Chapter 5: Quasistatic Processes with Irreversible Work Transfer and Non-Quasistatic Processes PDF

Classical thermodynamics, Srini Vanapalli CHAPTER 5: Quasistatic Processes with Irreversible Work Transfer and Non- Quasistatic Processes In this chapter, we delve into the complexities of quasistatic processes with irreversible work transfer and introduce the concept of...

Classical thermodynamics, Srini Vanapalli CHAPTER 5: Quasistatic Processes with Irreversible Work Transfer and Non- Quasistatic Processes In this chapter, we delve into the complexities of quasistatic processes with irreversible work transfer and introduce the concept of non-quasistatic processes. Traditional thermodynamic analysis often assumes idealized, reversible conditions, where changes occur slowly enough to maintain equilibrium at each infinitesimal step. However, real-world processes frequently involve irreversible effects—even within quasistatic conditions—that lead to energy dissipation and entropy generation. We begin by exploring how irreversible work transfer occurs within quasistatic processes, such as frictional and dissipative forces in mechanical systems, and analyze their effects on energy distribution. By separating the reversible and irreversible components of work, we reveal the impact of dissipative forces on internal energy and entropy. To illustrate these effects, we examine a range of ideal gas processes—isochoric, isobaric, isothermal, and adiabatic—that involve irreversibilities, providing insight into entropy generation and the limits of energy recoverability. The chapter then shifts to non-quasistatic processes, where equilibrium is not maintained, complicating the analysis. Here, classical thermodynamic relationships must be re-examined to account for rapid changes and non-equilibrium conditions. We provide examples, including gas expansion in a vacuum and expansion against constant external pressure, to demonstrate the unique characteristics of non-quasistatic processes and their implications for energy and entropy changes. Finally, we derive the thermodynamic identity, a fundamental relation that connects internal energy, entropy, temperature, pressure, and volume. This derivation emphasizes a comprehensive approach, respecting the contributions of both heat transfer and irreversibilities, distinguishing it from common simplified treatments. By establishing the thermodynamic identity in this broader context, we reinforce its universal applicability to both reversible and irreversible processes. Introduction to Quasistatic Processes with Irreversible Work Transfer In thermodynamics, a quasistatic process is one that occurs slowly enough for the system to remain in a state of equilibrium at each infinitesimal step. For a process to be reversible, however, the total entropy change of the combined system and surroundings must be zero, meaning there is no net entropy generation in the universe. In a reversible process, both the system and surroundings can, in principle, be returned to their initial states without any net impact on the universe. This reversibility is an idealization that assumes the absence of any dissipative effects, such as friction or unrestrained expansion, which would otherwise cause irreversibilities and increase the total entropy of the system and surroundings. In a purely reversible quasistatic process, the entropy of the system may increase, decrease, or remain constant, as long as any changes are exactly balanced by corresponding changes in the surroundings. Quasistatic, reversible processes are foundational in thermodynamics, serving as a benchmark for ideal energy transformations and allowing us to explore fundamental relationships and limits within thermodynamic systems. However, in real-world scenarios, processes often contain irreversibilities even when they are slow enough to be considered quasistatic. This introduces the concept of irreversible work transfer within a quasistatic process. Here, the process may still be slow enough to maintain equilibrium at each step, but certain irreversible effects—such as friction, inelastic deformations, or viscous dissipation— 60  Classical thermodynamics, Srini Vanapalli prevent the process from being fully reversible. These irreversibilities introduce additional energy losses, which cannot be recovered by simply reversing the process. Irreversible work transfer has a significant impact on energy distribution within these processes, primarily by causing temperature increases due to dissipative effects like friction. For example, friction within a piston-cylinder setup causes mechanical work to raise the temperature of the gas or material, which may lead to heat transfer if the system is not adiabatic. However, in an adiabatic system, this temperature increase would remain contained within the substance without necessarily converting mechanical work directly into thermal energy transferred to the surroundings. An illustrative example of irreversible work transfer can be seen in a mechanical shaft rotating within a liquid. In this scenario, the shaft's mechanical energy is dissipated as it encounters resistance from the fluid. This interaction increases the temperature of the liquid due to frictional forces, making the work done by the shaft irreversible; it cannot be fully recovered to drive the shaft in reverse. This irreversibility leads to an increase in the total entropy of the system and surroundings, reflecting the energy loss associated with irreversible work transfer. These examples highlight that even in quasistatic processes, irreversibilities are present, leading to entropy generation and irreversible energy loss within the system and surroundings. Energy Transfer in Quasistatic Processes with Irreversible Work In a quasistatic process with irreversible work transfer, energy distribution within the system consists of both recoverable and irrecoverable components. While the process is slow enough to assume that equilibrium is maintained at each step, certain irreversible effects—such as friction, resistance, and other non-conservative forces—dissipate a portion of the work. This dissipated work cannot be fully recovered and contributes to entropy generation, setting it apart from purely reversible work in an ideal quasistatic process. To clarify this, we separate the reversible and irreversible components of work. The reversible component represents the idealized, recoverable work done by the system that could, in principle, be fully recovered. In contrast, the irreversible component stems from dissipative forces like friction, viscosity, or