Questions Regarding the Phase Diagram

[Red_CCF]

Hello

I’m wondering if my understanding of how different processes relate to the phase diagram here is correct.

1) If a piston cylinder assembly maintains a constant 1atm pressure on some water and the water is cooled, we would be going horizontally along the red line from 373.15K to 273.15K at 1atm and the system will only have one phase at any point in the process?

2) If I have a cup of water exposed to air, assuming the air is saturated at all times, cooling the water from say 373.15K means we are following the saturation (black) line separating liquid-vapor until the triple point? What effect does total air pressure play in this?

3) For 2), what if the vapour pressure is not saturated (some sort of venting carrying it away)?

4) Is there a formal definition for what pressure in the phase diagram is? For a single phase closed system I see it as the pressure held by the container, but what about for a multiphase system?

Thank you very much

 

[Chestermiller]

Hello

I’m wondering if my understanding of how different processes relate to the phase diagram here is correct.

1) If a piston cylinder assembly maintains a constant 1atm pressure on some water and the water is cooled, we would be going horizontally along the red line from 373.15K to 273.15K at 1atm and the system will only have one phase at any point in the process?

Yes, liquid.

2) If I have a cup of water exposed to air, assuming the air is saturated at all times, cooling the water from say 373.15K means we are following the saturation (black) line separating liquid-vapor until the triple point? What effect does total air pressure play in this?

Not much.

3) For 2), what if the vapour pressure is not saturated (some sort of venting carrying it away)?

In this case, the system won't be at thermodynamic equilibrium. However, the interface will be very close to thermodynamic equilibirum, and the partial pressure of water vapor in the air at the interface will be equal to the equilibrium vapor pressure at the liquid temperature at the interface.

4) Is there a formal definition for what pressure in the phase diagram is?

Force per unit area?

For a single phase closed system I see it as the pressure held by the container, but what about for a multiphase system?

Thank you very much

Again, force per unit area.

Chet

 

[Red_CCF]

Not much.

What if the air pressure is exaggerated to the order of 100MPa? What I'm trying to get at is, on what order of magnitude does the air pressure have to increase to get to the freezing point in Case 1 (1atm constant applied pressure) or about a 0.01C decrease?

In this case, the system won't be at thermodynamic equilibrium. However, the interface will be very close to thermodynamic equilibirum, and the partial pressure of water vapor in the air at the interface will be equal to the equilibrium vapor pressure at the liquid temperature at the interface.

Had this scenario occurred and I continue to cool past the triple point, do I simply end up with solid and saturated vapour at the solid-vapour interface and the solid will sublimate until it disappears (air or no air wouldn't change a thing)?

Force per unit area?

Again, force per unit area.

Chet

For an adiabatic closed box with solid/liquid and saturated vapour, we always use the vapour pressure as the pressure of the system for calculations. Neglecting gravitational effects, is the vapour pressure equal and evenly distributed within the liquid/solid it is in equilibrium with (my feeling is that it is)?

Thank you very much

 

[Chestermiller]

What if the air pressure is exaggerated to the order of 100MPa? What I'm trying to get at is, on what order of magnitude does the air pressure have to increase to get to the freezing point in Case 1 (1atm constant applied pressure) or about a 0.01C decrease?

Do you know how to determine the change in partial molar free energy (chemical potential) of a component in a gas mixture and the change in free energy of a liquid with changes in pressure and temperature?

Had this scenario occurred and I continue to cool past the triple point, do I simply end up with solid and saturated vapour at the solid-vapour interface and the solid will sublimate until it disappears (air or no air wouldn't change a thing)?

Yes.

For an adiabatic closed box with solid/liquid and saturated vapour, we always use the vapour pressure as the pressure of the system for calculations. Neglecting gravitational effects, is the vapour pressure equal and evenly distributed within the liquid/solid it is in equilibrium with (my feeling is that it is)?

Yes, aside from statistical mechanical fluctuations.

Chet

 

[Red_CCF]

Do you know how to determine the change in partial molar free energy (chemical potential) of a component in a gas mixture and the change in free energy of a liquid with changes in pressure and temperature?

For an ideal gas mixture, this is what I found (I couldn't figure out how to do equations in this new format):

u_i = g_i + RTln(y_i*p/p_ref)

Taking the derivative with respect to T and p while holding one of them constant should give the change in chemical potential with respect to each variable. I'm not sure how to determine changes for liquid, is there some sort of table for this?

