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Thermodynamics - Change in the free energy

  1. Nov 24, 2012 #1
    1. The problem statement, all variables and given/known data

    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)


    2. Relevant equations

    F=U-TS

    U=[itex]\frac{3}{2}[/itex]NKbT - a[itex]\frac{N^2}{V}[/itex]

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

    3. 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([itex]\frac{2}{V}[/itex]-[itex]\frac{1}{V+ΔV}[/itex]-[itex]\frac{1}{V-ΔV}[/itex])+NKBT log([itex]\frac{(V-bN)^2}{(V+ΔV-bN)(V-ΔV-bN}[/itex])

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

    Thanks.
     
  2. jcsd
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