# Dividing the process of water vaporizing at 25°C and 1 atm into "steps" to calculate entropy of vaporization

On page number 297 and 298, Peter Atkins, in Chemical Principles writes, in essence, that to calculate phase change entropy at any temperature of a substance, we can use the fact that entropy is state function and hence can divide the the actual process into series of "steps" of which we know how to calculate the change in entropy, but these steps must led us to same final state as that of process we are interested,. Then the change in entropy we wanted will be just sum of the change in entropy of the steps.

To explain this idea he takes an example of finding change in entropy of water vaporizing at 25°C and 1 atm. He divides the actual process into the following steps:

• heat the liquid to its normal boiling point, 100°C,
• allow it to vaporize,
• then cool the vapor back to 25°C.

I have problem in the last step; how will the last step led us to same final conditions as when we vaporize the water at 25°C and 1 atm? It is clear than when we heat water the energy is used to do "phase-transformation", in other terms we provide enough energy to molecules of water so that they are just able to overcome the intermolecular forces. Clearly the heat energy is transformed into kinetic energy of vapours molecules. Hence the temperature of water vapour must be higher than that of water, which remains constant, throughout the process vaporization, i.e. 25°C. So, temperature of water vapours is not 25°C. Hence, the final states are not same in both the cases.

But this is just what I think, the author is a very well know chemist so he must be right and I must be wrong, but I don't see where I am wrong.

• Aren't they? In both cases you have a bunch of water vapour at 25 degrees C Commented Jul 15, 2021 at 10:14
• user253751 I don't understand you. Do you mean to say that in both cases water vapour will have same temperature i.e. 25 C? If yes, then please explain me how. I am not able to see how they will have same temperature. Thanks. Commented Jul 15, 2021 at 10:19
• oh I see, I misunderstood Commented Jul 15, 2021 at 10:27
• You had lots of doubts about this. One doubt you failed to express as the following: Is 25 C and 1 atm really a thermodynamic equilibrium state for water vapor, or is it a hypothetical ideal gas state? Commented Jul 15, 2021 at 12:48

## 2 Answers

The 25C atmospheric air acts as a thermal reservoir, i.e., a constant temperature source or sink, because of its large heat capacity. Thus the end state for the water vapor is 25C water vapor in air, not greater than 25C.

Evaporation occurs because the kinetic energies of the individual water molecules are distributed about the average kinetic energy which is the basis of the bulk temperature of the water. In other words, individual water molecules can have KE above or below the average. At the surface of the water those molecules having high enough kinetic energy to overcome intermolecular bonds can escape into the air as water vapor. That reduces the average KE of the remaining molecules near the surface thus reducing the local temperature of the water at the surface. This is the basis of evaporative cooling.

Heat then transfers to the surface from the higher temperature bulk water below the surface as well as from the air above. Assuming the heat capacity of the air is much greater than the water being evaporated, the air acts as a constant temperature reservoir. Thus the bulk of the water during evaporation as well as the water vapor escaping into the air, come into thermal equilibrium with the air at 25C.

Hope this helps.

Sorry for my poor english! It is not my native language.

During vaporization of a pure substance, at an certain pressure, liquid and its vapor are at the same temperature.

But I think that @Chet Miller has indicated the heart of the matter: this is a standard change of entropy. In France, the notation is very precise and conforms to what the IUPAC recommends : $$∆_r S°$$. This is not a real variation in entropy $$∆S$$ between two physical states. Standard states are hypothetical states, which cannot be shown in a test tube, but which can be assigned a numeric value theoratically.

And it is very useful to have the standard enthalpy of vaporization of water at $$25 ° C$$ and $$P ° = 1$$ bar if you want, for example, to compute the higher calorific value from the lower calorific value.