# Water / steam homogeneous mixture

• T C
In summary: Heated again and the valve will be reversed. What will happen?I don't want to go to a vivid description, but just want to remind that all ideal processes are reversible.
T C
We all here (I presume that members here have better understanding of physical processes than average person) know that it isn't a fact that water began to boil at 100°C but much before that. When being heated in an open pot, as the temperature rises, water began to boil and more and more water will be converted to steam. At 100°C, all the water will be converted into steam.
Now, suppose we have some amount of water in an enclosed container and that is being heated, if the heating is isobaric, then at around 80°C temperature; nearly about 46% of the water will then become steam. Now, question is, as the water is being heated in an enclosed container in isobaric way, what we will get at the end. Whether the steam will be separated from water inside the container or what we will get is an homogeneous mixture of steam and water inside the container.

T C said:
We all here (I presume that members here have better understanding of physical processes than average person) know that it isn't a fact that water began to boil at 100°C but much before that. When being heated in an open pot, as the temperature rises, water began to boil and more and more water will be converted to steam. At 100°C, all the water will be converted into steam.
I'm sorry, but no, neither of those are correct. Absent superheating (in which boiling is delayed), water boils at 100C, not below. And it doesn't all then spontaneously flash to steam at 100C, but reaches an equilibrium with the heat being added, staying at almost exactly 100C until finished boiling.

You may be confusing boiling with evaporation, though that only fixes part of it.
Now, suppose we have some amount of water in an enclosed container and that is being heated, if the heating is isobaric, then at around 80°C temperature; nearly about 46% of the water will then become steam.
Following from the above, this is wrong. It looks like here you are confusing the meaning of vapor pressure (which is about 46% of atmospheric at 80C).
Now, question is, as the water is being heated in an enclosed container in isobaric way, what we will get at the end. Whether the steam will be separated from water inside the container or what we will get is an homogeneous mixture of steam and water inside the container.
When you are "finished" boiling anything, what you get when you are done is 100% water vapor/steam. That's what it means to be finished.

You mat be confused by watching a pot on the stove. Bubbles form on the bottom long before general boiling, but that is because the temperature at the bottom is higher than at the top. @russ_watters told you what really happens.

Let's think about an imaginary experiment. We have an enclosed container having just one inlet fitted with an one way valve that will allow dry air to come in. The container contains boiling water and steam occupies the space above water. Now it has been left to cool slowly. What will happen? As more and more steam will be converted into water, dry air will come and occupy the space. At around 80°C, the empty space above the water will contain 46.12% steam and rest will be dry air. In short, as the temperature will fall, the % of steam above will be equal to the saturated steam pressure at that temperature. That's what Russ Waters has said in this thread.
Now, for the sake of simplicity, let's consider the process to be an ideal one. And after the fall of temperature to a certain level, the container will be heated again and the valve will be reversed. What will happen? I don't want to go to a vivid description, but just want to remind that all ideal processes are reversible.

T C said:
Let's think about an imaginary experiment. We have an enclosed container having just one inlet fitted with an one way valve that will allow dry air to come in. The container contains boiling water and steam occupies the space above water. Now it has been left to cool slowly. What will happen? As more and more steam will be converted into water, dry air will come and occupy the space.

Ok...
At around 80°C, the empty space above the water will contain 46.12% steam and rest will be dry air. In short, as the temperature will fall, the % of steam above will be equal to the saturated steam pressure at that temperature.
As measured by pressure, yes, but that is a bit of an odd way to say it (by mass or moles it won't be 46%).
And after the fall of temperature to a certain level, the container will be heated again and the valve will be reversed. What will happen? I don't want to go to a vivid description, but just want to remind that all ideal processes are reversible.
Yes, this process is reversible in the way you describe.

russ_watters said:
As measured by pressure, yes, but that is a bit of an odd way to say it (by mass or moles it won't be 46%).
By mass it would be greater than 46% as the molecular weight of steam is more than both Nitrogen and Oxygen, the main components of air.
Whatsoever, if % of steam above water rises with rise in temperature, that means with rise in temperature more and more water will be converted into steam. Not like as we have told that "water start to boil and becoming steam at standard pressure at 100°C.

T C said:
By mass it would be greater than 46% as the molecular weight of steam is more than both Nitrogen and Oxygen, the main components of air.
Whatsoever, if % of steam above water rises with rise in temperature, that means with rise in temperature more and more water will be converted into steam.
Yes, though I'm not sure what scenario we are working from at this point. There are potential caveats (one provided below).
Not like as we have told that "water start to boil and becoming steam at standard pressure at 100°C.
Sorry, but that still is not true. Your error appears to be that you think boiling and evaporation are the same thing. Evaporation happens at any temperature if the partial pressure of the vapor is lower than the vapor pressure of the water. Boiling only happens at 100C (at sea level and 1atm).

