuum, materials will eventually vaporize, making both solid and liquid states thermodynamically unstable. The melting point's dependence on pressure is negligible between 0 and 1 atmosphere, as indicated by the slope of the melting curve formula, ##d\ln T/dP = \Delta V/\Delta H##. Additionally, Teflon, being a polymer, does not have a distinct melting point; instead, it is defined by a viscosity threshold.

PREREQUISITES
  • Understanding of thermodynamics and phase transitions
  • Familiarity with the properties of polymers, specifically Teflon
  • Knowledge of pressure effects on melting points
  • Basic grasp of the Clausius-Clapeyron equation
NEXT STEPS
  • Research the Clausius-Clapeyron equation and its applications in phase transitions
  • Explore the properties of polymers and their thermal behavior
  • Investigate the effects of vacuum on melting points of various materials
  • Learn about viscosity and its relationship to melting points in polymers
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Researchers, materials scientists, and engineers involved in thermal analysis and polymer studies, particularly those working with Teflon and vacuum environments.

robousy
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Hey folks,

I'm trying to obtain the melting points of several materials in a vacuum.

It's pretty straightforward to look up the melting point in 1 atm, but I'm having difficulty doing this in a vacuum.

One material in particular I'm looking at is Teflon.

If anyone can provide any resources, formula or hints on how to do this I'd be grateful.

Thanks!
 
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The question bears some problems. First, at least in an ideal vacuum, any material will vapourize sooner or later, so both solid and liquid states are thermodynamically unstable. Hence melting becomes a non-equilibrium phase transition. However, from a practical point of view, I think that the changes in melting point with pressure are negligible between 0 and 1 atmosphere. The slope of the melting curve ##d\ln T/dP =\Delta V/\Delta H##, where ##\Delta V## is the volume change in melting and ##\Delta H## the heat of melting. The volume change being very small leads to the slope also being very small.

Secondly, teflon is a polymer and polymers don't have a sharp melting point. Rather ( I don't remember the details) one defines the melting point as the temperature where the viscosity becomes lower than a predefined (and quite arbitrary) level.
 

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