Irreversible Expansion of Ideal Gas into Vacuum in Insulated Vessel

To determine the correct conditions for the given expansion, we analyze the thermodynamics of the process step-by-step:

1. Evaluation of Heat Exchange ($q$): The container is thermally insulated from the surroundings, which means that no heat can enter or leave the system. Therefore, the process is adiabatic: $q = 0$ Hence, Option (A) is correct.

2. Evaluation of Work Done ($w$) and Internal Energy Change ($\Delta U$): The gas undergoes free expansion into a vacuum, meaning it expands against zero external pressure ($P_{\text{ext}} = 0$). The work done during the expansion is given by: $w = -\int P_{\text{ext}} \, dV = 0$ According to the First Law of Thermodynamics: $\Delta U = q + w$ Substituting $q = 0$ and $w = 0$: $\Delta U = 0$

3. Evaluation of Temperature ($T_2 = T_1$): For an ideal gas, the internal energy is solely a function of temperature ($U = U(T)$). Therefore: $\Delta U = n C_v (T_2 - T_1) = 0$ Since $n C_v \neq 0$, this implies: $T_2 - T_1 = 0 \implies T_2 = T_1$ Hence, Option (B) is correct.

4. Evaluation of the $PV$ Relationship: Using the ideal gas equation of state $PV = nRT$ for the initial and final states: $P_1 V_1 = n R T_1$ $P_2 V_2 = n R T_2$ Since $T_1 = T_2$, it follows directly that: $P_2 V_2 = P_1 V_1$ Hence, Option (C) is correct.

5. Evaluation of the Adiabatic Reversible Equation ($P_2 V_2^{\gamma} = P_1 V_1^{\gamma}$): The relation $P V^{\gamma} = \text{constant}$ is valid only for a reversible adiabatic process of an ideal gas. For the given irreversible free expansion: $P_2 V_2^{\gamma} = (P_2 V_2) V_2^{\gamma-1} = (P_1 V_1) V_2^{\gamma-1}$ Since the gas expands, $V_2 > V_1$, and for any ideal gas $\gamma > 1$, we have: $(P_1 V_1) V_2^{\gamma-1} \neq P_1 V_1^{\gamma}$ Thus, $P_2 V_2^{\gamma} \neq P_1 V_1^{\gamma}$. Hence, Option (D) is incorrect.

Therefore, the correct options are A, B, and C.