# Entropy of liquid water and water vapor

• jjand

## Homework Statement

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.

## Homework Equations

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.

#### Attachments

• Thermo table.pdf
413.9 KB · Views: 1,099
1. Δs = Δq/T = Δh/T since Δp = 0. Then s = s0 + Δs(T).

2. ΔG = 0 for phase change = ΔH - T*ΔS. So you use the G column to determine your enthalpy of vaporization ΔH. Then all you need is Cp for kquid water & vapor.

3. Should be able to handle this part now.

PS - I myself am confused about the reference temperature the table assumes for s = 0 and h = 0. I calculated very different values of T0.

Last edited:
Where did the s = s0 + Δs(T) equation come from?

s = 69.9 at T = 298.15K. You are to find s(T) at p = 1 atmosphere.
The change in entropy of water from any temperature between freezing and steam will be
Δs = ∫dq/T integrated from 298.15 to T, 273.15 < T < 373.15.
But dq = dh = cp*dT. So the integral becomes
s(T) = s0 + cp*ln(T/298.15) with cp = 75.3 and s0 = 69.9 per the table. s0 is the constant of integration. Note that at T = 298.15, s = 69.9 as the table says.

I still have no idea why they use s = 69.9 as the entropy at T = 298.15.