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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 relation between the molar specific heat `(C_(V))` and gas constant (R ) is

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
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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Knowledge Check

  • 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`

  • For a gas molecule with 6 degree of freedom , the law of equipartion of energy gives the following relations between the molecular speecific heat (C_V) and gas constant ( R )

    A
    ` C_V = ( R ) /(2) `
    B
    ` C_ V= R `
    C
    ` C_ V = 2R `
    D
    ` C_V = 3R `

  • Molar specific heat of a monatomic gas at constant pressure is

    A
    5/2 R
    B
    3/2 RT
    C
    7/2 R
    D
    R

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