# Understanding Vapor Pressure in Open Systems

• bubblewrap
In summary: The gas immediately adjacent to the interface with a liquid moves with the same velocity as the liquid. This is called the (experimentally observed) no-slip condition. So there is very thin layer of gas in contact with the liquid at the interface that is essentially stagnant relative to the liquid. It is therefore able to come to equilibrium with the liquid at the equilibrium vapor pressure corresponding to the interface temperature (even if the remainder of the system is not at equilibrium). (This is a slightly simplified picture, but it captures the essence of what is taking place).
bubblewrap
I learned that vapor pressure is defined in a closed system, but some questions ask me to consider vapor pressure in an open system. How is vapor pressure defined exactly? And where is this pressure acting against? Also why is it that at the boiling point, the vapor pressure and the atmospheric pressure equals? I mean is the vapor pressure acting against the atmosphere so that they cancel each other out at the boiling point? Thank you in advance.

bubblewrap said:
I learned that vapor pressure is defined in a closed system

That's not true. Pressure is a pressure, making it a "vapor pressure" doesn't change anything.

Sometimes it is easier to calculate things assuming a closed system, but that's another story.

In a closed system involving a single chemical species in which both liquid and vapor are present at thermodynamic equilibrium at a specified temperature, the pressure of the gas and the pressure in the liquid will be equal, and will equal the "equilibrium vapor pressure of the species at the specified temperature." There will be no net evaporation of liquid to form vapor and no net condensation of vapor to form liquid. The equilibrium vapor pressure is a unique function of the specified temperature.

In a closed system in which a second chemical species is also present in the gas phase (in addition to the chemical species of the above paragraph), and this second chemical species in virtually insoluble in the liquid (e.g., air in the gas phase), at thermodynamic equilibrium, the partial pressure of the original species in the gas phase will be equal to the same equilibrium vapor pressure at the specified temperature, but the total pressure in the container will be higher than the equilibrium vapor pressure, because of the presence of the second gaseous species. The key here is that the partial pressure of the original species is equal its equilibrium vapor pressure at the specified temperature.

In an open system, the partial pressure of the species under consideration is equal to its equilibrium vapor pressure at the interface between the liquid and the overlying gas. However, unless the entire room is at thermodynamic equilibrium, the partial pressure of the species further from the interface may be lower than the equilibrium vapor pressure. Under these circumstances, vapor formed at the interface will be transported away from the interface into the room, so that the liquid will be evaporating.

If the temperature of the liquid is raised to the point where the equilibrium vapor pressure matches the total pressure in the room, then bubbles can begin forming within the liquid, even below the interface with the room air. The gas within these bubble will consist of the pure species (no air), and the bubbles are able to grow in size because they can expand in the liquid, which pushes back the atmosphere. The process of forming bubbles under the liquid surface is what we call boiling.

Chet

lonely_nucleus, davidbenari and bubblewrap
So what is this pressure acting on?
And why doesn't the vapor.pressure change at the interface of gas and liquid?

bubblewrap said:
So what is this pressure acting on?
I don't understand this question. Pressure is present everywhere within a liquid or a gas. If it doesn't act on a wall, it is acting on its neighboring parcels of liquid or gas.
And why doesn't the vapor.pressure change at the interface of gas and liquid?
The gas immediately adjacent to the interface with a liquid moves with the same velocity as the liquid. This is called the (experimentally observed) no-slip condition. So there is very thin layer of gas in contact with the liquid at the interface that is essentially stagnant relative to the liquid. It is therefore able to come to equilibrium with the liquid at the equilibrium vapor pressure corresponding to the interface temperature (even if the remainder of the system is not at equilibrium). (This is a slightly simplified picture, but it captures the essence of what is taking place).

