# What is the mathematical equation showing that the quantity of heat absorbed by vaporizing is the same as the quantity of heat released when the vapor condenses?

Jun 9, 2017

...conservation of energy...?

Phase equilibria, in particular, are easily reversible in a thermodynamically-closed system... Thus, the process forwards requires the same amount of energy input as the energy the process backwards gives back.

At constant pressure:

${q}_{v a p} = n \Delta {\overline{H}}_{v a p}$,

$\text{X"(l) stackrel(Delta" ")(->) "X} \left(g\right)$

where $q$ is the heat flow in $\text{J}$, $n$ is of course mols, and $\Delta {\overline{H}}_{v a p}$ is the molar enthalpy in $\text{J/mol}$.

By definition, we must also have:

${q}_{c o n d} = n \Delta {\overline{H}}_{c o n d}$

$\text{X"(g) stackrel(Delta" ")(->) "X} \left(l\right)$

We know that $\Delta \overline{H}$ changes sign for the process of the opposite direction. Hence for any reversible process, $\Delta {\overline{H}}_{c o n d} = - \Delta {\overline{H}}_{v a p}$...

$\implies \textcolor{b l u e}{{q}_{c o n d} = - n \Delta {\overline{H}}_{v a p} = - {q}_{v a p}}$

Thus the heat flow that goes into the system for a vaporization process is equal in magnitude to the heat flow out of a system for a condensation process.