\(T_{2}=?\)
That the Final \(rms\) speed will be Double of inital.
\(\frac{V_{rms_{2}}}{V_{rms_{1}}}=\sqrt{\frac{T_{2}}{T_{1}}}=\sqrt{\frac{T_{2}}{273}}\)
\(V_{rms_{2}} =2 V_{rms_{1}}\)
\(\Rightarrow \quad 2 =\sqrt{\frac{T_{2}}{273}}\)
\(\Rightarrow \quad T_{2} =4 \times 273=1092 \; K.\)
\(\quad T_{2} =819^{\circ} C\)
The temperature raised \(=819^{\circ} C\)