Chet

davidbenari
What I meant was the pressure is a force(force divided by area) so what is this 'force' acting on. I thought that because at the boiling point the vapor pressure and the atmospheric pressure is the same, they'd cancel each other out and it would mean that the vapor pressure acts on the atmosphere, so how is the gas molecules acting on them. This concept is very confusing and every teacher I asked to couldn't come up with a satisfying answer, perhaps there isn't one?

bubblewrap said:
What I meant was the pressure is a force(force divided by area) so what is this 'force' acting on. I thought that because at the boiling point the vapor pressure and the atmospheric pressure is the same, they'd cancel each other out and it would mean that the vapor pressure acts on the atmosphere, so how is the gas molecules acting on them. This concept is very confusing and every teacher I asked to couldn't come up with a satisfying answer, perhaps there isn't one?
In a boiling liquid, the gas molecules inside a bubble do not act directly on the atmosphere. The bubbles are forming within the liquid, and the water vapor inside the bubbles is pushing back against the liquid around them. The pressure within the bubbles is only slightly higher than the pressure of the atmosphere. The formation of the bubbles within the liquid initially causes the combined volume of liquid and bubbles to expand against the atmosphere. After the bubbles form in the liquid, they rise to the surface of the liquid, and the water vapor in them is released there, while new bubbles are forming below. So the total expanded volume of liquid plus bubbles remains about the same during the boiling.

Chet

So what is vapor pressure acting on then?

Why does it have to act on anything? You are misunderstanding how the pressure works.

There is a point around a feet from my nose. There is nothing but air there. Air pressure doesn't act on anything (but air) there. Does it mean there is no pressure in this particular point?

Would it help if I tell you "put a test object there and the pressure will act on it"?

bubblewrap said:
So what is vapor pressure acting on then?
Think of the gas as consisting of a 3D array of small cubical parcels of gas stacked adjacent to one another (in all three dimensions). The gas in each parcel presses against the surrounding parcels at the boundary of the parcel. The surrounding parcels of gas press right back on the gas in the parcel.

Chet

In a closed container where there is only one chemical species, the pressure above the liquid is equal to the vapor pressure of the chemical species. Does that mean that in this scenario there is also bubble formation?

My guess is that bubbles won't form. But why?

The definition of boiling point seems to indicate that they should form!

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davidbenari said:
In a closed container where there is only one chemical species, the pressure above the liquid is equal to the vapor pressure of the chemical species. Does that mean that in this scenario there is also bubble formation?

My guess is that bubbles won't form. But why?

The definition of boiling point seems to indicate that they should form!
Well, at equilibrium, since you're not adding heat, there will be no more evaporation and no bubble formation. The liquid and vapor will just sit there.

Now, let's consider the next equilibrium state the mixture can achieve in the closed container if you have added heat. Will the internal energy of the mixture be higher? Will the temperature be higher? Will the pressure be higher? Will there be more liquid water or less liquid water? Will there be more water vapor or less water vapor?

Chet

Chestermiller said:
Well, at equilibrium, since you're not adding heat, there will be no more evaporation and no bubble formation. The liquid and vapor will just sit there.

Now, let's consider the next equilibrium state the mixture can achieve in the closed container if you have added heat. Will the internal energy of the mixture be higher? Will the temperature be higher? Will the pressure be higher? Will there be more liquid water or less liquid water? Will there be more water vapor or less water vapor?

Chet

I think that the internal energy would be higher, therefore the temperature higher, the pressure higher(because of higher vapor pressure) less liquid water (more would escape) and more water vapor. But I don't understand why the bubbles don't form here, is it something like water that doesn't freeze below zero but instantly freezes when you hit it? Something that needs a little push?

bubblewrap said:
therefore the temperature higher

Temperature doesn't change during phase changes.

In general - you are describing a static system at equilibrium, then you ask why it doesn't change (bubble formation is a change). Well, it doesn't change because it is static and at equilibrium, that's where you started.

bubblewrap said:
I think that the internal energy would be higher, therefore the temperature higher, the pressure higher(because of higher vapor pressure) less liquid water (more would escape) and more water vapor. But I don't understand why the bubbles don't form here?
Yes. This is all correct. And during the transition to the new state, bubbles can be forming. But, once you have stopped adding heat and the system has equilibrated to the new equilibrium state at the new higher pressure and temperature, boiling will stop.

Chet

Borek said:
Temperature doesn't change during phase changes.
I think you may have misunderstood the scenario: this is a closed container with a liquid-gas mixture in it and it is being heated. This process increases the temperature because it increases the pressure as the liquid boils/evaporates.

