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(D) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$,where $\Delta U$ is the change in internal energy and $\Delta W$ is

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$20$

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(D) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$,where $\Delta U$ is the change in internal energy and $\Delta W$ is the work done by the gas.For an isothermal process,the temperature remains constant,so the change in internal energy $\Delta U = 0$.For an adiabatic process,there is no heat exchange,so $\Delta Q = 0$,which implies $\Delta U = -\Delta W$.Let the total change in internal energy be $\Delta U_{total} = \Delta U_a + \Delta U_b + \Delta U_c = -20 \,J$.For the isothermal steps $(a)$ and $(c)$,$\Delta U_a = 0$ and $\Delta U_c = 0$.Therefore,the total change in internal energy is due to the adiabatic step $(b)$: $\Delta U_b = -20 \,J$.Since $\Delta U_b = -W$ for the adiabatic process,we have $-W = -20 \,J$.Thus,$W = 20 \,J$.

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