## Entropy Changes of a Physical State

$\Delta S = \frac{q_{rev}}{T}$

JuliaPark2H
Posts: 19
Joined: Fri Sep 25, 2015 3:00 am

### Entropy Changes of a Physical State

The book covers how to calculate the entropy of transition at a temperature other than the normal transition temperature. As an example, the entropy of vaporization of water at 25 degrees Celsius and 1bar would be: $\Delta Svap(25degrees)=\Delta S(heating liquid from 25 to 100)+\Delta S(vap100)+\Delta S(cooling vapor from 100 to 25)$
However, wouldn't the change in entropy from the heating and the change in entropy of the cooling cancel each other out, as the ln(T1/T2) is the negative value of ln(T2/T1). Then, the change in entropy would just be the entropy of vaporization, and temperature becomes insignificant..?

Ronald Yang 2F
Posts: 86
Joined: Fri Sep 25, 2015 3:00 am

### Re: Entropy Changes of a Physical State

The thing is, the molar heat capacities for water as a liquid and as a gas are different, so the cooling term and the heating term wouldn't cancel out. That's why I think for this kind of situation where you perform vaporization at a lower temperature than that of the boiling point that the total change in entropy would be positive, since the heating term includes a higher specific heat and it's overall positive (bigger, positive change in entropy), the cooling term includes a lower specific heat and it's overall negative (smaller, negative change in entropy), and the change in entropy of vaporization would be positive since the phase change is endothermic and the gas occupies a greater volume/more disorder. Thus, the total change in entropy for the vaporization of water at a lower temperature than that of the boiling point would be positive overall, not just equal to the change in entropy of the vaporization of water at 100oC.