# Chemical potential in thermodynamics to find latent heat of vaporization

1. Mar 10, 2012

### Hixy

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 $u_{0}$ in the liquid phase is -0.12 eV, what is the latent heat of vaporization in joules per mole?

My thoughts:
Since $u_{0}$ is -0.12 eV, then that must be the energy required to break one molecule away from the rest. This is $1.92 \cdot 10^{-20}$ J. Since they're asking for 1 mole, multiplying the above by $N_{A}$ (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 $\frac{N\nu}{2}kT$ overshoots the 15600 J/mol.