Evaporation, Vapor Pressure and Boiling
ฝัง
- เผยแพร่เมื่อ 22 ม.ค. 2014
- A physical description of the process of evaporation and its consequences.
Develop an understanding of the how and why of evaporation. Once you understand evaporation, you have all you need to understand vapor pressure and how boiling is just a specific vapor pressure.
0:24 Stability of Phases
1:22 Evaporation
5:14 Vapor Pressure
7:35 Boiling
This has been, by far, the best video explaining this topic, to date!
Best video I've seen on vapor pressure yet. This has seriously helped me in understanding the correlation between boiling and atmospheric pressure and vapor pressure!
Good seethank thsnkyou
such an awesome video which has cleared that how the phenomenon of boiling really occurs and even the difference between boiling and evaporation
Thank you very much sir, this the most essential description which I see on the process of evaporation.
I've watched about 10+ videos and finally understand it now!
Glad that your persistence has paid off and that my video played a role in your success!!
Great video! Really great explanation and visuals. Usually I have a lot of questions after a video but you explained all the "why's"!
Awesom explanation! It's so clear!
This is a much better explanation than my teacher's.. Thank you so much!
I am learning about cavitation in hydraulic pumps in school and could not wrap my head around WHY it happens. Thanks to you I now have a MUCH better understanding of this. This video helped me so much! Thank you!
Glad you found the video helpful!
Awesome video. I would like to leave more than one like!
You opened my eyes. Boiling "is" a specific vapor pressure. thank you
Glad the video helped!
Yep! You are at the top of your game with this explanation. On to Valhalla!
Thanks! Enjoy Asgard!
That was just amazing! I was having difficulty with psychometric charts, even thou I understood, I thought refrigeration pressures and temperatures, now it is so easy. I am going to recommend technicians learn this before they get into all the other HVAC concepts.
Well, your post made my day. Thanks a ton and I am so happy something I made is helping so many.
many thanks - very helpful!!
In the diagram when you have the dotted and solid line, if we were to lower the air pressure above the water, the vertical line would move to the left and therefore mean more molecules have the potential to escape - less velocity/energy is needed to escape. Very enlightening. I'm looking at this to get my head around dew points and re-watching is helping.
The vertical line in the diagram at about 5 minutes is dependent on the IMF, so that energy would not be affected by a pressure change. Instead, the vertical line would change to the left or right for different substances that have weaker IMFs (to the left, less energy needed to break the IMF) or stronger (to the right).
Changing the pressure will affect the vapor pressure to external pressure ratio and that in turn affect the boiling point. That is, boiling occurs when the gas produced in bubbles creates enough outward pressure to push back on the external pressure without popping. When the pressure is higher, you need more gas particles (more vaporization) to have bubbles with sufficient outward pressure. If the gas is ideal, then PV=nRT shows us why increasing the number of particles in the gas phase will increase the outward pressure the bubble creates.
So Clear, Thanks
One of the best video to explain boiling wrt vapour pressure..
Thanks 😊
Thanks. Glad it was helpful and worth your time.
that was amazing. Thank you
How would a pressure cooker change the cycle ?
It helped me a lot thank you
That is so good explanation.
Thank you!
Thats a freaky good video
thank you SOO much!
Thank you so muchhhh professor 🫡🫡🫡🫡🫡🫡
Thank you soooo much boss love ya
very nice explanation...
Thanks for the great video. It clarified a lot of things for me.
I have a question regarding a closed container initially filled with air and liquid water. Let's say that the total pressure within the container is 1.5 atm and the temperature is at 25C. In the beginning, air contributes to 1 atm and liquid water contributes to 0.5 atm. Vapor pressure of water at 25C is ~ 0.03 atm.
So, after some time,would the total pressure within the container remain the same, meaning air contributes 1 atm, liquid water exerts 0.47 atm, and water vapors exert 0.03atm?
Also, will the water vapors experience the same amount of pressure as the liquid water in this closed container? Meaning, they all experience 1.5atm.
Appreciate the help! thank you
The pressure discussed in the video is caused by gas particles colliding with the walls of the container and represents the pressure the walls of the container experience. Thus the liquid water does not contribute to the pressure that we are speaking of (only the gas particles do). So, if the initial pressure is 1.5 atm, the pressure is caused by the air is causing 1.5 atm by nature of their collisions with the containers walls.