Thank you

 

[Chestermiller]

For the liquid, you start with the free energy at the equilibrium vapor pressure of the liquid and temperature T, and you then integrate vdP from the equilibrium vapor pressure to the actual total pressure. This is called the Poynting correction. You then set the chemical potential (or fugacity) of the water in the liquid equal to the chemical potential (or fugacity) of the water vapor in the gas phase. To do high total pressures, of course, you also need to get the pressure correction for the fugacity of the water vapor in the gas.

Here is a nice link on how to do the whole thing in detail: http://web.mit.edu/10.213/www/handouts/vle.pdf

Chet

 

[Red_CCF]

For the liquid, you start with the free energy at the equilibrium vapor pressure of the liquid and temperature T, and you then integrate vdP from the equilibrium vapor pressure to the actual total pressure. This is called the Poynting correction. You then set the chemical potential (or fugacity) of the water in the liquid equal to the chemical potential (or fugacity) of the water vapor in the gas phase. To do high total pressures, of course, you also need to get the pressure correction for the fugacity of the water vapor in the gas.

Here is a nice link on how to do the whole thing in detail: http://web.mit.edu/10.213/www/handouts/vle.pdf

Chet

Hi Chet

Apologies, it look me a little while to look at this.

I'm confused about how the above method can be used to predict the melting temperature change. Assuming that my mixture of liquid and vapour is already at equilibrium at some total pressure (vapour is saturated), if I increase the total pressure without changing temperature, I see changes in the fugacity coefficient, activity coefficient, and Poynting Correction Factor (total pressure term) in the last equation but I can't draw the connection between that and phase change.

Am I correct to say that the effect on melting temperature in this case is a result of real gas effects only (due to the use of fugacity) and theoretically, if these effects did not exist (at any total pressure), melting point is independent of total pressure?

I am also wondering why the total pressure change must be several orders of magnitude higher to achieve the same melting point as that by a much smaller applied pressure (Case 1, horizontal red line of constant applied pressure of 1atm in the diagram in my OP), as the difference is essentially the presence of vapour phase.

Thank you very much

 

[Chestermiller]

I assume you are comfortable with the idea that, at melting equilibrium, the free energy per unit mass of liquid must equal the free energy per unit mass of the solid. So if the temperature changes, the pressure has to change in such a way that the difference in free energy between the liquid and solid must remain zero. Otherwise, all the solid will melt to form liquid, or vice versa. So, starting out from melting equilibrium at a specific melting temperature and pressure, the changes in free energy of the solid and liquid are:
$$dg_s=-s_sdT+v_sdP$$
$$dg_l=-s_ldT+v_ldP$$

So,
$$d(g_l-dg_s)=-(s_l-s_s)dT+(v_l-v_s)dP=0$$
This means that, in order to maintain melting equilibrium,
$$\frac{dP}{dT}=\frac{(s_l-s_s)}{(v_l-v_s)}$$
But, we also know that
$$(s_l-s_s)=\frac{h_l-h_s}{T}$$
where $h_l-h_s$ is the heat of fusion. If we substitute, we then get
$$\frac{dP}{dT}=\frac{(h_l-h_s)}{T(v_l-v_s)}$$
Now, we know that the specific volumes of the liquid and solid are very small, and the difference is ultra small. So the denominator on the right hand side is going to be very small, and dP/dT is going to be very large.

Chet

 

[Red_CCF]

I assume you are comfortable with the idea that, at melting equilibrium, the free energy per unit mass of liquid must equal the free energy per unit mass of the solid. So if the temperature changes, the pressure has to change in such a way that the difference in free energy between the liquid and solid must remain zero. Otherwise, all the solid will melt to form liquid, or vice versa. So, starting out from melting equilibrium at a specific melting temperature and pressure, the changes in free energy of the solid and liquid are:
$$dg_s=-s_sdT+v_sdP$$
$$dg_l=-s_ldT+v_ldP$$

So,
$$d(g_l-dg_s)=-(s_l-s_s)dT+(v_l-v_s)dP=0$$
This means that, in order to maintain melting equilibrium,
$$\frac{dP}{dT}=\frac{(s_l-s_s)}{(v_l-v_s)}$$
But, we also know that
$$(s_l-s_s)=\frac{h_l-h_s}{T}$$
where $h_l-h_s$ is the heat of fusion. If we substitute, we then get
$$\frac{dP}{dT}=\frac{(h_l-h_s)}{T(v_l-v_s)}$$
Now, we know that the specific volumes of the liquid and solid are very small, and the difference is ultra small. So the denominator on the right hand side is going to be very small, and dP/dT is going to be very large.

