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(a) Express the second law of thermodynamics as a mathematical equation. (b) In a particular spontaneous process the entropy of the system decreases. What can you conclude about the sign and magnitude of \(\Delta S_{\text {surr }}\) ? (c) During a certain reversible process, the surroundings undergo an entropy change, \(\Delta S_{\text {surr }}=-78 \mathrm{~J} / \mathrm{K} .\) What is the entropy change of the system for this process?

Short Answer

Expert verified
(a) The second law of thermodynamics is expressed mathematically as: \[\Delta S_\text{univ} = \Delta S_\text{sys} + \Delta S_\text{surr} \geq 0\] (b) If the system's entropy decreases in a spontaneous process, the surroundings' entropy change must be positive and have a magnitude greater than the system's entropy decrease. (c) For the reversible process with \(\Delta S_\text{surr} = -78 \mathrm{~J} / \mathrm{K}\), the system's entropy change is \(\Delta S_\text{sys} = 78 \mathrm{~J} / \mathrm{K}\).

Step by step solution

01

(a) Second Law of Thermodynamics

The second law of thermodynamics is expressed mathematically as: \[\Delta S_\text{univ} = \Delta S_\text{sys} + \Delta S_\text{surr} \geq 0\] Where \(\Delta S_\text{univ}\) is the change in entropy of the universe, \(\Delta S_\text{sys}\) is the change in entropy of the system, and \(\Delta S_\text{surr}\) is the change in entropy of the surroundings.
02

(b) Sign and Magnitude of \(\Delta S_{\text{surr}}\)

Given that the entropy of the system decreases in a spontaneous process, we have \[\Delta S_\text{sys} < 0\] According to the second law of thermodynamics, we know that \(\Delta S_\text{univ} \geq 0\Rightarrow \Delta S_\text{sys} + \Delta S_\text{surr} \geq 0\). Since \(\Delta S_\text{sys} < 0\), then to satisfy the second law: \[\Delta S_\text{surr} > - \Delta S_\text{sys}\] Thus, the magnitude of \(\Delta S_\text{surr}\) must be greater than the magnitude of the decrease in entropy of the system, and its sign must be positive.
03

(c) Entropy Change of the System

In a reversible process, the entropy of the universe remains constant, so \(\Delta S_\text{univ} = 0\). Given that the surroundings undergo an entropy change of \(\Delta S_\text{surr} = -78 \mathrm{~J} / \mathrm{K}\), we can find the entropy change of the system using the equation: \[\Delta S_\text{univ} = \Delta S_\text{sys} + \Delta S_\text{surr}\] Since \(\Delta S_\text{univ} = 0\) in a reversible process, \[0 = \Delta S_\text{sys} - 78 \mathrm{~J} / \mathrm{K}\] Solving for \(\Delta S_\text{sys}\), \[\Delta S_\text{sys} = 78 \mathrm{~J} / \mathrm{K}\]

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Key Concepts

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

Entropy Change
Entropy can be thought of as a measure of disorder or randomness in a system. When we discuss entropy change, symbolized as \( \Delta S \), we are referring to how much the disorder of the system has increased or decreased. According to the second law of thermodynamics, the total entropy of an isolated system can never decrease over time; it can remain constant for a reversible process or increase for an irreversible process.

To illustrate, if entropy decreases in the system (\( \Delta S_{\text{sys}} < 0 \) ), which might seem to violate the second law, we must consider the surroundings. For a spontaneous process, the entropy of the surroundings (\( \Delta S_{\text{surr}} \) ) must increase enough to offset the decrease in the system, ensuring that the total entropy of the universe (\( \Delta S_{\text{univ}} \) ) does not decrease.

Understanding this balance is crucial for analyzing any thermodynamic process and knowing how to calculate and interpret entropy changes helps us predict whether a process can occur spontaneously.
Spontaneous Process
A spontaneous process is one that occurs naturally without any external intervention. It can be challenging to predict spontaneity simply by looking at the initial and final states of a system. However, the second law of thermodynamics provides a valuable criterion: if the total entropy change of the universe \( \Delta S_{\text{univ}} \) ) is positive for a given process, that process can occur spontaneously.

