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(a) Express the second law of thermodynamics in words. (b) If the entropy of the system increases during a reversible process, what can you say about the entropy change of the surroundings? (c) In a certain spontaneous process the system undergoes an entropy change, \(\Delta S=42 \mathrm{~J} / \mathrm{K} .\) What can you conclude about \(\Delta S_{\text {surr }} ?\)

Short Answer

Expert verified
(a) The second law of thermodynamics states that the total entropy of a system and its surroundings never decreases; it either stays constant (in reversible processes) or increases (in irreversible processes). (b) In a reversible process, the entropy change of the surroundings is equal and opposite to the entropy change of the system. (c) In a spontaneous process with \(\Delta S_{system} = 42 \, \mathrm{J/K}\), the entropy change of the surroundings must be greater than \(-42 \, \mathrm{J/K}\).

Step by step solution

01

(a) The Second Law of Thermodynamics

The second law of thermodynamics states that in any natural process, the total entropy of a system and its surroundings never decreases; it either stays constant (in reversible processes) or increases (in irreversible processes). In other words, natural processes tend to increase the overall disorder (entropy) of the universe.
02

(b) Entropy Change in Surroundings during a Reversible Process

When the entropy of the system increases during a reversible process, we can say that the entropy change of the surroundings is equal and opposite to the entropy change of the system. In a reversible process, the total entropy change of the system and the surroundings is zero: \[ \Delta S_{total} = \Delta S_{system} + \Delta S_{surr} = 0\]
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(c) Analyzing Entropy Change in a Spontaneous Process

For a spontaneous process, we are given that the entropy change of the system is \(\Delta S = 42 \, \mathrm{J/K}\). According to the second law of thermodynamics, the entropy of the universe must increase for a spontaneous process, which means that the total entropy change must be positive: \[ \Delta S_{total} = \Delta S_{system} + \Delta S_{surr} > 0\] With the given \(\Delta S_{system}\), we can conclude for this spontaneous process that the entropy change of the surroundings must be: \[ \Delta S_{surr} > - \Delta S_{system}\] Hence, the entropy change of the surroundings must be greater than \(-42 \, \mathrm{J/K}\).

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Key Concepts

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

Entropy Change
Entropy is a measure of the disorder or randomness of a system, and it plays a crucial role in thermodynamics. When we talk about entropy change, \(\Delta S\), we are referring to how much the disorder of a system has changed. This can be caused by processes like heat transfer or matter exchange.
Often, in thermodynamic systems, the change in entropy can give us insight into whether a process can occur naturally.
  • For any process, the change in entropy of a system, \( \Delta S_{system} \), and its surroundings, \( \Delta S_{surr} \), is crucial for understanding the process's direction.
  • In reversible processes, there is a precise and balanced entropy change. In contrast, in irreversible processes, the entropy change tends to increase.
By calculating \( \Delta S \), we determine how energy is distributed within a system and to its environment, helping us predict natural occurrences.
Reversible Process
A reversible process is an idealized or theoretical concept in thermodynamics. It represents a process that can be reversed by an infinitely small change, leaving no net change in the system and surrounding.
In real-world applications, reversible processes are usually impossible due to practical limitations like friction and energy losses. However, they are essential for understanding thermodynamic efficiencies and serve as a benchmark.
  • In a reversible process, the total entropy change, \( \Delta S_{total} \), is zero.
  • This means \( \Delta S_{system} + \Delta S_{surr} = 0 \), indicating a perfect balance.
This balance illustrates that energy can be fully transferred without losses to randomness or disorder, making reversible processes very efficient hypothetically.
Spontaneous Process
Spontaneous processes occur naturally without the need for external energy input. They tend to increase the total entropy of the universe due to their irreversible nature.
For spontaneous processes to occur, the system must experience entropy changes that result in an overall entropy increase of the universe.
  • If \( \Delta S_{system} \) for a process is known, it can provide insights into the minimum entropy required for the surroundings.
  • For instance, with \( \Delta S_{system} = 42 \, \mathrm{J/K} \), the surroundings' entropy \( \Delta S_{surr} \) must be > \(-42 \, \mathrm{J/K}\).
Thus, these processes help illustrate how energy disperses in the universe, following the path of least resistance and maximizing entropy.
Entropy of the Universe
The entropy of the universe is a central concept in understanding the second law of thermodynamics. It refers to the sum total of entropy in all systems and their surroundings and acts as a gauge for the direction of natural processes.
In all natural processes, the second law states that the total entropy, \( \Delta S_{universe} \), will not decrease. It either increases, as seen in spontaneous processes, or remains constant, as seen in ideal reversible processes.
  • Thus, \( \Delta S_{universe} = \Delta S_{system} + \Delta S_{surr} \)
  • Understanding that this will either be zero or positive helps predict process feasibility.
This ensures a unidirectional flow of time, helping us interpret natural entropy as an arrow that points from order towards disorder.

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Most popular questions from this chapter

One way to derive Equation \(19.3\) depends on the observation that at constant \(T\) the number of ways, \(W\), of arranging \(m\) ideal-gas particles in a volume \(V\) is proportional to the volume raised to the \(m\) power: $$ W \propto V^{m} $$ Use this relationship and Boltzmann's relationship between entropy and number of arrangements (Equation 19.5) to derive the equation for the entropy change for the isothermal expansion or compression of \(n\) moles of an ideal gas.

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 \(13.5\).

Foreach of the following pairs, indicate which substance possesses the larger standard entropy: (a) 1 mol of \(\mathrm{P}_{4}(g)\) at \(300{ }^{\circ} \mathrm{C}, 0.01 \mathrm{~atm}\), or \(1 \mathrm{~mol}\) of \(\mathrm{As}_{4}(g)\) at \(300^{\circ} \mathrm{C}, 0.01 \mathrm{~atm} ;\) (b) \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(100^{\circ} \mathrm{C}, 1 \mathrm{~atm}\), or \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}(l)\) at \(100^{\circ} \mathrm{C}, 1 \mathrm{~atm} ;\) (c) \(0.5 \mathrm{~mol}\) of \(\mathrm{N}_{2}(\mathrm{~g})\) at \(298 \mathrm{~K}, 20\) - \(\mathrm{L}\) volume, or \(0.5 \mathrm{~mol} \mathrm{CH}_{4}(g)\) at \(298 \mathrm{~K}, 20\) -L volume; (d) \(100 \mathrm{~g}\), \(\mathrm{Na}_{2} \mathrm{SO}_{4}(s)\) at \(30^{\circ} \mathrm{C}\) or \(100 \mathrm{~g} \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) at \(30^{\circ} \mathrm{C}\).

(a) What is meant by calling a process irreversible? (b) After an irreversible process the system is restored to its original state. What can be said about the condition of the surroundings after the system is restored to its original state? (c) Under what conditions will the condensation of a liquid be an irreversible process?

Consider what happens when a sample of the explosive TNT (Section 8.8: "Chemistry Put to Work: Explosives and Alfred Nobel") is detonated. (a) Is the detonation a spontaneous process? (b) What is the sign of \(q\) for this process? (c) Can you determine whether \(w\) is positive, negative, or zero for the process? Explain. (d) Can you determine the sign of \(\Delta E\) for the process? Explain.

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