Given a themrmodynamical potential eg, the helmholtz free energy, $$F=-Nk_BT \ln(V-bN)+aN\ln V + k_BTN\left( \ln N! -\frac{3}{2}\ln T\right)$$ $a,b$ positive constants and $V \geq Nb$ and $N>>1$. How does one go about finding the equation of state?

My thinking is that the 1st law or 2nd laws of thermodynamics may help but I cannot see how.


You are on the right track. The first law of thermodynamics states that

$$\text{d} U = T \text{d} S - p \text{d}V + \mu \text{d} N$$

To get the free energy $F(T,V,N)$ you have to perform a Legendre Transformation with the respect to the variables $T\leftrightarrow S$. This will not affect the partial derivative with respect to $V$ and you get

$$ \left( \frac{\partial F}{\partial V} \right)_{T,N} = - p$$

This will be the equation of state. The left hand side represents a function of volume $V$ and particle number $N$, while the right hand side contains only the pressure $p$. This form is similar to the equation of state of the ideal gas or the Van der Waals equation.

  • $\begingroup$ Sorry I dont see where this goes $\endgroup$ – Permian May 7 '15 at 9:12
  • $\begingroup$ This will be you equation of state :) On the right hand side the pressure on the left hand side a function of volume and number of particles. Compare this for example with the ideal gas law or the van der waals gas . $\endgroup$ – sagittarius_a May 7 '15 at 9:14
  • $\begingroup$ I will include this clarification to my answer! $\endgroup$ – sagittarius_a May 7 '15 at 9:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.