Thermodynamics: vapor pressure water, given the gibbs energy

[nicolasdfz]

Homework Statement

Calculate the vapor pressure of water at 25°C, based on the Gibbs free energy when vaporising from liquid water to vapor (so at 1 atm and 25°C ).

Homework Equations

The Attempt at a Solution

After integrating d g/d p= RT/p. I get my formula p = p0*exp (-delta gm(p0, T)/RT). I know all these variables, except for delta gm. So the gibbs free energy at 1 atm and 25°C.

When differentiating G = H -TS:

dG = (dG/dp)T*dp + (dG/dT)p*dT
(partial derivative to p, when T is constant multiplied by change in pressure + partial derivative to T when p is constant multiplied by change in temperature)

Since there is no change in temperature only the partial derivative to p when T is constant is of importance. Which is equal to volume. I know the value should be 8560 J/mol, but no method I have used seems to give me the right answer.

I posted this on my phone. Sorry for not using proper symbols.

 

[Chestermiller]

Do you have a table of free energies of formation?

 

[nicolasdfz]

I have the enthalpy, vapor heat, volumes and entropy of water at 25°C and its respeptide vapor pressure in my steam table (3169 Pa, which I need to calculate. So I can't use this value). But I need to calculate these values for 1 atm and 25°C. Then I could calculate calculate the gibbs energy and solve it?

I had tried using the respective volumes in both liquid and gaseous phase to come to the value of 8560. But I can't multiply it by change in pressure, since I do not know the vapor pressure. And even when I did nonetheless the answer was wrong ..

 

[Chestermiller]

From your tables, what do you calculate for the free energy of liquid water and of water vapor at 25C and 1 atm?

 

[nicolasdfz]

I can calculate the free energy for liquid and gaseous at 3169 Pa. Values would be -83,71 J/mol for liquid water and -78,62 for the gaseous state at 25°C and 3169. I would need to calculate these values for standard conditions

 

[Chestermiller]

I can calculate the free energy for liquid and gaseous at 3169 Pa. Values would be -83,71 J/mol for liquid water and -78,62 for the gaseous state at 25°C and 3169. I would need to calculate these values for standard conditions

That does not seem to be what your problem statement is asking for. It seems to be asking for the vapor pressure of water at 25 C, given the free energies of liquid water and water vapor at 1 atm and 25C.

 

[nicolasdfz]

But if I could calculate the enthalpy and entropy under standard conditions I could calculate G = H- TS (the 8560 J/mol value). And then use the formula to calculate the vapor pressure. Or should I rather calculate using the respective volumes?

 

[Chestermiller]

But if I could calculate the enthalpy and entropy under standard conditions I could calculate G = H- TS (the 8560 J/mol value). And then use the formula to calculate the vapor pressure. Or should I rather calculate using the respective volumes?

You already know the free energies at standard conditions from the enthalpies and entropies at standard conditions. What values do you calculate? Please answer my question, so that I can lead you to the next step.

 

[nicolasdfz]

Yes, my bad. I was confused because of the vapour pressure. So my gibbs free energies for liquid and gaseous state would be -83,71 J/mol and -78,62 J/mol. But I don't see how that could be. It's no where near the values I could find on Internet.

I don't really see the mistage I made. I took the values in my table and multiplied them by 1000 and 0,018kg/mol and then G = H -TS

 

[Chestermiller]

Yes, my bad. I was confused because of the vapour pressure. So my gibbs free energies for liquid and gaseous state would be -83,71 J/mol and -78,62 J/mol. But I don't see how that could be. It's no where near the values I could find on Internet.

I don't really see the mistage I made. I took the values in my table and multiplied them by 1000 and 0,018kg/mol and then G = H -TS

These are steam tables. And this table is for the thermodynamic properties at saturation. So, the free energy of the liquid must be equal to the free energy of the vapor. If you didn't find that, you must have made a mistake.

You won't be able to solve this problem using the steam tables. You need to use a table of free energies of formation.