[diablo]

Hi folks,

can anyone please help me find a way to calculate this :

calculate the change of gibbs free energy at 337.8K ( boiling point of Methanol at 1.013bar) for following reactions

1)CH₃OH(l)--> CH₃OH(l) at p= 0.9bar
2)CH₃OH(l)--> CH₃OH(l) at p= 1.013 bar
3)CH₃OH(l)--> CH₃OH(l) at p= 1.100bar

Solution: -332.18kJ/mol, 0kJ/mol, 231.4kJ/mol

I dont have any idea what formula i should use... --> do i need additional values to solve this calculation?

[diablo]

i think i've got it, but still im not sure:

obviousely it got something to do with the chemical potential:

µ = G  (for pure substance)

µ = µ₀+RTln(p/p₀)

if i assume µ₀= 0 , i get the right solution...

but still i dont understand why i can assue µ₀ as zero..

[plankk]

In task you have to calculate "the change". So you can assume that G₀=µ₀ and G_e=µ₀+RTln(p/p0) so  :delta: G=G_e-G₀=RTln(p/p0)

[diablo]

In task you have to calculate "the change". So you can assume that G₀=µ₀ and G_e=µ₀+RTln(p/p0) so  :delta: G=G_e-G₀=RTln(p/p0)

  , thx alot for your answer !