Does Pressure in Phase Diagram refer to vapor pressure?

[Electric to be]

I am somewhat confused by what pressure refers to in a phase diagram? In a closed box it makes sense to me that the vapor pressure would eventually equilibriate at a pressure determined by the temperature. However, say you have an open box. It makes sense that the liquid would boil when the internal vapor pressure is equal to external pressure. However, at a lower pressure the liquid would still be evaporating. I'm basically wondering at what point would simple evaporation fall on the diagram? Since there is both liquid and gas? And what corresponding pressure would you use? Vapor + ambient pressure?

 

[russ_watters]

Phase diagrams show equilibrium states, they do not show processes. So "boiling" and "evaporation" do not appear on the diagram.

 

[Electric to be]

Phase diagrams show equilibrium states, they do not show processes. So "boiling" and "evaporation" do not appear on the diagram.

I've seen certain sources where going one direction across the "equilibrium line" is condensation and the other is evaporation. Is that simply wrong?

However regardless of this, I thought the phase diagram could be used for more than just equilibrium. For example, there is a "liquid" or "gas" only region. Can't the point we pick for our system be in this area, and therefore not in equilibrium? Also for a point that does fall on the equilibrium line, for what type of system does this apply to? Is it strictly a closed system like a closed beaker or container? And does the pressure refer strictly to the vapor pressure or is ambient air pressure inside the beaker included?

This is all somewhat confusing to me as I understand that boiling occurs when the vapor pressure within a liquid matches the ambient pressure. But apparently this can only occur in an open container as the ambient pressure will just become the vapor pressure and will reach equilibrium in a closed container? Just seems odd to include part of the system (the water vapor) as a contributing pressure acting on itself (the system).

Maybe I've just really mixed myself up. I would appreciate some help.

 

[SteamKing]

I've seen certain sources where going one direction across the "equilibrium line" is condensation and the other is evaporation. Is that simply wrong?

However regardless of this, I thought the phase diagram could be used for more than just equilibrium. For example, there is a "liquid" or "gas" only region. Can't the point we pick for our system be in this area, and therefore not in equilibrium? Also for a point that does fall on the equilibrium line, for what type of system does this apply to? Is it strictly a closed system like a closed beaker or container? And does the pressure refer strictly to the vapor pressure or is ambient air pressure inside the beaker included?

This is all somewhat confusing to me as I understand that boiling occurs when the vapor pressure within a liquid matches the ambient pressure. But apparently this can only occur in an open container as the ambient pressure will just become the vapor pressure and will reach equilibrium in a closed container? Just seems odd to include part of the system (the water vapor) as a contributing pressure acting on itself (the system).

Maybe I've just really mixed myself up. I would appreciate some help.

This is a phase diagram for water:

​

I assume by "equilibrium line" you are talking about the separation between the different phases of this substance.

If you boil water in a closed container, the boiling point will probably occur at a temperature slightly higher than if you were boiling water in an open container, because the pressure in the closed container will be slightly higher than ambient. This is why people use pressure cookers and our car radiators have pressure caps.

 

[russ_watters]

Ok, yes; moving from one state to another state is a process. Your initial wording threw me when you said "at what point would simple evaporation fall on the diagram?" (Still doesn't) But, you can and do diagram processes using that diagram (or the information in table form).

So here's how to do it: define your states and draw a line between them representing the process.

Example: If you have a bucket of water sitting on the floor of your house, in a steady evaporation, what are the two states? (phase, temp, pressure - pick values that make sense for your house).
Boiling is a special case of evaporation, so let's just do evaporation first.

 

[Khashishi]

The pressure refers to the environmental pressure, not the vapor pressure. The vapor pressure depends mostly on the temperature, whereas the environment pressure is orthogonal to the temperature.
They are equal along the vapor-liquid phase boundary.

 

[Khashishi]

Evaporation is not shown on this diagram. Evaporation depends on the relative humidity of the air. In the liquid zone of the phase diagram, there is actually an equilibrium between liquid and vapor. This equilibrium is called 100% humidity. If there is less than 100% humidity (only possible outside of equilibrium), then evaporation will take place. If more than 100% humidity, there will be dew or rain.

The equilibrium is where the vapor pressure of the water is balanced by the partial pressure in the air.

