# Vapor Pressure Curve on the PVT Diagram

1. Nov 5, 2013

### LunaFly

Hi all,

I am wondering about where the vapor pressure curve would be located on the PVT surface of a substance.

I know that as the temperature of a liquid increases, its vapor pressure increases. This continues until the vapor pressure is equal to the atmospheric pressure, at which point the substance begins to boil and goes through a phase transition.

What I am wondering is what kind of path does the vapor curve make on the PVT surface? I am using a closed flask being heated as an example. I would think that it would begin somewhere in the liquid region of the surface. Then as the temperature increases, the volume would decrease slightly as the water evaporates and the vapor pressure increases. Once the liquid reaches boiling point and the liquid-vapor region of the diagram, the volume would immediately increase (as vapor is now included as a part of the volume) then remain constant (as the size of the flask is constant). During this time the pressure would continue to increase until all of the liquid had turned into vapor.

I am very unsure if this is really how the vapor pressure curve would appear on the PVT surface or not. I am also curious as to how it would change if the flask was a different size but with the same amount of liquid initially. Any input would be much appreciated! Thanks.

-Luna

2. Nov 6, 2013

### Staff: Mentor

The vapor pressure curve doesn't stop at the boiling point. It goes all the way up to the critical point. It tells you the temperature and pressure of pure water within a closed container in which there is both water vapor and liquid present at equilibrium with one another. In your experiment, even if there is air in your closed vessel, the water won't boil. The pressure will just keep getting higher as you increase the temperature; from one temperature to the next, some liquid evaporates and forms more vapor at a higher pressure, and the pressure of the air also increases as it gets hotter.

Chet

3. Nov 7, 2013

### LunaFly

If the water in the flask could never boil, then what would be happening on the PVT surface? It would remain in the liquid phase region until the critical point is reached? Also, what would the volume be doing during this process? If the liquid never reaches boiling point it would seem that the volume would continue to decrease as the liquid evaporates, and if we are still in the liquid region of the PVT diagram then the overall volume would continue to decrease as well.

I am also not clear on why the water would not boil in the first place. Is it because it is not exposed to the atmosphere (i.e. vapor pressure could never overcome atmospheric pressure because atmospheric pressure isn't even present?), or is there a different reason?

Last question: if the vapor pressure curve continues past the boiling point, then past that point there is no liquid left and the data is just relating the temperature of the gas with the pressure of the gas, correct? Thanks for the input.

4. Nov 7, 2013

### Staff: Mentor

The easiest way to answer all these questions is to solve a specific problem. Are you willing to participate? Here is the problem: You have a 2 liter container containing 1 liter of liquid water and 1 liter of water vapor in equilibrium with the liquid water. We are going to assume that the density of the liquid water remains constant at 1 gm/cc, and that the vapor above the water behaves like an ideal gas. The container is rigid, and its volume can't change. Hydrostatic variations is pressure are to be assumed negligible. The initial condition of the container contents is 20 C. We are going to do calculations in which we gradually increase the temperature, and determine the pressure, the amount of liquid, and the amount of vapor as a function of the temperature (with the contents at equilibrium at each temperature). Temperatures above 100 C will be included in latter parts of the problem. Part 1 of the problem is to determine at 20 C the pressure in the container (mm Hg), the mass of liquid (gm) water in the container, the mass of water vapor (gm) in the container, and the total mass of water in the container. Also, how does the mass of water within the container change when we increase the temperature?

Chet

5. Nov 8, 2013

### LunaFly

Ok, I think that's a good idea.

So far, I have the initial pressure of the vapor at 17.5 mmHg (found from a vapor pressure table).

The mass of the liquid water is V*ρ= 1000g.

Using the ideal gas law, I found n, the number of moles of vapor, to be 9.58*10-4 moles, from which I found the mass by multiplying n*molar mass of H2O (18 g/mol) to get 0.017 g of water vapor.

From here the total mass is a simple addition resulting in 1000.017g of water present. Because we have a closed system, the mass is conserved throughout the processes.

So far, so good?

6. Nov 8, 2013

### Staff: Mentor

Excellent job!!!! Now let's heat it up to 60 C. When the system equilibrates at 60C, there will no longer be 1000 g of liquid present. Some of the liquid will have evaporated and entered the gas phase. There will also no longer be 1000 cc of liquid present. At 60 C, what is the new pressure of the system? What will be the new (a) grams of liquid water, (b) grams of water vapor, (c) volume of liquid water, and (d) volume of water vapor? To get you started with this, let G be the new number of grams of liquid water. Solve for G.

