1. Key facts for an ideal gas

   For an ideal gas:

   • Internal energy depends only on temperature.
   • In a reversible isothermal expansion, temperature remains constant.
   • In a reversible adiabatic expansion, temperature decreases during expansion.

2. Interpretation of the figure

   The standard $P$–$V$ diagram for reversible expansion shows:

   • The isothermal curve lies above the adiabatic curve during expansion from the same initial state, because pressure falls more slowly in the isothermal case.
   • Hence, for the same increase in volume, the area under the isothermal curve is greater than that under the adiabatic curve.

   Let the common initial state be at temperature $T_1$.

   • For the isothermal path, final temperature remains $T_1$.
   • For the adiabatic path, final temperature is lower, say $T_2$, so $T_2 < T_1$.

   Thus, if the figure labels the isothermal end temperature as $T_1$ and the adiabatic end temperature as $T_2$, then statement A corresponds to equality of the temperatures indicated on the same isotherm. From the usual interpretation of the figure, A is correct.

3. Check option B: $T_3 > T_1$

   In reversible adiabatic expansion of an ideal gas, temperature decreases: $TV^{\gamma-1}=\text{constant}$ So on expansion, $V$ increases and therefore $T$ decreases.

   Hence any final temperature on the adiabatic path must be less than the initial temperature. Therefore, the statement $T_3 > T_1$ is false.

4. Check option C: $w_{\text{isothermal}} > w_{\text{adiabatic}}$

   In chemistry sign convention, work done by the system during expansion is negative: $w=-\int P\,dV$

   Since the isothermal curve lies above the adiabatic curve, $\left|w_{\text{isothermal}}\right| > \left|w_{\text{adiabatic}}\right|$ but with sign convention, $w_{\text{isothermal}} < w_{\text{adiabatic}}$ because the isothermal work is more negative.

   Therefore the statement $w_{\text{isothermal}} > w_{\text{adiabatic}}$ is false.

5. Check option D: $\Delta U_{\text{isothermal}} > \Delta U_{\text{adiabatic}}$

   For an ideal gas, $\Delta U = nC_V\Delta T$

   • For isothermal expansion, $\Delta T=0$, so $\Delta U_{\text{isothermal}}=0$
   • For adiabatic expansion, temperature decreases, so $\Delta T<0 \implies \Delta U_{\text{adiabatic}}<0$

   Therefore, $\Delta U_{\text{isothermal}}=0 > \Delta U_{\text{adiabatic}}$

   So D is correct.

6. Final evaluation of options

   • A: Correct
   • B: Incorrect
   • C: Incorrect
   • D: Correct

Therefore, the correct options are A and D.