Question

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. $C_{V}=3R$

Answer 1

Answer: (D)

From $C_{V}=\frac{1}{2}fR=\frac{1}{2}\times 6R=3R $