Question

The average thermal energy for a mono-atomic gas is : ( \(k_{B}\) is Boltzmann constant and \(\mathrm{T}\), absolute temperature)
A \(\frac{1}{2} k_{B} T\)
B \(\frac{3}{2} k_{B} T\)
C \(\frac{5}{2} k_{B} T\)
D \(\frac{7}{2} k_{B} T\)

Answer 1

Solution:
Average thermal energy \(=\frac{3}{2} K_{B} T\)
(where 3 is translational degree of freedom)
For monoatomic gas total degree of freedom
\(f=3\) (translational degree of freedom)