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Chapter 19: Problem 40

Indicate whether each statement is true or false. (a) Unlike enthalpy, where we can only ever know changes in \(H\), we can know absolute values of \(S\). (b) If you heat a gas such as \(\mathrm{CO}_{2}\) you will increase its degrees of translational, rotational and vibrational motions. (c) \(\mathrm{CO}_{2}(g)\) and \(\mathrm{Ar}(\mathrm{g})\) have nearly the same molar mass. At a given temperature, they will have the same number of microstates.

Short Answer

Expert verified
Statement 1: True, because the Third Law of Thermodynamics allows us to know the absolute values of entropy (S). Statement 2: True, because heating a gas increases its kinetic energy, which in turn increases its translational, rotational, and vibrational motions. Statement 3: False, because the number of microstates depends not only on molar mass but also on factors such as molecular structure, internal complexity, and degrees of freedom.

Step by step solution

01

Statement 1: Entropy absolute values

The Third Law of Thermodynamics states that the entropy of a perfect crystal approaches zero as the temperature approaches absolute zero (0 K). Because there's a reference point (i.e., 0 K), we can indeed know the absolute values of entropy (S). Therefore, this statement is \( \boxed{\textbf{true}} \).
02

Statement 2: Heating a gas increases essential motions

When a gas is heated, it acquires more kinetic energy overall. This increase in kinetic energy will result in an increase in the translational, rotational, and vibrational motions of the gas particles. So, heating a gas like CO2 will increase its degrees of translational, rotational, and vibrational motions. This statement is \( \boxed{\textbf{true}} \).
03

Statement 3: Microstates of CO2(g) and Ar(g)

CO2(g) and Ar(g) have approximately the same molar mass. However, they have different molecular structures and complexities. CO2 is a linear molecule with 3 atoms, while Ar is an atomic gas (noble gas) with only one atom. The number of microstates is determined not only by molar mass but also by molecular structure, internal complexity, and degrees of freedom. Therefore, it is incorrect to say that they will have the same number of microstates at a given temperature, even if they have the same molar mass. This statement is \( \boxed{\textbf{false}} \).

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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 stands as a foundational principle in physics, particularly with regard to temperature and disorder.

It stipulates that as a system approaches absolute zero temperature (\(0 \text{ K}\)), the entropy鈥攐r measure of disorder鈥攐f a perfect crystalline structure will also approach zero. This assertion hinges on the presumption that at absolute zero, a perfect crystal exists in its state of minimum possible motion, implying that its atoms are arranged in a perfectly ordered state.

At this ultimate lower bound temperature, the system possesses only one possible microstate, meaning there is only one way the particles can be arranged. Since entropy is a measure of disorder or randomness within the system, having only one microstate translates to minimal entropy. As temperature increases, the number of microstates increases due to additional possible arrangements of particles, thus increasing the entropy.

Importantly, the third law provides a reference point for the absolute measure of entropy. This is critical for many calculations in thermodynamics since we can define the absolute entropy at zero Kelvin and then measure changes from this reference point.
Kinetic Energy in Gases
Delving into the behavior of gases, kinetic energy plays an integral role. Kinetic energy is the energy associated with the motion of particles.

In a gas, particles are in continuous, random motion and the kinetic energy of these particles is directly related to their temperature. Understanding the molecular theory of gases lets us know that when the temperature of a gas rises, the average kinetic energy of the molecules also increases. This is simply because temperature is a measure of the average kinetic energy of the particles in a substance.

As a result of kinetic energy increase, the molecules move more rapidly and this movement translates into greater translational (straight-line movement), rotational (spinning), and vibrational (oscillation) motions. Each molecule's range of motion contributes to the overall kinetic energy and therefore, additional amounts of heat鈥攅nergy transferred due to temperature difference鈥攚ill increase these motions in gases like carbon dioxide (\( \text{CO}_2 \)).
Molecular Structure Impact on Microstates
Microstates are specific ways in which a system may be arranged, especially in terms of the positions and energies of its molecules. They have profound implications for entropy and are largely influenced by molecular structure.

The molecular structure dictates the number of microstates available due to varying degrees of freedom such as translational, rotational, and vibrational movements. For example, a more complex molecule like carbon dioxide (\( \text{CO}_2 \)), with three atoms arranged linearly, will have more microstates than a simpler, single-atom molecule like argon (\( \text{Ar} \)) because of the additional rotational and vibrational possibilities.CO2 can rotate and vibrate in more ways than monatomic argon can, which allows CO2 to have more potential arrangements of energy and therefore more microstates.

Even if two substances, such as \( \text{CO}_2 \) and \( \text{Ar} \) have similar molar masses, their molecular structure, and complexity lead to different quantities of microstates. Understanding the distinct difference between how molecular complexity relates to potential microstate variety is a fundamental aspect of thermodynamics that extends into understanding properties like entropy and the distribution of energy levels within a substance.

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Most popular questions from this chapter

An ice cube with a mass of \(20 \mathrm{~g}\) at \(-20^{\circ} \mathrm{C}\) (typical freezer temperature) is dropped into a cup that holds \(500 \mathrm{~mL}\) of hot water, initially at \(83^{\circ} \mathrm{C}\). What is the final temperature in the cup? The density of liquid water is \(1.00 \mathrm{~g} / \mathrm{mL}\), the specific heat capacity of ice is \(2.03 \mathrm{~J} / \mathrm{g}-\mathrm{C}\); the specific heat capacity of liquid water is \(4.184 \mathrm{~J} / \mathrm{g}-\mathrm{C}\), the enthalpy of fusion of water is \(6.01 \mathrm{~kJ} / \mathrm{mol}\).

For a certain chemical reaction, \(\Delta H^{k}=-35.4 \mathrm{~kJ}\) and \(\Delta S^{n}=-85.5 \mathrm{~J} / \mathrm{K}\). (a) Is the reaction exothermic or endothermic? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (c) Calculate \(\Delta G^{\circ}\) for the reaction at \(298 \mathrm{~K}\). (d) Is the reaction spontaneous at \(298 \mathrm{~K}\) under standard conditions?

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) \(3 \mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(g)\) (d) \(\mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{~s})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{Al}(\mathrm{s})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

From the values given for \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\), calculate \(\Delta G^{*}\) 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? (a) \(2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g)\) \(\Delta H^{\circ}=-844 \mathrm{kl} ; \Delta S^{\circ}=-165 \mathrm{~J} / \mathrm{K}\) (b) \(2 \mathrm{POCl}_{3}(\mathrm{~g}) \longrightarrow 2 \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{O}_{\text {(a }}(\mathrm{a})\) \(\Delta H^{\circ}=572 \mathrm{~kJ} ; \Delta S^{\circ}=179 \mathrm{~J} / \mathrm{K}\)

A system goes from state 1 to state 2 and back to state 1 . (a) Is \(\Delta E\) the same in magnitude for both the forward and reverse processes? (b) Without further information, can you conclude that the amount of heat transferred to the system as it goes from state 1 to state 2 is the same or different as compared to that upon going from state 2 back to state 1? (c) Suppose the changes in state are reversible processes. Is the work done by the system upon going from state 1 to state 2 the same or different as compared to that upon going from state 2 back to state 1?
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