Questions Regarding the Phase Diagram

[Chestermiller]

For the case of a pressure of 1atm applied from a piston, in the phase diagram in my OP, I believe the melting point is 0C and 1atm. In the case of a fixed volume container, the melting/triple point is at 0.01C with the vapour pressure at 0.00611bar.

This has nothing to do with whether the container has fixed volume. The important parameters are the pressure and the temperature, and whether there is head space present. In the latter case, there is head space above the water/ice mixture, but with no air present. Here are the equilibrium cases of interest (please examine and compare them carefully):

1. Piston applying pressure of 1 atm at 0 C, no air present, no head space present.
2. Piston applying pressure of 0.00611 bar at 0.01C, no air present, head space present, water vapor present in head space at 0.00611 bar.
3. Piston applying pressure of 1 atm at 0 C, head space present, air present in head space, with total pressure of water vapor and air in head space equal to 1 atm.
4. Piston applying pressure between 0.00611 bar and 1 atm, temperature between 0.01 and 0 C, head space present, air present in head space, with total pressure of water vapor and air in head space between 0.00611 bar and 1 atm.

Essentially I see that the water is at two different thermodynamic states. My impression is that adding an ideal gas/air to increase the total pressure to 1atm for the fixed volume case does not alter the melting point and that only when the total pressure is extremely high does the melting point begin to converge to 0C as in the piston case. Is correct?

No. See my previous answer above.

I'm confused by the use of the equation. In my calculation I assumed a piston-cylinder at 50atm and 0C, which from the phase diagram is only liquid-solid (no vapour phase), so I'm confused on how the equilibrium vapor pressure comes in.

Please bear with me. This is only the first step in the calculation. The vapor pressure of ice at 0C will give you the fugacity of water at 0C and that pressure. We then take that number and continue the calculation to get the fugacity of ice and liquid water at 0C and 50 atm.

Chet

 

[Red_CCF]

This has nothing to do with whether the container has fixed volume. The important parameters are the pressure and the temperature, and whether there is head space present. In the latter case, there is head space above the water/ice mixture, but with no air present. Here are the equilibrium cases of interest (please examine and compare them carefully):

1. Piston applying pressure of 1 atm at 0 C, no air present, no head space present.
2. Piston applying pressure of 0.00611 bar at 0.01C, no air present, head space present, water vapor present in head space at 0.00611 bar.
3. Piston applying pressure of 1 atm at 0 C, head space present, air present in head space, with total pressure of water vapor and air in head space equal to 1 atm.
4. Piston applying pressure between 0.00611 bar and 1 atm, temperature between 0.01 and 0 C, head space present, air present in head space, with total pressure of water vapor and air in head space between 0.00611 bar and 1 atm.

Thank you, this summarizes the cases I was confused about perfectly. My original question is basically how and why freezing point vary between Case 1 and 3, since the system in both cases are under 1atm of pressure albeit the way the pressure is applied is different.

Please bear with me. This is only the first step in the calculation. The vapor pressure of ice at 0C will give you the fugacity of water at 0C and that pressure. We then take that number and continue the calculation to get the fugacity of ice and liquid water at 0C and 50 atm.

Chet

By vapour pressure of ice at 0C, are you referring to the 0.00611 bar value at the triple point (0.01C)?

Thank you

 

[Chestermiller]

Thank you, this summarizes the cases I was confused about perfectly. My original question is basically how and why freezing point vary between Case 1 and 3, since the system in both cases are under 1atm of pressure albeit the way the pressure is applied is different.

I'm confused by your response. Does your response mean that you now see why the freezing point is the same for Cases 1 and 3?

By vapour pressure of ice at 0C, are you referring to the 0.00611 bar value at the triple point (0.01C)?

No. I'm referring to the slightly lower vapor pressure of water vapor over ice at -0.357 C.

Chet

 

[Red_CCF]

I'm confused by your response. Does your response mean that you now see why the freezing point is the same for Cases 1 and 3?

Sorry, I am still confused about why there is a difference in melting point between Cases 1 and 3.

No. I'm referring to the slightly lower vapor pressure of water vapor over ice at -0.357 C.

Chet

I used the example here and the vapor pressure is calculated to be 0.00596bar (assuming heat of sublimation and vaporization is close enough).

Thank you

 

[Chestermiller]

Sorry, I am still confused about why there is a difference in melting point between Cases 1 and 3.

Maybe there's something wrong with my vision, but, in what I wrote, the melting points in cases 1 and 3 are both 0C.

I used the example here and the vapor pressure is calculated to be 0.00596bar (assuming heat of sublimation and vaporization is close enough).

You could also have looked this up in a table. Now if we treat all three phases in the system as ideal, this would be the partial pressure of water vapor in the gas phase (also its fugacity) at a total pressure of 50 atm. So, what would the mole fraction of water vapor in the gas phase be?

Now, if we want to get more accurate, we can start looking at the effect of non-idealities on all the phases.