Ah yes, I have missed the moment Chet changed the problem from an open system at equilibrium to a closed one heated.

The textbook definition of vapor pressure is the pressure exerted by a gas in contact with its pure condensed (liquid) phase. In a closed container at constant temperature that is partially filled with a given liquid, evaporation occurs until a dynamic equilibrium is achieved. That is, the rate of evaporation equals the rate of condensation. To measure a vapor pressure a simple laboratory manometer may be used. See link: http://www.csun.edu/~ml727939/coursework/695/vapor pressure/Vapor Pressure.htm.

To understand the relationship between vapor pressure and boiling one must understand the process of evaporation. To begin with, at temperatures below 'normal' boiling point molecules of a liquid can only evaporate from the surface of the liquid as they have sufficient energy to transition from liquid phase into the gas phase. Next, as the temperature is increased, the molecules have higher kinetic energy, overcome interactive electrostatic (+/-) attractions and escape faster producing higher vapor pressures. This will continue until reaching the boiling point at which ALL of the molecules of a given homogeneous, single component liquid have sufficient energy to escape into the vapor phase. Under these conditions, the kinetic energy content of the molecules of liquid and kinetic energy of the molecules in the vapor phase are equal. Such is why no temperature change occurs even with continued heating so long as the two phases are in contact. (Suggest reviewing the Heating Curve for Water. See link: http://chemistry.bd.psu.edu/jircitano/heatcurv.html ) For boiling water, the bubbles observed are mostly dissolved impurities ( e.g., N2 (g), O2(g), etc.) converting from liquid phase to gas phase as heating decreases solubility. It's interesting to note that a pure, uncontaminated liquid (with no impurities) would not, theoretically, produce bubbles. Evaporation would continue from the surface until all liquid was converted into gas phase.

Vapor pressure is also dependent upon molecular geometry, polar character and molecular size of the substance of interest. Typically, molecules that have a relatively high dipole character have stronger electrostatic attractive interactions (& higher boiling points) than molecules that have weak or non-polar character. Molecules with low particle-particle interactive forces require less energy (heat) to effect escape from the liquid phase and hence lower boiling points. Molecules with high particle-particle interactive forces require more heat energy to escape from the liquid phase and hence higher boiling points. Example, water is highly polar and boils at 100 Celsius (~760-mm Hg) has a mole weight of 18-amu. However, Hydrogen Sulfide liquid-form with molecular mass 34-amu boils at -60 Celsius (~760 mm Hg), Hydrogen Selenide ( mole wt. 81-amu ) at -40 Celsius and Hydrogen Tellurinide ( mole wt. 130-amu ) at ~0 Celsius. The high boiling point of water is due to electrostatic Hydrogen Bonding where as the other Group-VIA elements increase boiling points due to increasing molecular weights as their particle-particle interactions are much weaker than water. The same trends can be seen in Group-VA and Group-VIIA elements. Group-IVA (Carbon Group) hydrides are exclusively dependent upon molecular size as all have non-polar character.

Suggest further reading on how vapor pressure of solutions, and hence boiling points, are affected when a soluble solute is introduced into a liquid. Example, dissolving NaCl (ionic compound) vs a non-volatile polar solute (molecular compound). I also invite you to visit my website for lectures and additional info on many topics in chemistry, www.chemunlimited.com. (It's free and open for anyone interested).

James Pelezo said:
http://chemistry.bd.psu.edu/jircitano/heatcurv.html ) For boiling water, the bubbles observed are mostly dissolved impurities ( e.g., N2 (g), O2(g), etc.) converting from liquid phase to gas phase as heating decreases solubility. It's interesting to note that a pure, uncontaminated liquid (with no impurities) would not, theoretically, produce bubbles. Evaporation would continue from the surface until all liquid was converted into gas phase.
I don't know about anybody else, but this statement definitely does not seem correct to me. I looked up the solubility of air in 100 cc liquid water at 30 C in the Chemical Engineer's Handbook, and the value was < 5 cc (at 1 atm.). That means that, for half a liter of water, there would be < 25 cc of dissolved air. That would be enough to fill < 25 bubbles of volume 1 cc each. This would be a 1 shot deal. After that, there would be negligible dissolved air left.

