Relating molar volume to fugacity in solution thermodynamics

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 2K views
danago
Gold Member
Messages
1,118
Reaction score
4
Hey. I am currently studying chemical thermodynamics and have reached the section on solution thermodynamics.

For a pure liquid:

[tex] \left(\frac{\partial g}{\partial P}\right)_T=RT \left(\frac{\partial ln(f)}{\partial P}\right)_T[/tex]

Where g is the molar gibbs energy
P is pressure
T is temperature
R is the ideal gas constant
f is the fugacity

My first thought was to make use of one of the fundamental thermodynamic relations:

[tex]dg = v dP - s dT \Rightarrow \left(\frac{\partial g}{\partial P}\right)_T=v[/tex]

Where v is the molar volume and s is the molar entropy.

Anybody have any suggestions for a next step? :smile:

Thanks in advance,
Dan.
 
Physics news on Phys.org
Ok, just as i posted this i had an idea.

Fugacity is defined by the equation:

[tex] g-g^{o}=RT ln (\frac{f}{f^o})[/tex]

Since the reference state is fixed (i.e. [tex]dg^o=df^o=0[/tex]):

[tex] dg = RT d(ln f)[/tex]

Equating this with dg from the fundamental thermodynamic relation i mentioned in the first post (with dT=0 since temperature is being held constant):

[tex]RT d(ln f) = v dP \Rightarrow \frac{d(ln f)}{dP}=\frac{v}{RT}[/tex]

Substituting in (from the first post)

[tex] \left(\frac{\partial g}{\partial P}\right)_T=v[/tex]

will give the required relationship.

Does that look right?