Solving for Virial Coefficients: Find B2 & B3

[Clau]

Homework Statement

A gas obeys the equation of state

$$(P + \frac{a}{kTv^2})(v-v_{0})=kT$$.

Where a and v0 are constants and v=V/N is the volume per particle.

Find the second and third virial coefficients for this equation of state.

Homework Equations

$$B_{2}=V( 1/2 - Q_{2}/Q_{1}^2 )$$

$$B_{3}=V^2[ 2Q_{2}/Q_{1}^2 (2Q_{2}/Q_{1}^2 - 1) - 1/3(6Q_{3}/Q_{1}^3 - 1)$$

$$Q_{n}$$=canonical partition function of a subsystem of n particles.

The Attempt at a Solution

I was looking to the virial expansion:

$$PV/nRT = 1 + B(T)n/V + C(T)n^2/V^2 +...$$

In this expansion B(T) is the 2nd virial coeff., and C(T) is the 3rd virial coeff.
I was trying to find some relationship between this equation and the equation of state that was given in the problem.
My question is: how can I start this problem? What is the first thing that I have to do to find the virial coefficients?
Any hint will be apreciated.

 

[Clau]

I solve the problem this way...

Solving to P:
$$P=NkT/(V-Nv_{0}) - aN^2/(kTV^2)$$

The compressibility is:
Z=PV/NkT

Multilplying both sides by V and divide by NkT:

$$Z=PV/NkT=1/(1-Nv_{0}/V) - aN/(k^2T^2V)$$

For very low density
$$Nv_{0}/V << 1$$
Using approximation: 1/(1-x) ~ 1+x

$$Z= 1 + Nv_{0}/V - aN/(k^2T^2V) = 1 + (N/V)(v_{0} - a/k^2T^2)$$

So, the second virial coefficient is:

$$B_{2}(T)= v_{0} - a/k^2T^2$$

Is it right? And, how can I find the third virial coefficient?