Real gas, unable to reach the correct expression

[fluidistic]

Homework Statement

Consider a system of N particles contained in a volume V. The Hamiltonian of the system is $H=\sum{i=1}^N \frac{\vec p_i}{2m}+\sum _{i<j}u(|\vec r_i - \vec r_j|)$ where p_i and r_i are the momentum and position of the i-th's molecule.
1)Show that the state equation of the system is $\frac{Pv}{kT}=1+v\frac{\partial Z(v,T)}{\partial v }$ where v=V/N and $Z(v,T)=\frac{1}{N}\ln \left [ \frac{1}{V^N} \int d^3r_1... d^3 r _N \Pi _{i<j}(1+f_{ij}) \right ]$

Also $f_{ij}=f(|\vec r_i -\vec r_j|)$ with $f(r)=e^{-\beta u(r)}-1$.

Homework Equations

Relation between P and Z: $P=-\left ( \frac{\partial A}{\partial V} \right )_{\beta,N}$
Where $A=-\frac{1}{\beta}Z_N(\beta, V)$

The Attempt at a Solution

I used the relevant equations to get $A(\beta,V,N)=-\frac{1}{\beta} \ln [Z_N(\beta,V)]$ so that $P=\frac{1}{\beta}\frac{\partial}{\partial V} \{ \ln [Z_N(\beta,V)] \}$.
Hence $\frac{PV}{kT}=V\frac{\partial}{\partial V} \{ \ln [Z_N(\beta,V)] \}$.
Dividing by N I reach $\frac{Pv}{kT}=v\frac{\partial}{\partial V} \{ \ln [Z_N(\beta, V)] \}$
Now I believe that $\frac{\partial}{\partial V} \{ \ln [Z_N(\beta,V)] \}=N\frac{\partial}{\partial v} \{ \ln [Z_N(v,\beta)] \}$.
Which yields $\frac{Pv}{kT}=Nv \frac{\partial}{\partial v} \{ \ln Z_N (\beta ,v) \}=Nv\frac{1}{Z_N(v,\beta)}\cdot\frac{\partial}{\partial v}[Z_N(v,\beta)]$.
This differs from what I should have reached and I see no way to rewrite my expression into the desired one...
Any help on what's going on is appreciated.

 

[mark726]

Thank you for sharing your work on this problem. It seems like you have made some progress in understanding the equations involved, but there are a few errors in your solution that may be causing the discrepancy with the desired result.

Firstly, in the equation $P=\frac{1}{\beta}\frac{\partial}{\partial V} \{ \ln [Z_N(\beta,V)] \}$, the derivative should be taken with respect to the inverse temperature β, not the volume V. This is because the partition function Z depends on both the temperature and the volume, so we need to take the partial derivative with respect to each of these variables separately.

Secondly, when you divide by N to get $\frac{PV}{kT}=v\frac{\partial}{\partial V} \{ \ln [Z_N(\beta,V)] \}$, you should also divide the derivative term by N. This is because the derivative will also be affected by the number of particles N, and we want to keep everything in terms of the intensive variables v and β.

Finally, when you write $\frac{\partial}{\partial V} \{ \ln [Z_N(\beta,V)] \}=N\frac{\partial}{\partial v} \{ \ln [Z_N(v,\beta)] \}$, this is not correct. The derivative on the left side should be taken with respect to the inverse temperature β, not the intensive variable v. So the correct expression should be $\frac{\partial}{\partial \beta} \{ \ln [Z_N(\beta,V)] \}=N\frac{\partial}{\partial v} \{ \ln [Z_N(v,\beta)] \}$.

With these adjustments, you should be able to get to the desired result of $\frac{Pv}{kT}=1+v\frac{\partial}{\partial v} \{ \ln [Z_N(v,\beta)] \}$. I hope this helps clarify any confusion and leads you to the correct solution.