We were taught that the contents of the bubbles in boiling water are overwhelmingly water vapor, and that makes abundant sense to me. As far as nucleation sites are concerned for bubbles of water vapor to form, unless very special measures are taken, there are always tiny particles of dust present.

Chet

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From a practical point of view, Chet, you are right. It is virtually impossible to generate an absolutely pure sample of water. I did emphasized 'theoretical' in my post. Consider for a moment, that you 'do' have an absolutely pure, uncontaminated sample of water, which molecules would leave the liquid state 1st? I still support surface molecules 1st out. Without nucleation the molecules below the surface would have higher particle-particle interactions than surface molecules and ( in the absence of nucleation ) the surface molecules would, again 'theoretically', escape 1st followed by the next deepest layer. This then begs the question, what conditions would have to be present for water NOT to exist in liquid form. I think the standard answer is 'critical temperature' of the phase diagram (374oC, 218 atm). The conditions on the lab bench are way below that. So, that leads to the question, what is initiating the spontaneous conversion of liquid water into a bubble of water vapor? Since ALL of the molecules of water have the same kinetic energy and if spontaneous transition of water vapor occurs generating small bubbles (let's assume for a moment) without the benefit of nucleation, then what's to prevent the entire quantity of water ( half liter = 500 gms ) from 'exploding' into the vapor phase. In 50-yrs of studying chemistry, I have never seen that happen from just boiling water. So something has to be initiating the transition from vapor phase to gas phase; i.e., nucleation. Also your statement, "That means that, for half a liter of water, there would be < 25 cc of dissolved air". From the same 'Handbook of Chem Engineering', the %O2 is 20.976% of air by wt. If <25cc air (density ~ .001225 g/cc) would give < 0.031 g Air/cc water x 20% < 0.0061 g Oxygen/cc water. Since aquatic life thrives quite nicely it would appear that 'some' quantity of gas (especially Oxygen) is dissolved in water to support the life. A simple experiment can give some credence to the dissolved gases idea. Take a half-liter of water and bring it to boil and allow boiling to continue for ~15-minutes. Cover container and allow it to cool to room temperature and repeat boiling. You will see fewer bubbles generated during the second heating. A third and fourth will reveal even fewer. Have a good day.

James Pelezo said:
From a practical point of view, Chet, you are right.
Thanks James. I think that was all that bubblewrap (the OP) was interested in. He was obviously very confused to begin with. I just hope that our idealized theoretical discussion has not rekindled his confusion.

Chet

Chestermiller said:
Thanks James. I think that was all that bubblewrap (the OP) was interested in. He was obviously very confused to begin with. I just hope that our idealized theoretical discussion has not rekindled his confusion.

Chet
I hope not too... but, it's great to have someone to debate the science issues... Thanks for being there for all interested in science. Thanks for the reply. jp

James Pelezo said:
Since ALL of the molecules of water have the same kinetic energy

Nope. No idea what is the exact distribution of energies, but energies of water molecules are not the same. Think Maxwell–Boltzmann distribution (for gases, so won't work here, but it deals with exactly the same conceptual problem).

Chestermiller said:
I don't know about anybody else, but this statement definitely does not seem correct to me. I looked up the solubility of air in 100 cc liquid water at 30 C in the Chemical Engineer's Handbook, and the value was < 5 cc (at 1 atm.). That means that, for half a liter of water, there would be < 25 cc of dissolved air. That would be enough to fill < 25 bubbles of volume 1 cc each. This would be a 1 shot deal. After that, there would be negligible dissolved air left.

We were taught that the contents of the bubbles in boiling water are overwhelmingly water vapor, and that makes abundant sense to me. As far as nucleation sites are concerned for bubbles of water vapor to form, unless very special measures are taken, there are always tiny particles of dust present.

Chet

The bubbles on surfaces BEFORE boiling are mainly oxygen and nitrogen with some water vapour.
At 100C and atmospheric pressure, the amount of dissolved gases in water should be practically nil.

It should be important to note that oxygen is more soluble in water than nitrogen, so the bubbles forming before boiling have a greater O2/N2 ratio than the 20/80 in the air.

