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For a gas molecule with 6 degrees of fre...

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 )

A

`C_(V)=(R )/(2)`

B

`C_(V)=R`

C

`C_(V)=2R`

D

`C_(V)=3R`

Text Solution

Verified by Experts

The correct Answer is:
D
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`
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