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.
