# Chemical Thermodynamics Proof

1. May 11, 2010

### danago

Hey. Im currently studying chemical thermodynamics and have reached the section on solution thermodynamics.

For a pure liquid:

$$\left(\frac{\partial g}{\partial P}\right)_T=RT \left(\frac{\partial ln(f)}{\partial P}\right)_T$$

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:

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

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

Anybody have any suggestions for a next step?

Dan.

2. May 11, 2010

### danago

Ok, just as i posted this i had an idea.

Fugacity is defined by the equation:

$$g-g^{o}=RT ln (\frac{f}{f^o})$$

Since the reference state is fixed (i.e. $$dg^o=df^o=0$$):

$$dg = RT d(ln f)$$

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

$$RT d(ln f) = v dP \Rightarrow \frac{d(ln f)}{dP}=\frac{v}{RT}$$

Substituting in (from the first post)

$$\left(\frac{\partial g}{\partial P}\right)_T=v$$

will give the required relationship.

Does that look right?