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## Homework Statement

The latent heat of vaporization of a substance is defined as the amount of heat required to transform the substance from its liquid into its vapor phase. A certain material has a boiling point of 700K and its molecules have three degrees of freedom in the liquid phase but five in the gas phase. If the chemical potential, µ in the liquid phase is -0.12 eV, what is the latent heat of vaporization in J/mol?

## Homework Equations

L = J/mol

W = -ΔE = Nμ

_{o}+ [itex]\frac{NvkΔT}{2}[/itex]

k = 1.381x10

^{-23}

Avogrado's Constant = 6.022x10

^{23}

## The Attempt at a Solution

μ

_{o}= -0.12eV x 1.602x10

^{-20}= -1.9224x10

^{-20}J/atom

multiply by avogadro's constant to get J/mol

Nμ

_{o}= -1.9224x10

^{-20}x 6.022x10

^{23}= -11576.7 J/mol

Now for the [itex]\frac{NvkΔT}{2}[/itex] term:

[itex]\frac{NvkΔT}{2}[/itex] = [itex]\frac{N(3k)(700)}{2}[/itex] = 8732 for liquid

[itex]\frac{NvkΔT}{2}[/itex] = [itex]\frac{N(5k)(700)}{2}[/itex] = 14.6x10

^{6}for gas

How do I combine these to solve for L, the latent heat of vaporization?