JEE MAIN - Physics (2021 - 22th July Evening Shift - No. 15)
What will be the average value of energy for a monoatomic gas in thermal equilibrium at temperature T?
$${3 \over 2}{k_B}T$$
$${k_B}T$$
$${2 \over 3}{k_B}T$$
$${1 \over 2}{k_B}T$$
Explanation
For a monoatomic ideal gas, the average kinetic energy per molecule is determined by the equipartition theorem. This theorem states that the energy is equally distributed among all the available degrees of freedom.
A monoatomic gas has three translational degrees of freedom, corresponding to motion in the x, y, and z directions. Each degree of freedom contributes an average energy of $\frac{1}{2} k_BT$, where $k_B$ is the Boltzmann constant and $T$ is the absolute temperature.
So for a monoatomic gas with three translational degrees of freedom, the average energy per molecule is: $$\frac{3}{2} k_BT.$$
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