1

I am a high school student and I am very confused about what's actually happening at the microscopic scale in an ideal solution when it's boiling? Boiling as I understand at microscopic level is- suppose we are heating water at one atmospheric pressure, it's temperature will rise but only surface molecules will have enough energy to escape because the bubble formed didn't have enough to pressure inside them as compared to atmospheric pressure , so they collapse.

But as the temperature reaches 100 degrees Celsius the pressure inside the bubble will be sufficient enough to overcome the atmosphere pressure and it will rise and pops up at the surface, via this explanation we can Clearly see if we decrease the atmospheric pressure, the pressure which the bubble need inside them to not collapse will be less so the boiling point will be less.

But what matters here is the pressure inside those bubbles and not vapor pressure (which is the pressure exerted by liquid above it's surface at any temperature of its in equilibrium) . So why bubbles can't form inside the bulk at a point where the vapor pressure of liquid is less than atmospheric pressure? I mean there should be some high energy molecules in the bulk which should be able to form bubbles which can counter their surrounding pressure and rise, but why it doesn't happen? Why we put a condition that there has to be overall 0 pressure on liquid for bubbles to form?

Shyam
  • 29
  • Did you ask your high school teacher? – Bob D Feb 27 '23 at 11:03
  • Yes, but he just said one thing when V P=atm. Pressure there will be no force over the liquid and bubble will rise...but it doesn't make sense..as bubbles have some pressure too to overcome what's above them – Shyam Feb 27 '23 at 11:07
  • Also if all the vapors mixes with the atmosphere I don't think the pressure above liquid should change with time? – Shyam Feb 27 '23 at 11:10
  • Did you express your doubts to your teacher? – Bob D Feb 27 '23 at 11:25
  • I did, but I think he doesn't have any clear answer because he is saying just what written in the textbook ,he is not giving me any visual or logical answer and just confusing me with the statements – Shyam Feb 27 '23 at 11:51
  • If you can explain it me...please help...read the last part of my question specially which is the main thing I wanted to ask – Shyam Feb 27 '23 at 11:52
  • You can find a good visual explanation here: http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/vappre.html – Bob D Feb 27 '23 at 12:03
  • Here, also same thing is written that As VP= atm. Pressure bubbles can form.....but what about the pressure inside the bubble themselves? – Shyam Feb 27 '23 at 12:14
  • Think about it. It has has to be at least equal to the water pressure around it or it would collapse. – Bob D Feb 27 '23 at 13:28
  • So you mean the inside bubble pressure is not that much and is very less than atm. Pressure or V.P??It's just above the liquid pressure around it? – Shyam Feb 27 '23 at 15:19
  • But I still don't think it's logical because even if V.P is less than atm. Pressure...how can we say no bubble with enough pressure (i.e greater than atm pressure-V.P) at that moment can form and escape? – Shyam Feb 27 '23 at 15:29
  • I'm sorry, but I don't understand you. The bubble "escapes" (rises) because of buoyancy, i.e., the upward force on the bubble equal to the weight of the liquid water it displaces. – Bob D Feb 27 '23 at 15:44
  • It escapes because of buoyancy,I know that but I am saying .....why can't it form at the moment when say VP<Atm. Pressure ...why VP has to be equal to atm. Pressure, if you say it's because only at that moment there is no pressure collapsing the bubble...but what if the bubble pressure is greater than the difference of Atm. Pressure- VP...how can you say it only has to overcome the liquid pressure around it? – Shyam Feb 27 '23 at 15:47
  • Sorry, but I still don't understand what you are talking about, so I can't help you. Maybe someone else can. – Bob D Feb 27 '23 at 15:50
  • I think I have presented my point clearly, but no problem – Shyam Feb 27 '23 at 16:55

2 Answers2

2

Shorter answer: A higher temperature is needed to boil impure liquid water because its water concentration is less than 100%; therefore, we need to apply a greater driving force to attain the 100% water concentration of water vapor. (The same effect occurs with freezing impure liquid water into pure ice; some undercooling below 0°C is required. In this sense, impure liquid water seems to be a more stable phase than pure liquid water.)

Longer answer: Let's first address a misconception:

So why does it need more energy know for bubbles to form inside the water?

It doesn't; the latent heat of vaporization of water actually decreases slightly with increasing temperature. But it certainly does take a higher temperature to boil the water with the solute. Why?

