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(a) What is special about a reversible process? (b) Suppose a reversible process is reversed, restoring the system to its original state. What can be said about the surroundings after the process is reversed? (c) Under what circumstances will the vaporization of water to steam be a reversible process? (d) Are any of the processes that occur in the world around us reversible in nature? Explain.

Short Answer

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(a) A reversible process is one in which both the system and its surroundings can be returned to their original states, without any net change, by reversing the process. It occurs infinitely slowly, maintaining thermodynamic equilibrium with its surroundings. (b) When a reversible process is reversed, both the system and its surroundings return to their original states, with no net change. (c) Vaporization of water to steam can be a reversible process if it occurs isothermally and isobarically, with the system always in thermodynamic equilibrium with its surroundings, though practically achieving a truly reversible vaporization is impossible. (d) No processes in nature are truly reversible due to the finite rate of real processes and dissipative forces. However, some processes are considered approximately reversible for theoretical analysis in thermodynamics.

Step by step solution

01

Part (a): Definition of a Reversible Process

A reversible process is a thermodynamic process in which both the system and its surroundings can be returned to their original states, without any net change, by reversing the process. In a reversible process, the system is always in thermodynamic equilibrium with its surroundings, so the process occurs infinitely slowly through a series of infinitesimally small changes.
02

Part (b): Surroundings After Reversing a Reversible Process

When a reversible process is reversed, both the system and its surroundings are restored to their original states, with no net change. This means that all energy, entropy, and volume changes in the surroundings have been exactly reversed during the reversed process, returning the surroundings to their initial conditions.
03

Part (c): Conditions for Reversible Vaporization of Water

The vaporization of water to steam can be a reversible process under specific circumstances. For vaporization to be reversible, it must occur isothermally (constant temperature) and isobarically (constant pressure), and the system must always be in thermodynamic equilibrium with its surroundings. This can theoretically happen if the vaporization process occurs infinitely slowly, allowing for heat exchange with the surroundings at a constant temperature and pressure. However, in practice, achieving a truly reversible vaporization is impossible since it would require an infinitely long time.
04

Part (d): Reversible Processes in the World Around Us

In nature, there are no processes that are truly reversible. This is because reversible processes require thermodynamic equilibrium, which implies that the process must occur infinitely slowly through an infinite series of infinitesimal changes. Real processes always occur at a finite rate and are subject to dissipative forces (e.g. friction), making it impossible to completely reverse the process and restore both the system and its surroundings to their original states. However, some processes can be considered approximately reversible under specific conditions and are useful for theoretical analysis in the study of thermodynamics.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermodynamic Equilibrium
Thermodynamic equilibrium refers to a state where a system and its surroundings do not experience any changes over time. In this state, all the properties, such as temperature, pressure, and chemical potential, remain constant. This kind of balance is crucial for understanding how reversible processes work because
  • the system and surroundings need to be in perfect harmony;
  • no net flow of energy or matter across the boundaries;
  • any changes taking place should be infinitesimally slow.

When a system is in thermodynamic equilibrium, no driving force is present to cause any spontaneous change. This means all external influences are balanced, and the system is ideally prepared for running into a reversible path. This is often referred to as "quasi-static process," where changes occur so gradually, the system passes through continuous equilibrium states.
Vaporization
Vaporization is the process where a liquid turns into a gas. For example, water turning to steam. In an ideal world, vaporization can be reversible under certain circumstances:
  • It must happen at a constant temperature (isothermal) and pressure (isobaric).
  • The process must be infinitely slow.

For the process to be reversible, heat exchange with the surroundings should occur without any loss or gain in energy, maintaining equilibrium. In reality, achieving a completely reversible vaporization is virtually impossible because processes cannot be perfectly controlled to be this slow and lossless in real-world conditions.
Understanding vaporization is essential in thermodynamics, as it involves energy changes tied to the breaking of molecular bonds, which must also remain in equilibrium when reversed.
Entropy
Entropy is a measure of disorder or randomness in a system. In thermodynamics, entropy increases as systems tend toward disorder naturally. When a process is reversible, there is no net increase in entropy within the universe because any entropy generated in the process is used up to restore order.
  • A reversible process requires that entropy changes be perfectly compensated.
  • Entropy change is zero if the entire process can be reversed without any net change in energy.

