[cseil]

Hi,
I've got this problem from my book.

The vapor pressure of the ice at -3°C is 3.566mmHg.
The vapor pressure of the undercooled water at 3°C is 3.669mmHg.

I have to calculate the ΔG during the process "undercooled water -> ice (-3°C)".

I considered these transformations:

H2O(l) p=3.669mmHg -> H2O(s) p=3.669mmHg [ΔG=0]
H2O(s) p=3.667mmHg -> H2O(s) p2=3.566mmHg

but I don't get the physical meaning of what I've done and I'm really confused 

Can you give me a hint? Thank you

[cseil]

edit: sorry, I made a mistake editing the post

[mjc123]

Consider the process
water (-3°)  vapour (-3°, 3.669 mmHg)
vapour (-3°, 3.669 mm)  vapour (-3°, 3.566 mm)
vapour (-3°, 3.566 mm)  ice (-3°)
What is ΔG for steps 1 and 3?
What kind of process is step 2? What is ΔH? What is ΔS?

[cseil]

Consider the process
water (-3°)  vapour (-3°, 3.669 mmHg)
vapour (-3°, 3.669 mm)  vapour (-3°, 3.566 mm)
vapour (-3°, 3.566 mm)  ice (-3°)
What is ΔG for steps 1 and 3?
What kind of process is step 2? What is ΔH? What is ΔS?

dG is 0 for steps 1 and 3, because there's a phase transition within the equilibrium condition. But I'm not sure about it 
During the step 2 there's a change of pressure.

(dG/dP)=V
so I can say that ΔG = RTln(p2/p1).

(dS/dP)= -(dV/dT)
ΔS = -Rln(p2/p1)

[orthoformate]

The ΔG's for fusion and vaporization cancel out for 1, and 3?

[cseil]

I'm sorry but I haven't understood what you meant with that.

Anyway considering the process proposed by mcj123 I can get the right answer calculating the ΔG as RTln(p2/p1) [-15.27 cal].