/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 Comment on the difficulties of s... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Comment on the difficulties of solving environmental pollution problems from the standpoint of entropy changes associated with the formation of pollutants and with their removal from the environment.

Short Answer

Expert verified
Entropy increase in both the formation and removal of pollutants makes resolution of environmental pollution issues difficult. Sustainable solutions would have to concentrate on minimizing this entropy increase, such as by reducing high entropy–generating activities (like burning fossil fuels) and increasing the use of low entropy–producing energy sources (like solar or wind power).

Step by step solution

01

Understanding entropy

Entropy is a concept in thermodynamics that measures the level of disorder or randomness in a system. Technically, it is defined as the measure of energy dispersal. In most natural processes, entropy increases, making the system more disordered.
02

Reflecting on entropy and pollution creation

The formation of pollutants is a process that tends to increase entropy. This is because pollutants are normally byproducts of energy generation and industrial processes. Both involve the transformation and dispersion of energy, contributing to increased entropy. For example, burning fossil fuels releases carbon dioxide and other pollutants. This process increases total entropy because it involves the conversion of a concentrated fuel source into dispersed heat and gaseous waste
03

Discussing entropy and pollution control

Removing pollutants from the environment often requires energy input. This is due to the second law of thermodynamics, which states that in any energetic exchange, some energy will be lost as heat, which increases entropy. As removing pollution from the environment often means bringing a disordered system to a more ordered state, it tends to be energy intensive and increases the total entropy in the universe. As energy input usually comes from a source that generates some form of pollution, the overall process can still lead to an increase in entropy.
04

Conclusively understand the process

Therefore, both formation and removal of pollutants can contribute to entropy increase. This makes environmental pollution a difficult problem to solve. Any measures taken to control pollution could be counterproductive in terms of overall entropy change. Thus, it is necessary for sustainable solutions to pollution control to focus on minimizing the entropy increase as much as possible. This could mean reducing the use of high entropy processes (like burning fossil fuels), and increasing the use of lower entropy processes (like solar or wind power).

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Thermodynamics and Pollution
When it comes to understanding environmental pollution, thermodynamics plays a crucial role. The laws of thermodynamics provide insights into how energy is transformed and dispersed in various processes. Industrial activities and energy production often lead to pollution, as they increase the entropy of the system involved.
In simple terms, entropy measures the level of disorder within a system. Any process that transforms a concentrated form of energy, like fossil fuels, into dispersed energy forms like heat and gas usually leads to higher entropy. This transformation results in the generation of pollutants, which can significantly harm the environment.

When considering thermodynamics, we can emphasize:

  • Energy transformations often lead to pollution.
  • Higher entropy states are more disordered.
  • Pollutants are byproducts of these energy transformations.
Understanding these concepts can help us comprehend why reducing pollution is complex, as it often conflicts with nature's tendency to increase entropy.
Energy Dispersal
Energy dispersal is a fundamental concept linked to entropy. As energy is used in industrial and natural processes, it becomes less ordered and more spread out, thus increasing entropy. This dispersal is a significant contributor to the creation of pollutants.
For instance, burning fossil fuels is a primary method through which energy dispersal occurs. This process involves converting a stored energy form into a more chaotic form - i.e., the release of heat and gas. This transformation not only increases the overall entropy but also introduces pollutants like carbon dioxide into the atmosphere.
Important aspects of energy dispersal include:
  • It involves spreading out energy in processes.
  • It results in increased environmental disorder.
  • It can lead to the emergence of pollution.
By understanding energy dispersal, we gain insight into the challenges of managing pollution. Knowing how energy moves and transforms can inform strategies to reduce the entropy generated.
Entropy Increase
The concept of entropy increase is central to both the creation and mitigation of environmental pollution. At its core, entropy is about disorder. With almost all natural and artificial processes tending toward increased entropy, tackling pollution becomes inherently challenging.
Generating energy from traditional sources often results in entropy increase as concentrated resources turn into less ordered forms like heat and pollutants. Conversely, attempting to reorder or remove pollutants from the environment requires energy, which usually raises entropy even further.
Key points about entropy increase include:
  • Entropy naturally tends to increase as systems become more disordered.
  • Creating pollutants generally results in higher entropy.
  • Even reducing pollution often involves processes that increase entropy.
Understanding entropy increases helps us comprehend why both creating and removing pollutants can contribute to environmental challenges.
Sustainable Pollution Control
In tackling pollution, sustainable pollution control aims to minimize the entropy increase wherever possible. This is essential for creating long-lasting solutions that do not further exacerbate environmental problems.
Sustainable approaches focus on reducing reliance on high-entropy processes like fossil fuel combustion and increasing the adoption of low-entropy processes such as renewable energy sources - solar and wind.
Major elements of sustainable pollution control include:
  • Using sustainable, low-entropy energy sources.
  • Developing techniques that minimize energy dispersal.
  • Focusing on long-term environmental preservation.
By supporting sustainable methods, we work toward solutions that limit entropy increases, reducing the overall environmental impact and helping maintain ecological balance.

