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Let's consider an ideal gas that expands in vacuum and it is thermally isolated from the surroundings. Since it is isolated, it does not exchange heat with the exterior $Q=0$. Since it is expanding against the vacuum, it does no work $W=0$. From the first principle we get that the change in internal energy is zero.

$$\Delta U = Q - L = 0$$

Internal energy for an ideal gas depends only on the change in temperature, which will also be zero in this case. $$\Delta U = n C_{V}\Delta T=0$$

But since the expansion is adiabatic, we can write:

$$T_{1}V_{1}^{\gamma - 1} = T_{2}V_{2}^{\gamma - 1}$$ which, because we talk about expansion and $V_{2}>V_{1}$, would mean that:

$$\dfrac{T_{1}}{T_{2}}=\left(\dfrac{V_{2}}{V_{1}}\right)^{\gamma-1}$$ the temperature should decrease: $T_{2} < T_{1}$. What am I missing here?

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The error lies with the line

But since the expansion is adiabatic, we can write: $T_{1}V_{1}^{\gamma - 1} = T_{2}V_{2}^{\gamma - 1}$.

That equation is derived by assuming constant entropy. Specifically, infinitesimal heating $\delta Q(=0)$ is equated to $T\,dS$, which is valid for a reversible process. But expansion associated with a pressure difference is irreversible, so $\Delta S\neq 0$ here even though $Q=0$. The temperature remains constant in your scenario, as you showed in the first part of your analysis.

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  • $\begingroup$ Great! Thanks a lot! $\endgroup$ Commented Jan 11 at 19:18

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