electrical resistance, resulting in internal energy dissipation and entropy generation within the system. Examples of Reversible and Irreversible Work Consider a battery as an example of reversible work transfer. In an ideal scenario, the battery converts chemical energy into electrical work, which could be recovered by recharging, making it a reversible process. However, in real batteries, Joule heating due to internal resistance and electrode losses are irreversible components, as they result in entropy generation and energy loss. These effects cannot be fully reversed, even if the battery is recharged. Another example is a fuel cell that combines hydrogen and oxygen to produce electricity and water. Ideally, this process is reversible, as electricity could be used to split water back into hydrogen and oxygen, allowing complete recovery of the input energy. However, in practice, Joule heating, electrode degradation, and formation of gas bubbles within the electrolyte lead to irreversible energy losses that cannot be fully recovered, making part of the work irreversible. A third example is the expansion of a gas within a piston-cylinder setup. In an ideal scenario, the work done by the gas expanding against the piston could be recovered completely if there were no frictional losses. However, with a lossy piston, frictional effects dissipate some of the work, which increases the 61  Classical thermodynamics, Srini Vanapalli internal energy of the gas or piston material. This irreversibility results in an increase in the total entropy of the system and surroundings, and the dissipated work cannot be recovered. Mathematical Treatment of Reversible and Irreversible Work Using the first law of thermodynamics as defined in this book, we can write: Δ𝑈 = 𝑄 + 𝑊 where Δ𝑈 represents the change in internal energy of the system, 𝑄 is the heat added, and 𝑊 is the total work done on the system. In a quasistatic process with irreversibilities, we can separate 𝑊 into the reversible work 𝑊𝑟𝑒𝑣 and irreversible work 𝑊𝑖𝑟𝑟 components: 𝑊 = 𝑊𝑟𝑒𝑣 + 𝑊𝑖𝑟𝑟 The reversible work 𝑊𝑟𝑒𝑣 represents the ideal component of work that could, in principle, be fully recovered. The irreversible work 𝑊𝑖𝑟𝑟 , on the other hand, arises due to dissipative forces like friction or resistance, leading to energy dissipation within the system. In the modified energy equation, the presence of 𝑊𝑖𝑟𝑟 reflects the energy lost due to irreversibilities: Δ𝑈 = 𝑄 + 𝑊𝑟𝑒𝑣 + 𝑊𝑖𝑟𝑟 In a purely reversible process, 𝑊𝑖𝑟𝑟 would be zero, meaning there would be no net entropy generation and all the work could theoretically be recovered. However, when 𝑊𝑖𝑟𝑟 is non-zero due to dissipative forces, there is an irreversible loss of energy, contributing to an increase in total entropy. By separating 𝑊𝑟𝑒𝑣 and 𝑊𝑖𝑟𝑟 in the energy equation, we gain insight into how energy transfers are affected by irreversibilities in quasistatic processes. These examples and equations demonstrate the distinct roles of reversible and irreversible work, and underscore the impact of irreversibilities on entropy and energy transfer within thermodynamic systems. Entropy Generation in Quasistatic Processes with Irreversible Work In thermodynamic processes, entropy generation is a direct indicator of irreversibility. Whenever a process involves dissipative effects such as friction, electrical resistance, or inelastic deformation, it becomes irreversible, leading to a net increase in the total entropy of the system and surroundings. This increase in entropy, or entropy generation, is the hallmark of irreversible work within a thermodynamic process, including those that are quasistatic. Entropy Generation Due to Irreversibilities In an idealized reversible process, the total entropy change of the system and surroundings is zero, meaning no net entropy is generated, and the process could theoretically be reversed without any loss of energy or increase in entropy. However, in a quasistatic process with irreversible work, dissipative forces introduce irreversibilities that generate entropy. This entropy generation reflects the portion of the energy that cannot be recovered, as it has been dissipated in ways that increase the randomness or disorder of the system. Relationship Between Irreversible Work and Entropy Production The presence of irreversible work in a quasistatic process directly contributes to entropy generation. For instance, consider a gas expanding in a piston-cylinder arrangement with friction. The work done 62  Classical thermodynamics, Srini Vanapalli by the gas against the piston includes an irreversible component due to the frictional forces between the piston and cylinder wall. This irreversible work 𝑊𝑖𝑟𝑟 results in increased temperature and disorder within the system, which cannot be reversed without adding energy from an external source. Consequently, this irreversible work leads to a positive entropy generation term, representing the loss of available energy in the system. The relationship between entropy generation 𝑆𝑔𝑒𝑛 and irreversible work 𝑊𝑖𝑟𝑟 can be expressed as: 𝑆𝑔𝑒𝑛 ∝ 𝑊𝑖𝑟𝑟 ⁄𝑇 Where 𝑇 is the absolute temperature at which the irreversibility occurs. This relation implies that the greater the irreversible work, the more entropy is generated. The entropy generation term, 𝑆𝑔𝑒𝑛 , measures the degree of irreversibility in the process and signifies the portion of the energy transformation that is irrecoverable. Entropy in Reversible vs. Irreversible Quasistatic Processes In a purely reversible quasistatic process, there are no dissipative forces, and as a result, there is no entropy generation. Any entropy change within the system is exactly balanced by an equal and opposite change in the surroundings, leading to zero net entropy change in the universe. This reversibility implies that all work performed can be fully recovered, with no energy lost to irreversible effects. In contrast, in a quasistatic process with irreversible work, entropy is generated within the system or surroundings (or both), leading to a net positive entropy change in the universe. This irreversibility reduces the potential for energy recovery, as part of the work has been dissipated through mechanisms that increase entropy, such as frictional heating or resistive heating. The resulting entropy generation thus distinguishes quasistatic processes with irreversible work from ideal reversible processes, underscoring the fundamental difference in energy efficiency and recoverability. In summary, the presence of entropy generation in a quasistatic process with irreversible work signals the loss of usable energy and highlights the impact of