If the partition function for some system is given as $e^{\text{$\alpha $T}^3V}$; please note note that $\alpha$ is a constant.

I have computed $$\left[\frac{\text{$\delta $Z}}{\text{$\delta $T}}\right]_v=e^{\text{$\alpha $T}^3V}\text{(3}\text{$\alpha $T}^2\text{V)}$$

Could someone give me a hint to derive an expression for the pressure and internal energy U of the system?


1 Answer 1


You have to find out what kind of statistical ensemble you are dealing with.

As soon as you know that, you can get the corresponding thermodynamical potential from the knowledge of the partition function.

When you know the potential, you know everything!

EDIT: Since you don't know to which ensemble this partition function $\mathcal{Z}$ belongs to you can make a guess. The exponent does not depend on $N$, nor does it depend on $\mu$. All thermodynamical potentials are connected via Legendre transformations. You have

$$F(T,V,N)= U - TS $$ $$\Omega(T,V,\mu) = U - TS - \mu N = F - \mu N $$

You get $F$ from the canonical ensemble and $\Omega$ from the grand canonical ensemble via

$$F = - k T \ln\mathcal{Z}_c $$ $$\Omega = - k T \ln\mathcal{Z}_{gc} $$

Depending on which ensemble you have the ignorance of $\mathcal Z$ with respect to $\mu$ and $N$ leads to either $N=0$ or $\mu = 0$. This is a direct consequence from the first law of thermodynamics that gives you expressions for partial derivatives

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

And no matter which one you choose now you will get $F=\Omega$ from the above relations.

  • $\begingroup$ If I am not given the statistical ensemble to work with, do I make an assumption? $\endgroup$
    – Physkid
    May 8, 2015 at 11:04
  • $\begingroup$ You could make an assumption based on the variables that appear in the exponent. As there is no dependency of the chemical potential or the particle number I think you can assume that $\mu = 0$, from which follows that $\Omega = F$. So you can decide your favourite ensemble and the results will be the same $\endgroup$ May 8, 2015 at 11:13
  • $\begingroup$ I have to admit I'm lost. Could you elaborate? $\endgroup$
    – Physkid
    May 8, 2015 at 11:22
  • $\begingroup$ I edited the elaboration into my answer $\endgroup$ May 8, 2015 at 11:34
  • $\begingroup$ But how does pressure emerges from the given partition function in the OP? What are the areas I ought to be reading to acquaint myself with the relation between the thermodynamics potential and the partition function $\endgroup$
    – Physkid
    May 8, 2015 at 11:46

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