Q. For a gas molecule with 6 degrees of freedom the law of equipartition of energy gives the following relation between the molecular specific heat $(C_{V})$ and gas constant $(R)$
A) $C_{V} = \frac{R}{2}$
B) $C_{V} = R$
C) $C_{V} = 2R$
D) $C_{V} = 3R$
Explanation
(d) From $C_{V} = \frac{1}{2}fR = \frac{1}{2} \times 6R = 3R$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.