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19 March, 13:15

The temperature of a monatomic ideal gas remains constant during a process in which 7470 J of heat flows out of the gas. How much work (including the proper + or - sign) is done?

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  1. 19 March, 13:53
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    W = - 7470 J

    Explanation:

    Given that

    heat flow out of the system, Q = - 7470 J

    We know that internal energy of the ideal gas depends only on temperature and is given as

    ΔU = m Cv ΔT

    m=mass

    Cv=Specific heat at constant volume

    ΔT=Change in the temperature

    ΔU=Change in the internal energy

    Here given that temperature of the gas is constant, that is why

    ΔT = 0 ⇒ ΔU = 0

    We know that, from first law of thermodynamics

    Q = Δ U + W

    Q=Heat transfer

    W=Work transfer

    ΔU=Change in the internal energy

    Now by putting the values in the above equation

    - 7470 = 0 + W

    W = - 7470 J

    This applies that work is done on the system.
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