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

Two equations can be written for the dissolution of \(\mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{s})\) in acidic solution. $$\begin{aligned} \mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{s})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightleftharpoons & \mathrm{Mg}^{2+}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(1) \\ & \Delta G^{\circ}=-95.5 \mathrm{kJ} \mathrm{mol}^{-1} \\ (c) Will the solubilities of \(\mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{s})\) in a buffer solution at \(\mathrm{pH}=8.5\) depend on which of the two equations is used as the basis of the calculation? Explain. \frac{1}{2} \mathrm{Mg}(\mathrm{OH})_{2}(\mathrm{s})+\mathrm{H}^{+}(\mathrm{aq}) \rightleftharpoons & \frac{1}{2} \mathrm{Mg}^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1) \\ & \Delta G^{\circ}=-47.8 \mathrm{kJ} \mathrm{mol}^{-1} \end{aligned}$$ (a) Explain why these two equations have different \(\Delta G^{\circ}\) values. (b) Will \(K\) for these two equations be the same or different? Explain.

At \(298 \mathrm{K}, \Delta G_{\mathrm{f}}^{\mathrm{p}}[\mathrm{CO}(\mathrm{g})]=-137.2 \mathrm{kJ} / \mathrm{mol}\) and \(K_{\mathrm{p}}=\) \(6.5 \times 10^{11}\) for the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons\) \(\mathrm{COCl}_{2}(\mathrm{g}) . \quad\) Use these data to determine \(\Delta G_{f}\left[\mathrm{COCl}_{2}(\mathrm{g})\right],\) and compare your result with the value in Appendix D.

A tabulation of more precise thermodynamic data than are presented in Appendix D lists the following values for \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) at \(298.15 \mathrm{K},\) at a standard state pressure of 1 bar. $$\begin{array}{llll} \hline & \Delta H_{f}^{\circ}, & \Delta G_{f,}^{\circ} & S_{\prime}^{\circ} \\\ & \text { kJ mol }^{-1} & \text {kJ mol }^{-1} & \text {J mol }^{-1} \text {K }^{-1} \\ \hline \mathrm{H}_{2} \mathrm{O}(1) & -285.830 & -237.129 & 69.91 \\ \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) & -241.818 & -228.572 & 188.825 \\ \hline \end{array}$$ (a) Use these data to determine, in two different ways, \(\Delta G^{\circ}\) at \(298.15 \mathrm{K}\) for the vaporization: \(\mathrm{H}_{2} \mathrm{O}(1,1 \mathrm{bar}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g}, 1 \mathrm{bar}) .\) The value you obtain will differ slightly from that on page 838 because here, the standard state pressure is 1 bar, and there, it is 1 atm. (b) Use the result of part (a) to obtain the value of \(K\) for this vaporization and, hence, the vapor pressure of water at \(298.15 \mathrm{K}\) (c) The vapor pressure in part (b) is in the unit bar. Convert the pressure to millimeters of mercury. (d) Start with the value \(\Delta G^{\circ}=8.590 \mathrm{kJ}\), given on page 838 and calculate the vapor pressure of water at 298.15 K in a fashion similar to that in parts (b) and (c). In this way, demonstrate that the results obtained in a thermodynamic calculation do not depend on the convention we choose for the standard state pressure, as long as we use standard state thermodynamic data consistent with that choice.

Use thermodynamic data at \(298 \mathrm{K}\) to decide in which direction the reaction $$2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})$$ is spontaneous when the partial pressures of \(\mathrm{SO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{SO}_{3}\) are \(1.0 \times 10^{-4}, 0.20,\) and \(0.10 \mathrm{atm}\) respectively.

To establish the law of conservation of mass, Lavoisier carefully studied the decomposition of mercury(II) oxide: $$\mathrm{HgO}(\mathrm{s}) \longrightarrow \mathrm{Hg}(1)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ At \(25^{\circ} \mathrm{C}, \Delta H^{\circ}=+90.83 \mathrm{kJ}\) and \(\Delta G^{\circ}=+58.54 \mathrm{kJ}\) (a) Show that the partial pressure of \(\mathrm{O}_{2}(\mathrm{g})\) in equilibrium with \(\mathrm{HgO}(\mathrm{s})\) and \(\mathrm{Hg}(\mathrm{l})\) at \(25^{\circ} \mathrm{C}\) is extremely low. (b) What conditions do you suppose Lavoisier used to obtain significant quantities of oxygen?

See all solutions

Recommended explanations on Chemistry Textbooks

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.