The air would be at the pressure same on both sides of the container and thus the net result is no total pressure (the outside pushing down on the lid would be equal and opposite to the inside gas pushing up on the lid). It's this cancelling that allows us to just consider the evaporating liquid's effect when talking about boiling. That is, there is only the inside gas pushing upwards that affects the container, since we don't expect any of the liquid molecules to be over the container pushing downward on the lid.
Very helpful...
helpful!
nice man
vary helpful
but I am very much confused in some topics, plz help
1-explain vapour pressure in closed condition (fixe boundries) that if we continuely increase tempratute, what will hapen becouse air above it can not take more vapour
2-what is saturated vapour pressure is it fixed or vary with temperature and pressure and explain in both conditions open and closed system
3-what is boiling in closed system , is it a prosses or condition ,is it instentanious ?
4-what is roll of saturated vapour pressure in boiling or both have no relation
5- is all the liquid get completely vapourised when we put liquid into vacuum
Video is good..I have one qestion..
Is boiling possible in closed rigid container with liquid being heated?? or just pressure & temprtr would shoot up in close system,but no boiling??..please carify!
Yes, it is possible. However, to boil such that the only phase of matter in the container is a gas requires that the volume of the container be large enough to not only hold the gas but that the density of the matter in the container is a density that would be stable for a gas.
Imagine a small container holding only the liquid and say that it's 90% full of the liquid. There is not enough room in the container to hold the evaporated water at high temperature at a pressure that supports vapor. Thus, this system will initially be all liquid, will quickly become a liquid-vapor mixture (phase-wise), and then finally fall back into a completely liquid state with no vapor when the pressure is so high that only a liquid state is possible.
That same container with only 10% liquid water at the outset would undergo the same transitions, but in the end might be able to reach complete vapor without creating a pressure so high that only a liquid is stable. That is, since PV = nRT (approximately), the lower number of moles of vapor produced by complete evaporation never reaches a P that causes the molecules to be so close together that they condense.
To summarize, whether a liquid heated at constant V will boil depends on the V of liquid within the V of the container and the properties of the liquid molecules (IMFs, molecular mass, etc.). It's possible for the boiling to happen as we traditionally imagine it...pure liquid, then liqud-vapor mixture, then pure gas. It's also possible that no boiling occurs...pure liquid, then liquid-vapor mixture, then pure liquid.
You might imagine that there is a Volume of liquid in a particular container that might be right on the boundary between the two processes described above. That volume and temperature pair create what is called a critical point. At the critical point, the substance has properties that are not well defined as either liquid or gas. For example, no well formed meniscus will be seen between the liquid and vapor phase, but the density of the substance will be too high to be considered a gas.
Hope this helps.
Hello. I have a question. May I infer that vapor pressure ONLY occurs in a closed container? What happens when an open container has liquid inside. Will a vapor pressure can be measured? I am having some trouble understanding this particular issue. Thanks in advance.
Well, the evaporation still occurs but the vapor created doesn't reach an equilibrium pressure. Instead, the evaporated gas can go anywhere. We usually think of vapor pressures in terms of the equilibrium between the liquid and gas phase, so Pvap is usually measured in closed containers, where that equilibrium can occur.
When it comes to boiling, the amount of vapor created must still be sufficient to create bubbles in the liquid that can withstand the atmospheric pressure pushing down. So, an open container with an atmosphere will still show a distinct boiling point (as we see when boiling water on a stove and no lid on the pot). The boiling occurs when the vapor pressure for that temperature is equal to the atmospheric pressure.
Thanks a lot for replying and congrats for the video, it's excellent.
Your video is amazing....you just answered my burning question and you have the best explanation beautiful!!!!!
Great video but I have a question.
You said that when water is heated at the bottom some molecules vaporizes forming gas bubbles and then start they to grow .Is this really true? i mean, do these bubbles form as a consecuence of water being heated or were they there from the begining, trap on the surface of the pot?
In cavitation processes (which are in nature the same thing as boilling) is it asume that the liquid is populated with microbubbles (nuclei) of gas and that at certain level of external pressure they reach a point of unestable equilibrium and its radius start growing explosively (boilling).
