At what temperature will the root mean square speed of oxygen molecules be double its value at $27^{\circ }\text{C}$?

Solution

1. The root mean square speed is given by $v_{rms}=\sqrt{\frac{3RT}{M}}$, implying $v_{rms}\propto \sqrt{T}$.
2. Convert the initial temperature to Kelvin: $$\begin{array}{rl}{T}_1& =27^{\circ }\text{C}+273\\ & =300\text{K}\end{array}$$.
3. To double the speed, the temperature must increase by a factor of $2^2=4$, so $$\begin{array}{rl}{T}_2& =4\times 300\text{K}\\ & =1200\text{K}\end{array}$$.
4. Convert the final temperature back to Celsius: $$\begin{array}{rl}{T}_2& =1200\text{K}-273\\ & =927^{\circ }\text{C}\end{array}$$.
5. Hence the answer is (C).