Expansion and Compression Behavior of an Ideal Gas

To determine the correct statements regarding the expansion and compression of an ideal gas, let us analyze each option individually:

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Analysis of Option [A]

When the gas is compressed irreversibly from volume $V_2$ to $V_1$ ($V_2 > V_1$) in a single step against a constant external pressure equal to the final pressure $P_{\text{ext}} = p_1$:

The work done on the gas ($W_{\text{on}}$) is given by: $W_{\text{on}} = P_{\text{ext}} (V_2 - V_1) = p_1 (V_2 - V_1)$

Since $p_1$ is the maximum pressure reached at state $(p_1, V_1)$ (where $p_1 > p_2$), compressing in a single step against $p_1$ requires the largest possible boundary work compared to any multi-step or reversible path, where $P_{\text{ext}} \le p_1$. Thus, the work done on the gas is maximum in this single-step irreversible process.

Statement [A] is correct.

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Analysis of Option [B]

Free expansion refers to an expansion of a gas against zero external pressure ($P_{\text{ext}} = 0$).

1. Work done ($W$): $W = -\int P_{\text{ext}} \, dV = 0$

2. Heat transferred ($q$): By definition of free expansion in an isolated system, no heat exchange occurs with the surroundings, so $q = 0$. Hence, the process is adiabatic.

3. Internal Energy ($\Delta U$) and Temperature ($\Delta T$): According to the First Law of Thermodynamics: $\Delta U = q + W = 0 + 0 = 0$

   For an ideal gas, internal energy depends solely on temperature: $\Delta U = n C_v \Delta T = 0 \implies \Delta T = 0 \implies T_1 = T_2$

   Since $\Delta T = 0$, the process is also isothermal.

Therefore, free expansion of an ideal gas is simultaneously isothermal and adiabatic.

Statement [B] is correct.

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Analysis of Option [C]

Consider reversible expansion from volume $V_1$ to $V_2$ ($V_2 > V_1$):

• Isothermal Reversible Expansion: The pressure follows $P_{\text{iso}} = \frac{nRT_1}{V}$. The work done by the gas is: $W_{\text{by, iso}} = \int_{V_1}^{V_2} P_{\text{iso}} \, dV = n R T_1 \ln\left(\frac{V_2}{V_1}\right)$

• Adiabatic Reversible Expansion: The pressure follows $P_{\text{adi}} = \frac{p_1 V_1^\gamma}{V^\gamma}$, where $\gamma > 1$. Since $\gamma > 1$, for any volume $V > V_1$, we have: $P_{\text{adi}} < P_{\text{iso}}$

  Integrating $P \, dV$ from $V_1$ to $V_2$, the area under the adiabatic curve on a $P-V$ diagram is strictly smaller than the area under the isothermal curve: $W_{\text{by, adi}} = \int_{V_1}^{V_2} P_{\text{adi}} \, dV < \int_{V_1}^{V_2} P_{\text{iso}} \, dV = W_{\text{by, iso}}$

Thus, the work performed by the gas is smaller during reversible adiabatic expansion than during reversible isothermal expansion.

Statement [C] is correct.

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Analysis of Option [D]

For an ideal gas, the change in internal energy is given by: $\Delta U = n C_v (T_2 - T_1)$

1. Isothermal Reversible Expansion ($T_1 = T_2$): $\Delta U = n C_v (0) = 0 \quad \text{(Correct)}$

2. Adiabatic Reversible Expansion ($T_1 \neq T_2$): Since $q = 0$ during an adiabatic process, the First Law of Thermodynamics gives: $\Delta U = W = -W_{\text{by}}$

   During expansion ($V_2 > V_1$), work is done by the gas ($W_{\text{by}} > 0$). Consequently: $\Delta U = -W_{\text{by}} < 0$

   This means $\Delta U$ is negative (the gas cools down, $T_2 < T_1$), not positive.

Statement [D] is incorrect.

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Conclusion

The correct options are A, B, and C.