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.

 

[rude man]

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.

 

[jjand]

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

 

[rude man]

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.

 

[paranormal]

Your approach to finding the entropy of liquid water and water vapor using the thermodynamic table is correct. The equation dS = Q/T = Cp*dT/T can be used to calculate the change in entropy as long as the process is reversible and the temperature remains constant. In this case, since we are given a range of temperatures, we can use the integral form of the equation, as you have done, to find the change in entropy.

To find the entropy of vaporization, we can subtract the entropy of liquid water from the entropy of water vapor at the same temperature. This will give us the change in entropy during the phase change. Similarly, the enthalpy of vaporization can be found by subtracting the enthalpy of liquid water from the enthalpy of water vapor at the same temperature.

To calculate the total heat transfer, Qwater, we can use the enthalpy of vaporization and the given temperatures. We know that the enthalpy of vaporization is equal to the amount of heat required to vaporize one mole of water at a given temperature and pressure. Since we have 2 moles of water vapor at 100°C and 1 atm, we can use the enthalpy of vaporization to find the total heat transfer required to cool the water vapor to 50°C at constant pressure.

In summary, by using the thermodynamic table and the equations provided, we can find the entropy of liquid water and water vapor, as well as the entropy and enthalpy of vaporization. We can also use these values to calculate the total heat transfer required in a given process.