# Thermodynamics: vapor pressure water, given the gibbs energy

• nicolasdfz
In summary, the conversation was focused on calculating the vapor pressure of water at 25°C, based on the Gibbs free energy when vaporizing from liquid water to vapor at 1 atm and 25°C. The conversation involved using the equations for partial derivatives and integrating dG/dp= RT/p to get the formula p = p0*exp (-delta gm(p0, T)/RT). The person posting the question had all the necessary variables except for delta gm and was trying to calculate it using the respective volumes of liquid and gaseous water. However, the values obtained did not match the values found online and it was suggested that a table of free energies of formation be used instead.
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 ).

## 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.

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Do you have a table of free energies of formation?

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 ..

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

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

nicolasdfz said:
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.

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?

nicolasdfz said:
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.

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

nicolasdfz said:
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.

## What is the relationship between vapor pressure and water in thermodynamics?

The vapor pressure of water refers to the pressure exerted by water molecules in its gaseous state when in equilibrium with its liquid state. In thermodynamics, the vapor pressure of water is affected by factors such as temperature, atmospheric pressure, and humidity.

## How is the vapor pressure of water calculated using Gibbs energy?

The vapor pressure of water can be calculated using the Gibbs energy equation, which states that the natural logarithm of the vapor pressure is equal to the standard Gibbs energy of vaporization divided by the gas constant multiplied by the temperature.

## What is the significance of the vapor pressure of water in thermodynamics?

The vapor pressure of water is an important concept in thermodynamics as it is a measure of the tendency of water molecules to escape from its liquid phase and enter its gaseous phase. It also plays a role in various physical and chemical processes, such as evaporation and distillation.

## How does temperature affect the vapor pressure of water?

According to the Clausius-Clapeyron equation, the vapor pressure of water increases with increasing temperature. This means that as the temperature rises, the water molecules have more energy and are more likely to escape into the gas phase, resulting in a higher vapor pressure.

## What is the impact of atmospheric pressure on the vapor pressure of water?

The vapor pressure of water is inversely proportional to atmospheric pressure. This means that as atmospheric pressure increases, the vapor pressure decreases and vice versa. This is because a higher atmospheric pressure makes it more difficult for water molecules to escape into the gas phase.

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