Questions Regarding the Phase Diagram

[Red_CCF]

The water and the ice have the same state whether the piston is pushing down or whether the air is pushing down. The only difference is that some of the ice and liquid water can evaporate into the gas phase at the equilibrium vapor pressure of liquid water and ice at 0 C and 1 atm. if air is present. You already showed that with your calculation.

With regards to the first statement, I'm confused on what is meant by the same state. In the phase diagram for 1atm of pressure applied by the piston the freezing point is 0C (exactly) but for the case of cooling in an isolated fixed volume box with 1atm of air, the triple point/freezing point is 0.01C, so at least phase transition occurs at a different state.

Yes. The first step is to determine what the equilibrium temperature of a mixture of water and ice is (no air present) at a total pressure of 50 atm. Do you know how to do that?

Chet

I had thought some sort of table or data base existed that gives me the freezing point w.r.t pressure like the steam table but wasn't able to find one.

Thank you

 

[Chestermiller]

With regards to the first statement, I'm confused on what is meant by the same state. In the phase diagram for 1atm of pressure applied by the piston the freezing point is 0C (exactly) but for the case of cooling in an isolated fixed volume box with 1atm of air, the triple point/freezing point is 0.01C, so at least phase transition occurs at a different state.

This is not correct. The pressure at the triple point is way lower than 1 atm, and there is no air present. If air were present at a very low partial pressure << 1 atm, the equilibrium would be closer to 0.01 C than to 0 C. As you increased the total pressure, the equilibrium temperature would shift to 0 C.

I had thought some sort of table or data base existed that gives me the freezing point w.r.t pressure like the steam table but wasn't able to find one.

Do you know were the numbers in these tables come from? Do you remember the Clapeyron equation?

Chet

 

[Red_CCF]

This is not correct. The pressure at the triple point is way lower than 1 atm, and there is no air present. If air were present at a very low partial pressure << 1 atm, the equilibrium would be closer to 0.01 C than to 0 C. As you increased the total pressure, the equilibrium temperature would shift to 0 C.

Sorry, I was thinking of a different case (1atm pressure applied from by a piston); in that case it would be closer to 0C than 0.01C? Another way of asking my original question would be, why do we need a significantly higher total (air) pressure/force than an applied piston pressure to obtain the same melting temperature (0C)?

Do you know were the numbers in these tables come from? Do you remember the Clapeyron equation?

Chet

I actually never learned the Clapeyron Equation so I am likely doing something wrong. From this example on Wikipedia, assuming that 50atm is applied from a piston (no vapor phase), if starting from 1atm 0C (equilibrium point), using their numbers I get that at 50atm of piston pressure a -0.367C drop in melting point. I'm not sure how I would apply this equation when vapour is present and pressure is increased by adding more air to a fixed volume.

Thank you

 

[Chestermiller]

Sorry, I was thinking of a different case (1atm pressure applied from by a piston); in that case it would be closer to 0C than 0.01C?

Yes.

Another way of asking my original question would be, why do we need a significantly higher total (air) pressure/force than an applied piston pressure to obtain the same melting temperature (0C)?

I don't understand why you are saying this. Whether it is a piston applying the pressure or it is air applying the pressure, the total pressure for a melting temperature of 0C is 1 atm.

I actually never learned the Clapeyron Equation so I am likely doing something wrong. From this example on Wikipedia, assuming that 50atm is applied from a piston (no vapor phase), if starting from 1atm 0C (equilibrium point), using their numbers I get that at 50atm of piston pressure a -0.367C drop in melting point.

I'm assuming you did this calculation correctly. I'm guessing that you assumed a constant value for the heat of melting, and constant values for the specific volumes of ice and liquid water, correct? That's what I would have done.

I'm not sure how I would apply this equation when vapour is present and pressure is increased by adding more air to a fixed volume.

Let's start out by assuming that, even at 50 atm, everything is ideal for the gas phase. What is the equilibrium vapor pressure of ice at -0.367 C?

Chet

 

[Red_CCF]

I don't understand why you are saying this. Whether it is a piston applying the pressure or it is air applying the pressure, the total pressure for a melting temperature of 0C is 1 atm.

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. 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?

I'm assuming you did this calculation correctly. I'm guessing that you assumed a constant value for the heat of melting, and constant values for the specific volumes of ice and liquid water, correct? That's what I would have done.

Let's start out by assuming that, even at 50 atm, everything is ideal for the gas phase. What is the equilibrium vapor pressure of ice at -0.367 C?

Chet

I took the heat of melting and specific volumes as given in the Wikipedia page (for 0C) as constant for small temperature change.

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.

Thank you

 

[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