Well, I tried to solve a problem, where the pressure 1 atm is kept constant, and the water was condensed using the pressure give above.
I want to know the change in enthalpy, heat, work, change in entropy, and change in the internal energy
I immediately set internal energy change to zero, because I assume that the system has no internal energy
I note that
H = U - PV
H = -PV
U=q+W
q=-W
q=H
dH = q
S = q/T for a hypothetical entropy
my problem is what is the difference between "change in entropy" and change in "enthalpy of vaporization"
Vaporization is the inverse process of condensation
H-condens = - H_evap
I find the enthalpy of vaporization given and I find the
"enthapy of condensation" of water = 970 kcal/mol
What is the difference between "enthalpy change for condensation" and "standard enthalpy change of condensation"
Can you equate those quantities without making an error
I quess, I would like to discuss this matter with fellow chemists, and chemistry students.