# Van de Waals fluid in Free energy, Enthalpy representations

1. Feb 10, 2013

### Kidphysics

Compute the coefficient of expansion α in terms of P and V....

1. The problem statement, all variables and given/known data

Compute the coefficient of expansion α in terms of P and V for an ideal Van der Waals
gas

2. Relevant equations

(p+a/v^2)(v-b)=RT

3. The attempt at a solution

Is this as simple as solving for a? How would I go about eliminating T? I believe I have to take a derivative.

Last edited: Feb 10, 2013
2. Feb 11, 2013

### Staff: Mentor

Re: Compute the coefficient of expansion α in terms of P and V....

The coefficient of thermal expansion is defined as:

$$\alpha=\frac{1}{v}(\frac{\partial v}{\partial T})_p$$

Last edited: Feb 11, 2013
3. Feb 11, 2013

### Kidphysics

Re: Compute the coefficient of expansion α in terms of P and V....

ah yes, I should have looked that up myself I assumed it was the a in the formula. Since it seems I cannot isolate v in this equation I cannot explicitly find $$(\frac{\partial v}{\partial T})_p$$ I tried looking for some nifty maxwell's relations but I cannot find any that would be useful.. any helpful hints? and thank you for the reply!

4. Feb 11, 2013

### Staff: Mentor

Who says you have to do it explicitly?

5. Feb 11, 2013

### Kidphysics

Ok bare with me I'm not the brightest. So are you implying I should compute

∂/∂T(pv-pb+a/v-ba/v^2=RT)

and get something like p∂v/∂T-a/v^2(∂v/∂T)+ba/v^3(∂v/∂T)=R∂T/∂T

Then factor and get ∂v/∂T= R/(p-a/v^2+ba/v^3)

then

α=(1/v)(∂v∂T)p = (1/v)R/(p-a/v^2+ba/v^3)

it's in terms of p,v at least.. is this correct?

6. Feb 11, 2013

### Staff: Mentor

Looks OK, except for the omission of a factor of 2 in the ba term. If I were you, I would try playing with the final equation a little bit to see if I could combine it with the original equation in some way to manipulate it into a simpler form. If you don't feel like doing this, that's OK. Your answer is fine as it is. Nice job.

7. Feb 11, 2013

### Kidphysics

Pretty awesome stuff Chestermiller I appreciate it.