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(D) The change in internal energy for an ideal gas is given by $\Delta U = nC_{v}\Delta T$.Since $\Delta U \propto \Delta T$, we compare the temperatu

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$\Delta U_{1} > \Delta U_{2} > \Delta U_{3}$

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(D) The change in internal energy for an ideal gas is given by $\Delta U = nC_{v}\Delta T$.Since $\Delta U \propto \Delta T$, we compare the temperature changes for the three processes starting from the same initial state $(P_{0}, V_{0})$ to the same final volume $2V_{0}$.For process $1$ (isobaric): $T_{initial} = \frac{P_{0}V_{0}}{nR}$, $T_{final} = \frac{P_{0}(2V_{0})}{nR} = 2T_{0}$. Thus, $\Delta T_{1} = T_{0} > 0$.For process $2$ (isothermal): $T_{initial} = T_{0}$, $T_{final} = T_{0}$. Thus, $\Delta T_{2} = 0$.For process $3$ (adiabatic): Since the gas expands, the temperature decreases, so $T_{final} Comparing the changes: $\Delta T_{1} > \Delta T_{2} > \Delta T_{3}$.Therefore, $\Delta U_{1} > \Delta U_{2} > \Delta U_{3}$.

In the given figure, $1$ represents isobaric, $2$ represents isothermal, and $3$ represents adiabatic processes of an ideal gas. If $\Delta U_{1}, \Delta U_{2}, \Delta U_{3}$ are the changes in internal energy in these processes respectively, then:

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