NEET 2020 Physics Question 162 Solution
Question
The average thermal energy for a mono-atomic gas is: (kB is Boltzmann constant and T, absolute temperature)
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The average thermal energy for a mono-atomic gas is: (kB is Boltzmann constant and T, absolute temperature)
By the law of equipartition of energy, each quadratic degree of freedom contributes (1/2) kB T of thermal kinetic energy. A monoatomic gas molecule has exactly 3 translational degrees of freedom (x, y, z) and zero rotational/vibrational modes. Total average thermal energy = 3 × (1/2) kB T = (3/2) kB T. Mark Option (2) in 3 seconds!
A monoatomic gas molecule (such as Helium, Neon, or Argon) behaves as a point mass. It can move freely in 3 independent spatial directions (along the x, y, and z Cartesian axes), giving it 3 translational degrees of freedom (). It has negligible moment of inertia about its own center, so rotational degrees of freedom are absent.
According to Boltzmann's Law of Equipartition of Energy, in thermal equilibrium at absolute temperature , the average energy associated with each independent degree of freedom per molecule is: , where is Boltzmann's constant.
Multiplying the energy per mode by the total number of degrees of freedom (): .
Therefore, the average thermal energy of a monoatomic gas atom is , matching Option 2.
Diagnose every choice: Know exactly why the right option works and why distractors fail.
Incorrect; this is the thermal energy per single degree of freedom, not for all 3 translational axes.
Accurate value; 3 translational degrees of freedom each contributing (1/2) kB T.
Incorrect; this is the average thermal energy for a rigid diatomic gas molecule (f = 5: 3 translational + 2 rotational).
Incorrect; this is the average thermal energy for a non-rigid diatomic gas molecule with vibrational modes active at high temperature (f = 7).
Carelessly confusing monoatomic (f=3, E = 3/2 kB T) with diatomic (f=5, E = 5/2 kB T).
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