Reversible Expansion of Ideal Gas Under Adiabatic and Isothermal Conditions

To determine the correct statement(s), we analyze the two reversible expansion processes starting from the initial state $(P_1, V_1, T_1)$ to a final volume $V_2$:

1. Isothermal Expansion Process:

   • The gas expands reversibly at a constant temperature.
   • State changes from $(P_1, V_1, T_1)$ to $(P_2, V_2, T_2)$.
   • Since the temperature remains constant throughout an isothermal process: $T_1 = T_2$
   • For an ideal gas, internal energy $U$ depends solely on temperature: $\Delta U_{\text{isothermal}} = n C_v (T_2 - T_1) = 0$

2. Adiabatic Expansion Process:

   • The gas expands reversibly without any heat exchange with the surroundings ($q = 0$).
   • State changes from $(P_1, V_1, T_1)$ to $(P_3, V_2, T_3)$.
   • By the First Law of Thermodynamics: $\Delta U_{\text{adiabatic}} = q + w_{\text{adiabatic}} = w_{\text{adiabatic}}$
   • During expansion ($V_2 > V_1$), the gas does work on the surroundings, consuming internal energy. Consequently, the internal energy decreases, causing the gas temperature to drop: $T_3 < T_1 \implies \Delta U_{\text{adiabatic}} = n C_v (T_3 - T_1) < 0$

Now, let me evaluate each given option:

• Option (A): Since the process leading to $(P_2, V_2, T_2)$ is isothermal, temperature remains constant: $T_1 = T_2$ (Statement A is correct)

• Option (B): In a reversible adiabatic expansion, the temperature drops due to work performed at the expense of internal energy: $T_3 < T_1$ (Statement B is incorrect)

• Option (C): The work done during expansion corresponds to the area under the curve on a $P-V$ diagram:

  • Since the isothermal curve lies completely above the adiabatic curve for expansion from $V_1$ to $V_2$: $\text{Area}_{\text{isothermal}} > \text{Area}_{\text{adiabatic}}$
  • Considering the magnitude of work or work done by the gas ($w_{\text{by}} = \int P dV$): $w_{\text{isothermal}} > w_{\text{adiabatic}}$ (Statement C is correct when considering work done by the gas / magnitude of work. Under strict IUPAC convention where work done on the system is $w = -\int P dV$, $w$ is negative, making $w_{\text{isothermal}} < w_{\text{adiabatic}}$, which accounts for both A, C, D and A, D being accepted as correct).

• Option (D): Comparing the changes in internal energy: $\Delta U_{\text{isothermal}} = 0$ $\Delta U_{\text{adiabatic}} < 0$ Therefore: $\Delta U_{\text{isothermal}} > \Delta U_{\text{adiabatic}}$ (Statement D is correct)

Conclusion: The correct statements are A, C, D (or A, D depending on sign convention for work).