Question

A polyatomic gas with $n$ degrees of freedom has a mean energy per molecule given:-

• A. $\frac{nkT}{N}$
• B. $\frac{nkT}{2N}$
• C. $\frac{nkT}{2}$
• D. $\frac{3kT}{2}$

Answer 1

Answer: (C)

$K.E.$ per degree of freedom $=\frac{1}{2}kT$ For polyatomic gas, mean $K.E.=\frac{nkT}{2}$