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The temperature at which a molecule of nitrogen will have the same rms velocity as a molecule of oxygen at 127 ° C is

Options

  1. A457 ℃
  2. B273 ℃
  3. C350 ℃
  4. D77 ℃

Correct answer

D. 77 ℃

Step-by-step solution

Root mean square velocity v r m s = 3 R T M where R is gas constant, T the temperature and M molecular weight. Given, M N 2 = 28 , M O 2 = 32 , T O 2 = 127 ℃ = 127 + 273 = 400 K ∴ v O 2 v N 2 = T O 2 M O 2 × M N 2 T N 2 = 400 32 × 28 T N 2 = 1 ⇒ T N 2 = 350 K = 77 ℃ .

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