In all real processes, entropy typically increases, meaning the process takes the system and surroundings into more disorder. Reversible processes are theoretical models that help us understand the limitations and possibilities of energy transfer in thermodynamics.
Thermodynamics
Thermodynamics is the study of energy and its transformations. It looks at how energy moves through a system and how it affects matter. One of its core concepts is understanding the differences between reversible and irreversible processes.
  • Reversible processes are idealized because they require perfect balance and equilibrium.
  • Real-world processes are irreversible due to dissipative forces like friction.
  • Through thermodynamics, concepts like energy conservation and efficiency are explored.

Although perfect reversibility is unattainable, understanding these processes allows scientists and engineers to design more efficient energy systems. It teaches us the theoretical limits of energy use and sheds light on practical improvements we can make in various industrial applications. Thermodynamics aids in predicting how systems will behave under different conditions, helping in design optimization and enhancing resource utilization.

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Most popular questions from this chapter

The following data compare the standard enthalpies and free energies of formation of some crystalline ionic substances and aqueous solutions of the substances: \begin{tabular}{lrr} \hline Substance & \(\Delta H_{f}^{\circ}(\mathbf{k J} / \mathrm{mol})\) & \(\Delta G_{f}^{\circ}(\mathbf{k J} /\) moll \\ \hline \(\mathrm{AgNO}_{3}(s)\) & \(-124.4\) & \(-33.4\) \\ \(\mathrm{AgNO}_{3}(a q)\) & \(-101.7\) & \(-34.2\) \\ \(\mathrm{MgSO}_{4}(s)\) & \(-1283.7\) & \(-1169.6\) \\ \(\mathrm{MgSO}_{4}(a q)\) & \(-1374.8\) & \(-1198.4\) \\ \hline \end{tabular} (a) Write the formation reaction for \(\mathrm{AgNO}_{3}(s) .\) Based on this reaction, do you expect the entropy of the system to increase or decrease upon the formation of \(\mathrm{Ag} \mathrm{NO}_{3}(s)\) ? (b) Use \(\Delta H_{f}^{\circ}\) and \(\Delta G_{f}^{\circ}\) of \(\mathrm{AgNO}_{3}(s)\) to determine the entropy change upon formation of the substance. Is your answer consistent with your reasoning in part (a)? (c) Is dissolving \(\mathrm{AgNO}_{3}\) in water an exothermic or endothermic process? What about dissolving \(\mathrm{MgSO}_{4}\) in water? (d) For both \(\mathrm{AgNO}_{3}\) and \(\mathrm{MgSO}_{4}\), use the data to calculate the entropy change when the solid is dissolved in water. (e) Discuss the results from part (d) with reference to material presented in this chapter and in the second "Closer I onk" hox in Section \(13.5\).

(a) What do you expect for the sign of \(\Delta S\) in a chemical reaction in which two moles of gaseous reactants are converted to three moles of gaseous products? (b) For which of the processes in Exercise \(19.9\) does the entropy of the system increase?

Predict the sign of the entropy change of the system for each of the following reactions: (a) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (b) \(\mathrm{Ba}(\mathrm{OH})_{2}(s) \stackrel{\mathrm{L}}{\longrightarrow} \mathrm{BaO}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(l)\) (d) \(\mathrm{FeCl}_{2}(s)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{Fe}(s)+2 \mathrm{HCl}(g)\)

Using \(S^{\circ}\) values from Appendix \(C\), calculate \(\Delta S^{\circ}\) values for the following reactions. In each case account for the sign of \(\Delta S^{n}\). (a) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)\) (b) \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) (c) \(\mathrm{Be}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{BeO}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (d) \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\)

Use data in Appendix \(\mathrm{C}\) to calculate \(\Delta H^{\circ}, \Delta S^{\circ}\), and \(\Delta G^{\circ}\) at \(25^{\circ} \mathrm{C}\) for each of the following reactions. In each case show that \(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}\). (a) \(2 \mathrm{Cr}(s)+3 \mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{CrBr}_{3}(s)\) (b) \(\mathrm{BaCO}_{3}(s) \longrightarrow \mathrm{BaO}(s)+\mathrm{CO}_{2}(g)\) (c) \(2 \mathrm{P}(s)+10 \mathrm{HF}(g) \longrightarrow 2 \mathrm{PF}_{5}(g)+5 \mathrm{H}_{2}(g)\) (d) \(\mathrm{K}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{KO}_{2}(s)\)

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