Now the promised caveat: if the container is open and the relative humidity of the air is below 100%, all of the water will evaporate, regardless of its temperature.

I haven't started this thread to understand the difference between evaporation and boiling. My main concern is vapour formation, whether that by evaporation or by boiling doesn't matter. It's now clear that as temperature rises, more and more section of water will turn into steam. What I want to know is that if the pressure remain constant but the vapour formed has been forbidden to leave the water surface, what will happen then?

T C said:
I haven't started this thread to understand the difference between evaporation and boiling. My main concern is vapour formation, whether that by evaporation or by boiling doesn't matter.
Ok...it's tough for me to know what is important to you, so I answer and/or correct everything. And it is tough for me to know if you really know the concept behind what you are asking if you insist on describing it wrong...
It's now clear that as temperature rises, more and more section of water will turn into steam.
This remains false no matter how many times you say it and I correct it and you ignore the correction. Does that matter? I don't know, but I think probably yes.
What I want to know is that if the pressure remain constant but the vapour formed has been forbidden to leave the water surface, what will happen then?
How are you doing that? At what temperature? Are you adding heat? For any setup I can think of, what you have described is physically impossible; you have contradicting constraints.

For example, if you have a sealed vessel and heat it, The pressure has to rise.

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T C said:
I haven't started this thread to understand the difference between evaporation and boiling. My main concern is vapour formation, whether that by evaporation or by boiling doesn't matter. It's now clear that as temperature rises, more and more section of water will turn into steam. What I want to know is that if the pressure remain constant but the vapour formed has been forbidden to leave the water surface, what will happen then?
As I understand it...

Typically in a closed vessel containing dry air and water the water evaporates increasing the pressure. Once it reaches the saturation vapour pressure evaporation stops, or rather it's in balance, the number of molecules leaving and re-entering the water per second are the same.

If you keep the pressure constant (below saturation) by letting out some of the water vapour out of the container this doesn't happen. The water keeps evaporating until it's all gone. It's not "forbidden to leave the surface".

At what point are you "keeping the pressure constant"? Above or below the saturation vapour pressure?

If you increase the temperature then the saturation vapour pressure increases so a greater fraction of the water will need to evaporate to reach it.

russ_watters
russ_watters said:
How are you doing that? At what temperature? Are you adding heat? For any setup I can think of, what you have described is physically impossible; you have contradicting constraints.
For example, if you have a sealed vessel and heat it, The pressure has to rise.
In the Boiler of thermal power plants, pressurised water is heated at constant pressure. What I have clearly told in my starting thread that it's about heating water by keeping the pressure constant. In a totally sealed vessel, pressure will rise when heated because there is no place to add the increased volume (due to heating) of water. Think about scenario where there is piston fitted in your mentioned sealed vessel that can move to accommodate the increased volume of water. In that case, pressure wouldn't rise with rise in temperature.
CWatters said:
Typically in a closed vessel containing dry air and water the water evaporates increasing the pressure. Once it reaches the saturation vapour pressure evaporation stops, or rather it's in balance, the number of molecules leaving and re-entering the water per second are the same.
If you keep the pressure constant (below saturation) by letting out some of the water vapour out of the container this doesn't happen. The water keeps evaporating until it's all gone. It's not "forbidden to leave the surface".
At what point are you "keeping the pressure constant"? Above or below the saturation vapour pressure?
If you increase the temperature then the saturation vapour pressure increases so a greater fraction of the water will need to evaporate to reach it.
You have forgot that the temperature of water is rising and with that, the saturation pressure level too. What you have described is for a scenario where the temperature remains constant and my question is about what happens when temperature rises.

T C said:
In the Boiler of thermal power plants, pressurised water is heated at constant pressure.
True.
In a totally sealed vessel, pressure will rise when heated because there is no place to add the increased volume (due to heating) of water.
True, but that doesn't describe a boiler, which has inlet and outlet pipes.
Think about scenario where there is piston fitted in your mentioned sealed vessel that can move to accommodate the increased volume of water. In that case, pressure wouldn't rise with rise in temperature.
True. And that's yet a third scenario. And none of these fit the scenario you presented in post #8.
You have forgot that the temperature of water is rising and with that, the saturation pressure level too.
Since he said exactly that in his last sentence, no, he certainly didn't forget that.
What you have described is for a scenario where the temperature remains constant and my question is about what happens when temperature rises.
No, I specifically asked if heat was being added because I assumed the temperature was rising -- even though it presents a contradictory set of constraints.

