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JEE Mains · Physics · STD 11 - 12 . kinetic theory of gases
The temperature of an ideal gas is increased from \(200\,K\) to \(800\,K\). If r.m.s. speed of gas at \(200\,K\) is \(v_0\). Then, r.m.s. speed of the gas at \(800\,K\) will be:
- A \(v _0\)
- B \(4 v_0\)
- C \(\frac{v_0}{4}\)
- D \(2 v _0\)
Answer & Solution
Correct Answer
(D) \(2 v _0\)
Step-by-step Solution
Detailed explanation
\(V _{ rms }=\sqrt{\frac{3 RT }{ M }}\) \(\Rightarrow V _{ rms } \alpha \sqrt{ T }\) Increasing temperature \(4\) times, rms speed gets doubled.
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