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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.

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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.

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  • $\begingroup$ Sorry I dont see where this goes $\endgroup$
    – Trajan
    May 7, 2015 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 . $\endgroup$ May 7, 2015 at 9:14
  • $\begingroup$ I will include this clarification to my answer! $\endgroup$ May 7, 2015 at 9:18
  • $\begingroup$ @sagittarius_a Can thermodynamics predict the equation of state? I mean from the axioms of the theory there is no reason one can deduce that $U = \frac{3}{2}nRT$ for ideal gas or the van der Waals equation for real gases. $\endgroup$
    – Anton
    Jul 9, 2021 at 15:13
  • $\begingroup$ I think in the question, an "equation of state" was understood as something which relates the state variables (T,V,p). I am just showing a way to arrive at such a relation if you are handed some free or internal energy potential (which you might have obtained by other means) $\endgroup$ Jul 11, 2021 at 10:34

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