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Prove that for an adiabatic process TV^(...

Prove that for an adiabatic process `TV^(gamma-1)` = constant.

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In adiabatic expansion of monoatomic ideal gas, if volume increases by 24% then pressure decreases by 40% . Statement- 2 : For adiabatic process PV^(gamma) = constant

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Knowledge Check

  • The relation between pressure and temperature of a monoatomic gas in an adiabatic process is P prop T^(-C) , where C = ____

    A
    `2/5`
    B
    `5/2`
    C
    `3/5`
    D
    `5/3`
  • Assertion: The specific heat of a gas in an adiabatic process is zero and in an isothermal process is infinite. Reason: Specific heat of a gas is directly proportional to the heat in system and inversely proportional to change in temperature.

    A
    Both assertion and reason are true and the reason is the correct explanation of the assertion.
    B
    Both assertion and reason are true but reason is not the correct explanation of the assertion
    C
    The assertion is true but reason is false.
    D
    Assertion and reason both are false.
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    A box of negligible mass containing 2 moles of an ideal gas of molar mass M and adiabatic exponent gamma moves with constant speed v on a smooth horizontal surface. If the box suddenly stops, then change in temperature of gas will be :

    Molar heat capacity of an ideal gas in the process PV^(x) = constant , is given by : C = (R)/(gamma-1) + (R)/(1-x) . An ideal diatomic gas with C_(V) = (5R)/(2) occupies a volume V_(1) at a pressure P_(1) . The gas undergoes a process in which the pressure is proportional to the volume. At the end of the process the rms speed of the gas molecules has doubled from its initial value. The molar heat capacity of the gas in the given process is :-

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