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JEE Main 2023 (6 Apr, Shift 2) · Molecular Speeds & Maxwell Distribution

Molecular Speeds & Maxwell Distribution question from JEE Main 2023 (6 Apr, Shift 2)

Q39JEE Main 2023 · 6 APR · SHIFT 2Easy

Molecular Speeds & Maxwell Distribution

The temperature of an ideal gas is increased from 200 K200\text{ K} to 800 K800\text{ K}. If r.m.s. speed of gas at 200 K200\text{ K} is v0v_0. Then, r.m.s. speed of the gas at 800 K800\text{ K} will be:

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Answer: (b)

  1. vrms∝Tv_{rms} \propto \sqrt{T}
  2. vv0=800200=2\frac{v}{v_0} = \sqrt{\frac{800}{200}} = 2
  3. So the answer is (b) 2v02v_0.
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