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

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 relation between the molar specific heat (C_(V)) and gas constant (R ) is
Answer» `C_(V)=(R )/(2)` `C_(V)=R` `C_(V)=2R` `C_(V)=3R` Solution :As, `C_(v)=(F)/(2)R`, where f is the degree of FREEDOM. Here, `f=6 therefore C_(v)=(6)/(2)R=3R`

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

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