# Gibbs free energy problem

1. Apr 5, 2009

### pentazoid

1. The problem statement, all variables and given/known data

Suppose you have a mole of water at 198 K. and atmospheric pressure. use the data at the bac of this book to determine what happens to its gibbs free energy if you raise the temperature to 303 Kelvins. To compensate for this change, you could increase the pressure of the water. How much pressure would be require?

2. Relevant equations

V=(dG/dP), N and T are fixed
S=-(dG/dT), N and P are constant
delta(G)=delta(H)-Tdelta(S)

3. The attempt at a solution

H2O=> H2+.5*O2

delta(G)=-237.13 kJ/mol at 298 K(from thermodynamic reference table in back of my textbook).

I don't think I can used the equation delta(G)=delta(H)-Tdelta(S) since the temperature now changes ; therefore I turn to the equation S=-(dG)/dT. dT=303K-298 = 5 Kelvins. Not sure how to find

for the second part of the problem, I am now concerned with the amount of pressure I would add to compensate for the change in Gibbs free energy when the temperature increases. I think I would used the thermodynaimic identity equation , V=dG/dP? I would already have calculated dG from my equation S=-(dG/dT). How would I calculate my volume?

2. Apr 5, 2009

### Mapes

You don't know how to calculate the volume of a mole of water?
Also, doesn't your book have the specific or molar entropy of water?

3. Apr 5, 2009

### pentazoid

sure I know how to calculate the volume of 1 mole of water: I could either use NKT/P = V equation or used the knowledge that 1 mole of water is 18 grams which means the volume of water is 18 m^3 since density of water is 1.00 kg /m^3

4. Apr 5, 2009

### Mapes

Well, the first way only works with ideal gases...
The second way has some calculation errors or typos, but at least it's the right general approach. Does this help in calculating the necessary pressure for the desired change in G?

5. Apr 5, 2009

### pentazoid

yes since V=dG/dP