Chet

Is the P used here the vapour pressure or the total pressure?

Thank you

 

[Chestermiller]

Is the P used here the vapour pressure or the total pressure?

Thank you

This is a single component system, so the only place where you can have 3 phases present at equilibrium is at the triple point. Check this out using the phase rule: if the number of phases is 3 and the number of components is 1, there are zero degrees of freedom. This is also shown on your phase diagram. So, if you have solid and liquid present at equilibrium, there must be no vapor present (except at the triple point). So, in this equation P is the total pressure.

If you have three phases present in a closed container and set the temperature to any value other than the triple point temperature, when the system equilibrates at that temperature, there will be only one or two phases remaining.

Chet

 

[Red_CCF]

This is a single component system, so the only place where you can have 3 phases present at equilibrium is at the triple point. Check this out using the phase rule: if the number of phases is 3 and the number of components is 1, there are zero degrees of freedom. This is also shown on your phase diagram. So, if you have solid and liquid present at equilibrium, there must be no vapor present (except at the triple point). So, in this equation P is the total pressure.

If you have three phases present in a closed container and set the temperature to any value other than the triple point temperature, when the system equilibrates at that temperature, there will be only one or two phases remaining.

Chet

So in the above derivation we are essentially looking at something like a closed piston-cylinder assembly where we can apply pressure differently?

How does the analysis change if we have a closed, fixed volume container with a cup of water inside, and total pressure is varied by changing the air pressure?

Thank you

 

[Chestermiller]

So in the above derivation we are essentially looking at something like a closed piston-cylinder assembly where we can apply pressure differently?

How does the analysis change if we have a closed, fixed volume container with a cup of water inside, and total pressure is varied by changing the air pressure?

Thank you

Let me restate your question in a more precise way. You have W kg of water and A kg of air in a cylinder (no need for the cup) initially in equilibrium at a pressure of 1 atm pressure and 20 C. You can control the pressure in the cylinder using a piston and you can control the temperature in the cylinder using a constant temperature bath. You change the pressure to P and the temperature to T. What is the nature of the contents of the container when the system re-equilibrates? That is, how is the W kg of water distributed between ice, liquid water, and water vapor?

Is this a satisfactory statement of what you wish to analyze? (No need to hold the volume constant. You can answer what you want more easily by solving the problem as I have stated it.)

Chet

 

[Red_CCF]

Let me restate your question in a more precise way. You have W kg of water and A kg of air in a cylinder (no need for the cup) initially in equilibrium at a pressure of 1 atm pressure and 20 C. You can control the pressure in the cylinder using a piston and you can control the temperature in the cylinder using a constant temperature bath. You change the pressure to P and the temperature to T. What is the nature of the contents of the container when the system re-equilibrates? That is, how is the W kg of water distributed between ice, liquid water, and water vapor?

Is this a satisfactory statement of what you wish to analyze? (No need to hold the volume constant. You can answer what you want more easily by solving the problem as I have stated it.)

Chet

That is essentially what I'm looking for. However, in this setup, we cannot independently control the vapour pressure since the piston will raise the partial pressure of the vapour by the same proportion as the air pressure. Is there another possible setup where air pressure and vapour pressure are controlled independently (or a better question may be, does this even matter)?

Thank you

 

[Chestermiller]

That is essentially what I'm looking for. However, in this setup, we cannot independently control the vapour pressure since the piston will raise the partial pressure of the vapour by the same proportion as the air pressure.

This won't happen unless there is no liquid or solid present.

Is there another possible setup where air pressure and vapour pressure are controlled independently (or a better question may be, does this even matter)?

Thank you

It doesn't matter. So let's get started.

