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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 equipartion of energy gives the following relation between the molecular 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
Specific heat at constant volume (CV) and degreee freedom
` C_V = (fR)/( 2) = ( 6R)/(2) " " therefore f= 6`
` C_ V = 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 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`

  • If a gas has n degrees of freedom ratio of specific heats of gas is

    A
    `(1+n)/(2)`
    B
    `1+(1)/(n)`
    C
    `1+(n)/(2)`
    D
    `1+(2)/(n)`

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