# Why is temperature constant when an ideal gas expands into a vacuum?

An ideal gas in a box has pressure $$p$$ and temperature $$T$$. This box is kept in a vacuum, within a large container.

When the box is punctured, what happens to the temperature of the gas as it expands to fill the entire container?

I was taught that the temperature will remain constant. How is this possible? I understand that the pressure $$p$$ will decrease and the volume $$V$$ will increase, but for the temperature $$T$$ to remain constant, the decrease in $$p$$ must exactly counterbalance the increase in $$V$$. Is this the case?

## 1 Answer

This process is called an expansion into a vacuum. As you described, the volume of the gas increases and the pressure drops, such that $$pV$$ does not change at all. Since $$n$$ is constant, $$T = \frac{pV}{nR}$$ is also constant.

I think the fact that $$pV$$ is constant is not very intuitive if one only uses the ideal gas law. A more formal way of explaining that $$T$$ is constant would use thermodynamics. If you assume that the large container does not allow any transfer of heat (i.e. it is thermally insulating), then the large container is an isolated system: $$\mathrm{d}q = 0$$. Since the gas is expanding into a vacuum, $$p_{\text{ext}} = 0$$ and $$\mathrm{d}w = -p_{\mathrm{ext}}\,\mathrm{d}V = 0$$. By the First Law, $$\mathrm{d}U = \mathrm{d}q + \mathrm{d}w = 0$$ and temperature is therefore constant (since the temperature of an ideal gas depends fully on $$U$$).