identify derivations that span irreversible and reversible processes in Thermodynamics

[bhpayne]

As suggested on https://physicsderivationgraph.blogspot.com/2024/03/dichotomy-of-assumptions.html

[bhpayne]

prompt to Gemini 2.0 Flash Thinking Experimental 01-21, https://aistudio.google.com/prompts/

For the Ideal Gas Law and Thermodynamic Equations, derivations often assume reversible processes for simplicity, allowing for precise calculations of work and heat transfer (e.g., for isothermal or adiabatic reversible processes).

More realistic derivations account for irreversible processes. For example, the Second Law of Thermodynamics explicitly accounts for the increase in entropy in irreversible processes.

The reversible process is an idealized limit. The derivations for irreversible processes quantify the deviations from this ideal, showing how factors like friction introduce entropy generation.

Describe some derivations that bridge the equations used for reversible processes and irreversible processes. Explain what the initial equations are and what the ending equations are for each derivation.

thoughts

Here's a thinking process to construct the answer about bridging reversible and irreversible thermodynamic derivations:

1. Deconstruct the Prompt: The prompt asks for derivations that bridge reversible and irreversible processes in thermodynamics. This means starting with equations applicable to reversible processes and modifying them to incorporate irreversibility. It also emphasizes the Ideal Gas Law and Thermodynamic Equations as the starting point, and mentions entropy generation as a key aspect of irreversibility. The prompt wants examples of these derivations, outlining the starting and ending equations.

2. Brainstorm Key Concepts: To bridge reversible and irreversible, I need to think about what distinguishes them.

   • Reversible: Quasi-static, no entropy generation within the system, maximum work output (for expansion) or minimum work input (for compression).
   • Irreversible: Non-quasi-static, entropy generation within the system, less work output (for expansion) or more work input (for compression) compared to reversible.
   • Key Difference: Entropy generation. How do I incorporate entropy generation into reversible equations?

3. Identify Reversible Starting Points: What are common thermodynamic equations derived under reversible conditions?

   • Work: dW = -PdV (reversible PV work)
   • First Law: dU = dQ + dW (always true, but dQ and dW change meaning in irreversible)
   • Entropy Change: dS = dQ_rev / T (definition of entropy change for reversible processes)

4. Consider Irreversibility Mechanisms: What makes processes irreversible?

   • Friction: Mechanical irreversibility, dissipates work as heat.
   • Unrestrained Expansion: Pressure difference causes rapid expansion, not quasi-static.
   • Heat Transfer across a finite temperature difference: Heat flow from hot to cold is inherently irreversible.
   • Mixing: Diffusion and mixing are irreversible processes increasing entropy.

5. Choose Derivation Examples: Based on the above, select a few derivations that clearly illustrate the bridge. Good choices might be:

   • Work in Irreversible Expansion/Compression: Start with reversible work and modify it to account for pressure differences or friction.
   • Entropy Change in Irreversible Processes: Start with the reversible entropy change and relate it to the total entropy change including internal generation.
   • Heat Transfer Irreversibility: Focus on how finite temperature differences affect heat transfer and entropy.

6. Structure Each Derivation Example: For each chosen derivation, follow a consistent structure:

   • Process: Clearly define the thermodynamic process (isothermal, adiabatic, etc.) being considered.
   • Reversible Starting Equation: State the relevant equation for the reversible case.
   • Introduce Irreversibility: Explain how irreversibility is introduced (friction, pressure difference, etc.).
   • Modify the Equation: Show the steps to modify the reversible equation to account for irreversibility. This often involves adding terms related to entropy generation or accounting for deviations from equilibrium.
   • Irreversible Ending Equation: State the final equation that accounts for irreversibility.
   • Interpretation/Key Difference: Explain what the modified equation tells us compared to the reversible case, emphasizing the impact of irreversibility (e.g., less work, more entropy).

7. Refine and Organize: Review the examples and ensure they are clear, concise, and logically presented. Organize them with headings and bullet points for readability. Add an introductory and concluding paragraph to frame the discussion and summarize the key takeaways. Make sure to explicitly mention the Ideal Gas Law where relevant (e.g., for PV work).