Chet

 

[Red_CCF]

Maybe there's something wrong with my vision, but, in what I wrote, the melting points in cases 1 and 3 are both 0C.

So regardless of whether head space is present, the melting point when there is 1atm of pressure acting on the liquid (head space or not) is 0C, even if we are dealing with ideal gases in the head space?

You could also have looked this up in a table. Now if we treat all three phases in the system as ideal, this would be the partial pressure of water vapor in the gas phase (also its fugacity) at a total pressure of 50 atm. So, what would the mole fraction of water vapor in the gas phase be?

Now, if we want to get more accurate, we can start looking at the effect of non-idealities on all the phases.

Chet

My steam tables actually stop at 0.01C and I haven't been able to track down a water table that goes below it. Using the 0.00596bar value the mole fraction at 50atm should be 0.000118.

Thank you

 

[Chestermiller]

So regardless of whether head space is present, the melting point when there is 1atm of pressure acting on the liquid (head space or not) is 0C, even if we are dealing with ideal gases in the head space?

Yes.

My steam tables actually stop at 0.01C and I haven't been able to track down a water table that goes below it. Using the 0.00596bar value the mole fraction at 50atm should be 0.000118.

Good. Do you want to continue on to include non-idealities, or is this adequate for your present needs?

Chet

 

[Red_CCF]

Yes.

Good. Do you want to continue on to include non-idealities, or is this adequate for your present needs?

Chet

Taking a step back, this whole time I was under the (wrong) impression that for Case 2 (triple point, head space, no air), if I add air (ideal gas) until the total pressure in the head space is 1atm, the equilibrium state will not change; I did not realize that it actually does decrease by 0.01C (same as the piston case). Physically, what is causing the shift in melting point as air is added into the head space? How does the new equilibrium state fit on the phase diagram, which is for a pure system?

Thank you

 

[Chestermiller]

Taking a step back, this whole time I was under the (wrong) impression that for Case 2 (triple point, head space, no air), if I add air (ideal gas) until the total pressure in the head space is 1atm, the equilibrium state will not change; I did not realize that it actually does decrease by 0.01C (same as the piston case). Physically, what is causing the shift in melting point as air is added into the head space?

Well, I can't speak with authority over what is happening on the molecular level. But, when you increase the pressure to 1 atm, you are squeezing the molecules of the liquid and the solid closer together.

How does the new equilibrium state fit on the phase diagram, which is for a pure system?

Well, for the combination of ice and water (assuming no air dissolution), you are moving up the liquid/ice line on the phase diagram (to higher pressures). You already analyzed this.

Chet

 

[Red_CCF]

Well, I can't speak with authority over what is happening on the molecular level. But, when you increase the pressure to 1 atm, you are squeezing the molecules of the liquid and the solid closer together.

How does the squeezing effect reduce the melting point?

Well, for the combination of ice and water (assuming no air dissolution), you are moving up the liquid/ice line on the phase diagram (to higher pressures). You already analyzed this.

Chet

I think I'm reading the phase diagram wrong. Looking at the phase diagram here, if we start at the triple point with no air in the head space, and we add air until the head space pressure is 1atm, the vapour pressure should remain unchanged at 0.00611bar. However, what I'm confused by is how we "move up" in the liquid-ice line as we add pressure via air, since above the triple point (in terms of pressure) we can only have liquid-solid or liquid-vapour but not solid-vapour. I'm having trouble locating the new equilibrium state after air is added and the subsequent freezing point with air in the head space.

Thank you

 

[Chestermiller]

How does the squeezing effect reduce the melting point?

I really don't know how to explain this at the molecular level (which is what is required). The molecules are forced closer together, and this affects the potential energy of interaction.

I think I'm reading the phase diagram wrong. Looking at the phase diagram here, if we start at the triple point with no air in the head space, and we add air until the head space pressure is 1atm, the vapour pressure should remain unchanged at 0.00611bar.

No. The temperature is now lower (for water and ice to remain in equilibrium), so the vapor pressure is lower. You can't see this difference on the phase diagram because the changes are too small.

However, what I'm confused by is how we "move up" in the liquid-ice line as we add pressure via air, since above the triple point (in terms of pressure) we can only have liquid-solid or liquid-vapour but not solid-vapour. I'm having trouble locating the new equilibrium state after air is added and the subsequent freezing point with air in the head space.

If the air does not dissolve, then the combination of air and water vapor in the head space at 1 atm does the same thing as a piston. So the liquid water and ice are compressed slightly, and this changes the melting point to 0C. When you add the second component, what does the phase rule tell you? You now have one more degree of freedom. Without air present, you can have all three phases at equilibrium only at one point, the triple point (zero degrees of freed0m). But the additional component allows you to have all three phases present at equilibrium along an equilibrium line (one degree of freedom), which, if air doesn't dissolve, is the same temperature vs pressure locus as the solid ice liquid water equilibrium line on the pure water phase diagram.

Chet