Chet

7. Nov 10, 2013

### LunaFly

Now the system is at 60C = 333K. The mass of the liquid water can be found by solving:

1.000017 kg = G + n*M, where M is the molar mass of water, and n is the number of moles of vapor.

Using ideal gas law, n can be substituted with PV/(RT), in which V can be substituted with
V= (0.002m3 - G/ρ).

The expression becomes 1.000017 = G + [MP(0.002 - G/ρ)]/(RT).

Solving for G yields:

G= ρ(1.000017RT - 0.002MP)/(ρRT - MP)

Substituting in values of:

ρ= 1000 kg/m3
R= 8.31 J/(mol*K)
M= 0.018 kg/mol
P= 149.4 mmHg= 19,918 Pa
T= 60C= 333K

results in the mass of liquid water G= 0.9989 kg= 999.89 g.

The mass of the vapor (due to conservation of mass) is 1000.017 g - 999.89 g = 0.127 g of vapor.

The volume of the liquid is G/ρ= 0.99989 L.

The volume of the vapor is 2L - 0.999898L =1.00011 L

These answers seem reasonable. The temperature of the system increased, and so did the vapor pressure. Higher vapor pressure means more molecules in the vapor phase. The volume and mass of the liquid has slightly decreased, and the volume and mass of the vapor has increased.

8. Nov 10, 2013

### Staff: Mentor

OK. Nice work.

Next let's move up to 100 C, water's atmospheric boiling point. Does anything unusual happen there? We'll see that the answer is no.

We're going to do things a little differently this time. Instead of G representing the number of grams of liquid water, we are going to let G represent the number of grams of water vapor. Why are we making this change? Not for any particular physical reason. However, we will find that it will reduce the numerical roundoff error in the calculation.

We are also going to do the calculation two different ways, and compare the results:

Method 1: The same way we have been doing it up to now.
Method 2: Not assuming that the density of the liquid water stays constant, and not assuming that the water vapor behaves as an ideal gas. Look up the density of saturated liquid water and saturated water vapor at 100 C. This should be in the steam tables, or in some other set of tables that you can lay your hands on. Use these values in your calculations.

I think that the comparison of results between methods 1 and 2 will be pretty interesting, and method 2 will give you more practice in doing calculations of this type.

Chet

9. Aug 5, 2015

### chutonio

Dear Lunafly and Chestermiller,
first of all thank you for working on this topic. Phase diagrams teaching is always too far from real condition to be efficiently and safely applied to real world situations.
I hope you are already there and could help me with a development of this topic.
I reached to this post because I was trying to figure out how (and why) pressure builds up while heating a liquid in a sealed tank, with or without a second inert component (air). I am a chemist, and this affect very much the way one should properly use an autoclave to perform hydrothermal syntheses.
In such a process one typically fills with water (and reactants) a teflon lined reactor, seals everything up and starts heating up to a desired temperature.
Let´s forget about reactants and focus on the physics though

The teflon tube is a fixed volume reactor, in which are present a) a portion of water and b) a portion of gas (air + water vapour fraction).
It turns out experimentally that the pressure, that builds up upon heating, is linear with temperature. The steepness of this line is a function of the degree of filling of the reactor (ratio liquid/gas portions in the reactor). The higher the filling the higher the steepness and the pressure build-up. In practical cases one should avoid filling the reactor with more than 70%vol of liquid, otherwise a tremendous pressure will quickly build up with consequent rupture of the reactor.

I was wondering why this happens.
Is it because the liquid also expands considerably with temperature and constrains the expanding headspace gas containing air?
[At low filling the volume eaten up by the expanding gas is relatively small and expanding air is only building up pressure as it was alone (PV=nRT). On the contrary, when the initial volume of liquid is big, by expanding it eats up a relatively high portion of the initial gas volume, boosting up the compression of the headspace and multiplying the normal pressure increase produced by temperature (V' is forced to be much smaller)]

If this is true, I think this effect would not happen in presence of a single component system (air and its wapour, without air). Water vapour in equilibrium with liquid water would just be compensated by condensation. I suppose a critical point would be only reached if the final volume of water expanded at volumes higher than the tank.

