Gibbs Free Energy of Van Der Waals Gas

[HalfManHalfAmazing]

1. The problem statement, all variables and given/known data
Calculate the Gibbs free energy for a van der Waals gas (up to an integration constant), assuming a fixed amount of material and temperature.


2. Relevant equations
P = \frac{NkT}{V-Nb} - \frac{aN^2}{V^2}

dG = -SdT + VdP + udN (where u is the chemical potential)

3. The attempt at a solution

I am unsure of where to begin. I've tried looking at supplementary materials but everyone says stuff like "finding other free energies leads to 3rd order polynomials". If anyone can start me off I'd be grateful! Thanks!:!!)

[nrqed]

1. The problem statement, all variables and given/known data
Calculate the Gibbs free energy for a van der Waals gas (up to an integration constant), assuming a fixed amount of material and temperature.


2. Relevant equations
P = \frac{NkT}{V-Nb} - \frac{aN^2}{V^2}

dG = -SdT + VdP + udN (where u is the chemical potential)

3. The attempt at a solution

I am unsure of where to begin. I've tried looking at supplementary materials but everyone says stuff like "finding other free energies leads to 3rd order polynomials". If anyone can start me off I'd be grateful! Thanks!:!!)

It's the temperature is fixed, what does it tell you about dT? If the amount of material is fixed, what does it tell you?

[HalfManHalfAmazing]

Okay so because dT and dP are zero, we only have dG = udN. or G = uN. Thus we solve the equation of state for N and we're set?

[HalfManHalfAmazing]

If that's the case, I get: (ab/V^2)N^3 - (a/V)N^2 + (Pb +kT)N = PV. Solving for N is going to be annoying, unless there's a trick here?

[nrqed]

Okay so because dT and dP are zero, we only have dG = udN. or G = uN. Thus we solve the equation of state for N and we're set?

Why do you say that dP is zero??:confused:

The amount of material is fixed so the number of particles is not changing!

[HalfManHalfAmazing]

Oh wow. So now the only thing changing is pressure! dG = VdP. So now I solve the VdW equation of state for Volume. Plug that in and then integrate with respect to pressure? i'm going to give that a whirl! thanks!

[HalfManHalfAmazing]

Solving for V ALSO ends up being a third order polynomial which I have no idea on how to solve.

[jlragle]

This is an old chestnut of a (physical chemistry) problem. The nut is that attempting to solve for v (molal volume) gives a cubic. Use implicit differentiation or calculate dp in terms of dv to do the integration to give the chemical potential. You should be using Mathcad or Mathematica to help with the algebra and graphing. Depending on T and the vdW constants, the chemical potential may or may not contain a "loop" and the point at which the "loop" passes over to a "kink" in the curve (varying temperature) corresponds to the intersection (if you like to think of it this way) of two curves associated with two different molal volumes, i.e. a "phase transition" between two fluids of different densities. You can find this discussed in various physical chemistry books and in a couple of J. Chem. Ed. articles, q.v.

The nice thing about this "Maxwell Construction" is that it gives more insight into the nature of the phase change (crudely) represented by the vdW equation.

[Andrew Davies]

http://books.google.com/books?id=fSWEeTH1EeMC&pg=PA187&dq=calculate+Gibbs+free+energy+for+a+van+der+waals +gas&lr=&ei=Dv6NS-rZBYeuNYPX2dYM&client=firefox-a&cd=1#v=onepage&q=calculate%20Gibbs%20free%20energy%20for%20a%20va n%20der%20waals%20gas&f=false

Page 187 has a more elegant method.