You also said in your video that for a molecule of liquid to scape and form gas, one condition must be the presence of a boundary gas-liquid. Could it be posible for a liquid to vaporice in the absence of any gas-liquid boundary?
Thank you very much and good work!
Juan Maria Lopez The typical boiling process begins with energy transfer into the water from the surroundings. Usually, that energy source transfers the energy in the form of heat by direct contact at the bottom of the jar. In my example, that is where the energy is entering the water. Since there will now be more energy at the bottom, there is a slight temperature gradient between the top and bottom of the liquid. That is, the bottom increases temperature slightly faster than the top. That's why the bubbles tend to form in the bottom, then attempt to rise (less dense things float).
The "growth" is purely based upon more molecules that evaporate the larger the bubble will be. Just to rephrase, the bubble is not air or helium, it's a bubble consisting of the gaseous state molecules of water.
Cavitation is a far different process, and I am far from an expert. That said, I would rather an expert answer your well founded question...I'll see if I can find someone to chime in.
No liquid gas boundary means that the container's volume is the exact same as the liquid within, with no room for any gas. That's a tall order, but I bet someone has tried to do it. If a liquid molecule were to evaporate, then the actual, physical volume of that molecule would not change. But, it's now able to exert a pressure on the walls.
Let's assume the above continues for many molecules, and you would find a high pressure (due to the small volume) situation in the area where the gaseous molecules exist. If the container can handle the pressure, I see no reason why continued increase in the energy would not create a very high pressure container filled with a very hot gas. Well, there are critical points to deal with...points where the defined states of liquid and gas get kind of blurred. As I type this, I think those critical points would be very possible so that the stuff inside might be a very dense gassy-liquidy material.
Hope this helps!!
Eric Zuckerman Thank you very much for your answer Ericm although I wouldn't say that cavitation is far a different process, I think that they are both quite the same. The scientific definition of boiling is the point where the vapor pressure meets the ambient pressure. This is the same definition as for cavitation, the only difference is the path you follow to get to that point but the physics of boilling and cavitation are the same. Nevertheless I think there is a general misconception about this subject as many people think they are very different things.
I know my question may have some trick, because there is no real liquid with a 0% of gas content and thus there will be always microbubbles present in the fluid. I've read some articles where they measure the effect of gas content on cavitating water and it seems that the more "bubble-free" the more tension water can withstand (therefore the boilling point gets higher). I've read values up to -140MPa on extremely purified water, and in theory it seems that there is an exponential correlation so, in theory, for a pure liquid continium with 0% gas content there wouldn't be boilling. I just wonder what would happen to that fluid when heated.
I think there would not be any sings of boilling, just the molecules increasing its velocity and moving away from each other till the liquid swich softly to gas.
well indeed cavitation is related to boiling ..
as fluid energy is constant (assumption taken by Bernoulli ) In a pipe if at a point velocity is more there will be decrease in pressure of surroundings due to this, pressure of vapor of fluid is more than that of external therefore boiling starts (bubble forms). when this velocity is less at some other point these bubbles which are created here will burst which results in pits corrosion & hence pipes will fail due to cavitation.
In the graph thing shoun before 5:15 is temperature proportional to velocity?
Sarim Zulkifl The temperature of a substance is an interesting quantity. It is a single number that represents the average translational kinetic energy of the substance. Since the substance is composed of many, many particles (atoms, molecules, etc), then some of those molecules may move slowly and others quickly. The temperature, then, is not related to the velocity of any particular particle in the system. To specifically answer your question, it would be acceptable to say that the temperature is proportional to the average translational velocity of the particles that make up the system.
Eric Zuckerman Thanks that clears everything.
hi im josh thx for the video
i ve a question ...... when evaporation becomes equal to condensation, vapour pressure becomes constant.......it means that it was increasing before equilibrium point could reach and when equilibrium was attained, there was no increase in vapour pressure?????
+Zubair Sadiq You are correct. Prior to the equilibrium being established (evaporation rate and condensation rate being equal), the vapor pressure would gradually rise. Of course, this assumes that we start with the pure liquid in a closed container that has no (or little) gas over it.