Do you think this is a reasonable explanation or am I missing details?
Thank you

10. Aug 5, 2015

### Staff: Mentor

The volume of liquid water in the container is going to decrease with increasing temperature, not increase. So the explanation is not reasonable.

But all of this can be precisely calculated. In the case of only water being present in the container, the pressure is going to be equal to the equilibrium vapor pressure of water at the imposed temperature, as long as the liquid water lasts.

In the case of water and air in the container, as long as the total pressure is less than a few of atmospheres, the partial pressure of the water in the gas phase is going to be equal to the equilibrium vapor pressure of water at the imposed temperature, and the partial pressure of the air is going to be determined by the ideal gas law. The calculations to analyze this are pretty simple. The constraint is that the mass of water and air in the container are constant, and the total volume is constant. Part of the liquid water can evaporate to enter the gas phase as the temperature rises, so the volume of the gas phase is going to increase. It is reasonable to assume that none of the air dissolves in the liquid water and that the density of the liquid water is constant at 1000 kg/m^3. The game plan would be to set the initial conditions as: initial temperature (say 20 C) and initial fraction of liquid water. Then calculate what happens at equilibrium when the temperature is raised: new fraction liquid water, new partial pressure of water vapor, new partial pressure of air, new total pressure. Do this at a sequence of temperatures. Then repeat the calculations with a different initial fraction of liquid water.

Chet

11. Aug 5, 2015

### chutonio

I think your assumptions are not correct.
Under the boundaries condition you state the pressure variability vs. degree of filling would be relatively little.
What I´m talking about is the hundreds of bars variability well described by the picture 4.37 of this section

Extract from Schubert´s book Synthesis of Inorganic Materials

I think the author is right in saying that the liquid expands, it does not shrink.
Its density is decreasing upon temperature increase. At the same time the gas density is increasing (more pressure and molecules in the gas phase). Further increase in T follows the same trend up to the critical point, where the density of gas and liquid meet i.e. a single phase is formed.
It seems that the fact that higher T allows for more molecules to escape the liquid phase into the gas is completely overcome by the density change in the liquid. Still according to the author, and to give a practical example, density of water at 300 °C is only 0.75 g/ml.
Therefore if you fill your tank up to the 75% you are already going through the roof at that temperature (tank full, pressure exploding).

Do you feel my statements are correct and that you could reformulate your reply according to this?
Or I´m still missing something in the picture?

Thanks again

12. Aug 5, 2015

### Staff: Mentor

Why speculate? Let's do an actual calculation and see how it plays out. You pick the initial conditions. Then we'll do a sequence of calculations for increasing temperatures using the steam tables, and see what we come up with.

Chet

13. Aug 5, 2015

### Staff: Mentor

OK. So please specify the initial conditions for our calculations. I suggest using a tank volume of 1 m^3.

Chet

14. Sep 7, 2015

### chutonio

Hi Chet,
sorry for taking so long to reply.
I was thinking over it during my free time and I could not get a nice grip on it. The thing is density of liquid water varies substantially with temperature as I mentioned. Not only that. Density is somehow a function of what we want to calculate, that is pressure, since the latter affects the bulk modulus of elasticity (eventhough to a smaller extent apparently).
So basically P= f(T,ρ) but ρ is a function of both temperature and pressure.
That's why I guess practical people prefer to work with tables. Deriving a generally valid equation seems non-trivial to me.
Where do I start to clean up this self-biting snake mess? Do we have to work in a limited temperature range and start considering the bulk modulus invariant in that range?
Thanks

15. Sep 7, 2015

### Staff: Mentor

I still haven't seen a single indication from you as to a starting condition. Please define an initial condition for a constant volume container with a certain fraction of liquid water and fraction water vapor in the head space, and we can work the problem together. By doing modelling like this, all your uncertainties can be removed and we can stop with all the hand waving and speculation.

Using the steam tables will give the most accurate answer. Are you comfortable with using the steam tables? If not, we can model the liquid water using bulk compressibility and coefficient of thermal expansion. This is less accurate than the steam tables, particularly as the temperature increases. A still easier (but less accurate method) is to take the liquid density as constant and use the ideal gas law for the vapor. This method should be pretty good up to temperatures of about 200 C. So, what approach do you want to use? Whatever approach you choose, I will lead you through the calculations. Or, we can do all three approaches and compare the results.

