# Change in temperature due to adiabatic expansion against the vacuum

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?

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.

• Great! Thanks a lot! Commented Jan 11 at 19:18