For instance, if a system's entropy decreases, such as in our example from the original exercise, this does not automatically rule out spontaneity. The process can still be spontaneous if the surroundings experience an increase in entropy that more than compensates for the system's decrease. This interplay emphasizes that the direction of a spontaneous process is the direction that increases the universe's total entropy.
Reversible Process
A reversible process is a theoretical construct where a system undergoes changes so slowly that it is always in near-equilibrium with its surroundings. These processes are characterized by the fact that the system can be returned to its initial state without leaving any net change in either the system or the surroundings.

For a process to be reversible, the entropy change of the universe must be zero. In practical terms, this means that the entropy change of the system (\( \Delta S_{\text{sys}} \) ) is exactly opposite to the entropy change of the surroundings (\( \Delta S_{\text{surr}} \) ). Using the provided example, if the surroundings lose entropy (\( \Delta S_{\text{surr}} = -78 \text{~J/K} \) ), the system must gain the same amount of entropy to maintain reversibility. This concept challenges our understanding since in reality, all natural processes have some degree of irreversibility, where entropy is generated and the total entropy of the universe increases.
Thermodynamic Equations
Thermodynamic equations are mathematical expressions that describe the principles of thermodynamics in terms of measurable quantities like temperature, energy, volume, pressure, and entropy. The second law of thermodynamics, central to understanding entropy change, is encapsulated by the fundamental equation \( \Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} \) which can be set greater or equal to zero for spontaneous processes and exactly zero for reversible processes.

These equations allow us to quantify the entropy change of the system and surroundings. For example, if a process is spontaneous, the sum of the entropy changes of the system and its surroundings must be positive. Conversely, in a reversible process, the sum must be zero. Mastery of these thermodynamic equations enables students to analyze processes and predict behavior based on fundamental physical laws.

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Most popular questions from this chapter

In each of the following pairs, which compound would you expect to have the higher standard molar entropy: (a) \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\) or \(\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})\); (b) \(\mathrm{CO}_{2}(g)\) or \(\mathrm{CO}(g) ?\) Explain.

About \(86 \%\) of the world's electrical energy is produced by using steam turbines, a form of heat engine. In his analysis of an ideal heat engine, Sadi Carnot concluded that the maximum possible efficiency is defined by the total work that could be done by the engine, divided by the quantity of heat available to do the work (for example from hot steam produced by combustion of a fuel such as coal or methane). This efficiency is given by the ratio \(\left(T_{\text {high }}-T_{\text {low }}\right) / T_{\text {high }}\), where \(T_{\text {high }}\) is the temperature of the heat going into the engine and \(T_{\text {low }}\) is that of the heat leaving the engine. (a) What is the maximum possible efficiency of a heat engine operating between an input temperature of \(700 \mathrm{~K}\) and an exit temperature of \(288 \mathrm{~K} ?(\mathrm{~b})\) Why is it important that electrical power plants be located near bodies of relatively cool water? (c) Under what conditions could a heat engine operate at or near \(100 \%\) efficiency? (d) It is often said that if the energy of combustion of a fuel such as methane were captured in an electrical fuel cell instead of by burning the fuel in a heat engine, a greater fraction of the energy could be put to useful work. Make a qualitative drawing like that in Figure \(5.10\) that illustrates the fact that in principle the fuel cell route will produce more useful work than the heat engine route from combustion of methane.

Consider a system consisting of an ice cube. (a) Under what conditions can the ice cube melt reversibly? (b) If the ice cube melts reversibly, is \(\Delta E\) zero for the process? Explain.

(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?

(a) For each of the following reactions, predict the sign of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) and discuss briefly how these factors determine the magnitude of \(K .\) (b) Based on your general chemical knowledge, predict which of these reactions will have \(K>0 .\) (c) In each case indicate whether \(K\) should increase or decrease with increasing temperature. (i) \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{MgO}(s)\) (ii) \(2 \mathrm{KI}(s) \rightleftharpoons 2 \mathrm{~K}(g)+\mathrm{I}_{2}(g)\) (iii) \(\mathrm{Na}_{2}(g) \rightleftharpoons 2 \mathrm{Na}(g)\) (iv) \(2 \mathrm{~V}_{2} \mathrm{O}_{5}(s) \rightleftharpoons 4 \mathrm{~V}(s)+5 \mathrm{O}_{2}(g)\)

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