Why is temperature constant when an ideal gas expands into a vacuum?
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 $U$).