 

[Electric to be]

Evaporation is not shown on this diagram. Evaporation depends on the relative humidity of the air. In the liquid zone of the phase diagram, there is actually an equilibrium between liquid and vapor. This equilibrium is called 100% humidity. If there is less than 100% humidity (only possible outside of equilibrium), then evaporation will take place. If more than 100% humidity, there will be dew or rain.

The equilibrium is where the vapor pressure of the water is balanced by the partial pressure in the air.

This makes sense to me logically. However, on the phase diagram for water, take the point 1 ATM and 100 C. It lies on the equilibrium line. So if I have a bucket of water exposed to 1 ATM of ambient air (no water), the system obviously won't be under equilibrium because the air isn't a 100% humidity. Yet the system still lies on the line of equilibrium?

Because as you said, "The pressure refers to the environmental pressure, not vapor pressure"

 

[russ_watters]

This makes sense to me logically. However, on the phase diagram for water, take the point 1 ATM and 100 C. It lies on the equilibrium line. So if I have a bucket of water exposed to 1 ATM of ambient air (no water), the system obviously won't be under equilibrium because the air isn't a 100% humidity. Yet the system still lies on the line of equilibrium?

Because as you said, "The pressure refers to the environmental pressure, not vapor pressure"

I'm not sure that last bit was worded right: it is vapor pressure for the vapor. For the other phases, the environment provides the pressure.

So that means no, your bucket in air at 100C is not a single point, it is two separate points. The water vapor in the air is somewhere different from the water in the bucket.

 

[SteamKing]

This makes sense to me logically. However, on the phase diagram for water, take the point 1 ATM and 100 C. It lies on the equilibrium line. So if I have a bucket of water exposed to 1 ATM of ambient air (no water), the system obviously won't be under equilibrium because the air isn't a 100% humidity. Yet the system still lies on the line of equilibrium?

Because as you said, "The pressure refers to the environmental pressure, not vapor pressure"

You have a bucket of water at a pressure 1 atmosphere AND a temperature of 100 °C. The water in the bucket is going to be boiling, regardless of what the relative humidity is.

The atmosphere is able to have different amounts of water vapor suspended in it, but this does not alter what happens to a large amount of the substance which is relatively isolated from the atmosphere.

 

[Electric to be]

You have a bucket of water at a pressure 1 atmosphere AND a temperature of 100 °C. The water in the bucket is going to be boiling, regardless of what the relative humidity is.

The atmosphere is able to have different amounts of water vapor suspended in it, but this does not alter what happens to a large amount of the substance which is relatively isolated from the atmosphere.

Okay sure. However that is still a point on the equilibrium line. So what exactly does that mean if it's boiling.

I essentially understand several things and correct me if I'm wrong:

1. Water will evaporate until it reaches the saturated vapor pressure given by the specific temperature. (This is independent of any external pressure, as far as I understand)

2. If the vapor pressure within the water (which would be the vapor pressure in the air, if it was able to reach equilibrium) exceeds the external pressure the water will boil at the boiling point as additional heat is added.

So basically my question is, what exactly does a phase diagram apply to? An open system, closed system, the phases seperately, together? It seems like the equilibrium line isn't an actual point of equilibrium since the system reaches equilibrium when the vapor pressure reaches a certain level, and on the actual equilibrium line the water is boiling.

Ehhh haha

 

[Khashishi]

At the boiling point, the water has enough vapor pressure to spawn bubbles of pure water vapor anywhere in the liquid where there is a nucleation site. Raising the pressure raises the boiling point because it requires a higher vapor pressure (which increases with temperature) to push the water out of the way to spawn a bubble. Below the boiling point, but above the dew point, water only evaporates at the surface. The water doesn't have enough vapor pressure to lift the water out of the way form a vapor bubble.

The difference between dew point and boiling point exists because air a mixture of many gases. Things will become even more complicated if your liquid is also a mixture.

1. (Edit: actually, your statement looks ok). Saturated vapor pressure depends on the external pressure. Relative humidity of 80% at 30 degrees C has a lot more absolute water content than relative humidity of 80% at 20 degrees C.

2. Yes. I would replace the term vapor pressure with partial pressure in the air.

 

[SteamKing]

So basically my question is, what exactly does a phase diagram apply to? An open system, closed system, the phases seperately, together? It seems like the equilibrium line isn't an actual point of equilibrium since the system reaches equilibrium when the vapor pressure reaches a certain level, and on the actual equilibrium line the water is boiling.

