- #1

Hixy

- 8

- 0

**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.