We are going to assume that the air is insoluable in liquid water and ice. As an initial state, the temperature is going to be 20 C, the pressure is going to be 1 atm, and the piston is situated so that the volume of material in the cylinder is 1 liter. The bottom half of the cylinder is filled with liquid water, and the top half of the cylinder is filled with a mixture of air and water vapor, with water vapor at its equilibrium vapor pressure at 20 C. Determine:

• the mass of liquid water in the bottom half
• the partial pressure of water vapor in the top half
• the partial pressure of air in the top half
• the mass of water vapor in the top half
• the total mass of water in the cylinder
• the total mass of air in the cylinder

We are going to change the temperature and/or the pressure to new values. I want you to specify what these new values should be. I suggest that you begin by choosing conditions that are easy to get answers for, so that we can get some experience. Then, we can start to move to more complicated combinations of temperature and pressure. So please, specify the first combination of pressure and temperature that you want to move to.

Chet

 

[Red_CCF]

This won't happen unless there is no liquid or solid present.

It doesn't matter. So let's get started.

We are going to assume that the air is insoluable in liquid water and ice. As an initial state, the temperature is going to be 20 C, the pressure is going to be 1 atm, and the piston is situated so that the volume of material in the cylinder is 1 liter. The bottom half of the cylinder is filled with liquid water, and the top half of the cylinder is filled with a mixture of air and water vapor, with water vapor at its equilibrium vapor pressure at 20 C. Determine:

• the mass of liquid water in the bottom half
• the partial pressure of water vapor in the top half
• the partial pressure of air in the top half
• the mass of water vapor in the top half
• the total mass of water in the cylinder
• the total mass of air in the cylinder

We are going to change the temperature and/or the pressure to new values. I want you to specify what these new values should be. I suggest that you begin by choosing conditions that are easy to get answers for, so that we can get some experience. Then, we can start to move to more complicated combinations of temperature and pressure. So please, specify the first combination of pressure and temperature that you want to move to.

Chet

• the mass of liquid water in the bottom half: 0.499kg (assuming it is 50% by volume)
• the partial pressure of water vapor in the top half: 2.339 kPa
• the partial pressure of air in the top half: 98.96 kPa (assuming 1atm = 101.3kPa)
• the mass of water vapor in the top half: 8.65mg
• the total mass of water in the cylinder: 0.499kg (neglecting mass of vapour)
• the total mass of air in the cylinder: 0.588g

I think a easy new condition would be 0C, 1atm (same pressure).

Thank you

 

[Chestermiller]

• the mass of liquid water in the bottom half: 0.499kg (assuming it is 50% by volume)
• the partial pressure of water vapor in the top half: 2.339 kPa
• the partial pressure of air in the top half: 98.96 kPa (assuming 1atm = 101.3kPa)
• the mass of water vapor in the top half: 8.65mg
• the total mass of water in the cylinder: 0.499kg (neglecting mass of vapour)
• the total mass of air in the cylinder: 0.588g

I think a easy new condition would be 0C, 1atm (same pressure).

Thank you

Nice job.

OK. Now 0 C is slightly below the triple point of water, 0.01 C, so you will have no liquid water in the container. Let's see what you can do with this case. Please do the calculation for the same amounts of water and air in the container, and, of course, also determine the new volume. What is the specific volume of ice under these conditions?

Chet

 

[Red_CCF]

Nice job.

OK. Now 0 C is slightly below the triple point of water, 0.01 C, so you will have no liquid water in the container. Let's see what you can do with this case. Please do the calculation for the same amounts of water and air in the container, and, of course, also determine the new volume. What is the specific volume of ice under these conditions?

Chet

Actually I will do 0.01C (right at the triple point) since that's the lowest temperature my book's steam table goes to. I'm going to assume that we only have liquid and vapour here (I don't have the properties for solid but I think density at least is close to the liquid)

The new vapour/air volume is: V = mRT/P = 0.000588*0.287*273.25/100.64 = 0.458L

• the mass of liquid water in the bottom half: 0.499kg (negligble amount of vapour condensed)
• the partial pressure of water vapor in the top half: 0.661 kPa
• the partial pressure of air in the top half: 100.64 kPa (assuming piston moves to maintain 1atm)
• the mass of water vapor in the top half: 2.22mg
• the total mass of water in the cylinder: 0.499kg (neglecting mass of vapour)
• the total mass of air in the cylinder: 0.588g

 

[Chestermiller]

Actually I will do 0.01C (right at the triple point) since that's the lowest temperature my book's steam table goes to. I'm going to assume that we only have liquid and vapour here (I don't have the properties for solid but I think density at least is close to the liquid)

The new vapour/air volume is: V = mRT/P = 0.000588*0.287*273.25/100.64 = 0.458L

• the mass of liquid water in the bottom half: 0.499kg (negligble amount of vapour condensed)
• the partial pressure of water vapor in the top half: 0.661 kPa
• the partial pressure of air in the top half: 100.64 kPa (assuming piston moves to maintain 1atm)
• the mass of water vapor in the top half: 2.22mg
• the total mass of water in the cylinder: 0.499kg (neglecting mass of vapour)
• the total mass of air in the cylinder: 0.588g

OK. What temperature and pressure do you want to try next?