Broadly, we seek a way to determine what phase of matter is the most stable, or—even better—a general rule for how matter moves and transforms.

That rule is energy minimization, a consequence of the Second Law of Thermodynamics. (In familiar examples, stretched or compressed springs holding strain energy tend to relax when the load is removed; liquid surfaces holding surface energy tend to curve to reduce their surface area; suspended objects with gravitational potential energy tend to fall when released.)

The relevant energy here when we're controlling temperatures and pressures is the Gibbs free energy. An open liquid–vapor system is at equilibrium when its Gibbs free energy is minimized.

Considering individual phases, the partial molar Gibbs free energy is called the chemical potential. Using this new term, matter moves to where the chemical potential is lowest.

The chemical potential of a substances decreases with increasing temperature, and the rate of decrease is just the entropy of the substance. The gas phase has a higher entropy than the liquid phase, and so the curve of the chemical potential or Gibbs free energy is always steeper for the gas phase:

Again, the lowest curve corresponds to the most stable equilibrium phase at that temperature. The Gibbs free energy decreases with stronger bonding (as with solids) but also with more configurations (as with gases), and the latter tendency wins out with increasing temperature. That's why solids melt and liquids boil with increasing temperature.

The chemical potential is related to a quantity called the activity. The activity of pure condensed matter is 1; the activity of impure condensed matter is lower. The activity of a component in an ideal mixture is just its concentration.

This gives us a framework for understanding boiling of pure and impure water: With increasing temperature, the Gibbs free energy decreases for both the liquid and the gas phases, but the decrease is stronger for the gas phase. At 100°C, the Gibbs free energy or chemical potential of liquid and gaseous water is identical. This corresponds to a vapor pressure of 1 atm, enough for nucleated vapor bubbles to begin to push liquid water out of the way.

If the water is impure, then its concentration is less than 100%, and its Gibbs free energy or chemical potential is therefore lower. But the impurity is nonvolatile—meaning it doesn't evaporate—and so the vapor phase is still 100% water, and its chemical potential remains unchanged. The mismatch favors the liquid state! Its curve is now lower, and so a higher temperature is required for the curve crossover associated with boiling. The effect scales up with increasing solute concentration because the water (solvent) concentration continues to decrease.

On a molecular level, you could imagine fewer water molecules per unit surface area of the impure liquid (because the solute molecules occupy some space), and so the rate of evaporation is lower. However, the gas phase contains the same concentration of water, and so condensation occurs at the same rate regardless of the liquid water's purity. This is another way to look at the shifted equilibrium: If evaporation is suppressed whereas condensation is unaltered, the liquid phase is favored somewhat, and we need to move to a higher temperature to reach the rate of evaporation that corresponds to boiling. We need to compensate for the solute reducing the water's concentration.

Does this all make sense?

