Work in reversible expansion of Van der Waals gas with mobile wall

[Jalo]

Homework Statement

Consider a Van der Waals gas.
Consider a recipient of volume 2V, with a mobile wall (with no friction) that divides the recipient in two, each part having exactally N particles. The system is at equilibrium and the mobile wall is exactally in the middle of the recipient.

Consider that the system is in contact with a heat reservoir of temperature T.
Now imagine reversible work is realized on the system so that the volume of oe of the parts increases ΔV.Find the work done by the system. (The work given in a reversible process to a system is equal to the decrease of the Helmholtz free energy)

Homework Equations

F=U-TS

U=$\frac{3}{2}$NKbT - a$\frac{N^2}{V}$

F(T,V,N)=-$\frac{aN^2}{V}$-NKbT [ log(V-bN) + $\frac{3}{2}$log($\frac{3}{2}$KbT)-log(N)+log(c)-$\frac{3}{2}$ ]

The Attempt at a Solution

I found the function F(T,V,N) in the first exercise. Since the work done by the system is equal to the decrease of the Helmholtz free energy ( dW = -dF ) I just calculated

W=ΔF=F(T,V+ΔV,N) - F(T,V,N)

The answer I got was incorrect tho... The correct answer is:

W=aN²($\frac{2}{V}$-$\frac{1}{V+ΔV}$-$\frac{1}{V-ΔV}$)+NK_BT log($\frac{(V-bN)^2}{(V+ΔV-bN)(V-ΔV-bN}$)

I'm a little confused as to why my resolution isn't correct.. If anyone could give me a hand I'd appreciate.

Thanks.

 

[Chestermiller]

The OP had the right idea, but he didn't consider the work done on both compartments; he only considered one compartment. The equation for the change in Helmholtz free energy for both compartments should have been:
$$\Delta F=F(T, V+\Delta V,N)+F(T,V-\Delta V,N)-2F(T,V,N)$$