Borek said:
Nope. No idea what is the exact distribution of energies, but energies of water molecules are not the same. Think Maxwell–Boltzmann distribution (for gases, so won't work here, but it deals with exactly the same conceptual problem).

Borek you are right if describing the distribution of Kinetic Energy of particles in the gas phase as the Maxwell-Boltzman distribution does deal with distribution of kinetic energy of gas phase particles. However, for water particles in the liquid phase at boiling temp, all molecules in the liquid phase do have the same kinetic energy. Such gives all of the molecules in a sample of water equal chance of transitioning into the gas phase. As they transition into the gas phase they do lose energy on expansion. As described by the Maxwell-Boltzman distribution. Particles at the Vaporization Energy transition into the gas phase. At 1 atm, this energy has been determined to be 40.7 Kj/mole water;i.e., Heat of Vaporization. To sustain boiling, a continuous application of energy is applied to maintain sufficient kinetic energy to sustain the transition into the gas phase. Remove the heat and the boiling stops as the kinetic energy decreases with cooling. A simple test that indicates that all of the particles have the same Kinetic Energy is that during boiling the measured temperature remains constant through out the liquid so long as two phases are in contact during boiling. That is, measuring the temperature with a thermometer finds that it is the same at all parts of the liquid. If the Kinetic Energy of different points in the liquid were not the same, different temperature values would be noted.

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James Pelezo said:
However, for water particles in the liquid phase at boiling temp, all molecules in the liquid phase do have the same kinetic energy.

Do you mean that energy of water molecules in the liquid phase is always identical for all molecules (no distribution at all), or do you mean the distribution disappears at the boiling point? Neither looks correct to me.

James Pelezo said:
If the Kinetic Energy of different points in the liquid were not the same, different temperature values would be noted.

No. Molecules of gas follow Maxwell -Boltzmann distribution, yet nobody suggests different temperatures in different points.

Chestermiller
Borek said:
Do you mean that energy of water molecules in the liquid phase is always identical for all molecules (no distribution at all), or do you mean the distribution disappears at the boiling point? Neither looks correct to me.
No. Molecules of gas follow Maxwell -Boltzmann distribution, yet nobody suggests different temperatures in different points.
I'm not talking about gas phase particles, I'm referring to all liquid phase particles at the normal boiling point. I'd venture to guess that the gas particles at the interface of the l/g phase transition might be the same, but the kinetic energy is equal in all particles in the liquid phase.

You have not answered my questions:

Do you mean that energy of water molecules in the liquid phase is always identical for all molecules (no distribution at all), or do you mean the distribution disappears at the boiling point?

Gas was mentioned only to show that your earlier notion (about different temperatures at different points) is wrong. Fact that kinetic energies of molecules in a phase are different doesn't imply that the temperatures in different places are different. It clearly doesn't work this way for gases, why should it be true for liquids?

Let me ask a different way: if all of the molecules of water have the same kinetic energy, which molecules are going to boil when and why?

The reality is as Borek said: at all times, there is a Maxwell -Boltzmann and in addition there can be othere differences throughout the liquid, such as higher temperature closer to the heat source and further below the surface where the pressure is higher.

Borek said:
Gas was mentioned only to show that your earlier notion (about different temperatures at different points) is wrong. Fact that kinetic energies of molecules in a phase are different doesn't imply that the temperatures in different places are different. It clearly doesn't work this way for gases, why should it be true for liquids?
This is a common mistake that you are noting with regard to the term "velocity distribution." At the macroscale, velocity distribution refers to the variation of average molecular velocity with spatial position. At the molecular scale, velocity distribution refers to the distribution of molecular speeds at a given spatial location (e.g., the Boltzmann distribution).

A similar error is made by people new to polymer science with regard to the term "molecular weight distribution." They think it refers to the variation of average molecular weight with respect to spatial position, when actually it refers to the distribution of polymer chain molecular weight at a given spatial location.

Chet

Borek said:
You have not answered my questions:
Gas was mentioned only to show that your earlier notion (about different temperatures at different points) is wrong. Fact that kinetic energies of molecules in a phase are different doesn't imply that the temperatures in different places are different. It clearly doesn't work this way for gases, why should it be true for liquids?

