# Why does temperature remain constant at the boiling point of the liquid? And what kind of an equilibrium is boiling?

When we were in school, the teachers used to tell us that heat increases the kinetic energy of the particles (reflected in the temperature) until the boiling point is reached. Afterwards, all the heat goes into breaking the intermolecular bonds between the liquid particles until all of them change phase. I realize that this is the answer, but I cannot conceptually relate to it for several reasons:

1) I do not understand if the temperature being measured is that of the liquid or of the entire system. 2) Doesn't breaking the intermolecular bond automatically mean that the particles have become faster?

I have another explanation for this phenomenon but this assumes that the temperature being measured is that of the liquid only. Please let me know if it is correct:

Since evaporation has a cooling effect, when the boiling point is reached, the rate at which the liquid cools down because of evaporation becomes equal to the rate at which heat is added to the container, thus keeping the temperature of the liquid constant.

Correct or not?

1) I do not understand if the temperature being measured is that of the liquid or of the entire system.

You have thermometer in liquid. I can't see why it will measure system's (Referring to the Latent heat of vaporization experiment we all must have witnessed in 10th grade or so)

2) Doesn't breaking the intermolecular bond automatically mean that the particles have become faster

I may not agree with you on this (see end)

Since evaporation has a cooling effect, when the boiling point is reached, the rate at which the liquid cools down because of evaporation becomes equal to the rate at which heat is added to the container, thus keeping the temperature of the liquid constant.

Now see, Evaporation is phase transition from liquid to vapors while Boiling is a phase transition from liquid to gases. More over. Evaporation may occur when the partial pressure of vapor of a substance is less than the equilibrium vapour pressure while Boiling, as opposed to evaporation, occurs below the surface. Boiling occurs when the equilibrium vapour pressure of the substance is greater than or equal to the environmental pressure. I can''t see how you are correlating them

If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance. The example we will use here is ice melting into water. Immediately after the molecular bonds in the ice are broken the molecules are moving (vibrating) at the same average speed as before, so their average kinetic energy remains the same, and, thus, their Kelvin temperature remains the same.

Note: The above explanation is borrowed from this Link

• On: "Immediately after the molecular bonds in the ice are broken the molecules are moving (vibrating) at the same average speed as before". Now I see what you are saying. May 14, 2017 at 10:00
• @dimyak sorry for the hurried typing You can follow the same logic for vaporisation model also. Please see the link May 14, 2017 at 10:03
• ...well the reason I am asking is the following: It can either be that the energy provided does work by breaking the intermolecular bond (which is doing work by applying a force over distance -meaning that the particles are as fast as before but are far apart from each other) and in this sense the system is the liquid and the gas OR that the system is both the liquid and since particles that escape take energy away with them from the system then this is how temperature remain constant. One of the two...The first one seems to be a better explanation. May 14, 2017 at 10:05

I think the description in your first paragraph is perfectly reasonable, but I'll try to give a slightly more technical explanation.

If you hold a substance at temperature $T$, the phase that it adopts is determined by a trade-off between energy $E$, which it tries to minimize, and entropy $S$, which it tries to maximize. The actual phase at a given temperature is the one that minimizes the free energy $F = E - TS$.

Entropy is a measure of the disorder of the system, and is high in the gas phase, where the molecules are essentially moving freely and independently of each other, and lower in the liquid, where the density is high and the molecules have less freedom to move. But the energy is also higher in the gas phase, where the molecules have lots of kinetic energy and roughly zero potential energy, than the liquid phase, where they have less kinetic energy and negative potential energy (because of attractive interactions between the molecules, the “intermolecular bonds”).

At high $T$, the best way to minimize $F$ is for the substance to form a gas and hence maximize $S$. At lower $T$, it instead minimizes $F$ by forming a liquid, which reduces $E$. (At still lower temperature, it forms a solid, which has even lower energy and entropy.) At some intermediate temperature, which we call the boiling point $T_b$, the two effects balance out: the energy is lower by $\Delta E$ in the liquid phase and the entropy is higher by $\Delta S$ in the gas; if $\Delta E = T_b \Delta S$ then the two phases have the same free energy, and there is nothing to choose between the two. ($\Delta E$ is called the latent heat.) At this temperature, the two phases are in equilibrium.

So imagine starting with a liquid at $T < T_b$ and heating it up. As you put energy in, the molecules start to move a bit faster and to get a little further apart, increasing both their kinetic and potential energy. But the substance is still a liquid as long as you stay below $T_b$. Once $T$ reaches the boiling point, the extra energy goes into converting liquid into gas, which costs energy $\Delta E$, instead of increasing the temperature. Both liquid and gas are then at the same temperature (as they must be, to be in equilibrium), but the molecules in the gas phase have more energy. Finally, when all of the substance has become gas, the extra energy again goes into increasing the temperature.