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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
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