I could imagine the opposite set up occurring, too. Imagine liquid water with all of the gas pumped out. Then, I add a whole bunch of water vapor over the liquid. If the pressure of the vapor is greater than the vapor pressure, as determined by the temperature, then the vapor pressure would decrease until the equilibrium vapor pressure is established.
I have a question . How humidity can affect evaporation??
Aditya Bhosale Humidity is inversely proportional to evaporstion because humidity means presence of water vapours in the air and if there is water vapours in the air they wont allow the liquid molecules to turn into vapour
so in space will the water boil without any heat source as the external pressure is close to zero?
You are correct. In a vacuum, the water would boil as long as there is sufficient energy in the system for some molecules to have enough kinetic energy to overcome the intermolecular forces (hydrogen bonds, dipole-dipole, dispersion) that hold water molecules together. In other words, at room temperature we would see the water boil very quickly if opened to a vacuum. On the other hand, in interstellar space, where the temperature is below 3 K, there isn't much energy available in the system and few, if any, molecules have the ability to evaporate...well, it would be sublimation since the water at that temperature is in solid form.
Comets are known for their tails that point away from the Sun. The tails are composed of dust and other molecules from the comet that have been sublimated away due to the extra energy obtained by being closer to the Sun. It is not uncommon for that tail to have some water, ammonia, carbon dioxide, etc. in it along with the dust.
Eric Zuckerman thanks for your response 😃
Are molecules closely packed in liquids, as u have shown?
Yes. They don't "touch" each other physically as the simple model shows, but they are quite close. Both condensed phases (solid and liquid) have species very close to each other.
When does liquid evaporate easily? In a open or closed container?
In general, evaporation is not dependent on the container. That is, the rate of evaporation is dependent upon the nature of the molecule (type of IMF) and temperature. A molecule in the liquid but at the surface does not “know” of the lid on the container, so there is no difference in evaporation rate due to be open or closed.
Why vapor pressure is only a function of temperature?? I mean if we had increased the volume between the beaker and the lid, wouldn't more molecules of the liquid have evaporated?
Note that the vapor pressure is defined when an equilibrium is established between the evaporating molecules (liquid to gas) and the condensing molecule (gas to liquid). These two transitions are opposites of each other, energetically. That is, the amount of energy required to break the IMF's holding the liquid together is equal and opposite to the energy released when a gas particle establishes an IMF with the liquid. That's where the clear energetics appear and make it obvious that temperature will affect the vapor pressure. I'm assuming this part is pretty clear to you.
How about increasing the volume that the gas can move in, you ask? At first glance, it appears that there is more space and thus a lower pressure in that space (just look at Boyle's law, right)?
The problem with the above Boyle's law explanation is that it doesn't take into account that equilibrium condition. That the two processes must be at equilibrium to talk about the vapor pressure. And that equilibrium is defined as when the rate of evaporation is equal to the rate of condensation.
Let's imagine a system at equilibrium and having a constant vapor pressure. For every molecule at the liquid surface that becomes gas phase (evaporates), a gas molecule collides with the surface of the liquid and condenses. It's in this mechanism that the same number of gas molecules exist above the liquid and exert a constant pressure (the vapor pressure).
Now imagine the volume of the container increasing (maybe our system is a mechanical piston and we can control this). At this instance, the number of gas molecules hasn't changed and the temperature hasn't changed...only the volume. But the pressure would now be lower. Why? Well, the pressure is caused by the gas molecules colliding with the walls and the walls just got further apart. There is more distance to move between collisions for the average gas molecule in our system. Some of those collisions would have been with the surface of the liquid, where condensation occurs. At the instance of expansion, the condensation rate would likely decrease (less collisions with surface means less chances that a collision will happen that results in IMF's holding the gas molecule to the liquid).
The evaporation rate has not changed a bit, though. Why would it? The liquid's surface area is the same and the energy of the liquid hasn't changed due to the volume increase. So, while a decrease in condensation rate is instantly a result of the larger volume, the evaporation rate goes unchanged. That means the system is NOT at equilibrium the instance the volume increases.