 

[Red_CCF]

OK. What temperature and pressure do you want to try next?

I was thinking 90C

The new vapour/air volume is: V = mRT/P = 0.000588*0.287*363.15/31.36 = 1.95L

• the mass of liquid water in the bottom half: 0.499kg
• the partial pressure of water vapor in the top half: 70.14kPa
• the partial pressure of air in the top half: 31.16kPa
• the mass of water vapor in the top half: 0.826g
• the total mass of water in the cylinder: 0.499kg (neglecting mass of vapour)
• the total mass of air in the cylinder: 0.588g

 

[Chestermiller]

Actually, the mass of liquid water in the cylinder would now be only 0.498 kg. What do you want to try next?

Chet

 

[Red_CCF]

I was thinking 90C

The new vapour/air volume is: V = mRT/P = 0.000588*0.287*363.15/31.36 = 1.95L

• the mass of liquid water in the bottom half: 0.499kg
• the partial pressure of water vapor in the top half: 70.14kPa
• the partial pressure of air in the top half: 31.16kPa
• the mass of water vapor in the top half: 0.826g
• the total mass of water in the cylinder: 0.499kg (neglecting mass of vapour)
• the total mass of air in the cylinder: 0.588g

I think now I will vary the pressure to say 10atm (i think we are still in the ideal gas range?) at 90C. New volume is 0.065L

• the mass of liquid water in the bottom half: 0.499kg
• the partial pressure of water vapor in the top half: 70.14kPa
• the partial pressure of air in the top half: 942.86kPa
• the mass of water vapor in the top half: 27.5mg
• the total mass of water in the cylinder: 0.499kg
• the total mass of air in the cylinder: 0.588g

 

[Chestermiller]

I think now I will vary the pressure to say 10atm (i think we are still in the ideal gas range?) at 90C. New volume is 0.065L

• the mass of liquid water in the bottom half: 0.499kg
• the partial pressure of water vapor in the top half: 70.14kPa
• the partial pressure of air in the top half: 942.86kPa
• the mass of water vapor in the top half: 27.5mg
• the total mass of water in the cylinder: 0.499kg
• the total mass of air in the cylinder: 0.588g

Okay. What next?

 

[Red_CCF]

Okay. What next?

I think I will increase pressure to 50atm:

• the mass of liquid water in the bottom half: 0.499kg
• the partial pressure of water vapor in the top half: 70.14kPa
• the partial pressure of air in the top half: 4994.86kPa
• the mass of water vapor in the top half: 5.2mg
• the total mass of water in the cylinder: 0.499kg
• the total mass of air in the cylinder: 0.588g

I'm thinking about increasing the pressure again, at what point does the ideal gas assumption not hold? Is it possible to lower the triple point to 0C via increasing the pressure?

 

[Chestermiller]

I think I will increase pressure to 50atm:

• the mass of liquid water in the bottom half: 0.499kg
• the partial pressure of water vapor in the top half: 70.14kPa
• the partial pressure of air in the top half: 4994.86kPa
• the mass of water vapor in the top half: 5.2mg
• the total mass of water in the cylinder: 0.499kg
• the total mass of air in the cylinder: 0.588g

I'm thinking about increasing the pressure again, at what point does the ideal gas assumption not hold?

50 atm is already beyond ideal gas assumption accuracy.

Is it possible to lower the triple point to 0C via increasing the pressure?

In order for ice and liquid water to be in equilibrium, the fugacity of the ice must be equal to the fugacity of the liquid water. Since we are assuming that no air dissolves in either the liquid water or the ice, the fugacity of each must be the same as for pure ice and pure liquid water at temperature T and pressure P. Therefore, under our assumption, the only T-P path under which they can be in equilibrium is the path on the pure water phase diagram at the ice/liquid water boundary. Is there a pressure at which the fugacity of ice is equal to the fugacity of liquid water at 0 C?

