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 $C_V=\frac{1}{2}fR=\frac{1}{2}\times 6R=3R$