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Indicate whether each statement is true or false. (a) The entropy of the universe increases for any spontaneous process. (b) The entropy change of the system is equal and opposite that of the surroundings for any irreversible process. (c) The entropy of the system must increase in any spontaneous process. (d) The entropy change for an isothermal process depends on both the absolute temperature and the amount of heat reversibly transferred.

Short Answer

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a) True. The entropy of the universe increases for any spontaneous process. b) False. The entropy change of the system is not equal and opposite that of the surroundings for any irreversible process. c) False. The entropy of the system doesn't always increase in a spontaneous process. d) True. The entropy change for an isothermal process depends on both the absolute temperature and the amount of heat reversibly transferred.

Step by step solution

01

Statement a) Entropy for spontaneous processes

The second law of thermodynamics states that the total entropy of the universe tends to increase in spontaneous processes. Based on this principle, statement a) is true. The entropy of the universe increases for any spontaneous process.
02

Statement b) Entropy change for irreversible processes

For an irreversible process, the entropy change of the system is not equal and opposite to that of the surroundings. In fact, the total entropy change of the universe (system and surroundings combined) is greater than zero in an irreversible process, which means that the entropy of the system and surroundings both increase. Statement b) is false.
03

Statement c) Entropy increase in spontaneous processes

For a process to be spontaneous, the total entropy change of the universe (system and surroundings combined) must be greater than zero. However, it doesn't mean that the entropy of the system itself must always increase. In some spontaneous processes, the entropy of the system may decrease, but the entropy of the surroundings increases enough such that the total entropy change remains positive. Therefore, statement c) is false.
04

Statement d) Entropy change for isothermal processes

For an isothermal process, the temperature remains constant, and the entropy change is given by \(\Delta S = \int \frac{dq}{T}\), where dq is the infinitesimal heat exchange, and T is the constant absolute temperature. So, the entropy change for an isothermal process indeed depends on both the absolute temperature and the amount of heat reversibly transferred. Statement d) is true.

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Key Concepts

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

Entropy
Entropy is a fascinating concept in thermodynamics. It measures the amount of disorder or randomness in a system. In simple terms, it tells us how spread out energy is within a system. Generally, higher entropy means more disorder and lower energy concentration.

Several key points help to clarify the concept of entropy:
  • Entropy tends to increase in isolated systems, helping us understand the natural tendency for systems to move towards disorder.
  • Entropy is central to understanding why certain processes are spontaneous. This is tied to the second law of thermodynamics, which states that the total entropy of the universe increases in spontaneous processes.
  • When dealing with entropy changes, always consider both the system and its surroundings. Even if a system's entropy decreases, the increase in its surroundings' entropy can result in an overall increase in the universe's total entropy.
Overall, entropy is a pivotal concept in grasping how thermodynamic systems behave and evolve.
Isothermal Process
An isothermal process is a type of thermodynamic transformation in which the temperature of the system remains constant throughout the process. This constancy makes isothermal processes distinct and somewhat easier to analyze.

What makes an isothermal process unique?
  • In an isothermal process, any heat added to the system is used to do work, keeping the temperature constant by balancing energy input and output.
  • The entropy change during an isothermal process can be calculated using the formula: \[\Delta S = \int \frac{dq}{T}\]This equation implies that the entropy change depends on the heat transfer \(dq\) and the constant temperature \(T\).
  • These processes frequently occur in systems in thermal contact with a large reservoir ensuring the temperature remains unchanged even with energy transfers.
Isothermal processes are often seen in idealized scenarios such as the slow compression or expansion of gases where the temperature is kept consistent by implementing energy balance.
Second Law of Thermodynamics
The second law of thermodynamics is a fundamental principle governing energy transformations and directionality of processes. It sets a definitive rule about the direction of spontaneous processes, stating that the total entropy of an isolated system can only increase over time. Here's how it unfolds:
  • Spontaneous processes are those that occur naturally, without any external influence.
  • According to the second law, these processes often lead to an increase in the universe's total entropy, driving systems toward states of higher disorder.
  • This law helps us understand why certain processes are irreversible: once nature finds a path leading to higher entropy, 'rolling back the clock' to a lower entropy state requires added energy or work, often impossible or highly inefficient.
This law plays a critical role in several real-world applications, such as understanding the efficiency limits of engines and predicting the spontaneous nature of chemical reactions.