8. Self-Correction/Review: Read through the answer and ask:

   • Does it directly address the prompt? Yes, it provides derivations bridging reversible and irreversible processes.
   • Are the examples relevant and clear? Yes, work, entropy, and heat transfer are central concepts.
   • Are the starting and ending equations clearly identified? Yes, for each example.
   • Is the explanation of irreversibility mechanisms sufficient? Yes, friction, pressure differences, etc., are mentioned.
   • Is the language precise and thermodynamically sound? Review terms and definitions. (e.g., Be careful about dQ in irreversible processes. It's not directly related to TdS anymore for internal entropy generation, but dQ/T can still represent entropy change due to heat transfer across the system boundary. Clarify this subtly if needed, but focus on the broader picture.)

By following these steps, I can construct a comprehensive and accurate answer that addresses all aspects of the prompt and provides valuable insights into the bridge between reversible and irreversible thermodynamics.

response

Let's explore how derivations bridge the gap between reversible and irreversible thermodynamic processes, focusing on key examples and equations.

Core Concept: Entropy Generation

The central difference between reversible and irreversible processes lies in entropy generation. Reversible processes are idealized scenarios where entropy remains constant for the system plus surroundings (isentropic for adiabatic reversible processes within the system, but overall entropy of the universe is constant). Irreversible processes, however, always lead to an increase in the total entropy of the system plus surroundings. This entropy generation arises from factors like friction, unrestrained expansion, heat transfer across finite temperature differences, and chemical reactions far from equilibrium.

Example 1: Work in Isothermal Expansion of an Ideal Gas

• Process: Isothermal expansion of an ideal gas.

• Reversible Process (Starting Point):

  • Initial Equation: For a reversible isothermal expansion, the work done by the system is derived by integrating the pressure with respect to volume, knowing that for an ideal gas P = nRT/V:

    W_{rev} = -∫_{V_1}^{V_2} P_{ext} dV
    

    For a reversible process, the external pressure P_{ext} is infinitesimally smaller than the system pressure P_{sys} at each step, and in the limit, P_{ext} ≈ P_{sys} = P = nRT/V.

  • Derivation: Substituting P = nRT/V and integrating:

    W_{rev} = -∫_{V_1}^{V_2} (nRT/V) dV
    W_{rev} = -nRT ∫_{V_1}^{V_2} (1/V) dV
    W_{rev} = -nRT [ln(V)]_{V_1}^{V_2}
    W_{rev} = -nRT ln(V_2/V_1)
    

    Since V_2 > V_1 for expansion, ln(V_2/V_1) is positive, and W_{rev} is negative, indicating work done by the system. This is the maximum work obtainable from an isothermal expansion between V_1 and V_2.

  • Ending Equation (Reversible Work):

    W_{rev} = -nRT ln(V_2/V_1)
    

• Irreversible Process (Bridging to Irreversible):

  • Introducing Irreversibility: Let's consider an irreversible isothermal expansion where the gas expands against a constant external pressure P_{ext}, which is significantly smaller than the initial gas pressure P_1. This simulates a sudden expansion, like a piston moving rapidly against a constant opposing force.

  • Modifying the Equation: In this irreversible case, the external pressure P_{ext} is constant during the expansion. Therefore, it can be taken out of the integral:

    W_{irr} = -∫_{V_1}^{V_2} P_{ext} dV
    W_{irr} = -P_{ext} ∫_{V_1}^{V_2} dV
    W_{irr} = -P_{ext} [V]_{V_1}^{V_2}
    W_{irr} = -P_{ext} (V_2 - V_1)
    

  • Ending Equation (Irreversible Work):

    W_{irr} = -P_{ext} (V_2 - V_1)
    

  • Bridging and Interpretation:

