Question

An ideal gas is expanded from $(P_1, V_1, T_1)$ to $(P_2, V_2, T_2)$ under different conditions. Correct statement(s) among the following is(are):

• A. The work done by the gas is maximum when it is expanded reversibly from $(P_1, V_1)$ to $(P_2, V_2)$.
• B. If the expansion is carried out in vacuum then $\Delta T = 0$ but $\Delta S > 0$
• C. The final temperature of the gas is less when it is expanded reversibly from $V_1$ to $V_2$ under adiabatic conditions as compared to that when expanded irreversibly from $V_1$ to $V_2$ against constant pressure $P_1$ under adiabatic conditions.
• D. The change in entropy of system is maximum when it is expanded reversibly from $V_1$ to $V_2$ under isothermal conditions as compared to that when expanded reversibly from $V_1$ to $V_2$ under adiabatic conditions

Answer 1

Answer: (A), (B), (C), (D)

A, B, C, D

Answer 2

The question asks to identify the correct statement(s) among the given options regarding the expansion of an ideal gas.

Statement A: The work done by the gas is maximum when it is expanded reversibly from $(P_1, V_1)$ to $(P_2, V_2)$.

For a given initial state $(P_1, V_1)$ and final state $(P_2, V_2)$, the work done by the gas is given by $W = \int P dV$. In a reversible process, the pressure of the gas is always infinitesimally greater than the external pressure during expansion, and the process follows a well-defined path on the P-V diagram. In an irreversible expansion between the same initial and final states, the external pressure is less than the internal pressure throughout the expansion. On a P-V diagram, the work done is the area under the process curve. For expansion, $V_2 > V_1$. For a reversible process, the pressure at each volume is determined by the equation of state and the process path (e.g., isothermal or adiabatic). For an irreversible process between the same states, the path is different. However, if we consider expansion from $(P_1, V_1)$ to $(P_2, V_2)$, the reversible path gives the maximum work done compared to any irreversible path between the same states. This is a fundamental concept in thermodynamics.

Consider expansion from $V_1$ to $V_2$. The work done is $W = \int_{V_1}^{V_2} P_{ext} dV$. For reversible expansion, $P_{ext} = P_{int}$. For irreversible expansion, $P_{ext} < P_{int}$. Since $P_{int}$ follows the reversible path, the reversible work $W_{rev} = \int_{V_1}^{V_2} P_{int, rev} dV$ is greater than the irreversible work $W_{irr} = \int_{V_1}^{V_2} P_{ext, irr} dV$ because $P_{int, rev} > P_{ext, irr}$ at each step of expansion.

Thus, the work done by the gas is maximum when expanded reversibly between the same initial and final states. Statement A is correct.

Statement B: If the expansion is carried out in vacuum then $\Delta T = 0$ but $\Delta S > 0$.

Expansion in vacuum is a free expansion, where the external pressure $P_{ext} = 0$. The work done by the gas is $W = \int P_{ext} dV = 0$. For an ideal gas, the internal energy depends only on temperature, $\Delta U = n C_v \Delta T$. According to the first law of thermodynamics, $\Delta U = Q - W$. In free expansion, $W=0$. If the process is adiabatic ($Q=0$), then $\Delta U = 0$, which implies $\Delta T = 0$ for an ideal gas. Even if it's not explicitly stated as adiabatic, free expansion is usually considered to be adiabatic or occurring in an isolated system where $Q=0$. So, $\Delta T = 0$ is correct for free expansion of an ideal gas.

Free expansion is an irreversible process. For any irreversible process in an isolated system, the entropy of the system increases. The change in entropy of the system for an isothermal process from $V_1$ to $V_2$ is given by $\Delta S = n R \ln(\frac{V_2}{V_1})$. Since it is an expansion, $V_2 > V_1$, so $\ln(\frac{V_2}{V_1}) > 0$, which means $\Delta S > 0$. Also, for an irreversible process in an isolated system, $\Delta S_{system} > 0$. Free expansion into vacuum is an irreversible process. The system is the gas, and the surroundings are the vacuum and the container walls. Since there is no work done and no heat exchange, $\Delta S_{surroundings} = 0$. For an irreversible process, $\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings} > 0$. Since $\Delta S_{surroundings} = 0$, $\Delta S_{system} > 0$.

Thus, for free expansion of an ideal gas, $\Delta T = 0$ and $\Delta S > 0$. Statement B is correct.

Statement C: The final temperature of the gas is less when it is expanded reversibly from $V_1$ to $V_2$ under adiabatic conditions as compared to that when expanded irreversibly from $V_1$ to $V_2$ against constant pressure $P_1$ under adiabatic conditions.

Let's consider two adiabatic expansions from initial state $(P_1, V_1, T_1)$ to final volume $V_2 > V_1$.

Case 1: Reversible adiabatic expansion to volume $V_2$. The process follows $T V^{\gamma-1} = constant$. Let the final temperature be $T_{2,rev}$. Then $T_1 V_1^{\gamma-1} = T_{2,rev} V_2^{\gamma-1}$, so $T_{2,rev} = T_1 (\frac{V_1}{V_2})^{\gamma-1}$. Since $V_2 > V_1$ and $\gamma > 1$, $T_{2,rev} < T_1$. The work done is $W_{rev} = n C_v (T_1 - T_{2,rev})$. Since $W_{rev} > 0$, $T_1 > T_{2,rev}$.

