# Thermodynamics (entropy) of boiling a liquid

1. Aug 12, 2010

### Moogie

Hi

I would like to consider the thermodynamics of boiling a liquid to a gas.

I am assuming there are no intermolecular forces in a gas. I am also presuming intermolecular forces in a liquid are exothermic, though people hardly ever talk about the energy of IMF

Lets assume the the boiling point of the liquid is X. I am assuming that at temperatures lower than X, the increase in the entropy of the surroundings from the heat released by the intermolecular forces is the thermodynamical reason why the substance exists as a liquid. As the temperature of the system and surroundings increases, the IMF forces are broken because the increase in positional disorder from the substance existing as a gas now outweighs the 'thermal disorder' from the energy released from the IMF (as heat energy is less significant at higher temperatures because$$\Delta$$s=q/T). The substance then exists as a gas.

Is this correct?

2. Aug 12, 2010

### Ygggdrasil

Yes, that's a good way of looking at it. Boiling a liquid requires the transfer of heat from the surroundings to the system as the heat is needed to break the intermolecular bonds holding the liquid together. If the loss of entropy of the surroundings due to heat transfer is outweighed by the gain in entropy from going from a liquid to a gas, the process becomes favorable.

An alternative way of looking at it is to consider the free energy of the system. From the second law of thermodynamics, we can derive the expression that ΔG <= 0 (for a system at constant pressure). That is, the free energy of a system is always decreasing (unless non-PV work is performed on the system, for example, electrical work is done). Now, based on the definition of Gibbs free energy, we can break down ΔG into two components, a component based on the enthalpy change of the system and a component based on the entropy change of the system. The exact relation is:

ΔG = ΔH - TΔS

Now, we know that, for the transition from liquid to gas, ΔHvap > 0 (it is endothermic) and ΔSvap > 0 (a gas has more entropy than a liquid). Here we can see that a crossover point from an unfavorable reaction (ΔG > 0) to a favorable reaction (ΔG < 0) occurs at a specific temperature Tvap = ΔHvap/ΔSvap. At temperatures above Tvap, boiling will occur spontaneously while at temperatures below Tvap, boiling will not occur.