Solving for Virial Coefficients: Find B2 & B3

  • Thread starter Thread starter Clau
  • Start date Start date
Clau
Messages
9
Reaction score
0

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.
 
Last edited:
Physics news on Phys.org
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?
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top