irreversibilities. This entropy generation reflects the inherent inefficiencies in real-world processes, contrasting sharply with the idealized behavior of purely reversible systems. Quasistatic Irreversible Ideal Gas Processes This section explores various types of ideal gas processes that are quasistatic but involve irreversibilities. In each case, the process is slow enough to maintain equilibrium at every step, yet friction, heat transfer resistance, or other dissipative effects introduce irreversibilities, resulting in entropy generation. Isochoric Process with Irreversibility (Constant Volume) In an isochoric process, the volume of the gas remains constant, so no boundary work is done (𝑊 = 0). We’ll analyze two cases where the system undergoes a change in internal energy due to heat transfer, and in one case, additional work through a rotating shaft. In both cases, we assume the gas is ideal, and the system goes from the same initial state to the same final state. Let 𝑇𝑠𝑢𝑟 be the temperature of the surroundings, which remains constant during the process. Case 1: Isochoric Process with Only Heat Transfer In this first case, the gas undergoes an isochoric process where heat 𝑄1 is transferred from the surroundings to the system. According to the first law of thermodynamics as defined in this book: 63  Classical thermodynamics, Srini Vanapalli 𝛥𝑈 = 𝑄1 where 𝛥𝑈 is the change in internal energy of the gas. Given that volume is constant, all the heat transfer goes into changing the internal energy. For an ideal gas undergoing an isochoric process, the entropy change of the system is: 𝑇𝑓 𝛥𝑆𝑠𝑦𝑠𝑡𝑒𝑚 = 𝑛𝑐𝑣 𝑙𝑛 ( ) 𝑇𝑖 where 𝑇𝑖 and 𝑇𝑓 are the initial and final temperatures of the gas, 𝑛 is the number of moles, and 𝑐𝑣 is the molar heat capacity at constant volume. Since this process is irreversible (likely due to a temperature gradient between the surroundings and system), there is entropy generation in the surroundings, denoted 𝑆𝑔𝑒𝑛,1. For the surroundings, the entropy change due to heat transfer 𝑄1 at temperature 𝑇𝑠𝑢𝑟 is: 𝑄1 Δ𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠,1 = − 𝑇𝑠𝑢𝑟 The total entropy generation for Case 1, combining the system and surroundings, is: 𝑇𝑓 𝑄1 𝑆𝑔𝑒𝑛,1 = 𝛥𝑆𝑠𝑦𝑠𝑡𝑒𝑚 + Δ𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠,1 = 𝑛𝑐𝑣 𝑙𝑛 ( ) − 𝑇𝑖 𝑇𝑠𝑢𝑟 Case 2: Isochoric Process with Heat Transfer and Shaft Work In the second scenario, in addition to heat transfer, a shaft rotates within the gas, performing mechanical work 𝑊𝑠ℎ𝑎𝑓𝑡 on the gas. This work is dissipated entirely as heat within the system, contributing to the internal energy change and reducing the amount of heat 𝑄2 needed from the surroundings. The total change in internal energy is the same as in Case 1: 𝛥𝑈 = 𝑄2 + 𝑊𝑠ℎ𝑎𝑓𝑡 Since both cases start and end at the same states, we have: 𝑄1 = 𝑄2 + 𝑊𝑠ℎ𝑎𝑓𝑡 Therefore, 𝑄2 < 𝑄1 due to the additional energy input from the shaft work. For the system, the entropy change remains the same as in Case 1, as it only depends on the initial and final states: 𝑇𝑓 𝛥𝑆𝑠𝑦𝑠𝑡𝑒𝑚 = 𝑛𝑐𝑣 𝑙𝑛 ( ) 𝑇𝑖 For the surroundings, the entropy change due to the heat transfer 𝑄2 is: 𝑄2 Δ𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠,2 = − 𝑇𝑠𝑢𝑟 64  Classical thermodynamics, Srini Vanapalli Comparison of Entropy Generation The total entropy generation for Case 2, combining the system and surroundings, is: 𝑇𝑓 𝑄2 𝑆𝑔𝑒𝑛,2 = 𝛥𝑆𝑠𝑦𝑠𝑡𝑒𝑚 + Δ𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠,2 = 𝑛𝑐𝑣 𝑙𝑛 ( ) − 𝑇𝑖 𝑇𝑠𝑢𝑟 Since 𝑄2 < 𝑄1 , we have: 𝑄2 𝑄1 − >− 𝑇𝑠𝑢𝑟 𝑇𝑠𝑢𝑟 Therefore, 𝑆𝑔𝑒𝑛,2 > 𝑆𝑔𝑒𝑛,1 This shows that the additional shaft work in Case 2 introduces more irreversibility, leading to a greater entropy generation than in Case 1. Although less heat is needed from the surroundings in Case 2, the total irreversibility of the process is higher due to the dissipative work done by the shaft. This example highlights how work dissipation can increase entropy generation, even when reducing external heat transfer. Isobaric Process with Irreversibility (Constant Pressure) In an isobaric process, the pressure of the gas remains constant, which allows for boundary work as the gas expands or contracts. We’ll consider two cases where the system undergoes an internal energy change due to heat transfer, and in one case, additional work is provided through a rotating shaft. The gas is assumed to be ideal, and both cases go from the same initial to the same final state. Let 𝑇𝑠𝑢𝑟 represent the temperature of the surroundings, which remains constant throughout the process. Case 1: Isobaric Process with Only Heat Transfer In the first scenario, heat 𝑄1 is transferred to the system from the surroundings at constant pressure, causing the gas to expand. According to the first law of thermodynamics as defined in this book: 𝛥𝑈 = 𝑄1 + 𝑊 where Δ𝑈 is the change in internal energy, 𝑄1 is the heat transfer, and 𝑊 is the work done by the system on the surroundings. For an isobaric process, the work done by the gas on the surroundings is given by: 𝑊 = −𝑝Δ𝑉 Since 𝛥𝑈 depends only on the initial and final states, the heat transfer 𝑄1 is required to account for both the internal energy change and the work done: 𝑄1 = 𝛥𝑈 + 𝑝𝑑𝑉 The entropy change of the system, given it’s an ideal gas, can be expressed as: 𝑇𝑓 𝛥𝑆𝑠𝑦𝑠𝑡𝑒𝑚 = 𝑛𝑐𝑝 𝑙𝑛 ( ) 𝑇𝑖 65  Classical thermodynamics, Srini Vanapalli where 𝑇𝑖 and 𝑇𝑓 are the initial and final temperatures, 𝑛 is the number of moles, and 𝑐𝑝 is the molar heat capacity at constant pressure. Because this process is irreversible (possibly due to a temperature gradient between the surroundings and the system), there is entropy generation in the surroundings, which we denote as 𝑆𝑔𝑒𝑛,1. For the surroundings, the entropy change due to heat transfer Q1Q_1Q1 at temperature 𝑇𝑠𝑢𝑟 is: 𝑄1 Δ𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠,1 = − 𝑇𝑠𝑢𝑟 The total entropy generation for Case 1, combining system and surroundings, is: 𝑇𝑓 𝑄1 𝑆𝑔𝑒𝑛,1 = 𝛥𝑆𝑠𝑦𝑠𝑡𝑒𝑚 + Δ𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠,1 = 𝑛𝑐𝑝 𝑙𝑛 ( ) − 𝑇𝑖 𝑇𝑠𝑢𝑟 Case 2: Isobaric Process with Heat Transfer and Shaft Work In the second scenario, in addition to heat transfer, a shaft is rotating within the gas, performing mechanical work 𝑊𝑠ℎ𝑎𝑓𝑡 on the gas. This work is fully dissipated as frictional heat within the system, supplementing the heat input required from the surroundings. The total change in internal energy remains the same: 𝛥𝑈 = 𝑄2 + 𝑊𝑠ℎ𝑎𝑓𝑡 + 𝑊 Since both cases start and end at the same states, we have: 𝑄1 = 𝑄2 + 𝑊𝑠ℎ𝑎𝑓𝑡 Thus, in this case, the heat transfer Q2Q_2Q2 from the surroundings is lower than 𝑄1 , as part of the energy input is provided by the shaft work 𝑊𝑠ℎ𝑎𝑓𝑡. The entropy