How Do You Calculate the Triple Point of Uranium Hexafluoride?

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SUMMARY

The discussion focuses on calculating the triple point of uranium hexafluoride (UF6) using the natural logarithm equations for pressure in solid and liquid states. The equations provided are lnp(solid) = 29.411 - (5893.5K/T) and lnp(liquid) = 22.254 - (3479.9K/T). The user successfully determined a temperature of 337.2 K by equating the two equations, resulting in pressures of approximately 1.522443 x 10^5 Pa and 1.523616 x 10^5 Pa, which are sufficiently close to be considered equivalent. The key takeaway is that the triple point occurs where the solid and liquid states coexist at the same temperature and pressure.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically the concept of the triple point.
  • Familiarity with natural logarithm functions and their application in pressure calculations.
  • Knowledge of the properties of uranium hexafluoride (UF6) including its molecular weight (MW = 352.02).
  • Basic skills in algebra for solving equations and manipulating variables.
NEXT STEPS
  • Research the concept of the triple point in thermodynamics for various substances.
  • Learn about the properties and behavior of uranium hexafluoride under different temperature and pressure conditions.
  • Explore advanced mathematical techniques for solving logarithmic equations in physical chemistry.
  • Investigate the significance of pressure units in scientific calculations, particularly in Pascal (Pa) and atmospheres (atm).
USEFUL FOR

This discussion is beneficial for chemistry students, particularly those studying physical chemistry and thermodynamics, as well as professionals working with uranium hexafluoride in nuclear science and engineering.

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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


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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Is there any reason you put P=1atm?

In my opinion, you should equate those two equations since the triple point is where they meet each other.
 
I just assumed it to try and work through the problem. I also tried setting the two equal to each other after posting to determine the temp. The temp was 337.2, which i put into each respective equation to get the pressures. It worked well, but there was a small difference in the answer I got for the pressure. one was 1.522443*10^5, and the other was 1.523616*10^5. I believe these values are close enough to consider them the same, what do you think?
 

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