Chet

 

[Red_CCF]

50 atm is already beyond ideal gas assumption accuracy.

In order for ice and liquid water to be in equilibrium, the fugacity of the ice must be equal to the fugacity of the liquid water. Since we are assuming that no air dissolves in either the liquid water or the ice, the fugacity of each must be the same as for pure ice and pure liquid water at temperature T and pressure P. Therefore, under our assumption, the only T-P path under which they can be in equilibrium is the path on the pure water phase diagram at the ice/liquid water boundary. Is there a pressure at which the fugacity of ice is equal to the fugacity of liquid water at 0 C?

Chet

In the above examples I assumed that air was an ideal gas and that changing the air pressure did not affect the phase equilibrium (if our volume was constant adding air wouldn't have changed anything). Base on the link you recommended, when real gas effects are considered, increasing the total pressure would increase the Poynting correction factor; I'm not sure how this affects the activity and fugacity coefficients but I read that increasing total pressure lowers the melting temperature, which implies that at 0.01C, with sufficiently high total pressure, the fugacity of liquid-solid (that otherwise would be equal) now shifts such that the liquid has a lower fugacity coefficient than of solid. Is this correct?

For ideal gases, I'm wondering why the assumption made in the previous examples are true. Physically, why does adding say 1atm of ideal gas not affect the equilibrium, since the liquid is being pounded with a much higher force than when it was at equilibrium with its vapour whose saturation pressure is a fraction of the total pressure?

Thank you

 

[Chestermiller]

Let's do the case of having the system at 1 atm. total pressure at 0.01C with and without air present in the head space.

Without air present in the head space, we are dealing with the exact situation shown in your phase diagram. There is no vapor phase, and only ice and liquid water are present at equilibrium. There can be no head space. For the pure water situation, the only point at which you can have all 3 phases present at equilibrium is at the triple point.

Now for the case in which air is present in the head space. If the air cannot dissolve in the liquid and ice (our assumption), then as far as the liquid water and ice are concerned, it is the same as if there were no gas phase present. They don't know what is causing the total pressure to be 1 atm. However, a head space can now exist, and the gas phase can now have water vapor in it. Let's temporarily treat the gas phase as an ideal gas. What is the equilibrium vapor pressure of water at 0.01C? This will be the partial pressure of the water vapor in the gas phase when air is present and the total pressure is 1 atm. From this you can calculate the partial pressure of the air in the gas phase, and the mole fractions of air and water vapor in the gas phase. What values do you get?

Introducing the air into the container has given you one more degree of freedom for your system. There is now not just one combination of temperature and pressure in which you can have all three phases present at equilibrium (i.e., the triple point). You now have the ability to vary the temperature and pressure (along the liquid/ice equilibrium line) and still have all 3 phases present.

Including non-ideal gas effects at higher pressures is do-able, but is, of course, more complicated. Going to much higher pressures does not affect the liquid and solid phases (assuming air does not dissolve), but does affect how we handle the gas phase and the equilibrium calculation.

Questions?? Where do you want to go from here?

Chet

 

[Red_CCF]

Now for the case in which air is present in the head space. If the air cannot dissolve in the liquid and ice (our assumption), then as far as the liquid water and ice are concerned, it is the same as if there were no gas phase present. They don't know what is causing the total pressure to be 1 atm. However, a head space can now exist, and the gas phase can now have water vapor in it. Let's temporarily treat the gas phase as an ideal gas. What is the equilibrium vapor pressure of water at 0.01C? This will be the partial pressure of the water vapor in the gas phase when air is present and the total pressure is 1 atm. From this you can calculate the partial pressure of the air in the gas phase, and the mole fractions of air and water vapor in the gas phase. What values do you get?

I can't get my head around why ideal gas has no effect on equilibrium state or in a sense, hold the water down. The pressure on the liquid (and in the head space) applied by the air at 1atm has a drastically different effect than 1atm applied by a piston even though the pressure felt by the liquid surface is the same.

For water at 0.01C the partial vapor pressure is 0.611kPa so the partial pressure of air is 100.6kPa. Mole fractions are 0.00603 and 0.993 for water and air respectively.