change of the system, given it’s an ideal gas and depends only on the initial and final states, remains the same as in Case 1: 𝑇𝑓 𝛥𝑆𝑠𝑦𝑠𝑡𝑒𝑚 = 𝑛𝑐𝑝 𝑙𝑛 ( ) 𝑇𝑖 The entropy change of the surroundings, due to the heat transfer 𝑄2 , is: 𝑄2 Δ𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠,2 = − 𝑇𝑠𝑢𝑟 Comparison of Entropy Generation The total entropy generation for Case 2, combining system and surroundings, is: 𝑇𝑓 𝑄2 𝑆𝑔𝑒𝑛,2 = 𝛥𝑆𝑠𝑦𝑠𝑡𝑒𝑚 + Δ𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠,2 = 𝑛𝑐𝑝 𝑙𝑛 ( ) − 𝑇𝑖 𝑇𝑠𝑢𝑟 Since 𝑄2 < 𝑄1 , we have: 𝑄2 𝑄1 − >− 𝑇𝑠𝑢𝑟 𝑇𝑠𝑢𝑟 66  Classical thermodynamics, Srini Vanapalli Therefore, 𝑆𝑔𝑒𝑛,2 > 𝑆𝑔𝑒𝑛,1 This analysis shows that the additional shaft work in Case 2 increases the total entropy generation. While less heat is required from the surroundings, the irreversibility associated with the dissipative shaft work results in greater entropy generation, highlighting the impact of energy dissipation on entropy production in an isobaric process. Isothermal Process with Irreversibility (Constant Temperature) In an isothermal process, the temperature of the gas remains constant, so the internal energy change is zero for an ideal gas (𝛥𝑈 = 0). For this analysis, we assume that the surroundings are at a slightly higher temperature 𝑇𝑠𝑢𝑟 = 𝑇 + 𝛥𝑇 than the system's temperature 𝑇. This slight temperature difference drives heat transfer from the surroundings to the system, allowing the gas to perform work while maintaining an isothermal state. Case 1: Isothermal Process with Only Heat Transfer In the first case, the gas undergoes an isothermal expansion where the work done on the surroundings is entirely supplied by heat transfer Q1Q_1Q1 from the surroundings. According to the first law of thermodynamics: 𝛥𝑈 = 𝑄1 + 𝑊 Since 𝛥𝑈 = 0 for an isothermal process, we have: 𝑄1 = −𝑊 The work done by the system during an isothermal expansion from an initial volume 𝑉𝑖 to a final volume 𝑉𝑓 at temperature 𝑇 is given by: 𝑊 = −𝑛𝑅𝑇𝑙𝑛 (𝑉𝑓 /𝑉𝑖 ) Therefore, the heat transfer required is: 𝑄1 = 𝑛𝑅𝑇𝑙𝑛 (𝑉𝑓 /𝑉𝑖 ) The entropy change for the system, due to the isothermal heat transfer, is: 𝑄1 𝛥𝑆𝑠𝑦𝑠𝑡𝑒𝑚 = = 𝑛𝑅𝑙𝑛(𝑉𝑓 /𝑉𝑖 ) 𝑇 Since this process is irreversible due to the temperature difference 𝛥𝑇 between the system and surroundings, entropy is generated. For the surroundings, the entropy change due to the heat transfer 𝑄1 at temperature 𝑇𝑠𝑢𝑟 : 𝑄1 𝑛𝑅𝑇𝑙𝑛(𝑉𝑓 /𝑉𝑖 ) 𝛥𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠,1 = − =− 𝑇𝑠𝑢𝑟 𝑇 + Δ𝑇 The total entropy generation for Case 1, combining the system and surroundings, is: 67  Classical thermodynamics, Srini Vanapalli 𝑛𝑅𝑇𝑙𝑛(𝑉𝑓 /𝑉𝑖 ) 𝑆𝑔𝑒𝑛,1 = Δ𝑆𝑠𝑦𝑠𝑡𝑒𝑚 + Δ𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠,1 = 𝑛𝑅𝑙𝑛(𝑉𝑓 /𝑉𝑖 ) − 𝑇 + Δ𝑇 Since 𝑇 < 𝑇 + 𝛥𝑇, the entropy generation 𝑆𝑔𝑒𝑛,1 is positive, reflecting the irreversibility due to the temperature difference. Case 2: Isothermal Process with Heat Transfer and Shaft Work In the second scenario, in addition to heat transfer, a shaft rotates within the gas, performing mechanical work 𝑊𝑠ℎ𝑎𝑓𝑡 on the gas, which is fully dissipated as frictional heat within the system. This dissipative work supplements the heat required from the surroundings. According to the first law of thermodynamics, we have: 𝛥𝑈 = 𝑄2 + 𝑊 + 𝑊𝑠ℎ𝑎𝑓𝑡 = 0 Thus: 𝑉𝑓 𝑄2 = −(𝑊 + 𝑊𝑠ℎ𝑎𝑓𝑡 ) = 𝑛𝑅𝑇𝑙 𝑛 ( ) − 𝑊𝑠ℎ𝑎𝑓𝑡 𝑉𝑖 Since 𝑊𝑠ℎ𝑎𝑓𝑡 is positive, 𝑄2 < 𝑄1 , meaning less heat is required from the surroundings due to the additional work from the shaft. The entropy change for the system remains the same: 𝛥𝑆𝑠𝑦𝑠𝑡𝑒𝑚 = 𝑛𝑅𝑙𝑛(𝑉𝑓 /𝑉𝑖 ) For the surroundings, the entropy change due to the reduced heat transfer 𝑄2 is: 𝑉𝑓 𝑄2 𝑛𝑅𝑇𝑙𝑛 ( ) − 𝑊𝑠ℎ𝑎𝑓𝑡 𝑉𝑖 𝛥𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠,2 =− =− 𝑇𝑠𝑢𝑟 𝑇 + Δ𝑇 Comparison of Entropy Generation The total entropy generation for Case 2 is: 𝑉𝑓 𝑛𝑅𝑇𝑙𝑛 ( ) − 𝑊𝑠ℎ𝑎𝑓𝑡 𝑉𝑖 𝑆𝑔𝑒𝑛,2 = Δ𝑆𝑠𝑦𝑠𝑡𝑒𝑚 + Δ𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠,2 = 𝑛𝑅𝑙𝑛(𝑉𝑓 /𝑉𝑖 ) − 𝑇 + Δ𝑇 Since 𝑄2 < 𝑄1 , we find: 𝑄2 𝑄1 − >− 𝑇𝑠𝑢𝑟 𝑇𝑠𝑢𝑟 Thus, 𝑆𝑔𝑒𝑛,2 > 𝑆𝑔𝑒𝑛,1 68  Classical thermodynamics, Srini Vanapalli This analysis demonstrates that the inclusion of shaft work in Case 2 results in greater total entropy generation compared to the heat-only scenario in Case 1, due to additional irreversibilities introduced by the dissipative work. By assuming the surroundings are at a slightly higher temperature than the system, the entropy generation is more accurately represented. Adiabatic Process with Irreversibility (No Heat Transfer) In an adiabatic process, there is no heat exchange between the system and its surroundings (𝑄 = 0). Any change in the system’s internal energy must come solely from the work done on or by the system. We will examine two cases involving an ideal gas: (1) an ideal, reversible adiabatic process with no entropy generation, and (2) an adiabatic process with irreversibilities introduced by a dissipative piston. Case 1: Ideal (Reversible) Adiabatic Process with Only Expansion/Compression Work In the first case, the gas undergoes an ideal, reversible adiabatic process where the work 𝑊1 done on or by the system is the sole contributor to the change in internal energy. According to the first law of thermodynamics: 𝛥𝑈 = 𝑄 + 𝑊1 Since the process is adiabatic (𝑄 = 0), this simplifies to: 𝛥𝑈 = 𝑊1 For an ideal gas undergoing an adiabatic process, the relationship between pressure and volume can be described by: 𝑝𝑉 𝛾 = constant where 𝛾 = 𝑐𝑝 /𝑐𝑣 is the heat capacity ratio. For an adiabatic expansion, the work 𝑊1 done by the gas is negative, leading to a decrease in internal energy and a drop in temperature. For an adiabatic compression, the work 𝑊1 done on the gas is positive, increasing the internal energy and raising the gas temperature. Since this process is reversible, there are no irreversibilities, and the total entropy change for the system and surroundings is zero: 𝛥𝑆𝑠𝑦𝑠𝑡𝑒𝑚 = 0 and 𝛥𝑆𝑡𝑜𝑡𝑎𝑙 = 0 With no heat exchange, the surroundings experience no entropy change, and the total entropy generation for this process is zero. Case 2: Adiabatic Process with a Dissipative Piston (Irreversible Work) In the second scenario, the gas undergoes an adiabatic expansion or compression, but with a dissipative piston that introduces friction as it moves. This frictional force dissipates part of the work 𝑊𝑑𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑖𝑣𝑒 done by or on the gas, causing an increase in the internal energy of the gas, which leads to entropy generation within the system and represents an irreversible process. According to the first law of thermodynamics: 69  Classical thermodynamics, Srini Vanapalli 𝛥𝑈 = 𝑄 + 𝑊2 + 𝑊𝑑𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑖𝑣𝑒 