# Questions Regarding the Phase Diagram

1. Sep 26, 2014

### 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

2. Sep 26, 2014

### Staff: Mentor

Yes, liquid.
Not much.
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.
Force per unit area?
Again, force per unit area.

Chet

3. Sep 26, 2014

### Red_CCF

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?

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)?

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

4. Sep 27, 2014

### Staff: Mentor

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?

Yes.

Yes, aside from statistical mechanical fluctuations.

Chet

5. Sep 28, 2014

### Red_CCF

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

6. Sep 28, 2014

### Staff: Mentor

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

Last edited: Sep 28, 2014
7. Oct 4, 2014

### Red_CCF

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

8. Oct 5, 2014

### Staff: Mentor

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

Last edited: Oct 5, 2014
9. Oct 5, 2014

### Red_CCF

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

Thank you

10. Oct 5, 2014

### Staff: Mentor

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

11. Oct 5, 2014

### Red_CCF

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

12. Oct 6, 2014

### Staff: Mentor

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

13. Oct 6, 2014

### Red_CCF

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

14. Oct 6, 2014

### Staff: Mentor

This won't happen unless there is no liquid or solid present.
It doesn't matter. So lets 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

15. Oct 7, 2014

### Red_CCF

• 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

16. Oct 8, 2014

### Staff: Mentor

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

17. Oct 9, 2014

### Red_CCF

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

18. Oct 9, 2014

### Staff: Mentor

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

19. Oct 10, 2014

### 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

20. Oct 10, 2014

### Staff: Mentor

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

Chet