Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 65

Comment on the statement: "Just talking about entropy increases its value in the universe."

Short Answer

Expert verified
No, just talking about entropy does not increase its value in the universe. The statement is a metaphor indicating that discussing such complex topics could lead to confusion or misunderstanding, metaphorically 'increasing entropy' in the sense of information disorder.

Step by step solution

01

Understand the Concept of Entropy

The entropy in thermodynamics is a physical quantity that represents the degree of randomness or disorder of a system. It's often denoted by the letter 'S'. According to the second law of thermodynamics, in any energy exchange, if no energy enters or leaves the system, the potential energy of the state will always be less than that of the initial state, which implies that the entropy of an isolated system always increases or remains constant.
02

Interpret the Phrase

The statement 'Just talking about entropy increases its value in the universe' is more of a metaphoric representation rather than a scientific fact. The act of talking about entropy or any other topic doesn't physically contribute to the total entropy of the universe. This statement may refer metaphorically to the idea that discussing complex or abstract topics, such as entropy, can lead to a state of 'disorder' or 'confusion', especially for those not familiar with such subjects. The 'increased entropy' is, therefore, on the level of information understood or misunderstood by the listener.
03

Conclude the Discussion

Based on the analysis and interpretation, we conclude that speaking about entropy, or anything else, doesn't impact the total entropy of the universe in the context of the second law of thermodynamics. The statement should be understood as a metaphor about communication and comprehension rather than a factual scientific assertion.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Predict whether the entropy change is positive or negative for each of these reactions: (a) \(\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)\) (b) \(\mathrm{O}(g)+\mathrm{O}(g) \longrightarrow \mathrm{O}_{2}(g)\) (c) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(2 \mathrm{H}_{2} \mathrm{O}_{2}(l) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\)

The internal combustion engine of a \(1200-\mathrm{kg}\) car is designed to run on octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right),\) whose enthalpy of combustion is \(5510 \mathrm{~kJ} / \mathrm{mol}\). If the car is moving up a slope, calculate the maximum height (in meters) to which the car can be driven on 1.0 gallon of the fuel. Assume that the engine cylinder temperature is \(2200^{\circ} \mathrm{C}\) and the exit temperature is \(760^{\circ} \mathrm{C},\) and neglect all forms of friction. The mass of 1 gallon of fuel is \(3.1 \mathrm{~kg} .\) [Hint: The efficiency of the internal combustion engine, defined as work performed by the engine divided by the energy input, is given by \(\left(T_{2}-T_{1}\right) / T_{2},\) where \(T_{2}\) and \(T_{1}\) are the engine's operating temperature and exit temperature (in kelvins). The work done in moving the car over a vertical distance is \(m g h,\) where \(m\) is the mass of the car in \(\mathrm{kg}, g\) the acceleration due to gravity \(\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right),\) and \(h\) the height in meters. \(]\)

What is a coupled reaction? What is its importance in biological reactions?

Use the following data to determine the normal boiling point, in kelvins, of mercury. What assumptions must you make in order to do the calculation? $$ \begin{aligned} \mathrm{Hg}(l): & \Delta H_{\mathrm{f}}^{\circ} &=0 \text { (by definition) } \\\ & S^{\circ} &=77.4 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol} \\ \mathrm{Hg}(g): & \Delta H_{\mathrm{f}}^{\circ} &=60.78 \mathrm{~kJ} / \mathrm{mol} \\ & S^{\circ} &=174.7 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol} \end{aligned} $$

State whether the sign of the entropy change expected for each of the following processes will be positive or negative, and explain your predictions. (a) \(\mathrm{PCl}_{3}(l)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(s)\) (b) \(2 \mathrm{HgO}(s) \longrightarrow 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}(g)\) (d) \(\mathrm{U}(s)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{UF}_{6}(s)\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Kinetics

Read Explanation

Making Measurements

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free