    • W_{irr} is always less negative (i.e., less work done by the system) than W_{rev} for the same volume change (V_2 > V_1), because P_{ext} < P_{sys} throughout the irreversible process.
    • The "lost work" in the irreversible process is dissipated as heat within the system and surroundings, leading to entropy generation.
    • To quantify entropy generation, we can use the First Law and Second Law. For an isothermal process, ΔU = 0 for an ideal gas. Therefore, Q = -W.
    • For the reversible case, Q_{rev} = -W_{rev} = nRT ln(V_2/V_1). Entropy change of the system in the reversible case is ΔS_{sys, rev} = Q_{rev}/T = nR ln(V_2/V_1).
    • For the irreversible case, Q_{irr} = -W_{irr} = P_{ext} (V_2 - V_1). The entropy change of the system is still ΔS_{sys, irr} = nR ln(V_2/V_1) because entropy is a state function and depends only on the initial and final states, which are the same in both reversible and irreversible isothermal expansions between V_1 and V_2.
    • Entropy generation occurs in the surroundings. In the irreversible case, less heat is absorbed from the surroundings (Q_{irr} < Q_{rev}), so the entropy decrease of the surroundings is smaller in magnitude. The total entropy change (system + surroundings) is positive for the irreversible process, indicating entropy generation.

Example 2: Entropy Change in Irreversible Heat Transfer

• Process: Heat transfer between two reservoirs at different temperatures.

• Reversible Process (Conceptual Starting Point):

  • Idealized Reversible Heat Transfer: Reversible heat transfer would require an infinitesimal temperature difference. In this idealized limit, we can imagine a series of reservoirs with infinitesimally varying temperatures between T_H and T_C. Heat transfer would occur reversibly at each step. In this ideal case, the total entropy change of the universe would be zero.

  • Entropy Change for Reversible Heat Transfer (within a system): For a system undergoing reversible heat transfer at temperature T, the entropy change is given by:

    dS = dQ_{rev} / T
    

• Irreversible Process (Bridging to Irreversible):

  • Introducing Irreversibility: Consider heat transfer directly from a hot reservoir at temperature T_H to a cold reservoir at temperature T_C (where T_H > T_C). This is a classic irreversible process.

  • Modifying the Equation (Focusing on Entropy Change):

    • Let's say an amount of heat Q is transferred from the hot reservoir to the cold reservoir.
    • Entropy change of the hot reservoir: Since heat is leaving the hot reservoir, dQ_{hot} = -Q. Assuming the reservoir is large enough that its temperature remains constant, the entropy change of the hot reservoir is:

      ΔS_{hot} = Q_{hot} / T_H = -Q / T_H
      

    • Entropy change of the cold reservoir: Since heat is entering the cold reservoir, dQ_{cold} = +Q. The entropy change of the cold reservoir is:

      ΔS_{cold} = Q_{cold} / T_C = +Q / T_C
      

    • Total entropy change (universe, in this case, hot + cold reservoirs):

      ΔS_{total} = ΔS_{hot} + ΔS_{cold} = -Q / T_H + Q / T_C = Q (1/T_C - 1/T_H) = Q (T_H - T_C) / (T_H T_C)
      

  • Ending Equation (Irreversible Entropy Change):

    ΔS_{total} = Q (1/T_C - 1/T_H)  = Q (T_H - T_C) / (T_H T_C)
    

  • Bridging and Interpretation:

    • Since T_H > T_C, ΔS_{total} > 0. This confirms that irreversible heat transfer leads to an increase in entropy of the universe.
    • In a reversible heat transfer, T_H ≈ T_C, and in the limit as T_H → T_C, ΔS_{total} → 0, recovering the reversible condition of no net entropy change.
    • The entropy generation is directly related to the temperature difference. The larger the temperature difference, the more irreversible the process and the greater the entropy generation for the same amount of heat transferred.

Example 3: Adiabatic Irreversible Expansion (Free Expansion)

• Process: Adiabatic expansion of an ideal gas into a vacuum (free expansion).

