Analysis of the Process

The problem describes an irreversible expansion of an ideal gas in a thermally insulated vessel against zero external pressure. This specific process is known as a free expansion.

Let's analyze the thermodynamic properties of this process step-by-step and evaluate each option.

Step 1: Evaluate Option A (q = 0)

The gas is in a thermally insulated vessel. This means there is no heat exchange between the system (the gas) and the surroundings.

By definition, a process with zero heat transfer is an adiabatic process.

$q = 0$

Therefore, option A is correct.

Step 2: Evaluate Option B (T2 = T1)

We apply the First Law of Thermodynamics:

$\Delta U = q + w$

where $\Delta U$ is the change in internal energy, $q$ is the heat added to the system, and $w$ is the work done on the system.

From Step 1, we know that $q = 0$ because the vessel is thermally insulated.

The work done ($w$) during an expansion against a constant external pressure ($p_{ext}$) is given by:

$w = -p_{ext} \Delta V = -p_{ext} (V_2 - V_1)$

The problem states that the gas expands against zero external pressure, so $p_{ext} = 0$.

Therefore, the work done on the system is:

$w = -0 \times (V_2 - V_1) = 0$

Now, substituting $q=0$ and $w=0$ into the First Law of Thermodynamics:

$\Delta U = 0 + 0 = 0$

The change in internal energy of the gas is zero.

For an ideal gas, the internal energy is a function of temperature only:

$\Delta U = nC_v \Delta T = nC_v (T_2 - T_1)$

where $n$ is the number of moles and $C_v$ is the molar heat capacity at constant volume.

Since $\Delta U = 0$, and both $n$ and $C_v$ are non-zero, we have:

$T_2 - T_1 = 0 \implies T_2 = T_1$

The final temperature is equal to the initial temperature. The process is isothermal.

Therefore, option B is correct.

Step 3: Evaluate Option C (p2V2 = p1V1)

We use the ideal gas equation, $pV = nRT$.

For the initial state:

$p_1V_1 = nRT_1$

For the final state:

$p_2V_2 = nRT_2$

From Step 2, we found that $T_2 = T_1$. Substituting this into the equation for the final state:

$p_2V_2 = nRT_1$

Comparing the equations for the initial and final states, we see that:

$p_2V_2 = p_1V_1$

This is consistent with Boyle's Law, which applies to isothermal processes for a fixed amount of gas.

Therefore, option C is correct.

Step 4: Evaluate Option D (p2V2^γ = p1V1^γ)

The relation $pV^\gamma = \text{constant}$ (where $\gamma = C_p/C_v$) is valid for a reversible adiabatic expansion of an ideal gas.

However, the process described here is:

1. Irreversible: The expansion is rapid against a zero external pressure, not a slow, quasi-static process.
2. Isothermal, not just adiabatic: As shown in Step 2, the temperature remains constant ($T_2=T_1$). In a reversible adiabatic expansion, the gas does work, its internal energy decreases, and its temperature drops ($T_2 < T_1$).

Since the given process is an irreversible free expansion (which happens to be both adiabatic and isothermal), the equation for a reversible adiabatic process does not apply.

Therefore, option D is incorrect.

Conclusion

Based on the analysis, options A, B, and C are correct, while option D is incorrect.