Question

The average thermal energy for a mono-atomic gas is :
($k_B$ is Boltzmann constant and $T$ is absolute temperature)

• A. $\frac{5}{2}{k}_{B}T$
• B. $\frac{7}{2}{k}_{B}T$
• C. $\frac{1}{2}{k}_{B}T$
• D. $\frac{3}{2}{k}_{B}T$

Answer 1

The correct option is D $\frac{3}{2}{k}_{B}T$

For a monatomic gas, the total degree of freedom is $3$.
Thus, the energy of monatomic gas is on account of its translational energy along $x,y,z$ directions respectively.

$\Rightarrow$ The translational kinetic energy per degree of freedom per gas molecule is $\frac{1}{2}{k}_{B}T$

Thus, the average thermal energy of monatomic gas per molecule is,
$E_T=3\times \frac{1}{2}{k}_{B}T$
$\therefore E_T=\frac{3}{2}{k}_{B}T$