Why does bubble formation only happens at the point when vapor pressure becomes equal to atmospheric pressure during boiling? 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 suppose we add a non volatile ideal solute to water at one atmospheric pressure, we know an ideal solute doesn't change any molecular interactions. So why does it need more energy know for bubbles to form inside the water? I know according to raoult's law as surface water molecules decreases on adding ideal solute the vapor pressure also decreases, but how does vapor pressure play the role here, if it's defined for surface water molecules only? The pressure inside the bubbles should matter and the bubbles formed at 100 degrees should still have the pressure to overcome the the atmospheric pressure right? I mean to say, even though the vapor pressure due to surface water molecules is less than atmospheric pressure at 100degress, the pressure inside the bubble at 100 degrees should still be sufficient to overcome the atmospheric pressure because ideal solute doesn't changes any molecular interaction so bubble pressure should remain the same. Why do we say that when V.P(Vapor pressure)=atm. Pressure then there will be no pressure over the liquid and bubble will rise, why don't we talk about inside bubble pressure as they have some pressure too to overcome the Net pressure(i.e atm pressure-V.P)?Also as all the water vapors spread out into atmosphere ,I don't think the pressure above liquid should change at all with the time during boiling?so the confusion is that 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? Please try to explain it to me in visual manner and try to avoid using complex terminologies
 A: 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?