Introducing the air into the container has given you one more degree of freedom for your system. There is now not just one combination of temperature and pressure in which you can have all three phases present at equilibrium (i.e., the triple point). You now have the ability to vary the temperature and pressure (along the liquid/ice equilibrium line) and still have all 3 phases present.

Including non-ideal gas effects at higher pressures is do-able, but is, of course, more complicated. Going to much higher pressures does not affect the liquid and solid phases (assuming air does not dissolve), but does affect how we handle the gas phase and the equilibrium calculation.

Questions?? Where do you want to go from here?

Chet

So adding air will essentially change the phase diagram, as opposed to just shifting it, such that for different air pressure we have a different phase diagram? I'm basically wondering, if we have a closed box at the triple point (0.01C), how would one find out how much air to add into melt the water (essentially shifting the melting point downwards).

Thank you

 

[Chestermiller]

I can't get my head around why ideal gas has no effect on equilibrium state or in a sense, hold the water down. The pressure on the liquid (and in the head space) applied by the air at 1atm has a drastically different effect than 1atm applied by a piston even though the pressure felt by the liquid surface is the same.

This is quite a bold statement. What makes you feel this way? How does the ice/liquid water mixture know what is causing the pressure on its surface?

For water at 0.01C the partial vapor pressure is 0.611kPa so the partial pressure of air is 100.6kPa. Mole fractions are 0.00603 and 0.993 for water and air respectively.

Yes. that's about it. Problem solved.

So adding air will essentially change the phase diagram, as opposed to just shifting it, such that for different air pressure we have a different phase diagram?

For the ice/liquid water mixture, varying the pressure just moves you to a different point on the phase diagram. The pressure-temperature relationship for the ice/liquid water mixture does not change (assuming negligible air dissolves). But, if you don't vary the temperature and pressure along the pure water equilibrium line so that the temperature corresponds to the new total pressure, one of the phases will disappear.

I'm basically wondering, if we have a closed box at the triple point (0.01C), how would one find out how much air to add into melt the water (essentially shifting the melting point downwards).

Do you want to solve this specific problem? Tell me what specific combination of temperature and air partial pressure you want to look at?

Chet

 

[Red_CCF]

This is quite a bold statement. What makes you feel this way? How does the ice/liquid water mixture know what is causing the pressure on its surface?

I phrased that poorly, it was meant as a question not a statement. I'm was thinking, from the liquid's perspective, if I have air at 1atm, or a piston applying a pressure at 1atm, the liquid "feels" the same force, but the equilibrium state is very different. I get that the liquid has no idea what's causing the force and that the air cannot hold the liquid down like the piston can, but how come the same force has two different effects on equilibrium and cause the liquid itself to be at two different states despite being under the same pressure and at the same temperature?

Yes. that's about it. Problem solved.For the ice/liquid water mixture, varying the pressure just moves you to a different point on the phase diagram. The pressure-temperature relationship for the ice/liquid water mixture does not change (assuming negligible air dissolves). But, if you don't vary the temperature and pressure along the pure water equilibrium line so that the temperature corresponds to the new total pressure, one of the phases will disappear.


Do you want to solve this specific problem? Tell me what specific combination of temperature and air partial pressure you want to look at?

Chet

I'm not too sure if there is a relatively simple way to do this, but if I start at the triple point equilibrium in a fixed volume container (no air), is there a way to see how changing the total pressure (maybe going from air partial pressure = 0 to some quantity like 50atm) "shifts" the equilibrium position on the phase diagram; I just want to get a qualitative feel about how pressure affects equilibrium position?

Thank you

 

[Chestermiller]

I phrased that poorly, it was meant as a question not a statement. I'm was thinking, from the liquid's perspective, if I have air at 1atm, or a piston applying a pressure at 1atm, the liquid "feels" the same force, but the equilibrium state is very different. I get that the liquid has no idea what's causing the force and that the air cannot hold the liquid down like the piston can, but how come the same force has two different effects on equilibrium and cause the liquid itself to be at two different states despite being under the same pressure and at the same temperature?

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.

I'm not too sure if there is a relatively simple way to do this, but if I start at the triple point equilibrium in a fixed volume container (no air), is there a way to see how changing the total pressure (maybe going from air partial pressure = 0 to some quantity like 50atm) "shifts" the equilibrium position on the phase diagram; I just want to get a qualitative feel about how pressure affects equilibrium position?

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

 

[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