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Given that C(p)-C(V)=R and gamma=C(p)//C...

Given that `C_(p)-C_(V)=R and gamma=C_(p)//C_(V)`, where `C_p`=Molar specific heat at constant pressure, `C_(V)`=molar specific heat at constant volume. Then `C_(V)`=

A

`(gammaR)/(gamma-1)`

B

`(R)/(gamma-1)`

C

`(gamma-1)/(R)`

D

`(gamma-1)/(gammaR)`

Text Solution

Verified by Experts

The correct Answer is:
B
`C_(P)-C_(V)=R`
`(C_(V))/(C_(V))=(C_(P))/(C_(V))=(R)/(C_(V))`
`1=gamma-(R)/(C_(V))`
`1-gamma=-(R)/(C_(V))`
`1-gamma=-(R)/(C_(V))`
`(R)/(C_(V))=gamma-1`
`C_(V)=(R)/(gamma-1)`
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Knowledge Check

  • Given the , C_p-C_v=R and gamma=(C_p)/(V_v) where C_p =molar specific heat at constant pressure C_v =molar specific heat at constant volume. Then C_v =

    A
    `(gammaR)/(gamma-1)`
    B
    `R/(gamma-1)`
    C
    `(gamma-1)/R`
    D
    `(gamma-1)/(gammaR)`

  • Specific heat at constant pressure C_P of a gas

    A
    more than the specific heat at constant volume `(C_(V))`
    B
    less then the specific heat at constant volume `(C_(V))`
    C
    equal to the specific heat at constant volume `(C_(V))`
    D
    may be more or less than specific heat at constant volume `(C_(V))`

  • The ratio of specific heat at constant pressure to specific heat at constant volume (i.e. C_p//C_V) for noble gases is

    A
    1.66
    B
    1.33
    C
    1.42
    D
    1.83

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