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One mole of an ideal gas undergoes a pro...

One mole of an ideal gas undergoes a process `p=(p_(0))/(1+((V_(0))/(V))^(2))`. Here, `p_(0)` and `V_(0)` are constants. Change in temperature of the gas when volume is changed from `V=V_(0)` to `V=2V_(0)` is

Text Solution

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The correct Answer is:
`(11P_0V_0)/(10R)`
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One mole. of an ideal gas undergoes a process P=(P_(0))/(1+(V_(0) / V)^(2)) . Here P_(0) and V_(0) are constants. The change in temperature of the gas when volume is changed from V=V_(0) to V=2 V_(0) is (x P_(0) V_(0))/(10 R) . Find x . (Here R= Uniyersal gas constant)

One mole of an ideal gas undergoes a process p=(p_(0))/(1+((V)/(V_(0)))^(2)) where p_(0) and V_(0) are constants. Find temperature of the gaas when V=V_(0) .

Knowledge Check

  • One mole of an ideal gas undergoes a process P = P_(0) [1 + ((2 V_(0))/(V))^(2)]^(-1) , where P_(0) V_(0) are constants. Change in temperature of the gas when volume is changed from V = V_(0) to V = 2 V_(0) is:

    A
    `(4)/(5) (P_(0) V_(0))/(nR)`
    B
    `(3)/(4) (P_(0) V_(0))/(nR)`
    C
    `(2)/(3) (P_(0) V_(0))/(nR)`
    D
    `(9)/(7) (P_(0) V_(0))/(nR)`

  • For one mole of ideal gas if P=(P_(0))/(1+((V)/(V_(0)))) where P_(0) and V_(0) are constant, then temperature of gas when V=V_(0) is:

    A
    `(P_(0)V_(0))/(R)`
    B
    `(P_(0)V_(0))/(4R)`
    C
    `(P_(0)V_(0))/(2R)`
    D
    `(P_(0)V_(0))/(3R)`

  • One mole of a diatomic gas undergoes a process P = P_(0)//[1 + (V//V_(0)^(3))] where P_(0) and V_(0) are constant. The translational kinetic energy of the gas when V = V_(0) is given by

    A
    `5 P_(0) V_(0)//4`
    B
    `3 P_(0) V_(0)//4`
    C
    `3 P_(0) V_(0)//2`
    D
    `5 P_(0) V_(0)//2`

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    One mole of a diatomic gas undergoes 'a process P=(P_(0))/(1+((V)/(V_(0)))^(s),) where P_(0), V_(0) are constants. The translational kinetic energy of the gas when V=V_(0) is given by (alpha P_(0) V_(0))/(beta) , Find (alpha+beta) .

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