The phase diagram for any substance is obtained by testing samples of the pure article under a variety of combinations of pressure and temperature, primarily, but not exclusively, to establish the boundaries of the different phases of that substance. Using the phase diagram in an actual environment will be subject to certain limitations and cautions about its applicability.

https://en.wikipedia.org/wiki/Phase_diagram

There are a variety of different diagrams employed in science and engineering, and it takes some training to be able to use them properly. Phase diagrams can also be prepared for mixtures of different substances, depending on the purpose for which they are needed.

The vapor pressure of a substance is defined as the pressure exerted by the vapor phase of a substance which is in equilibrium with the solid or liquid phases of that substance in a closed system:

https://en.wikipedia.org/wiki/Vapor_pressure

The vapor pressure of a substance changes with temperature, given a constant ambient pressure. When the vapor pressure = ambient pressure, then the boiling point of the substance has been achieved at that temperature.

 

[Electric to be]

1. Saturated vapor pressure depends on the external pressure. Relative humidity of 80% at 30 degrees C has a lot more absolute water content than relative humidity of 80% at 20 degrees C.

2. Yes. I would replace the term vapor pressure with partial pressure in the air.

In your response to my first point, I'm not sure if your edit is taking your statement back but how exactly does external pressure affect the saturated vapor pressure. Isn't that only a function of temperature? I see how external pressure matters in regards to boiling but how is this true?My guess at it would be that external pressure limits how much water can escape so the equilibrium saturated vapor pressure would therefore lower? But would that mean a certain pressure that the vapor pressure in the air is zero?

 

[Khashishi]

I edited it because I think you understood it correctly. The saturated vapor pressure only depends on the temperature.

 

[Electric to be]

I edited it because I think you understood it correctly. The saturated vapor pressure only depends on the temperature.

Okay thanks for all the help so far. However, doesn't the external pressure affect how much water is escaping the liquid state in an interval of time? Wouldn't this therefore affect the equilibrium vapor pressure?

And this could therefore correspond to the liquid areas of the phase diagram where essentially no vapor is able to escape due to external pressures

 

[russ_watters]

So basically my question is, what exactly does a phase diagram apply to? An open system, closed system, the phases seperately, together?

States. I'm wondering if the problem is that you don't understand what a "state" is. In thermodynamics, a "state" is a list of properties that describe a substance. You can grab as much or as little of that substance as you want and as long as that's all at the same conditions, it is all at the same state. It doesn't matter if the system is open or closed: but phase is a property, so clearly different phases are at different states.

It seems like the equilibrium line isn't an actual point of equilibrium since the system reaches equilibrium when the vapor pressure reaches a certain level, and on the actual equilibrium line the water is boiling.

What equilibrium? If the water is boiling, it is changing state: it is not at equilibrium.

You haven't taken my advice from earlier and defined the properties at the states. It will become much, much clearer if you do that because then you will find that the states you are suggesting are at equilibrium with each other are actually very far apart. In other words: the states you are describing do not both fall on that line.

 

[Electric to be]

States. I'm wondering if the problem is that you don't understand what a "state" is. In thermodynamics, a "state" is a list of properties that describe a substance. You can grab as much or as little of that substance as you want and as long as that's all at the same conditions, it is all at the same state. It doesn't matter if the system is open or closed: but phase is a property, so clearly different phases are at different states.

What equilibrium? If the water is boiling, it is changing state: it is not at equilibrium.

You haven't taken my advice from earlier and defined the properties at the states. It will become much, much clearer if you do that because then you will find that the states you are suggesting are at equilibrium with each other are actually very far apart. In other words: the states you are describing do not both fall on that line.

Could you possibly explain a specific example, for example with a bucket or cup of water, both closed and open?

And also, why is that line called an equilibrium line then?

Sorry for being difficult

 

[russ_watters]

Could you possibly explain a specific example, for example with a bucket or cup of water, both closed and open?

Sure. You've already started describing one: a bucket of water at 100C and atmospheric pressure. That's a state. The system is open and the water evaporates into the room. That's another state. Define the conditions of that state (pressure, temperature) and see where it would be plotted on the phase diagram.

And also, why is that line called an equilibrium line then?

The equilibrium line is an equilibrium line. You're just not describing/using it properly yet.

Sorry for being difficult

Not a problem. But the methods we try to use to teach do matter. You'll get it better if you follow our steps/instructions than if we just spoon feed you an answer.

