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16 September, 05:00

onsider the following cyclic process carried out in two steps on a gas: Step 1: 42 J of heat is added to the gas, and 14 J of expansion work is performed. Step 2: 53 J of heat is removed from the gas as the gas is compressed back to the initial state. Calculate the work for the gas compression in Step 2.

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  1. 16 September, 05:25
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    W = - 25 J

    Explanation:

    For a cyclic process, the variation of internal energy (ΔU) must be 0. It means that ΔU from step 1 must be equal to ΔU from step 2. For the first law of the thermodynamics, the energy must be conserved, so:

    ΔU = Q - W, from each step. Q is the heat and W the work

    When the heat is gained by the system, the process is endothermic, so Q >0, when the system loses heat, the process is exothermic, and Q0, and when it is compressing, W<0.

    From step 1:

    ΔU = 42 - 14

    ΔU = 28 J

    From step 2:

    28 = - 53 - W

    W = - 53 + 28

    W = - 25 J
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