Coefficient of thermal expansion for van der Waals gas

[Kidphysics]

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

Homework Statement

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

Homework Equations

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

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.

 

[Chestermiller]

Homework Statement

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

Homework Equations

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

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.

The coefficient of thermal expansion is defined as:

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

 

[Kidphysics]

The coefficient of thermal expansion is defined as:

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

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!

 

[Chestermiller]

Who says you have to do it explicitly?

 

[Kidphysics]

Who says you have to do it explicitly?

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?

 

[Chestermiller]

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.

 

[Kidphysics]

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.

Pretty awesome stuff Chestermiller I appreciate it.