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Most popular questions from this chapter

Predict which member of each of the following pairs has the greater standard entropy at \(25^{\circ} \mathrm{C}:\) (a) \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\) or \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\), (b) \(\mathrm{CO}(g)\) or \(\mathrm{CO}_{2}(g)\) (c) \(1 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{4}(g)\) or \(2 \mathrm{~mol} \mathrm{NO}_{2}(g)\) (d) \(\mathrm{HCl}(g)\) or \(\mathrm{HCl}(a q) .\) Use Appendix \(\mathrm{C}\) to find the standard entropy of each substance.

Isomersare moleculesthat havethesamechemical formula but different arrangements of atoms, as shown here for two isomers of pentane, \(\mathrm{C}_{5} \mathrm{H}_{12} .\) (a) Do you expect a significant difference in the enthalpy of combustion of the two isomers? Explain. (b) Which isomer do you expect to have the higher standard molar entropy? Explain. \([\) Section 19.4\(]\)

Predict the sign of \(\Delta S_{s y s}\) for each of the following processes: (a) Gaseous \(\mathrm{H}_{2}\) reacts with liquid palmitoleic acid \(\left(\mathrm{C}_{16} \mathrm{H}_{30} \mathrm{O}_{2},\right.\) unsaturated fatty acid) to form liquid palmitic acid \(\left(\mathrm{C}_{16} \mathrm{H}_{32} \mathrm{O}_{2}\right.\) saturated fatty acid). (b) Liquid palmitic acid solidifies at \(1^{\circ} \mathrm{C}\) to solid palmitic acid. (c) Silver chloride precipitates upon mixing \(\mathrm{AgNO}_{3}(a q)\) and \(\mathrm{NaCl}(a q) .\) (d) Gaseous \(\mathrm{H}_{2}\) dissociates in an electric arc to form gaseous H atoms (used in atomic hydrogen welding).

From the values given for \(\Delta H^{\circ}\) and \(\Delta S^{\circ},\) calculate \(\Delta G^{\circ}\) for each of the following reactions at \(298 \mathrm{~K}\). If the reaction is not spontaneous under standard conditions at \(298 \mathrm{~K}\), at what temperature (if any) would the reaction become spontaneous? $$ \begin{array}{l} \text { (a) } 2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g) \\ \Delta H^{\circ}=-844 \mathrm{~kJ} ; \Delta S^{\circ}=-165 \mathrm{~J} / \mathrm{K} \\ \text { (b) } 2 \mathrm{POCl}_{3}(g) \longrightarrow 2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g) \\ \Delta H^{\circ}=572 \mathrm{~kJ} ; \Delta S^{\circ}=179 \mathrm{~J} / \mathrm{K} \end{array} $$

The standard entropies at \(298 \mathrm{~K}\) for certain group 14 elements are: \(\mathrm{C}(s,\) diamond \()=2.43 \mathrm{~J} / \mathrm{mol}-\mathrm{K}, \mathrm{Si}(s)=18.81 \mathrm{~J} /\) \(\mathrm{mol}-\mathrm{K}, \mathrm{Ge}(s)=31.09 \mathrm{~J} / \mathrm{mol}-\mathrm{K}, \quad\) a n d \(\quad \mathrm{Sn}(s)=51.818 \mathrm{~J} /\) mol-K. All but \(S\) n have the same (diamond) structure. How do you account for the trend in the \(S^{\circ}\) values?

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