What will be the average value of energy along one degree of freedom for an ideal gas in thermal equilibrium at a temperature T? (k B is Boltzmann constant)

Solution

Physics sir · 3 steps

Equipartition hands out energy equally: every degree of freedom gets exactly half of k_BT.

1. Equipartition law. In thermal equilibrium at temperature T, the energy of a molecule is shared equally among all its degrees of freedom. Each one gets the same average share.

   E_(per degree of freedom) = 1/2k_BT

2. Quick check. A monatomic molecule has 3 translational degrees of freedom. Three shares of 1/2k_BT give its familiar average kinetic energy, 3/2k_BT.
3. So. The average energy along one degree of freedom is 1/2k_BT, which is option (a).

The idea to keep. Count the degrees of freedom f, then the average energy per molecule is f/2k_BT. Each degree of freedom always carries 1/2k_BT, whatever the gas.

Answer from the NTA final answer key. Solution by PrepWiser.