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I am trying to determine the equation of state and see if $PV = nRT$ is satisfied. For an ultra-relativistic gas of identical particles in a volume $V$ the energy (I am assume there is no potential) is $E = c\sum_{i = 1}^Np_i$ where $p_i = \sqrt{p_{xi}^2 + p_{yi}^2 + p_{zi}^2}$. Hence I estimate the number of microstates with energy $E$ is

$\Omega(E) = \bigg(\frac{1}{\Delta p}\bigg)^{3N} \int_{0}^{E/c}\Pi\bigg(4\pi p_i^2 dp_i\bigg) = \bigg(\frac{E}{c\Delta p} \bigg)^{3N} \frac{(8\pi)^{N}}{(3N)!}$

I used the fact that $\int_{0}^{E/c}\Pi p_i^2dp_i = \frac{2^N}{(3N)!}\bigg(\frac{E}{c} \bigg)^{3N}$

Now using the equation $\frac{1}{T} = \frac{\partial S}{\partial E}$ I obtain $E = 3Nk_BT$. However I am not sure how to find an equation for pressure and volume. Any help will be appreciated

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  • $\begingroup$ There should be an $V^N$ in you expression of $\Omega$. $\endgroup$ Commented Mar 18, 2021 at 16:33

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Hint: You have to use $$dE=TdS-PdV+\mu dN$$ $$\frac{P}{T}= \left. \frac{\partial S}{\partial V}\right|_{N,E}$$ Where $$S=k_B\ln\Omega$$


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  • $\begingroup$ Slightly confused about how (dS/dE at const N,V) = P/V? Should have been 1/T... $\endgroup$ Commented Mar 18, 2021 at 12:18
  • $\begingroup$ Ah! Sorry, it should be $\partial S/\partial V$. I should confirm it from the book. It's now corrected. $\endgroup$ Commented Mar 18, 2021 at 12:47
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    $\begingroup$ The first equality implies that entropy $S$ can be measured in the same units as pressure $P$. How is that possible? In addition, it is impossible to differentiate $S$ with respect to $V$, provided $V = const$. $\endgroup$
    – Gec
    Commented Mar 18, 2021 at 15:30
  • $\begingroup$ It's should be $E$, I have corrected it! $\endgroup$ Commented Mar 18, 2021 at 16:12
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    $\begingroup$ Could be P/T = (dS/dV, at const T), since then TdS-pdV=dU=0, for constant T. $\endgroup$ Commented Mar 18, 2021 at 16:16
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The calculation is easier in the canonical ensemble. Use $$P=-{\partial F\over\partial V}$$ where the free energy is given by $$F=-k_BT\ln{\cal Z}$$ Note that for free particles, the partition function factorizes $${\cal Z}=z^N$$

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  • $\begingroup$ Is the no. of particles varying, for it to be in cannonical ensemble? If so, it should become a Grand-Cannonical ensemble. $\endgroup$ Commented Mar 18, 2021 at 16:32

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