CHEM: Triple point Temp and Pressure

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SUMMARY

The discussion focuses on calculating the temperature and pressure at the triple point of uranium hexafluoride using the equations for the natural logarithm of pressure for both solid and liquid phases. The equations provided are lnp(solid) = 29.411 - (5893.5K/T) and lnp(liquid) = 22.254 - (3479.9K/T). The user initially attempted to solve for temperature using a pressure of 1 atm (101.325 kPa) but encountered difficulties in relating the two equations to find the triple point. The key takeaway is that the intersection of the two equations indicates the conditions at the triple point.

PREREQUISITES
  • Understanding of thermodynamics and phase diagrams
  • Familiarity with natural logarithmic functions
  • Knowledge of uranium hexafluoride properties
  • Basic algebra for solving equations
NEXT STEPS
  • Study the concept of triple points in phase diagrams
  • Learn how to graph equations to find intersections
  • Research the properties of uranium hexafluoride and its phase behavior
  • Explore advanced thermodynamic equations related to phase transitions
USEFUL FOR

Chemistry students, thermodynamics enthusiasts, and professionals involved in material science or chemical engineering who are studying phase transitions and the behavior of substances at their triple points.

ktmtalker
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Homework Statement


I need to calculate the temperature and pressure at the triple point. natural log of pressure of the liquid/solid: lnp(solid)= 29.411-(5893.5K/T) lnp(liquid)=22.254-(3479.9K/T). The compound is uranium hexafluoride, MW= 352.02. I do not believe the actual molecule matters to much, just the weight. pressure is in Pa.








Homework Equations


lnp(solid)= 29.411-(5893.5K/T)
lnp(liquid)=22.254-(3479.9K/T)

The Attempt at a Solution


3. The Attempt at a Solution
I used P= 1 atm -->101.325kPa, and using the lnp(solid) equation determined a temperature. I then used that same temp in the lnp(liquid) to solve the other lnp equation. It did not go well. I am unsure on how to relate these two equations to the triple point. I should be able to grind through the math when I know how to relate it all.
 
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ktmtalker said:
I used P= 1 atm -->101.325kPa, and using the lnp(solid) equation determined a temperature. I then used that same temp in the lnp(liquid) to solve the other lnp equation. It did not go well. I am unsure on how to relate these two equations to the triple point. I should be able to grind through the math when I know how to relate it all.
When one "assumes" one atmosphere, one makes a big mistake. You have two equations. Do they intersect? What is the relationship of that intersection to the triple-point?
 

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