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The volume of one mode of an ideal gas w...

The volume of one mode of an ideal gas with adiabatic exponent `gamma` is varied according to the law `V = a//T`, where a is constant . Find the amount of heat obtained by the gas in this process, if the temperature is increased by `Delta T`.

Text Solution

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The correct Answer is:
`(2-gamma)/(gamma-1)RDeltaT`
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Knowledge Check

  • The volume of 1 mole of an ideal gas with the adiabatic exponenty gamma is changed according to the relation V=b/T Where b= constant The amount of hear absorbed by the pas in the proces if the temperature is increased by triangleT will be

    A
    `((1-gamma)/(gamma+1)) R triangleT`
    B
    `(R )/(gamma-1) triangleT`
    C
    `((2-gamma)/(gamma-1)) R triangleT`
    D
    `(R triangleT)/(gamma-1)`

  • The volume of one mole of ideal gas with adiabatic exponent is varied according to law V = 1/T . Find amount of heat obtained by gas in this process if gas temperature is increased by 100 K.

    A
    `100 R`
    B
    `200 R`
    C
    `300 R`
    D
    `400 R`

  • The efficiency of an ideal gas with adiabatic exponent 'gamma' for the shown cyclic process would be

    A
    `((gamma-1)("2 ln 2"-1))/(1+(gamma-1)"2 ln 2")`
    B
    `((gamma-1)(1-"2 ln 2"))/((gamma-1)"2 ln 2"-1)`
    C
    `(("2 ln 2"+1)(gamma-1))/((gamma-1)"2 ln 2"+1)`
    D
    `(("2 ln 2"-1))/(gamma//(gamma-1))`

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