Since 𝑄 = 0 for an adiabatic process, this simplifies to: 𝛥𝑈 = 𝑊2 + 𝑊𝑑𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑖𝑣𝑒 The dissipative work 𝑊𝑑𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑖𝑣𝑒 effectively adds to the internal energy change of the gas. For an adiabatic expansion with a dissipative piston, the temperature of the gas does not drop as much as it would in the reversible case because the dissipative work partially offsets the cooling effect of the expansion. Conversely, for an adiabatic compression with a dissipative piston, the gas temperature increases more than in the reversible case due to the additional frictional contribution to internal energy. The key point here is that, because of the irreversibility introduced by the dissipative piston, the end state in Case 2 cannot match the end state in Case 1 for the same amount of work done by or on the gas. The presence of friction changes the internal energy distribution, resulting in a different temperature and pressure compared to the reversible adiabatic process. The entropy change of the system in Case 2 is now positive, as the dissipative piston introduces irreversibilities: 𝛥𝑆𝑠𝑦𝑠𝑡𝑒𝑚 = 𝑆𝑔𝑒𝑛,2 > 0 Since there is no heat exchange, the surroundings experience no entropy change, and the total entropy generation is entirely within the system: 𝛥𝑆𝑡𝑜𝑡𝑎𝑙 = 𝑆𝑔𝑒𝑛,2 > 0 Comparison of Entropy Generation In Case 1, the total entropy change 𝛥𝑆𝑡𝑜𝑡𝑎𝑙 is zero, as it represents an ideal, reversible adiabatic process with no irreversibilities. In Case 2, the additional dissipative work due to friction in the piston results in greater entropy generation within the system, represented by 𝑆𝑔𝑒𝑛,2. Thus: 𝑆𝑔𝑒𝑛,2 > 0 Furthermore, because of this dissipative work, the final state in Case 2 differs from that in Case 1. The irreversibilities prevent reaching the same pressure, volume, and temperature as in the ideal, reversible case, even if the initial states were identical. This example illustrates how introducing dissipative work in an adiabatic process increases total entropy generation and results in a different final state than in the ideal, reversible case. Readers are encouraged to plot the adiabatic expansion and compression processes on 𝑝𝑉 and 𝑇𝑆 diagrams to visualize the impact of irreversibilities on the process path and final state. Additionally, in the isobaric and isothermal examples discussed earlier, a similar analysis can be conducted by considering a lossy piston instead of shaft work, as the friction in the piston would also introduce irreversibilities. While we have primarily analyzed expansion processes, readers are encouraged to apply a similar approach to compression processes to explore how irreversibilities affect entropy generation and final states in those scenarios as well. 70  Classical thermodynamics, Srini Vanapalli Transition to Non-Quasistatic Processes In the study of thermodynamics, we have focused primarily on quasistatic processes—those that proceed slowly enough for the system to remain in equilibrium at every step. In a quasistatic process, the state variables (pressure, volume, temperature) are well-defined at each instant, allowing the system to follow a continuous path of equilibrium states. This assumption of equilibrium simplifies the analysis, as we can use precise thermodynamic relationships to describe the process. However, in real-world applications, many processes are inherently non-quasistatic, or non- equilibrium, as they occur too rapidly or involve too many internal interactions for the system to maintain equilibrium throughout. In a non-quasistatic process, variables such as pressure and temperature can vary significantly within different parts of the system, making it impossible to describe the system with a single, uniform state at any given moment. These processes introduce new complexities: they require us to account for internal gradients, rapid changes, and potentially chaotic interactions, and they often result in irreversible energy dissipation and entropy generation. In non-quasistatic processes, the departure from equilibrium means that classical thermodynamic relationships—such as those for work, heat, and entropy—cannot always be applied in their standard forms. Instead, analyzing non-quasistatic processes often requires more advanced approaches, such as non-equilibrium thermodynamics. As we transition to examining non-quasistatic processes, we’ll explore the unique challenges they pose and investigate the principles that govern thermodynamic behavior in far-from-equilibrium conditions. This shift offers a broader, more realistic view of how thermodynamic systems operate in nature and technology, expanding our understanding beyond the idealized, reversible framework of quasistatic processes. Introduction to Non-Quasistatic Processes and Moving Boundary Work Transfer for an Ideal Gas In previous chapters, we discussed quasistatic processes—those in which the system remains in mechanical equilibrium with its surroundings at each step. This equilibrium allows us to define the system’s pressure throughout the process. However, in a non-quasistatic process, the system is not in mechanical equilibrium with its surroundings, and, as a result, the pressure within the system varies in ways that are not well-defined at each instant. Consequently, non-quasistatic processes cannot be represented on standard thermodynamic property diagrams, such as the pressure-volume (𝑝𝑉) or temperature-entropy (𝑇𝑆) diagrams. However, we can still determine changes in state variables, such as internal energy and entropy, by comparing the initial and final states, as these are state functions and do not depend on the process path. Moving Boundary Work Transfer in a Non-Quasistatic Process In a non-quasistatic process, moving boundary work occurs when an external force acts to change the system’s volume (i.e., moves the system boundary). The work done on the system due to this external force is: 𝑉 𝑊 = 𝐹⃗𝑒𝑥𝑡 ⋅ 𝑑𝑠⃗ = (𝑝𝑒𝑥𝑡 ⋅ 𝐴)𝑑( ) = −𝑝𝑒𝑥𝑡 𝑑𝑉 𝐴 where: 𝑝𝑒𝑥𝑡 is the external pressure, defined as 𝐹𝑒𝑥𝑡 /𝐴, 𝑑𝑉 is the volume change equal to 𝑠⃗ ⋅ 𝐴, and the minus sign accounts for the convention that work done on the system is positive when the volume decreases. 