/*! 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 40 (a) If you are told that the ent... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 19: Problem 40

(a) If you are told that the entropy of a certain system is zero, what do you know about the system and the temperature? (b) The energy of a gas is increased by heating it. Using \(\mathrm{CO}_{2}\) as an example, illustrate the different ways in which additional energy can be distributed among the molecules of the gas. (c) \(\mathrm{CO}_{2}(g)\) and \(\mathrm{Ar}(g)\) have nearly the same molar mass. At a given temperature, will they have the same number of microstates? Explain.

Short Answer

Expert verified
(a) A system with zero entropy is a perfect crystal at absolute zero (0 K) temperature, indicating perfect order and no uncertainty in energy distribution. (b) In a \(\mathrm{CO}_2\) gas, additional energy can be distributed as translational (linear motion), rotational (rotation around axes), and vibrational (molecular vibrations) energy. (c) \(\mathrm{CO}_2(g)\) and \(\mathrm{Ar}(g)\) won't have the same number of microstates at a given temperature as \(\mathrm{CO}_2(g)\) has additional rotational and vibrational energy levels, providing more ways to distribute energy among the molecules.

Step by step solution

01

(a) Interpreting the Entropy of a System

In this case, the entropy of the system is zero. Entropy is a measure of the disorder or randomness of a system. When a system has zero entropy, it means that the system is perfectly ordered, and there is no uncertainty about its internal energy distribution. If we consider the Third Law of Thermodynamics, it states that the entropy of a perfect crystal at absolute zero (0 kelvin) is zero. Therefore, we can conclude that the system is a perfect crystal and the temperature of the system is absolute zero (0 K).
02

(b) Energy Distribution in CO2 Molecules

When the energy of a \(\mathrm{CO}_2\) gas is increased by heating it, the additional energy can be distributed among the molecules of the gas in several ways: 1. Translational energy - As the gas is heated, the molecules will move faster in their linear paths. This increase in translational energy will result from an increase in their kinetic energy. 2. Rotational energy - The molecules of \(\mathrm{CO}_2\) can also rotate around their axes. By heating the gas, some of the energy will be used to increase the rate of rotation of the molecules. 3. Vibrational energy - The \(\mathrm{CO}_2\) molecules can vibrate in different modes, such as asymmetric stretch, symmetric stretch, and bending vibrations. With the increase in energy, the amplitude and frequency of these vibrations will also increase.
03

(c) Microstates of CO2(g) and Ar(g)

At a given temperature, \(\mathrm{CO}_2(g)\) and \(\mathrm{Ar}(g)\) will not have the same number of microstates. Here's why: Microstates are the different possible arrangements of energy among the molecules in a system. While both \(\mathrm{CO}_2(g)\) and \(\mathrm{Ar}(g)\) have nearly the same molar mass, their structures are different and have different degrees of freedom for energy distribution. \(\mathrm{Ar}(g)\) is a monatomic gas and can only distribute energy in translational and electronic energy levels. On the other hand, \(\mathrm{CO}_2(g)\) is a linear polyatomic molecule and has more degrees of freedom due to additional rotational and vibrational energy levels. As a result, at a given temperature, \(\mathrm{CO}_2(g)\) will have more microstates (ways to distribute energy among the molecules) than \(\mathrm{Ar}(g)\).

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 91影视!

