1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Chemical potential in thermodynamics to find latent heat of vaporization

  1. Mar 10, 2012 #1
    Doing some fun problems in Keith Stowe's 'An Introduction to Thermodynamics and Statistical Mechanics'. Good book.

    Problem statement:
    A certain material vaporizes from the liquid phase at 700 K. In both phases, the molecules have three degrees of freedom. If [itex]u_{0}[/itex] in the liquid phase is -0.12 eV, what is the latent heat of vaporization in joules per mole?

    My thoughts:
    Since [itex]u_{0}[/itex] is -0.12 eV, then that must be the energy required to break one molecule away from the rest. This is [itex]1.92 \cdot 10^{-20}[/itex] J. Since they're asking for 1 mole, multiplying the above by [itex]N_{A}[/itex] (Avogadro's constant) gives 11577 J/mol. But the correct answer is 15600 J/mol. What am I missing?

    My next consideration was that some thermal energy also would be added, but since the molecules have the same degrees of freedom in both the liquid phase and the vapor phase, I was hesitant to include this. Doesn't that mean the thermal energy is the same in both states? Adding the thermal energy [itex]\frac{N\nu}{2}kT[/itex] overshoots the 15600 J/mol.
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted



Similar Discussions: Chemical potential in thermodynamics to find latent heat of vaporization
Loading...