## ΔS of Phase Changes

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

Ariel Fern 2B
Posts: 105
Joined: Fri Aug 30, 2019 12:17 am

### ΔS of Phase Changes

When a question asks to calculate the entropy of a phase change at a certain temperature, why do we bring the temperature back down to the specified temperature? Does it stay in that phase when you bring the temperature back down?

For example, for calculating the ΔS of vaporization of water at 85°C, we have three steps:
1. Change Temperature to 100°C
2. Entropy of Vaporization
3. Change Temperature back to 85°C

Why do we do this last step, and does the water stay in its gaseous state? Thanks so much!

Andrew Pfeiffer 2E
Posts: 101
Joined: Sat Sep 28, 2019 12:16 am

### Re: ΔS of Phase Changes

We do this last step because we want to calculate the entropy of vaporization at that specific temperature. If we didn't do the last step, we would only be calculating the normal entropy of vaporization after heating up the liquid.

Jielena_Bragasin2G
Posts: 104
Joined: Sat Aug 24, 2019 12:18 am
Been upvoted: 1 time

### Re: ΔS of Phase Changes

We have to add all these changes in enthalpy because we cannot do q=n*C*deltaT because there is no delta T, since there us no change in temperature. We have to do the last step because this will give us the change in enthalpy at 85 degrees C. Since enthalpy is a state function, adding all these enthalpies gives us the enthalpy needed for this reaction to occur at 85 degrees C.

Rida Ismail 2E
Posts: 139
Joined: Sat Sep 07, 2019 12:16 am

### Re: ΔS of Phase Changes

This method is used when you want to calculate the entropy of vaporization for a value other than the known or given. In this case, the temp we want to find is at 85 degrees Celcius. So we need to bring up the temp to 100, use the known entropy of vaporization of 100-degree Celcius, and then cool it back to 85. this works because entropy is a state function and it doesn't matter what path was taken to get the value.

Gabriella Bates 2L
Posts: 113
Joined: Thu Jul 11, 2019 12:15 am

### Re: ΔS of Phase Changes

We complete this last step because we want the entropy change at that specific temperature. When we cool the species back down to the original temperature, it completes the cycle, giving up the total entropy required to change the phase at the given temperature. If we don't cool it back down, then we would simply calculate the phase change after raising the species to 100 degrees, opposed to the phase change when the temperature stays at 85 degrees.

Philip
Posts: 100
Joined: Sat Sep 07, 2019 12:16 am

### Re: ΔS of Phase Changes

We calculate the last step because the change of entropy is a state function and we only care about the initial and final state of the reaction. In this case, we want the entropy change of the water evaporating at 85 degrees so we need the water to heat up and change phase and then bring the temp of the vapor down so we end up with water evaporating with 85 degree temperature.