• Reversible Process (Starting Point - Conceptual):

  • Reversible Adiabatic Expansion: For a reversible adiabatic expansion, dQ = 0 and dS = 0 for the system. The work is done at the expense of internal energy, leading to a temperature decrease. The relationship between pressure and volume is given by PV^γ = constant, where γ = C_p/C_v.

• Irreversible Process (Free Expansion):

  • Introducing Irreversibility: Free expansion occurs when a gas expands into a vacuum (P_{ext} = 0). This is highly irreversible and unrestrained.

  • Modifying the Equation (Focusing on Work and Entropy Change):

    • Work: Since P_{ext} = 0, the work done in free expansion is:

      W_{free} = -∫_{V_1}^{V_2} P_{ext} dV = -∫_{V_1}^{V_2} 0 dV = 0
      

      So, no work is done by or on the system in free expansion.

    • First Law: For an adiabatic process, dQ = 0. Therefore, from the First Law (ΔU = dQ + dW), we have ΔU = 0 + 0 = 0. For an ideal gas, internal energy U depends only on temperature. Thus, ΔU = 0 implies ΔT = 0. The temperature remains constant in the free expansion of an ideal gas.

    • Entropy Change: Even though it's adiabatic (dQ = 0), it's not reversible, so we cannot use dS = dQ/T. Instead, we need to calculate the entropy change using a reversible path that connects the same initial and final states. Since the temperature is constant and volume increases from V_1 to V_2, we can imagine a reversible isothermal expansion between the same volumes. For an isothermal expansion of an ideal gas, we derived earlier:

      ΔS_{sys} = nR ln(V_2/V_1)
      

      Since V_2 > V_1 for expansion, ΔS_{sys} > 0.

  • Ending Equations (Irreversible Free Expansion):

    • W_{free} = 0
    • ΔT = 0 (for ideal gas)
    • ΔS_{sys} = nR ln(V_2/V_1)

  • Bridging and Interpretation:

    • In the reversible adiabatic expansion, work is done, and temperature decreases. In the irreversible free expansion, no work is done, and temperature remains constant (for an ideal gas).
    • Despite being adiabatic (dQ = 0), the free expansion is highly irreversible, resulting in a significant increase in entropy of the system. This entropy generation is due to the unrestrained expansion, which is far from equilibrium.
    • The entropy change in the irreversible process is calculated by considering a reversible path between the same initial and final states, highlighting that entropy is a state function.

General Approach to Bridging Reversible and Irreversible Equations:

1. Start with Reversible Equations: Begin with the thermodynamic equations derived under reversible conditions (e.g., dW = -PdV for reversible work, dS = dQ_{rev}/T for reversible entropy change).

2. Identify the Source of Irreversibility: Pinpoint the factor causing irreversibility (friction, pressure difference, temperature difference, etc.).

3. Modify Equations to Account for Irreversibility:

   • For work, replace the reversible pressure P with the actual external pressure P_{ext} or consider frictional forces.
   • For heat transfer, acknowledge finite temperature differences and their impact on entropy generation.
   • For entropy change, realize that dS ≠ dQ/T for irreversible processes within the system if entropy is being generated internally. However, dS = dQ/T + dS_{gen} where dS_{gen} is the entropy generated within the system. For calculating the change in state, we can still use reversible paths to find ΔS because entropy is a state function. The total entropy change (system + surroundings) will always increase in irreversible processes.

4. Focus on Entropy Generation: Irreversible processes are fundamentally linked to entropy generation. Quantify this entropy generation by considering the deviations from reversible conditions and using the Second Law of Thermodynamics.

5. State Functions vs. Path Functions: Remember that state functions like internal energy U, enthalpy H, and entropy S depend only on the initial and final states. Their changes (ΔU, ΔH, ΔS) can be calculated using reversible paths, even for irreversible processes. Path functions like work W and heat Q do depend on the path, and their values are different for reversible and irreversible processes between the same initial and final states.

By understanding these principles, we can bridge the idealized world of reversible thermodynamics with the reality of irreversible processes, allowing for more accurate descriptions and analyses of real-world phenomena.