# 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?

##### 1 Answer

...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_(vap) = nDeltabarH_(vap)# ,

#"X"(l) stackrel(Delta" ")(->) "X"(g)# where

#q# is the heat flow in#"J"# ,#n# is of course mols, and#DeltabarH_(vap)# is the molar enthalpy in#"J/mol"# .

By definition, we must also have:

#q_(cond) = nDeltabarH_(cond)#

#"X"(g) stackrel(Delta" ")(->) "X"(l)#

We know that

#=> color(blue)(q_(cond) = -nDeltabarH_(vap) = -q_(vap))#

*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.*