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    Question

    An ideal gas expands into vacuum in an insulated

    container. Then:
    A ΔU = 0,Q = 0,W = 0 Correct Answer Incorrect Answer
    B ΔU = 0,Q = 0,W > 0 Correct Answer Incorrect Answer
    C ΔU > 0,W < 0, Correct Answer Incorrect Answer
    D Q>0,W> 0 Correct Answer Incorrect Answer

    Solution

    The process described is the free expansion of an ideal gas into a vacuum in an insulated container.

    1. Work Done (W): The gas expands into a vacuum. This means there is no external pressure opposing the expansion. The work done by the gas is given by W=∫Pext ​dV. Since the external pressure Pext is zero (expansion into a vacuum), the work done by the gas is: W=∫0 ⋅ dV=0 So, W=0.
    2. Heat Exchange (Q): The container is insulated, which means there is no heat exchange between the system (the gas) and its surroundings. Therefore, the heat added to or removed from the system is zero: Q=0
    3. Change in Internal Energy (ΔU): According to the first law of thermodynamics, the change in internal energy of a system is given by: ΔU=Q−W Substituting the values we found for Q and W: ΔU=0−0=0 So, ΔU=0.
    For an ideal gas, the internal energy depends only on its temperature. Since the internal energy remains constant (ΔU=0), the temperature of the ideal gas also remains constant during free expansion. Therefore, ΔU=0, Q=0, and W=0. This corresponds to option (A).

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