Chemomechanics
  • 25,546
  • Sorry, I didn't fully understand your answer as I mentioned I am in High school and there are some terms in your answer which I don't know about.....but on reading the last para of your answer..you said on molecular level no. Of water molecules per unit area dec. On adding non volatile solute and So does the rate of evaporation, I know this and it's raoult's law ...but as I understand boiling is characterized by bubble formation in bulk liquid...but as the solute is ideal and is not changing any net molecular bonding...how does it take more energy for bubbles to form now? – Shyam Mar 02 '23 at 17:46
  • Also if you'll say that bubbles formation doesn't take more energy...it's just that..because at 100 degrees now(on adding solute) VP is less than atm. Pressure so the net pressure(i.e atm. Pressure -VP) will try to collapse the bubble and so it doesn't get form now at 100 degrees....but bubbles have some inside pressure too...so how do we know if they can overcome the net pressure around them or no? Mainly I want to ask why do we have the condition that at the moment when VP becomes exactly equal to atm. Pressure we have boiling? Why bubbles can't form below it and..... – Shyam Mar 02 '23 at 17:50
  • why doesn't temp. Change further than this point? – Shyam Mar 02 '23 at 17:51
  • Vapor bubbles don't form below the boiling temperature because at any point in the liquid, the molecules don't have the energy for a large number of them to detach from the liquid at the same time. – Chemomechanics Mar 02 '23 at 19:07
  • Upon boiling, the temperature stays the nearly same even with strong heating because the latent heat is so large. Even a slightly higher temperature than the boiling temperature would accelerate the boiling process tremendously, removing energy and imparting cooling. It's a negative feedback effect. See here for much more discussion. – Chemomechanics Mar 02 '23 at 19:08
  • As you said...vapor bubbles don't form below the point than VP= atm. Pressure because they don't have much energy, then adding ideal solute doesn't change any molecular interactions with water molecules so why do they form at a high temperature than normal? Why do they require more energy now? – Shyam Mar 03 '23 at 17:57
  • See...the confusion is ...there is a discrepancy in boiling....on one hand we say ..that boiling temperature is when the water molecules get enough energy to move apart and form bubbles and on the other hand...we say boiling is when VP= atm. Pressure......I don't see any single point on how these two statements are related? – Shyam Mar 03 '23 at 17:59
  • Boiling/condensation is a race between the evaporation rate and the condensation rate. If you add impurities to the liquid, there aren't as many water molecules at the surface that can evaporate. This changes the race, and a larger temperature is then needed for the evaporation rate to beat the condensation rate. This is equivalent to what I wrote in my answer, but perhaps the wording was unclear. – Chemomechanics Mar 03 '23 at 18:13
  • To connect the vapor pressure to bubble formation, do a force balance on the water above the vapor bubble. As the pressure in the bubble exceeds atmospheric pressure, it begins to be capable of lifting the water out of the way. – Chemomechanics Mar 03 '23 at 18:15
  • That's what I am saying...for bubble to form..the bubble pressure has to exceed its surrounding pressure but...I don't think it's equal to VP..because we define that only from surface molecules as per raoult's law...and we say as on adding ideal impurities surface molecules decreases and hence the VP...but it shouldn't affect the pressure inside the bubble...then why it needs more temperature now to form and exceed the surrounding pressure around it? – Shyam Mar 05 '23 at 04:06
  • The vapor pressure can be defined anywhere in the bulk liquid, not just at the surface. It is $P_\text{vapor}=P_0\exp(-\mu/RT)$, where $P_0$ is a reference pressure, $\mu$ is the chemical potential, $R$ is the gas constant, and $T$ is the temperature. This is discussed here, for example. – Chemomechanics Mar 05 '23 at 05:25
  • Ok,,,,it seems clear..but just one confusion is there....how we can say that the decrease in VP (acc. To raoult's law) that arise when the surface water molecules density get decrease...how can we say that the VP inside the bubble is exactly equal to that...I mean if the impurities are suspended in bulk and it's ideal..than the molecular bonding is not changing and no. Of water molecules are also same (if we consider the whole liquid) then why it's pressure is decreasing for the same temperature and it's not forming? – Shyam Mar 05 '23 at 07:23
  • Also(not related to the question just asking) from what I understand now...as the temperature of a certain part of bulk liquid gets to the boiling point ..the vapors start forming there....the temperature of whole liquid is not same right? Otherwise the whole liquid would simultaneously turn into vapors right? – Shyam Mar 05 '23 at 07:26
  • The total number of water molecules doesn't change when you add an impurity, but the concentration goes down. The entire liquid doesn't turn to vapor instantaneously because evaporation/boiling is a cooling process; heat must be constantly delivered from elsewhere. – Chemomechanics Mar 05 '23 at 08:31
  • The picture I have in my mind is..a non volatile solute molecule is surrounded by many water molecules and the bonding between them is same as it's between two water molecules...the why I am saying still the energy require to separate them will be same as if there was no impurity...so to generate the same bubble pressure it should have taken the same energy and it should form at the same temperature.? What's missing – Shyam Mar 05 '23 at 08:56
  • The water concentration is lower in the liquid state—but not the gas state—so there’s now an imbalance between evaporation and condensation that requires a higher temperature to achieve net boiling. – Chemomechanics Mar 05 '23 at 16:50
0

When a bubble is forming in a boiling liquid, it's filled with vapor from that liquid. The bubble is in direct contact with the liquid, allowing molecules to rapidly move between the liquid and vapor phases.

Over time, the rate at which molecules evaporate from the liquid into the bubble and the rate at which they condense back into the liquid become equal. This is a state of dynamic equilibrium.

At equilibrium, the pressure of the vapor inside the bubble (the internal pressure) is effectively the same as the vapor pressure above the liquid. This is because the same process of evaporation and condensation is happening both inside the bubble and at the liquid's surface.

Aakash Bagga
  • 1
  • 2