Entropy/enthelpy/gibbs for vaporization of water

[Puchinita5]

This is a question from a quiz in my physical chemistry class

For 1 mole of H20, calculate \Delta{S_{sys}}, \Delta{S_{surr}},
\Delta{S_{univ}}, and \Delta{G_{sys}} for the following transition at 1 atm pressure and 95 degrees Celcius

H20 (l) ---------> H20 (g) Right now, what I'm confused about is...liquid water doesn't turn into a gas at 95 degrees at 1 atm (it's 100 degrees no?), so why would this reaction occur at all?

Is it perhaps just poorly worded? Maybe he meant to say that it STARTS at 95 degrees but will be heated to 100 degrees?

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Puchinita5]

Oh, and looking at the solution to this, there are three steps, 1 from 95-100, the transition, and one from 100 to 95! I would never have thought to have those steps because I thought T was constant.

Is it just a poorly worded question, or does this reaction actually happen at 95 degrees and I'm just totally missing the concept?

 

[Borek]

Right now, what I'm confused about is...liquid water doesn't turn into a gas at 95 degrees at 1 atm (it's 100 degrees no?),

100 deg C is a boiling point, but water evaporates at lower temperatures as well.

 

[Timo]

Oh, and looking at the solution to this, there are three steps, 1 from 95-100, the transition, and one from 100 to 95! I would never have thought to have those steps because I thought T was constant.

Is it just a poorly worded question, or does this reaction actually happen at 95 degrees and I'm just totally missing the concept?

I think the question means the process at 95 Celsius. But since you are (supposedly) not given the evaporation contribution at 95 Celsius you must calculate a different path from the initial state (liquid water at 95 Celsius) to the final state (gaseous water at 95 Celsius). This actually only makes sense if the values you are supposed to calculate are path-independent, and I strongly suggest you try to understand if and possibly why they are. Just our of curiosity: what are the results?

 

[Puchinita5]

Here is the solution...I just don't understand why all these things were done. If you could explain the whole "path" thing because that is a concept I definitely don't really get and might be why I don't understand this solution.

There are some missing parts to the solution where I couldn't read what the professor wrote. But for the most part this is all of it...

These were given also:
\Delta{H_{fH20(l)}}=-285.8 kJ/mol
\Delta{H_{fH20(g)}}=-241.8 kJ/mol
C_{pm}H20(l)=75.29 J/Kmol
C_{pm}H20(g)=33.58 J/Kmol

The solution was given in three steps:
1)
\Delta{H1}=Cp\Delta{T}=75.29 J/Kmol*(373K-368K)=376.5J

\Delta{S1}=Cp*ln(T2/T1)=75.29 J/Kmol*ln(373K/368K)=1.02J

2)

\Delta{H2}=\Delta{H_{fH20(g)_{373}}} - \Delta{H_{fH20(l)_{373}}}=?

\Delta{H_{f(373)}}=Cp(373-298)+ \Delta{H_{f(298)}}=?

\Delta{H2}= -24.8 kJ+285.8 kJ=44,000J

\Delta{S2}=\Delta{H2}/373 =44,000J/373=117.96 J/K

3)

\Delta{H3}=33.58 J/K*(368-373)=-168 J

\Delta{S3}=33.58 J/K*ln(368/373)= -.453 J

\Delta{HTotal}=44.208 J

\Delta{STotal}=118.5 J

\Delta{GTotal}=600 J

\Delta{S_{surr}}=-\Delta{HTotal}/368=-120.13

\Delta{S_{univ}}=-1.63 J/K


Yea, so if you can just provide me with some rational on why these steps are taken, I would appreciate it! I'm really just trying to understand.