The rms speed of oxygen molecules (O₂) at a certain temperature T (degree absolute) is v. If the temperature is doubled and oxygen gas dissociates into atomic oxygen, what is its rms speed in atomic form?

The rms speed is given by $\mathrm{v}=\sqrt{\frac{3\mathrm{RT}}{{\mathrm{M}}_{\mathrm{w}}}}.$
When the temperature is doubled, the new temperature T’ = 2T and oxygen (O₂) dissociates into atomic oxygen, then molecular weight $\mathrm{M}{\text{'}}_{\mathrm{w}}=\frac{{\mathrm{M}}_{\mathrm{w}}}{2}$
Therefore, 
$$\begin{gathered} \mathrm{v}\text{'}=\sqrt{\frac{3\mathrm{RT}\text{'}}{\mathrm{M}{\text{'}}_{\mathrm{w}}}}=\sqrt{\frac{3\mathrm{R}\left(2\mathrm{T}\right)}{\left(\frac{{\mathrm{M}}_{\mathrm{w}}}{2}\right)}}=\sqrt{4\left(\frac{3\mathrm{RT}}{{\mathrm{M}}_{\mathrm{w}}}\right)} \\ \Rightarrow \mathrm{v}\text{'}=2\sqrt{\frac{3\mathrm{RT}}{{\mathrm{M}}_{\mathrm{w}}}}=2\mathrm{v} \\ \mathrm{Hence}, {\left({\mathrm{v}}_{\mathrm{O}}\right)}_{\mathrm{rms}}=2{\left({\mathrm{v}}_{{\mathrm{O}}_2}\right)}_{\mathrm{rms}} \end{gathered}$$