71  Classical thermodynamics, Srini Vanapalli In a quasistatic process, the system is always in equilibrium with its surroundings, meaning that the system pressure 𝑝 equals the external pressure 𝑝𝑒𝑥𝑡 at all points during the process. Thus, the work expression simplifies to: Quasistatic process: 𝑊 = −𝑝𝑒𝑥𝑡 𝑑𝑉 = −𝑝𝑑𝑉 However, in a non-quasistatic process, the system is not in mechanical equilibrium with its surroundings, so the external pressure 𝑝𝑒𝑥𝑡 differs from the system pressure 𝑝. Therefore, for non- quasistatic processes, the work transferred to the system is: Non − quasistatic process: 𝑊 = −𝑝𝑒𝑥𝑡 𝑑𝑉 In this case, we must use the external pressure 𝑝𝑒𝑥𝑡 rather than the system pressure 𝑝 to calculate the boundary work, as the system’s pressure is not well-defined during the non-equilibrium process. Examples of Moving Boundary Work in Non-Quasistatic Processes 1. Gas Expansion in a Vacuum (External Pressure Zero) Consider an ideal gas enclosed in a cylinder with a weightless piston, initially held in place by stops. The initial pressure and volume of the gas are 𝑝1 and 𝑉1, respectively. When the stops are removed, the gas expands until it reaches a second set of stops, increasing to volume 𝑉2 with a final pressure 𝑝2. Since the external pressure 𝑝𝑒𝑥𝑡 = 0 (a vacuum): 𝑊 = −∫ 𝑝𝑒𝑥𝑡 𝑑𝑉 = 0 Although the gas volume increases, no work is done by the system to expand, as there is no opposing external force. 2. Gas Expansion Against a Constant External Pressure In this scenario, a mass 𝑚 is placed on a weightless piston, creating a constant external pressure 𝑝𝑒𝑥𝑡 = 𝑚𝑔/𝐴, where 𝐴 is the piston area. When the gas expands, the moving boundary work is: 𝑉2 𝑉2 𝑚𝑔 𝑚𝑔 𝑊 = − ∫ 𝑝𝑒𝑥𝑡 𝑑𝑉 = − ∫ 𝑑𝑉 = − (𝑉 − 𝑉1 ) 𝑉1 𝑉1 𝐴 𝐴 2 If the mass 𝑚 is chosen such that 𝑚𝑔 = 𝑝2 𝐴, the piston reaches the second set of stops, with the work done by the gas given by: 𝑊 = −𝑝2 (𝑉2 − 𝑉1 ) This area on a 𝑝𝑉 diagram represents the work performed by the gas. 72  Classical thermodynamics, Srini Vanapalli V1 V2 3. Multi-Step Expansion with Incremental Mass Changes Now, suppose we add another mass 𝑚 on the piston, creating a higher external pressure 𝑝3. After removing the stops, the gas expands to volume 𝑉3. Then, we remove one mass, allowing the gas to further expand to volume 𝑉2. The work done during this process can be broken down as follows: 𝑉3 𝑉2 𝑉3 𝑉2 𝑊 = − ∫ 𝑝𝑒𝑥𝑡 𝑑𝑉 − ∫ 𝑝𝑒𝑥𝑡 𝑑𝑉 = − ∫ 𝑝3 𝑑𝑉 − ∫ 𝑝2 𝑑𝑉 𝑉1 𝑉3 𝑉1 𝑉3 = −𝑝3 (𝑉3 − 𝑉1 ) − 𝑝2 (𝑉2 − 𝑉3 ) The shaded areas represent the work performed by the gas, which is greater than that with a single mass. 73  Classical thermodynamics, Srini Vanapalli P3 V1 V2 4. Continuous Expansion with Incrementally Smaller Masses If we place incrementally smaller masses on the piston and remove them one by one, the total work done by the gas approaches that of a quasistatic process, represented by the bold line in the 𝑝𝑉 diagram. P3 V1 V2 To Conclude: In the limit of small changes to the external force, the work performed by the gas converges to that of a quasistatic process. Suggested Analysis for Compression Processes The above examples illustrate moving boundary work during gas expansion under various external conditions. Readers are encouraged to conduct a similar analysis for compression processes, starting from volume 𝑉2 to 𝑉1. This analysis could involve a single mass, two masses, or a series of smaller masses to observe the effects of different external pressures during compression. By applying the same systematic approach, readers can explore how external forces influence work in non-quasistatic compression processes. Non-Quasistatic Processes This section introduces a methodology for calculating changes in internal energy and entropy in non- quasistatic processes, focusing on ideal gas behavior without additional forms of work transfer. The first and second laws of thermodynamics still apply to non-quasistatic processes. For clarity, we restate them here: First Law (with only pV work): 𝛥𝑈 = 𝑄 + 𝑊 Second Law: 74  Classical thermodynamics, Srini Vanapalli 𝑄 Δ𝑆 = ∫ + 𝑆𝑖𝑟𝑟 𝑇 where 𝑆𝑖𝑟𝑟 represents the irreversible entropy generated within the system due to irreversibilities. We will analyze two types of non-quasistatic processes for an ideal gas: an adiabatic expansion and an isothermal compression, and we will compare each with its quasistatic counterpart. Example 1: Adiabatic Expansion - Quasistatic vs. Non-Quasistatic Consider an ideal gas initially at pressure 𝑝1 and volume 𝑉1, enclosed in a cylinder with a weightless piston. Quasistatic Adiabatic Expansion In a quasistatic adiabatic expansion, an external force 𝐹𝑒𝑥𝑡 on the piston is gradually reduced, allowing the gas to expand along an adiabat. At the final state 𝑛, the gas pressure equals the external pressure. For a quasistatic adiabatic process: 𝛾 𝛾 𝑝𝑉 𝛾 = 𝑝1 𝑉1 = 𝑝𝑛 𝑉𝑛 = constant 𝑐𝑝 Where 𝛾 is the heat capacity ratio 𝑐. The ideal gas law applies throughout: 𝑣 𝑝𝑉 𝑝1 𝑉1 𝑝𝑛 𝑉𝑛 = = = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑇 𝑇1 𝑇𝑛 Since this is an adiabatic process, 𝑄 = 0. Applying the first law: 𝛥𝑈 = 𝑊 With no irreversible entropy generation (𝑆𝑖𝑟𝑟 = 0), the entropy change is: 𝑄 Δ𝑆 = ∫ + 𝑆𝑖𝑟𝑟 = 0 𝑇 Non-Quasistatic Adiabatic Expansion Now consider a non-quasistatic expansion starting from the same initial state. In this process, the gas expands rapidly against the external pressure 𝑝𝑒𝑥𝑡 , so the system is not in equilibrium with its surroundings, and intermediate states are not well-defined. At the final state 𝑛′, the gas pressure equals the external pressure, but the volume differs from the quasistatic case. The moving boundary work in the non-quasistatic process is: 𝑊 = −𝑝𝑒𝑥𝑡 (𝑉𝑛′ − 𝑉1 ) With 𝑄 = 0, applying the first law yields: Δ𝑈 = 𝑊 = 𝑈𝑛′ − 𝑈1 = −𝑝𝑒𝑥𝑡 (𝑉𝑛′ − 𝑉1 ) Since entropy is a function of internal energy and volume, and with 𝑆𝑖𝑟𝑟 > 0 in a non-quasistatic process, the entropy at 𝑛′ is greater than at 𝑛. This implies that 𝑉𝑛′ > 𝑉𝑛 due to additional irreversible entropy generation in the non-quasistatic expansion. 75  Classical thermodynamics, Srini Vanapalli Consider an ideal diatomic gas initially at pressure 𝑝1 = 10 bar, volume 𝑉1 = 1 𝐿, and temperature 𝑇1 = 300 K. We will examine both a quasistatic and a non-quasistatic adiabatic expansion process, beginning from the same initial state. Given Values and Constants (for an ideal diatomic gas): Degrees of freedom, 𝑓 = 5 Heat capacity at constant volume, 𝑐𝑣 = 20.8 J/(mol ⋅ K) Heat capacity ratio, 𝛾 = 1.4 Gas constant, 𝑅 = 8.315 J/(mol ⋅ K) Molar quantity, 𝑛 = 0.4 mol State State 1 State n State n′ State x State x′ p (bar) 10 1 1 10 10 V (L) 1 5.18 7.43 5.18 7.43 U (J) 2500 1295 1857 12949 18571 T (K) 300 155 223 1554 2229 76  Classical thermodynamics, Srini Vanapalli Quasistatic Adiabatic Expansion For this quasistatic expansion from State 1 to State 𝑛, we calculate: Change in internal energy: 𝛥𝑈 = −1205 J Work done: 𝑊 = −1205 J Entropy change: 𝛥𝑆 = 0 To determine the entropy change, we can conceptually follow an alternative path involving two steps: (1) an isobaric process