1- Use the thermodynamic table to establish an equation giving the molar entropy of liquid water as a function of temperature. Do the same for water vapor.
2- Use these equations to calculate the entropy of vaporization and the enthalpy of vaporization.
3- From the enthalpy of vaporization of water calculate the total heat transfer, Qwater, when 2 moles of water vapor at 100°C and 1 atm is cooled down to 50°C at constant pressure.
dS = Q/T = Cp*dT/T
dG = dH - TdS
Thermodynamic Constants table is attached
The Attempt at a Solution
I figured I could just put into the dS equation the values for each respective Cp, then integrate and get deltaS = 75.291*ln(T2/298) for example with liquid water, using 298K for T1 as that is what the Cp value is quoted at. The question doesn't give any other temperatures and it doesn't seem to make sense to me to find a specific entropy, but instead a change in entropy, but maybe I'm thinking this through wrong.
Once I get the first question I know the entropy of vaporization would just be the difference of the individual entropies and the enthalpy of vaporization would be just using the dG equation with dG = 0, but I can't get to this one without figuring out the first one.
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