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The change in entropy of a system can be calculated because a) it depends only on the c) entropy always increases. initial and final states. d) none of the above. b) any process is reversible.

Short Answer

Expert verified
Answer: The change in entropy of a system can be calculated because it depends only on the initial and final states, making it a state function.

Step by step solution

01

Understanding Entropy

Entropy is a thermodynamic concept that measures the level of disorder in a system. In any process, the total entropy of the system and its surroundings can only increase or stay the same, as stated by the second law of thermodynamics.
02

Analyzing Option A

Option A states that the change in entropy of a system can be calculated because it depends only on the initial and final states. This statement is correct; the change in entropy (∆S) is a state function, meaning it depends only on the initial and final states of a system. It does not matter which path the system takes between its initial and final states.
03

Analyzing Option B

Option B states that the change in entropy of a system can be calculated because any process is reversible. This statement is incorrect. Although reversible processes have a maximum entropy change, not all processes are reversible. Some, like most real-world processes, are irreversible and have a greater increase in entropy than reversible processes.
04

Analyzing Option C

Option C states that the change in entropy of a system can be calculated because entropy always increases. While it is true that entropy tends to increase (or stay the same) for any process, this does not explain why we can calculate the change in entropy.
05

Analyzing Option D

Option D states that the change in entropy of a system can be calculated because of none of the above. Since we've already determined that option A is correct, option D is incorrect.
06

Choosing the Correct Answer

Based on the analysis of each option, the correct answer to the question is option A, stating that the change in entropy of a system can be calculated because it depends only on the initial and final states.

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Key Concepts

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

Second Law of Thermodynamics
Understanding the second law of thermodynamics is crucial for grasping the behavior of entropy in any system. It postulates that the total entropy, which is a measure of disorder or randomness in a system, can never decrease over time in an isolated system. It's this law that implies the natural tendency for systems to progress from order to disordered states.
Implications in Real-World Systems: The second law has profound consequences in our everyday life, from engines running to ice melting. It ensures that energy efficiency, such as in a heat engine, is always limited, and it also dictates the direction of heat flow, naturally from a hotter to a cooler body.
Cosmic Perspective: On a larger scale, it plays a pivotal role in the thermal death of the universe concept, where eventually the universe would reach a state of maximum entropy, with no available energy to do work.
State Function
The concept of a state function is vital in thermodynamics because it relates to the state of a system, independent of the path taken to reach that state. Entropy is an example of a state function. Specifically, this means that the difference in entropy between two states of a system only depends on those states and not on the process used to go from one to the other.
Importance of State Functions:
  • They simplify the analysis of thermodynamic systems.
  • Calculations are easier, as they do not require knowledge of the system's history.
  • Many thermodynamic properties, like temperature, pressure, and volume, are state functions.
When considering the exercise, this is why the correct answer emphasizes the irrelevance of the path taken between initial and final states when calculating entropy changes.
Irreversible Process
In contrast to idealistic reversible processes, an irreversible process is one that cannot be reversed without leaving a change in the universe. These processes are common in the real world, often accompanied by friction, turbulence, and inelastic deformation which contribute to the overall increase in entropy.
Features of Irreversible Processes:
  • They are non-equilibrium and cannot be reversed exactly.
  • They produce entropy, contributing to the irreversibility.
  • Examples include natural processes like mixing of substances and spontaneous chemical reactions.
In practical scenarios, understanding that most processes are irreversible helps anticipate the inevitable increase in entropy, which is notably absent in the idealized reversible processes.
Reversible Process
A reversible process in thermodynamics is a theoretical concept where a system changes in such a manner that the system and environment can be returned to their original states without leaving any trace on the universe. While no process is truly reversible, this concept is useful for calculating the maximum possible efficiency of thermodynamic systems, often serving as a benchmark.
Characteristics of Reversible Processes:
  • They maintain thermodynamic equilibrium at all stages.
  • Changes occur infinitesimally slowly to ensure reversibility.
  • Reversible processes like isothermal and adiabatic expansions are often used in thermodynamic cycle analyses.
This idealized concept provides a way to understand the upper limit of engine efficiency, but it's also a reminder of the difference between theoretical and practical scenarios in thermodynamics.

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Most popular questions from this chapter

A Carnot refrigerator is operating between thermal reservoirs with temperatures of \(27.0^{\circ} \mathrm{C}\) and \(0.00^{\circ} \mathrm{C}\) a) How much work will need to be input to extract \(10.0 \mathrm{~J}\) of heat from the colder reservoir? b) How much work will be needed if the colder reservoir is at \(-20.0^{\circ} \mathrm{C}^{2}\)

A coal-burning power plant produces \(3000 .\) MW of thermal energy, which is used to boil water and produce supersaturated steam at \(300 .{ }^{\circ} \mathrm{C}\). This high-pressure steam turns a turbine producing \(1000 .\) MW of electrical power. At the end of the process, the steam is cooled to \(30.0^{\circ} \mathrm{C}\) and recycled. a) What is the maximum possible efficiency of the plant? b) What is the actual efficiency of the plant? c) To cool the steam, river water runs through a condenser at a rate of \(4.00 \cdot 10^{7} \mathrm{gal} / \mathrm{h}\). If the water enters the condenser at \(20.0^{\circ} \mathrm{C}\), what is its exit temperature?

What is the minimum amount of work that must be done to extract \(500.0 \mathrm{~J}\) of heat from a massive object at a temperature of \(27.0^{\circ} \mathrm{C}\) while releasing heat to a high temperature reservoir with a temperature of \(100.0^{\circ} \mathrm{C} ?\)

A volume of \(6.00 \mathrm{~L}\) of a monatomic ideal gas, originally at \(400 . \mathrm{K}\) and a pressure of \(3.00 \mathrm{~atm}\) (called state 1 ), undergoes the following processes, all done reversibly: \(1 \rightarrow 2\) isothermal expansion to \(V_{2}=4 V_{1}\) \(2 \rightarrow 3\) isobaric compression \(3 \rightarrow 1\) adiabatic compression to its original state Find the entropy change for each process.

Suppose a person metabolizes \(2000 .\) kcal/day. a) With a core body temperature of \(37.0^{\circ} \mathrm{C}\) and an ambient temperature of \(20.0^{\circ} \mathrm{C}\), what is the maximum (Carnot) efficiency with which the person can perform work? b) If the person could work with that efficiency, at what rate, in watts, would they have to shed waste heat to the surroundings? c) With a skin area of \(1.50 \mathrm{~m}^{2}\), a skin temperature of \(27.0^{\circ} \mathrm{C}\) and an effective emissivity of \(e=0.600,\) at what net rate does this person radiate heat to the \(20.0^{\circ} \mathrm{C}\) surroundings? d) The rest of the waste heat must be removed by evaporating water, either as perspiration or from the lungs At body temperature, the latent heat of vaporization of water is \(575 \mathrm{cal} / \mathrm{g}\). At what rate, in grams per hour, does this person lose water? e) Estimate the rate at which the person gains entropy. Assume that all the required evaporation of water takes place in the lungs, at the core body temperature of \(37.0^{\circ} \mathrm{C}\).

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