I did an experiment and need help with the questions.?

Question

In the experiment, we were to find the boiling point using an one open ended microbell capillary tube attached to a thermometer which was put open end down in a melting point capillary tube. A liquid was placed in the melting point capillary tube. The melting point apparatus was heated, and the melting point was when the stream of bubbles ceased and when the liquid begins to rise in the bell.

Q1: When the stream of bubbles in the capillary tube stops, we define this as the boiling point. Explain why this is and what is happening on the molecular level.

Q2: Why does the alcohol rise into the capillary tube? Explain what is happening on the molecular level?

Answer 1

it has something to do with vapor pressure

Answer 2

Lancenigo di Villorba (TV), Italy

In some manners, you repeated experimental works executed by several english scientists (J. Black, in XVIII century and J. Daniell, XIX century) in thermometric determinations.
You are interested by experimental "boiling point" determinations.

BOILING CONDITIONS
Once you determine pressure on liquid stuff, one lonely temperature's value permit phase's equilibrium between liquid phase and vapour bubbles leaving up from liquid one. This temperature value is measured by your thermometer, e.g. it is called as BOILING POINT.
In these novel conditions, the heat's amounts you fournish to this two-phase's system permits bubble's forming. In this manner, you refer to "latent heat" (e.g. Lambda) as appears in this relation :
Q = Lambda * V
where Q and V are respectively heat and volume's amounts exchanged in "ebullition phenomena".
How do vapour bubbles use this heat amounts?
Since I write about pure substance, I refer to its molecules...
....I overcome to MOLECULAR SCALE.
Liquid molecules of a pure liquid (remember, Boiling Conditions) may crush intermolecular bonds so any molecule may become "free". This process needs many energies.
In effects, when you apply "First Principle of Thermodynamics", the following relationship :
Delta E = Q + W
where Q, W and Delta E are respectively heat and work exchanges, finally the changes of "Inner Energy", e.g. a "state's function" of thermodynamical system.
As you may see, there is none changes of liquid temperature hence none changes of "Inner Energy",
e.g. Q + W = 0.
Heat fournished to boiling liquid is spent completely as "mechanical work", e.g. bubble's expansion.
At boiling conditions, some liquid portions crush intermolecular bonds and they expand in a novel form, e.g. vapour state.
Bubble derived by local collection of small vapour amount in a greater one, e.g. the bubble which is attourned by surrounding liquid mass. Thus, you can think vapour bubbles as spread collection's points of vapour amounts. Why this happens? Vapour has a VERY LOWER mass density than liquid mass, hence vapour result as a LIGHTER SPECIES than surrounding liquids. In a thin pipe, the vapour and liquid are placed in a vertical space where lighter substances get "buoyancy force" that lifts up vapour phase. The buoyancy forces determine fast rising of vapour bubbles.
Futhermore, a specified amount of liquid at boiling conditions overcome to a diminuted value of overall density, e.g. diminuted overall density once a vapour amounts have vey low density values. Finally, since overall denisty value is diminuted then you overcome to increased overall volume occupied by two-phase's system....liquid system rises in the pipe.

I hope this helps you.