Key Concepts

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

Third Law of Thermodynamics
The Third Law of Thermodynamics is a fundamental principle in physics. It states that the entropy of a perfect crystal approaches zero as the temperature approaches absolute zero (0 Kelvin). Entropy measures the disorder in a system, and a perfectly ordered crystal at absolute zero has no disorder. This is because the molecules are in their most organized state.
Understanding this concept helps explain why systems can't reach absolute zero. As temperature decreases, less thermal energy is available, and molecular motion reduces. However, reaching absolute zero is impossible because it would require removing all entropy, a theoretical challenge.
  • Perfect crystals naturally have zero entropy at 0 Kelvin due to maximum order.
  • Real systems can't achieve absolute zero, as removing all disorder is practically impossible.
  • This principle explains why cooling becomes challenging as temperatures approach absolute zero.
Energy Distribution
When you heat a gas like CO_2, energy is distributed among molecules in different ways. This increase in energy leads to various changes within the system, enhancing both molecular motion and activity.
  • Translational Energy: Molecules move faster along a path, increasing kinetic energy. This movement is typically straight-line motion.
  • Rotational Energy: CO_2 molecules spin around their axes. Heating increases rotation rates, distributing additional energy.
  • Vibrational Energy: Molecules vibrate in different modes, like stretching and bending. Extra energy increases these vibration rates and amplitudes.
Each way of energy distribution enhances the activity level of molecules differently. It's fascinating how heating changes molecular dynamics in gases!
Microstates
Microstates are the different ways in which energy can be arranged in a system at a given temperature. While it might sound complex, it's a core concept in understanding molecular behavior.
Imagine comparing CO_2(g) and Ar(g). Although they have similar molar masses, they differ in structure. Ar is a monatomic gas, allowing energy distribution only through translation. However, CO_2 is polyatomic, with additional vibrational and rotational modes.
  • Ar(g): Simplistic energy distribution through translational moves only.
  • CO_2(g): Rich energy distribution possibilities through extra rotations and vibrations.
  • More complex molecules like CO_2 have far more microstates than simpler gases like Ar.
This variance in microstates explains why different gases have unique energy behaviors even under similar conditions. Unique structures lead to different energy arrangements and levels of molecular activity.

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

When most elastomeric polymers (e.g., a rubber band) are stretched, the molecules become more ordered, as illustrated here: Suppose you stretch a rubber band. (a) Do you expect the entropy of the system to increase or decrease? (b) If the rubber band were stretched isothermally, would heat need to be absorbed or emitted to maintain constant temperature? (c) Try this experiment: Stretch a rubber band and wait a moment. Then place the stretched rubber band on your upper lip, and let it return suddenly to its unstretched state (remember to keep holding on). What do you observe? Are your observations consistent with your answer to part (b)?

Using data in Appendix C, calculate \(\Delta H^{\circ}, \Delta S^{\circ},\) and \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) for each of the following reactions. In each case show that \(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} .\) (a) \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)\) (b) \(\mathrm{C}(s,\) graphite \()+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(g)\) (c) \(2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{POCl}_{3}(g)\) (d) \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\)

The \(K_{b}\) for methylamine \(\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is given in Appendix D. (a) Write the chemical equation for the equilibrium that corresponds to \(K_{b} .\) (b) By using the value of \(K_{b},\) calculate \(\Delta G^{\circ}\) for the equilibrium in part (a). (c) What is the value of \(\Delta G\) at equilibrium? (d) What is the value of \(\Delta G\) when \(\left[\mathrm{H}^{+}\right]=6.7 \times 10^{-9} \mathrm{M},\left[\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\right]=2.4 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{CH}_{3} \mathrm{NH}_{2}\right]=0.098 \mathrm{M} ?\)

Predict the sign of the entropy change of the system for each of the following reactions: (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) \(\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\) (c) \(3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(g)\) (d) \(\mathrm{Al}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Al}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)\)

Consider the following equilibrium: $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ Thermodynamic data on these gases are given in Appendix C. You may assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not vary with temperature. (a) At what temperature will an equilibrium mixture contain equal amounts of the two gases? (b) At what temperature will an equilibrium mixture of 1 atm total pressure contain twice as much \(\mathrm{NO}_{2}\) as \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (c) At what temperature will an equilibrium mixture of 10 atm total pressure contain twice as much \(\mathrm{NO}_{2}\) as \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (d) Rationalize the results from parts (b) and (c) by using Le Ch芒telier's principle. [Section 15.7]
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

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.