In your mind, you can now imagine that the larger evaporation rate is producing more gas molecules. Maybe for every five evaporating molecules only one condenses. Increasing the number of gas molecules above the liquid. A little later, 3 molecules evaporate for every 1 that condenses. A little later, 2 molecules evaporate for every 1 that condenses. A little later, 3 molecules evaporate for every 2 that condense. Until finally there is enough gas in the container to get our two rates equal again...for every 1 evaporating liquid molecule there is 1 condensing gas molecule.
The end result is that more gas molecules now exist. The extra volume created a condition where the system would respond to the change by creating just enough gas to reestablish this physical equilibrium between evaporation and condensation. Amazingly, the amount of extra gas created to fill the empty space created is exactly the amount of gas needed to reach the same vapor pressure as we saw before the system expanded.
The only way, it turns out, to change the vapor pressure of a pure liquid is to alter the temperature. It's because the temperature of the system dictates the energetics of the two opposite processes: evaporation and condensation.
I sure hope this helps. Let me know if any of the above explanation requires further clarifying.
Eric Zuckerman
Now it makes sense... Excellent explanation! Thank You!😇
Hi Sir.GREAT VIDEO.Sir will the Vapour Pressure be the same for any amount of liquid???
Chandrachud Mishra The amount of liquid will not impact vapor pressure.
@@EricZuckerman1 than you sir
@Eric Zuckerman
Is the liquid in equilibrium with vapour at boiling point?
Yes, they are in equilibrium as long as the external pressure does not change. As more energy is added, the equilibrium shifts more to the vapor side until all of the liquid becomes vapor. Until that point, the system is at equilibrium.
Now it gets challenging to describe this for open systems, like a pot of water on a stove.
@@EricZuckerman1
I am not able to infer what's my book trying to imply. There are two statements given :
1. "At the saturation temperature, the boiling liquid is in equilibrium with its own vapour."
2. "The steam is called saturated when the molecules escaping from liquid become equal to the molecules returning to it."
In case of first statement, I think that definition is not right. The liquid can be in equilibrium with its own vapour at a temperature lower than boiling point too.
Same argument holds for second statement too.
@@keshavbassi6501 The book is telling you the story in reverse, as I read your description.
A "saturated steam" (#2) is one in which the number of moles of gas are constant. So, for every new evaporated molecule there is a different molecule that is condensing. In other words, the idea of saturation is that no more gas molecules can exist as steam at this temperature. When "saturated", the pressure caused by the vapor is constant....it is the vapor pressure. I've not heard this use of saturated before.
That brings us to #1. At some temperature, the vapor pressure caused by the "saturated vapor" will be equal to the external pressure. That temperature is the boiling point.
Remember that the liquid to gas equilibrium can occur at temperatures lower than the boiling point. Only at the boiling point does a system have enough pressure to overcome the external pressure.
To give an example, consider a pot of water on your stove at room temperature. The external pressure is caused by the air if no lid is present. If you put a lid on the pot, then the external pressure is the pressure needed to lift that lid off the pot.
I hope this helps.
@@EricZuckerman1
So basically the vapour generated at boiling point in a closed container is saturated steam and it is in equilibrium with liquid water, i.e. the molecules escaping from liquid become equal to the molecules returning to it. And, as we add more energy the amount of steam generated ( or number of moles) goes on increasing. Is it right?
@@keshavbassi6501 Be careful not to always imagine the liquid is water. In general, in a closed container any liquid will evaporate some and create a vapor pressure in the container. Even when not boiling, the system is at equilibrium.
At the boiling point (temperature), it just so happens that the vapor pressure is equal to the external pressure. The boiling point also represents the temperature and pressure combination where both the liquid and gas states are equally stable, which is why a phase transition can happen for the substance.
Otherwise, I think you have a good handle on this topic. Best of luck!!
05:14 Vapor Pressure
SARVESH SIR OPPP
it means container without lid will boil quicker than container with lid
????
Comments are back here again uyosssssssssss
Soooooo…the lighter the lid on a pot on a flame, the quicker the water will boil.
The answer to this is "yes and no". Assuming that the same amount of energy flows in per unit time, then we can discuss how fast the boiling will occur. The real takeaway is the temperature of the water when it boils will be different. So, it's not a race to boiling at a specific temperature, but rather the temperature increases until the criteria for boiling is met. With a lower external pressure, the criteria is met at a lower T.
1 FOKU$