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A rigid diatomic ideal gas undergoes an ...

A rigid diatomic ideal gas undergoes an adiabatic process at room temperature .The relation between temperature and volume for this process in `TV^x`= constant ,then x is :

A

`3/5`

B

`2/5`

C

`2/5`

D

`5/3`

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Knowledge Check

  • The ideal gas equation for an adiabatic process is

    A
    `PV^(gamma)`=constant
    B
    `TV^(gamma+1)`=constant
    C
    `P^(gamma-1)`=constant
    D
    `P^(gamma+1)`T=constant

  • For a monoatomic ideal gas undergoing an adiabatic change, the relation between temperature and volume TV^(x) = constant, where x is

    A
    `(7)/(5)`
    B
    `(2)/(5)`
    C
    `(2)/(3)`
    D
    `(1)/(3)`

  • For a monoatomic ideal gas undergoing an adiabatic change, the relation between temperature and volume is TV^(x) = constant, where x is

    A
    `7//5`
    B
    `2//5`
    C
    `2//3`
    D
    `1//3`

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