 

[Electric to be]

Sure. You've already started describing one: a bucket of water at 100C and atmospheric pressure. That's a state. The system is open and the water evaporates into the room. That's another state. Define the conditions of that state (pressure, temperature) and see where it would be plotted on the phase diagram.

The equilibrium line is an equilibrium line. You're just not describing/using it properly yet.

Not a problem. But the methods we try to use to teach do matter. You'll get it better if you follow our steps/instructions than if we just spoon feed you an answer.

Okay so the two states are plotted on different areas. So what would be a situation in which a system would fall on the equilibrium line?(in regards to this water bucket)

 

[russ_watters]

Okay so the two states are plotted on different areas. So what would be a situation in which a system would fall on the equilibrium line?(in regards to this water bucket)

Put a lid on the bucket and pump out the air. It will then be a water/vapor mixture at 100C and atmospheric pressure.

 

[Khashishi]

Put a lid on the bucket and pump out the air. It will then be a water/vapor mixture at 100C and atmospheric pressure.

It will be somewhere on the liquid/vapor boundary, but would be colder than 100C, and lower than atmospheric pressure.

For the pure chemical case (e.g. water, with no air), it might help to see the 3D phase diagram, as shown here:
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/pvtexp.html#c1
We see that there is a 2D surface which you can regard as pressure as a function of temperature and volume. The state of the chemical is given by a point sitting somewhere along this surface. The surface is broken up into several faces which correspond to different phases. The formulae for the faces are called equations of state. (For example, the ideal gas law PV=nRT is an equation of state which is approximately valid in the "gas" part of the diagram above the critical temperature.)
If you put water in a closed container, this will constrain the density and volume, and the state will lie somewhere along the 2D cross section which looks like a traditional P vs T phase diagram. If the volume is large, that means your density is low, and you have only gas. If we half fill a bucket of water and pump out the remaining air, we will end up somewhere on the "liquid and vapor" face. That means there is an equilibrium between the liquid water, and the water vapor in the container. The pressure will be the vapor pressure of the chemical at whatever temperature it ends up at.
Therefore, it will be at the boiling "point", but as we can see on the 3D diagram, boiling is possible over a whole face (the liquid and vapor face), not a point.

In a closed container, we can't really control the pressure, unless we have some kind of piston.

 

[russ_watters]

It will be somewhere on the liquid/vapor boundary, but would be colder than 100C, and lower than atmospheric pressure.

You are assuming different constraints than what was provided: 100C is what was provided, so I am assuming it stays at 100C.

 

[Khashishi]

Well, it does cool a little if you pump out the air.

 

[russ_watters]

Well, it does cool a little if you pump out the air.

It doesn't need to if you keep heating it with the same heater that brought it to 100C in the first place. I'm trying to keep the problem simple: I think trying to deal with too many state changes before the OP has an understanding of what states are will make it harder for him to learn this.

 

[Electric to be]

It doesn't need to if you keep heating it with the same heater that brought it to 100C in the first place. I'm trying to keep the problem simple: I think trying to deal with too many state changes before the OP has an understanding of what states are will make it harder for him to learn this.

So I've got the basic idea but I've kind of come up with a contradiction. Say I have a closed box, and the water has reached equilibrium with its vapor pressure at 100C, with a vapor pressure of 1 ATM. Now I pump in some external gas, which will increase the total pressure inside the box but the vapor pressure will not change since the water is in equilibrium and adding some extra gas doesn't change anything. I also assume adding this gas doesn't change the temperature.

Now I will plot both states on the phase diagram as you have told me, the vapor and liquid. Initially, the vapor and liquid both lied on the same point on the equilibrium line. However now, hasn't the pressure in both states increased by the same amount? Won't the vapor and liquid states somehow now both lie in the liquid area of the phase diagram?

The contradiction being: how can a vapor's state be located in the liquid only region of the phase diagram if it's clearly a vapor?So basically considering the scenario you mentioned earlier, without pumping out the air.

 

[russ_watters]

If you pump dry air back in and pressurize it while keeping it at 100C, you haven't changed the vapor's state because its pressure is still 1ATM, but you have changed the water's state by raising its pressure.

 

[Khashishi]

Before you pump in air, the dew point and boiling point are the same, 100C. If you pump in air, while keeping the temperature the same, then the pressure increases, so the boiling point will increase, but the dew point will stay the same. The water is at its dew point because the vapor pressure equals the partial pressure of water in the gas, so it is in equilibrium.