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

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$13$

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(D) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$,where $\Delta Q$ is the heat supplied,$\Delta U$ is the change in internal energy,and $\Delta W$ is the work done.For the first expansion:$\Delta W_1 = 20 \, kJ$$\Delta Q_1 = 16 \, kJ$Using $\Delta U = \Delta Q_1 - \Delta W_1 = 16 \, kJ - 20 \, kJ = -4 \, kJ$.Since internal energy $U$ is a state function,the change in internal energy $\Delta U$ depends only on the initial and final states. Therefore,$\Delta U$ remains the same for the second expansion.For the second expansion:$\Delta U = -4 \, kJ$$\Delta Q_2 = 9 \, kJ$Using $\Delta W_2 = \Delta Q_2 - \Delta U = 9 \, kJ - (-4 \, kJ) = 13 \, kJ$.Thus,the work done in the second expansion is $13 \, kJ$.

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