ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J.

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ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J.

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Vân Khánh 6 months 2021-07-29T15:59:26+00:00 1 Answers 3 views 0

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    2021-07-29T16:00:34+00:00

    Complete question:

    ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.

    Answer:

    The magnitude of q for the process 568 J.

    Explanation:

    Given;

    change in internal energy of the gas, ΔU = 475 J

    work done by the gas, w = 93 J

    heat added to the system, = q

    During gas expansion process, heat is added to the gas.

    Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.

    ΔU = q – w

    q = ΔU +  w

    q = 475 J  +  93 J

    q = 568 J

    Therefore, the magnitude of q for the process 568 J.

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