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For monoatomic, ideal gases, $PV^{\frac{5}{3}}$ is constant. I have been told that this does not hold for diatomic, or other non-monoatomic gases. However, I derived the relationship using only the ideal gas law and the assumption that $Q=0$.

Since $Q = \Delta U+W$,

$$dU = -W$$ $$dU=-PdV$$ $$\frac{dU}{dV}=-P$$ The internal energy of an ideal gas is $\frac{3}{2} nRT = \frac{3}{2} PV$: $$\frac{3}{2} \frac{d(PV)}{dV}=-P$$ By product rule: $$\frac{3}{2}(\frac{dP}{dV}V+P)=-P$$ $$\frac{dP}{dV}V= -\frac{5}{3} P$$

This is a seperable differential equation which we can easily solve to get $PV^{5/3} = const$.

Nowhere did I assume that the gas was monotomic. I suspect the error might lie in the step setting $U=\frac{3}{2} PV$, but I'm not sure if this is the case and, if so, why.

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  • $\begingroup$ There are other ways a molecule can hake up energy other than by translation - Equipartition of energy $\endgroup$
    – Farcher
    Commented Apr 9 at 21:52
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    $\begingroup$ The monoatomic gas assumption is implicit when you used $U=\frac{3}{2}k_B T$ $\endgroup$
    – Lucky Charms
    Commented Apr 9 at 21:57
  • $\begingroup$ Thank you, I was not previously aware of the derivation of the formula but this makes sense. I will accept the answer once the cooldown is complete. $\endgroup$
    – QWERTYL
    Commented Apr 9 at 22:00

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The expression for molar internal energy is not always $U = 3RT/2$. That is only true for monatomic gases. The general expression is, by the equipartition theorem, $$\frac{N}{2}RT$$ where $N$ is the number of degrees of freedom. This leads to the correct formula $\gamma = (N+2)/N$.

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