OK, two points seem to be in this issue of discussion, energy distribution in gas phase particles and energy distribution in liquid phase particles. Gases 1st: While it's true the Boltzman-Maxwell distribution of particle energies vs numbers of particles is an illustration of particle energy distributions, it is a common understanding that the Kinetic Energy of a mole of any gas equals (3/2RT). This suggests that Kinetic Energy (homogeneous gas-phase system) is directly proportional to Temperature and changes in Kinetic Energy would, in fact, demonstrate different temperature values. The average kinetic energy of a molecule, which by definition is (1/2mv2), is obtained by dividing (3/2RT) by Avogadro's number, NA. This gives gas phase kinetic energy as (1/2mv2 = (3/2RT/2NA) that is, according to this equation, temperature dependent.

2nd, liquid phase: In a sample of homogeneous a single component system in liquid phase, the energy content is described by (mc∆T) and has a temperature change factor included. However, when a system is in phase transition;i.e., liquid phase => vapor phase, energy input goes exclusively to providing the kinetic energy to effect vaporization and temperature change is not observed under such circumstance. The results of a common experimental investigation of this issue is verified in the 'Heating Curve' for a given mass of water. Tracing, say energy change as a function of temperature change of a finite mass of water from below freezing to a few degrees above boiling (liquid-gas transition) reveals that single phase components demonstrate a temperature change with the input or extraction of heat. However, when two phases are in contact, no temperature change is observed and heat content (Kinetic Energy) for the entire mass quantity of interest is given by (m∆Ht), where ∆Ht is the phase change transition constant. This energy is uniform and distributed homogeneously throughout the liquid-phase system so long as energy input continues and no changes in atmospheric pressure are imposed. Here is a link that may better clarify energy changes observed experimentally. http://chemistry.bd.psu.edu/jircitano/heatcurv.html I specifically direct your attention to 'Step 2' statement, "The heat does not increase the kinetic energy of the molecules, so the temperature remains constant. As long as there are H-bonds to break (as long as there is solid present), the temperature cannot increase". Also, here's another link that applies the Boltzman-Maxwellean Distribution to factors required for reaction initiation including phase transition issures. http://www.chemunlimited.com/The Arrhenius Equation.html
Hope this helps, jp

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I don't think this helps because we all agree with this. What we don't agree with is that the kinetic energies of all the molecules in a thermally homogeneous liquid are identical to one another, particularly at the boiling point. This was our (Chet, Borek, Russ) understanding of what you were saying. If it wasn't what your were saying, please excuse us for our misinterpretation. Please clarify.

Chet

## 1. What is vapor pressure in open systems?

Vapor pressure is the pressure exerted by a vapor in equilibrium with its condensed phase (liquid or solid) in a closed container at a specific temperature. In open systems, the pressure is affected by external factors such as temperature and volume, and the vapor can escape into the surrounding environment.

## 2. How does temperature affect vapor pressure in open systems?

As temperature increases, the vapor pressure in an open system also increases. This is because higher temperatures provide more energy to the molecules of the liquid or solid, allowing them to escape into the vapor phase more easily.

## 3. What is the relationship between volume and vapor pressure in open systems?

In open systems, an increase in volume leads to a decrease in vapor pressure. This is because when the volume increases, there is more space for the vapor molecules to spread out and escape into the surrounding environment, resulting in a decrease in pressure.

## 4. How does the type of substance affect vapor pressure in open systems?

The type of substance can greatly affect the vapor pressure in open systems. Substances with weaker intermolecular forces, such as nonpolar molecules, tend to have higher vapor pressures because their molecules can escape into the vapor phase more easily. On the other hand, substances with stronger intermolecular forces, such as polar molecules, have lower vapor pressures.

## 5. What are some real-life applications of understanding vapor pressure in open systems?

Understanding vapor pressure in open systems is important in various industries, such as in the production of food and beverages, pharmaceuticals, and chemicals. It is also crucial in weather forecasting, as vapor pressure plays a role in the formation of clouds and precipitation. Additionally, it is essential in the design and operation of equipment that involves the use of liquids or gases, such as refrigerators and boilers.

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