from State 1 to State 𝑥, and (2) an isochoric process from State 𝑥 to State 𝑛. This approach allows us to integrate along known paths, as the entropy change across the actual adiabatic path from State 1 to nnn would be zero for this quasistatic, reversible process. Non-Quasistatic Adiabatic Expansion In the non-quasistatic adiabatic expansion, the gas expands rapidly against a constant external pressure 𝑝𝑒𝑥𝑡 = 1 bar. The system is not in equilibrium with its surroundings, so intermediate states are undefined. The gas eventually reaches a final state 𝑛′ where the pressure equals 𝑝𝑒𝑥𝑡 , but the volume and internal energy differ from the quasistatic case due to irreversibilities. Since the external pressure is constant: 𝑊 = −𝑝𝑒𝑥𝑡 (𝑉𝑛′ − 𝑉1 ) With 𝑄 = 0, the first law becomes: Δ𝑈 = 𝑊 = 𝑈𝑛′ − 𝑈1 = −𝑝𝑒𝑥𝑡 (𝑉𝑛′ − 𝑉1 ) Calculations yield the following values for the process from State 1 to State 𝑛′: Change in internal energy: 𝛥𝑈 = −643 J Work done: 𝑊 = −643 J Entropy change: 𝛥𝑆 = 4.2 J/K For the non-quasistatic process, we can also determine the entropy change by conceptualizing an alternate two-step path, moving first from State 1 to State 𝑥′ along an isobar, then from 𝑥′ to 𝑛′ along 77  Classical thermodynamics, Srini Vanapalli an isochoric path. This approach helps identify the total entropy generated due to irreversibility in the system, which results in a net positive entropy change: 𝛥𝑆 = 4.2 J/K Since 𝑆𝑖𝑟𝑟 > 0, the entropy of the final state 𝑛′ is greater than that of state 𝑛 in the quasistatic process. Additionally, 𝑉𝑛′ > 𝑉𝑛 , illustrating that the final volume in the non-quasistatic process is greater than in the quasistatic case due to irreversible entropy generation. Main Findings for Adiabatic Expansion Comparison Work Comparison: The magnitude of 𝑊 in the non-quasistatic expansion process is lower than in the quasistatic process. Volume Difference: In the non-quasistatic process, the final volume 𝑉𝑛′ is greater than the quasistatic final volume 𝑉𝑛 , as irreversible entropy generation requires a larger volume change. Entropy Generation: The entropy change in the non-quasistatic process is positive due to irreversible entropy generation. In contrast, the entropy change for the quasistatic process is zero, making it reversible. This example clarifies how entropy generation and work differ in quasistatic versus non-quasistatic adiabatic expansions. By folloswing alternative paths (involving states 𝑥 and 𝑥′) for entropy calculations, we can compare these processes with a more systematic approach. To deepen your understanding, you are encouraged to conduct a similar analysis for an adiabatic compression process, considering both quasistatic and non-quasistatic scenarios. Start with the same initial conditions and examine how changes in internal energy, work done, and entropy differ between the two cases. By following a similar methodology—perhaps using an alternate path with intermediate states as we did with states 𝑥 and 𝑥′—you’ll gain insights into the effects of irreversibility on entropy generation and final state differences during compression. This exercise will further illustrate the unique characteristics of non-quasistatic processes. Example 2: Isothermal Compression - Quasistatic vs. Non-Quasistatic Now let’s examine an ideal gas compression process under isothermal conditions, contrasting a quasistatic process with a non-quasistatic one. Quasistatic Isothermal Compression The gas is initially at a pressure of 1 bar, a volume of 1.0 liter, and a temperature of 300 K. It is compressed isothermally to a final volume of 0.1 liter, with the surroundings as a constant- temperature reservoir at 290 K. The work done in the quasistatic isothermal process is: 𝑛 𝑉𝑓 𝑉𝑓 𝑊 = − ∫ 𝑝𝑑𝑉 = −𝑛𝑅𝑇0 ln = −𝑝𝑖 𝑉𝑖 ln = 100 ln 10 J 1 𝑉𝑖 𝑉𝑖 With no change in internal energy (𝛥𝑈 = 0) for an isothermal process, the heat transfer is: 𝑄 = −𝑊 = −100 ln 10 J The entropy change of the system is: 𝑄 −100 ln 10 ln 10 Δ𝑆𝑠𝑦𝑠𝑡𝑒𝑚 = ∫ = =− J/K 𝑇 300 3 78  Classical thermodynamics, Srini Vanapalli Since the surroundings act as a heat reservoir, the entropy change is: −𝑄 100 ln 10 300 ln 10 Δ𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 = = = J/K 𝑇𝑟𝑒𝑠𝑒𝑟𝑣𝑜𝑖𝑟 290 3 × 290 Thus, the total entropy change is: ln 10 300 ΔS𝑡𝑜𝑡𝑎𝑙 = Δ𝑆𝑠𝑦𝑠𝑡𝑒𝑚 + Δ𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 = ( − 1) > 0 3 290 Non-Quasistatic Isothermal Compression In this case, an external force exerts a constant pressure of 10 bar on the gas, causing an instantaneous compression to 0.1 liter. The final temperature is maintained at 300 K. The work done in the non-quasistatic process is: 𝑊 = −𝑝𝑒𝑥𝑡 (𝑉𝑓 − 𝑉𝑖 ) = −10 bar × (0.9 liter) = 900 J With no change in internal energy (ΔU=0\Delta U = 0ΔU=0), the heat transfer is: 𝑄 = −𝑊 = −900 J The entropy change of the system is the same as in the quasistatic process: ln 10 Δ𝑆𝑠𝑦𝑠𝑡𝑒𝑚 = − J⁄K 3 However, the entropy change of the surroundings differs due to the larger heat transfer: −𝑄 900 Δ𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 = = J⁄K 𝑇𝑟𝑒𝑠𝑒𝑟𝑣𝑜𝑖𝑟 290 The total entropy change for the non-quasistatic process is: ln 10 900 ΔS𝑡𝑜𝑡𝑎𝑙 = Δ𝑆𝑠𝑦𝑠𝑡𝑒𝑚 + Δ𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 = − + >0 3 290 This larger total entropy change in the non-quasistatic process indicates higher irreversibility than in the quasistatic case. Main Findings Work Comparison: The magnitude of 𝑊 in non-quasistatic expansion processes is generally lower than in quasistatic processes. Entropy Generation: Entropy change in the non-quasistatic process is positive due to irreversible entropy generation, whereas the total entropy change for the quasistatic process is zero, making it a reversible process. In summary, both examples demonstrate that non-quasistatic processes lead to higher entropy generation, indicating greater irreversibility compared to their quasistatic counterparts. Readers are 79  Classical thermodynamics, Srini Vanapalli encouraged to apply this approach to other scenarios, such as non-quasistatic adiabatic compression, as well as to explore arbitrary variations in external conditions. It is important to note that there are countless ways in which a non-quasistatic process can unfold, as the system is not constrained to maintain equilibrium with its surroundings. Because of this, analyzing a non-quasistatic process requires specific information about how the process occurred or detailed data about the states involved. Only with this information—whether through measurement, experimental data, or clear process definitions—can an accurate analysis of the system’s behavior and energy transformations be performed. Thermodynamic