Solution:

The root mean square (r.m.s.) speed \( v_{\text{rms}} \) of gas molecules is given by:

\[
v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}}
\]

Since \( v_{\text{rms}} \propto \sqrt{T} \), if the temperature changes, the r.m.s. speed changes proportionally to the square root of the temperature.

Let the initial r.m.s. speed at \( T = 120 \, \text{K} \) be \( v \). Then, at \( T = 480 \, \text{K} \), the r.m.s. speed \( v' \) is:

\[
v' = v \cdot \sqrt{\frac{480}{120}} = v \cdot \sqrt{4} = 2v
\]

So, the r.m.s. speed at \( 480 \, \text{K} \) will be \( 2v \).