For a gas molecule with 6 degrees of freedom the law of equipartition of energy gives the following relation between the molar specific heat (C_(V)) and gas constant (R )

For a gas molecule with 6 degrees of freedom the law of equipartition

1.	For a gas molecule with 6 degrees of freedom the law of equipartition of energy gives the following relation between the molar specific heat (C_(V)) and gas constant (R )
Answer» `C_(V)=(R )/(2)` `C_(V)=R` `C_(V)=2R` `C_(V)=3R` Solution :Here, `f=6 therefore gamma=1+(2)/(f)=1+(2)/(6)=(4)/(3)` As `C_(P)-C_(V)=Rand(C_(P))/(C_(V))=gamma` `therefore gammaC_(V)-C_(V)=Ror(4)/(3)C_(V)-C_(V)=RorC_(V)=3R`