Identity In this final section, we introduce the thermodynamic identity, a fundamental relation that links internal energy, entropy, temperature, pressure, and volume in a concise mathematical form. This identity encapsulates the changes in internal energy in terms of entropy and volume, providing a powerful tool for analyzing thermodynamic processes. Derivation of the Thermodynamic Identity We begin by considering entropy 𝑆 as a function of internal energy 𝑈, volume 𝑉, and particle number 𝑁. For simplicity, we’ll focus on a fixed number of particles, so entropy becomes a function of 𝑈 and 𝑉 alone: 𝑆 = 𝑆(𝑈, 𝑉) Taking the differential of entropy 𝑆, we have: 𝜕𝑆 𝜕𝑆 𝑑𝑆 = ( ) 𝑑𝑈 + ( ) 𝑑𝑉 𝜕𝑈 𝑉 𝜕𝑉 𝑈 From our previous discussions on thermal and mechanical equilibrium, we defined temperature 𝑇 and pressure 𝑝 in terms of these partial derivatives: 1 𝜕𝑆 =( ) 𝑇 𝜕𝑈 𝑉 𝜕𝑆 𝑝 = 𝑇( ) 𝜕𝑉 𝑈 Substituting these definitions into the differential of entropy, we obtain: 1 𝑝 𝑑𝑆 = 𝑑𝑈 + 𝑑𝑉 𝑇 𝑇 Rearranging, we can multiply through by 𝑇 to isolate 𝑑𝑈: 𝑇𝑑𝑆 = 𝑑𝑈 + 𝑝𝑑𝑉 Rearranging further, we arrive at the thermodynamic identity: 𝑑𝑈 = 𝑇𝑑𝑆 − 𝑝𝑑𝑉 80  Classical thermodynamics, Srini Vanapalli This identity provides an essential link between state variables, relating changes in internal energy 𝑈 to changes in entropy 𝑆 and volume 𝑉. It reveals that, for a small change in a system's internal energy, the amount of energy is partitioned between the change in entropy (multiplied by temperature) and the change in volume (multiplied by pressure). Significance of the Thermodynamic Identity The thermodynamic identity captures the fundamental interplay of energy, entropy, and volume in thermodynamic systems. Each term in the identity—𝑇𝑑𝑆 and −𝑝𝑑𝑉—represents a component of the system’s internal energy change: Entropy-Related Energy Change: The term 𝑇𝑑𝑆 represents the portion of internal energy change associated with entropy change. This includes contributions from both heat transfer and irreversible processes, such as frictional effects or dissipative work, which increase entropy within the system. Work: The term −𝑝𝑑𝑉 corresponds to the work associated with changes in the system’s volume. This is often called boundary work, as it reflects the energy required for the system to expand or compress against an external pressure. Together, these terms describe how internal energy changes in response to both entropy and volume variations, making the thermodynamic identity a versatile tool for analyzing any thermodynamic process, whether reversible or irreversible. This identity is foundational because it holds for any infinitesimal process, whether it’s reversible or irreversible, as long as it involves small changes. It allows us to analyze complex processes by breaking them down into contributions from heat and work, thereby deepening our understanding of how energy is conserved and transformed in thermodynamic systems. Thermodynamic Identity Clarification In deriving the thermodynamic identity, many authors take a simplified approach, starting directly from the first law of thermodynamics 𝑑𝑈 = 𝑄 + 𝑊 and substituting 𝑄 = 𝑇𝑑𝑆 and 𝑊 = −𝑝𝑑𝑉. While this approach is often convenient, it overlooks an important aspect of real thermodynamic systems— namely, that other forms of work transfer, such as friction and shaft work, can occur, especially in irreversible processes. Our derivation, however, takes a more comprehensive route by incorporating the idea that entropy change results from both heat transfer and irreversibilities. Specifically, we start with the expression 𝑄 𝑑𝑆 = 𝑇 + 𝑆𝑖𝑟𝑟 , where 𝑆𝑖𝑟𝑟 represents the entropy generated due to irreversible effects within the system. Thus, the internal energy change can be more completely expressed as: 𝑑𝑈 = 𝑄 + 𝑊𝑝𝑉 + 𝑊𝑜𝑡ℎ𝑒𝑟 where: 𝑄 represents heat transfer, leading to a change in system entropy, 𝑊𝑝𝑉 = −𝑝𝑑𝑉 is the work associated with volume change (boundary work), 𝑊𝑜𝑡ℎ𝑒𝑟 represents additional forms of work, such as frictional or shaft work, which can contribute to entropy generation. This perspective respects the contributions of all energy transfers and entropy changes in real-world systems, offering a more complete picture. The entropy change in this approach accounts for both the reversible entropy change from heat transfer and the irreversible contribution from other forms of work. 81  Classical thermodynamics, Srini Vanapalli The Thermodynamic Identity in Any Process Through this derivation, we obtain the thermodynamic identity: 𝑑𝑈 = 𝑇𝑑𝑆 − 𝑝𝑑𝑉 This identity holds universally for any infinitesimal process, regardless of whether it’s reversible or irreversible, because it shows the interplay of state variables—internal energy 𝑈, entropy 𝑆, temperature 𝑇, pressure 𝑝, and volume 𝑉—that define the system. By deriving it through the fundamental principles of entropy and energy transfer, we ensure that it remains applicable even when additional irreversibilities and forms of work transfer are present. This approach not only honors the first law but also provides a fuller understanding of energy and entropy changes, making it a powerful tool for analyzing any thermodynamic process, no matter the complexity. Summary Chapter 5 provides a detailed examination of thermodynamic processes that extend beyond idealized, reversible assumptions. Through an analysis of quasistatic processes with irreversible work transfer, we highlight the essential role of dissipative forces—such as friction and inelastic deformation—in generating entropy and causing energy losses that cannot be fully recovered. By distinguishing between reversible and irreversible work components, we gain a clearer understanding of how energy distribution is affected by irreversibilities in real-world systems. In discussing non-quasistatic processes, the chapter underscores the limitations of equilibrium-based thermodynamic analysis and introduces methodologies for evaluating systems far from equilibrium. Non-quasistatic processes, such as rapid gas expansion, showcase the challenges in defining state variables and calculating work and heat under non-equilibrium conditions. The chapter concludes with the derivation of the thermodynamic identity, incorporating both heat transfer and irreversibilities. This identity provides a unified framework for analyzing energy changes in thermodynamic systems, applicable to any infinitesimal process, whether reversible or irreversible. Through this comprehensive approach, we establish the thermodynamic identity as a powerful tool for understanding the interplay of state variables, reinforcing its foundational role in thermodynamics. 82