Case 2: Irreversible adiabatic expansion from $V_1$ to $V_2$ against constant external pressure $P_{ext}$. Let the initial pressure be $P_1$. For expansion to occur, $P_{ext} < P_1$. The statement says expansion against constant pressure $P_1$. This is not possible for expansion from $P_1$. Assuming the statement meant against a constant external pressure $P_{ext}$ which is less than $P_1$. Let's assume the expansion is against a constant external pressure $P_f$, where $P_f$ is the final pressure in the reversible expansion. Or let's assume the statement meant expansion against constant pressure $P_{ext}$.

Let's re-examine the statement: "expanded irreversibly from $V_1$ to $V_2$ against constant pressure $P_1$". This is likely a typo. It should be against a constant external pressure $P_{ext}$. Let's assume the expansion is against a constant external pressure $P_{ext}$. The work done is $W_{irr} = P_{ext}(V_2 - V_1)$. For an adiabatic process, $\Delta U = -W$. $n C_v (T_{2,irr} - T_1) = -P_{ext}(V_2 - V_1)$. So $T_{2,irr} = T_1 - \frac{P_{ext}(V_2 - V_1)}{n C_v}$.

Comparing the work done in reversible and irreversible adiabatic expansion from $V_1$ to $V_2$: $W_{rev} > W_{irr}$ for expansion between the same initial and final states. However, here the final states are not the same (only the final volume is the same).

Let's compare $T_{2,rev}$ and $T_{2,irr}$.

$T_{2,rev} = T_1 (\frac{V_1}{V_2})^{\gamma-1}$.

$T_{2,irr} = T_1 - \frac{P_{ext}(V_2 - V_1)}{n C_v}$.

We know that for expansion from $V_1$ to $V_2$, the work done is greater in reversible adiabatic process than in irreversible adiabatic process against a constant external pressure $P_{ext}$. $W_{rev} > W_{irr} = P_{ext}(V_2 - V_1)$.

Since $W = n C_v (T_1 - T_2)$ for adiabatic process, $T_1 - T_{2,rev} = \frac{W_{rev}}{n C_v}$ and $T_1 - T_{2,irr} = \frac{W_{irr}}{n C_v}$.

Since $W_{rev} > W_{irr}$, it follows that $T_1 - T_{2,rev} > T_1 - T_{2,irr}$. This implies $T_{2,rev} < T_{2,irr}$.

So, the final temperature of the gas is less when it is expanded reversibly from $V_1$ to $V_2$ under adiabatic conditions as compared to that when expanded irreversibly from $V_1$ to $V_2$ against constant pressure $P_{ext}$ under adiabatic conditions.

Assuming the statement meant expansion against constant external pressure $P_{ext}$ where $P_{ext}$ is less than $P_1$. The statement says "against constant pressure $P_1$". If this means $P_{ext} = P_1$, then expansion would not occur. If it means the initial pressure is $P_1$ and the expansion is against some constant $P_{ext} < P_1$, then the comparison $T_{2,rev} < T_{2,irr}$ holds.

Let's interpret "against constant pressure $P_1$" as the constant external pressure is $P_1$. For expansion from $V_1$ to $V_2$ to occur, the initial pressure $P_1$ must be greater than the external pressure. So, this interpretation is problematic.

Let's interpret "against constant pressure $P_1$" as the external pressure is constant and equal to the initial pressure $P_1$. This would lead to no expansion, or infinite expansion depending on the initial conditions.

Let's assume the statement meant irreversible expansion from $V_1$ to $V_2$ against a constant external pressure $P_{ext}$ where $P_{ext}$ is related to the initial state. Perhaps it means against a constant pressure equal to the initial pressure $P_1$, which doesn't make sense for expansion.

Let's assume the statement meant irreversible expansion from $V_1$ to $V_2$ against a constant external pressure $P_{ext}$ which is less than $P_1$. In this case, $T_{2,rev} < T_{2,irr}$. So the final temperature is less in reversible adiabatic expansion. Thus, statement C is correct under this interpretation.

Statement D: The change in entropy of system is maximum when it is expanded reversibly from $V_1$ to $V_2$ under isothermal conditions as compared to that when expanded reversibly from $V_1$ to $V_2$ under adiabatic conditions.

Consider reversible expansion from $V_1$ to $V_2$.

Case 1: Isothermal reversible expansion. Temperature is constant, $T_1$. The change in entropy is $\Delta S_{iso} = \int \frac{dQ_{rev}}{T} = \frac{Q_{rev}}{T_1}$. For isothermal expansion of ideal gas, $\Delta U = 0$, so $Q_{rev} = W_{rev} = n R T_1 \ln(\frac{V_2}{V_1})$. So $\Delta S_{iso} = n R \ln(\frac{V_2}{V_1})$. Since $V_2 > V_1$, $\Delta S_{iso} > 0$.

Case 2: Adiabatic reversible expansion. For a reversible adiabatic process, $dQ_{rev} = 0$. Therefore, the change in entropy is $\Delta S_{adia} = \int \frac{dQ_{rev}}{T} = 0$. This is consistent with the definition of a reversible adiabatic process as an isentropic process.

Comparing the two cases, $\Delta S_{iso} = n R \ln(\frac{V_2}{V_1})$ and $\Delta S_{adia} = 0$. Since $V_2 > V_1$, $\ln(\frac{V_2}{V_1}) > 0$, so $\Delta S_{iso} > 0$. Thus, $\Delta S_{iso} > \Delta S_{adia}$.

The change in entropy of the system is maximum when it is expanded reversibly from $V_1$ to $V_2$ under isothermal conditions as compared to that when expanded reversibly from $V_1$ to $V_2$ under adiabatic conditions. Statement D is correct.

All four statements A, B, C, and D are correct.