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NTA Abhyas NEET2020ChemistryChemical KineticsPractice

The average kinetic energy of one molecule of an ideal gas at 27 ℃ and 1 atm pressure is [Avogadro number N A = 6.023 × 10 23 ]

Options

  1. A900 cal K -1 mol -1
  2. B6.21 × 10 - 2 1 J K -1 molecule -1
  3. C336.7 J K -1 mol -1
  4. D3741.3 J K -1 mol -1

Correct answer

B. 6.21 × 10 - 2 1 J K -1 molecule -1

Step-by-step solution

The average kinetic energy per molecule, = 3 2 kT Or = 3 2 R N A T = 3 2 × 8.314 6.023 × 1 0 2 3 × 3 0 0 = 6.21 × 1 0 - 2 1 J K -1 molecule -1

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