To find the change in internal energy of the gas, we use the formula:$\\ \Delta U = nC_V\Delta T \\$For a diatomic ideal gas, $C_V = \frac{5}{2}R. \\$Given:$\\$• The transition is from point A to point B.$\\$• Initial state A: $P_A = 5$ kPa$, V_A = 4$ m$^3. \\$• Final state B: $P_B = 2$ kPa$, V_B = 6$ m$^3. \\$Using the ideal gas equation $PV = nRT,$ we find the temperatures:$\\ T_A = \frac{P_A V_A}{nR} = \frac{5 \times 10^3 \times 4}{R} = \frac{20 \times 10^3}{R} \\ T_B = \frac{P_B V_B}{nR} = \frac{2 \times 10^3 \times 6}{R} = \frac{12 \times 10^3}{R} \\$The change in temperature $\Delta T$ is:$\\ \Delta T = T_B - T_A = \frac{12 \times 10^3}{R} - \frac{20 \times 10^3}{R} \\ \Delta T = \frac{-8 \times 10^3}{R} \\$Now, calculate $\Delta U: \\ \Delta U = nC_V\Delta T = 1 \times \frac{5}{2}R \times \frac{-8 \times 10^3}{R} \\ \Delta U = \frac{5}{2} \times (-8 \times 10^3) \\ \Delta U = -20 \times 10^3$ J$\\ \Delta U = -20$ kJ$\\$Therefore, the change in internal energy is $-20$ kJ$. \\$This corresponds to Option 1.