The mean kinetic energy of translation per gram molecule of a gas is

Physics · NEET · NTA ExamsKINETIC THEORY OF GASES

The mean kinetic energy of translation per gram molecule of a gas is
  1. \(\frac{1}{2} \mathrm{RT}\)
  2. \(\frac{3}{2} \mathrm{RT}\)
  3. \(\frac{5}{2} \text { RT }\)
  4. \(\frac{9}{2} \mathrm{RT}\)

Answer

(B) 3 2 RT

Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.

Related practice questions

More KINETIC THEORY OF GASES questions · Browse all practice questions