I've really had enough of this: you need to lock-in on one specific scenario, describe it in detail (better yet, with a diagram) and discuss only that scenario. Throwing out several different scenarios in one post and picking and choosing bits and pieces of each to assemble into an unknown master scenario just doesn't produce anything fruitful. You need to stop this random, meandering, idle speculation or the thread will be closed.

+1

Just going back to the OP...

T C said:
Now, question is, as the water is being heated in an enclosed container in isobaric way, what we will get at the end. Whether the steam will be separated from water inside the container or what we will get is an homogeneous mixture of steam and water inside the container.

That bit seems easy to answer..

If you keep adding heat, in the "end" all the water is turned to steam.

Before that (when you still have both water and water vapour) the water stays at the bottom of the container because it's denser than both dry air and water vapour. If there is no gravity surface tension will prevent a uniform mixture.

russ_watters
CWatters said:
If you keep adding heat, in the "end" all the water is turned to steam.
What I am asking is what will happen in between. Anybody can understand that when the temperature will reach 100°C. What I want to know is what will happen before temperature reaching 100°C.
CWatters said:
Before that (when you still have both water and water vapour) the water stays at the bottom of the container because it's denser than both dry air and water vapour. If there is no gravity surface tension will prevent a uniform mixture.
To do that, the pressure of the vapour should be 1 barA or more and before reaching 100°C, that's not possible. What I want to know is whether the steam will leave water or we will get a mixture of steam and water. When the temperature will reach 100°C, then the steam will start leaving water as the pressure then will be 1 barA and above then.

T C said:
What I am asking is what will happen in between. Anybody can understand that when the temperature will reach 100°C. What I want to know is what will happen before temperature reaching 100°C.
This has been fully addressed, several times.

But if this is really what you care about:
What I want to know is whether the steam will leave water or we will get a mixture of steam and water.

[or from the OP]
...Whether the steam will be separated from water inside the container or what we will get is an homogeneous mixture [emphasis added] of steam and water inside the container.]
Regardless of the impossible constraints you are using to try to achieve it, no, you will never get a homogeneous mixture of steam and water. As said above, water is much heavier than steam, so they separate.

Note though that while eventually all water will evaporate from an uncovered pot, you can prevent evaporation by putting an air-tight but non-rigid cover on the container (such as plastic wrap on a pot). If there is a bit of air in the container, that air will quickly reach saturation and evaporation will stop in the case of constant temperature -- or with a rising temperature will slowly continue in order to match the saturation point in the air. If there is no air in the container (the plastic wrap is tight to the water), there will be no evaporation at any temperature.

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I have searched google with "water steam mixture" and "steam water mixture" and found a lot of links. Here is just one.

T C said:
I have searched google with "water steam mixture" and "steam water mixture" and found a lot of links. Here is just one.
[clicks link] Ok, so you found a lot of papers that contain the words you want to see, but not assembled in the way you want to see them, nor describing scenarios anywhere close to what you described? Ok...

This is just silly. I think we've run this to ground. Thread locked.

phinds, Bystander and CWatters

## 1. What is a water/steam homogeneous mixture?

A water/steam homogeneous mixture is a uniform combination of water and steam molecules. This means that both water and steam are evenly distributed throughout the mixture, resulting in a single phase system.

## 2. How is a water/steam homogeneous mixture different from a heterogeneous mixture?

A heterogeneous mixture is a combination of substances that are not evenly distributed, resulting in a mixture with more than one distinct phase. In contrast, a water/steam homogeneous mixture has a single phase due to the uniform distribution of water and steam molecules.

## 3. What causes a water/steam homogeneous mixture to form?

A water/steam homogeneous mixture forms when water is heated to its boiling point and transforms into steam. The steam molecules then mix with the remaining water molecules, resulting in a homogeneous mixture.

## 4. What are some properties of a water/steam homogeneous mixture?

A water/steam homogeneous mixture has properties that are a combination of both water and steam. It has a uniform composition and boiling point, and the temperature remains constant during the phase transition. Additionally, it is a good conductor of heat and has a high vapor pressure.

## 5. How is a water/steam homogeneous mixture used in scientific research?

A water/steam homogeneous mixture is commonly used in experiments and studies related to phase transitions, thermodynamics, and heat transfer. It is also used in the production of electricity through steam turbines and in various industrial processes such as sterilization and heating.

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