Chet

16. Sep 8, 2015

### chutonio

Alright. I am not confident with steam tables, but it sounds like the perfect occasion to learn.

As to conditions I would do as follows. 1 m3 container is fine.

The container is filled at T1 = 293 K, sealed, and heated up to T2 = 586K (to make it nasty)

The question is, what is the final pressure inside the tank at 586 K varying the following conditions:

- Initial degree of filling of water in the tank a) 10% b) 50% c) 80%
- Composition: i) only water in equilibrium with its vapour (ideal condition); ii) Air in the headspace (realistic condition of sealing a tank, headspace is a mix of air and water vapour in equilibrium )

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Do you agree on the selected points?

I read from steam table the saturated water pressure at 586 K ===> 103 bar , finished
It feels to me that points i,b and i,c are exactly the same , as the equilibrium liquid-vapour is an intensive property (as long as there's enough initial liquid water to inject into the headspace, but this is not the point of the discussion at the moment).
I suppose the tank will blast only if the expansion of the liquid fills completely the tank. At 586 K water density is 0.68
So the final volume of
i,a ) 0.1 / 0.68 = 0.15 m3 SAFE
i,b) 0.5 / 0.68 = 0.73 m3 SAFE
i,c) 0.8 / 0.68 = 1.17 m3 BLAAST (or containment of water expansion if the container is super resistant)
[I approximated density of water to 1 at 293 K ]

let´s consider now cases ii), in presence of air headspace.
How can I use correctly the steam tables here? It is possibly my ignorance here stopping me from going forward. I don´t like to wave speculating, I just don´t know how to proceed.
Can I treat independently the variation of physical properties of air and water upon heating?
Here air pressurizes upon heating, and even more because of water expansion. But water expansion and air pressurization it´s an equilibrium (I think): are there models describing it or can we look up a separate set of steam tables?
Do we approximate density of water as only dependent on temperature?
In this case pressure depens dramatically on degree of filling I think, but I don´t know where to start from do it correctly.

Thank
Cheers

17. Sep 8, 2015

### Staff: Mentor

yes.
I thought you were also interested in getting the volumes and masses of liquid water and water vapor at 586. Were you?
Let's wait on the air in head space cases until we are sure we are really through with the pure water cases.

Chet

18. Sep 8, 2015

### chutonio

Not really.
But should we? Don´t steam tables already account for it?
Was my pressure derivation directly form the steam tables correct?

In case we were interested...
Initial conditions
pressure : 0.023392147667769 [ bar ]
density water : 998.16080927879 [ kg / m3 ]
density steam : 0.017312574945658 [ kg / m3 ]

starting with 0.1 m3 of liquid water in equilibrium with its vapour.
23.4 mbar in V0 = 0.9 m3 of headspace,
n = RT/PV0=8.314E-5*293 / (0.0234*0.9) = 1.156 mol
m = 18 * 1.156 = 20.8 g of water vapour (0.2% of liquid phase mass)

Now I heat up.
At 586 K I would do the same starting from the 100 bar pressure given from the table, but how do I know now the residual headspace volume after some water evaporated/expanded? I should anyway staple on the new density value given by the table shouldn't I?

Chut

19. Sep 8, 2015

### Staff: Mentor

Yes.

If you are interested in the actual pressure at 586K of compressed liquid water at a density of 1.25 gm/cc, you can find it in steam tables that include compressed (subcooled) liquid water. The pressure under these conditions is about 800 bars.
This result does not look correct. (0.9)(0.0173126)=0.0156 kg = 15.6 g. Applying the ideal gas law, I get 15.8 g. It looks like you got your gas law relation inverted.

So the total mass of water is 99.82+.02=99.84 kg
Yes. The total mass is constant and the total volume is constant. What is the density of saturated liquid and saturated vapor at 586? Let m = mass of liquid remaining at 586 and let (99.82-m) = mass of vapor at 586. In terms of m, what is the volume of liquid? In terms of m, what is the volume of vapor? These have to add up to the total volume. What is the value of m?

Chet

20. Sep 8, 2015

### insightful

I start with a 1m3 rigid vessel with 0.8m3 liquid water and 0.2m3 vapor in equilibrium at 20oC giving 798.4kg liquid and 0.00346kg vapor. Then heat to 100oC, I get 0.0961kg liquid water transfers to the vapor space giving 0.167m3 vapor and 0.833m3 liquid.