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Erik Lindahl

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Free Energy<br />

• Use V=exp{ln(V)}<br />

• This gives us:<br />

pVA/pVB =<br />

exp{-EA/kT+ln VA} / exp{-EB/kT+ln VB}=<br />

exp{-(EA-T*k ln VA)/kT}/exp{-(EB-T*k ln VB)/kT}<br />

• Looks just like a Boltzmann distribution?<br />

But now it says (E-T*k ln V) instead of E?

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