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For each of the following pairs, choose the substance with the higher entropy per mole at a given temperature: (a) \(\operatorname{Ar}(l)\) or \(\operatorname{Ar}(g)\), (b) \(\mathrm{He}(g)\) at 3 atm pressure or \(\mathrm{He}(\mathrm{g})\) at \(1.5\) atm pressure, (c) \(1 \mathrm{~mol}\) of \(\mathrm{Ne}(\mathrm{g})\) in \(15.0 \mathrm{~L}\) or 1 mol of \(\mathrm{Ne}(\mathrm{g})\) in \(1.50 \mathrm{~L},(\mathrm{~d}) \mathrm{CO}_{2}(g)\) or \(\mathrm{CO}_{2}(s)\)

Short Answer

Expert verified
(a) \(\operatorname{Ar}(g)\) has higher entropy than \(\operatorname{Ar}(l)\). (b) \(\mathrm{He}(\mathrm{g})\) at \(1.5\) atm pressure has higher entropy than that at 3 atm pressure. (c) \(1 \mathrm{~mol}\) of \(\mathrm{Ne}(\mathrm{g})\) in \(15.0 \mathrm{~L}\) has higher entropy than that in \(1.50 \mathrm{~L}\). (d) \(\mathrm{CO}_{2}(g)\) has higher entropy than \(\mathrm{CO}_{2}(s)\).

Step by step solution

01

(a) Comparing entropy of liquid and gaseous Argon

In the case of Argon, we need to compare the entropy for liquid \((l)\) and gaseous \((g)\) states at the same temperature. Generally, entropy increases from solid to liquid to gas. Since gaseous Argon has more disorder and randomness than liquid Argon, the entropy of \(\operatorname{Ar}(g)\) is higher than the entropy of \(\operatorname{Ar}(l)\).
02

(b) Comparing entropy of gaseous Helium at different pressures

In this case, we need to compare the entropy for gaseous Helium at two different pressures: 3 atm and 1.5 atm. The entropy of a gas is inversely proportional to its pressure since an increase in pressure leads to a decrease in volume and a more ordered state. Therefore, the entropy of \(\mathrm{He}(\mathrm{g})\) at \(1.5\) atm pressure is higher than that at 3 atm pressure.
03

(c) Comparing entropy of gaseous Neon at different volumes

Here, we need to compare the entropy for gaseous Neon at two different volumes: \(15.0 \mathrm{~L}\) and \(1.50 \mathrm{~L}\). Entropy is directly proportional to the volume of a gas, as an increase in volume allows more disorder and randomness in the gas molecules. Consequently, the entropy of \(1 \mathrm{~mol}\) of \(\mathrm{Ne}(\mathrm{g})\) in \(15.0 \mathrm{~L}\) is higher than that in \(1.50 \mathrm{~L}\).
04

(d) Comparing entropy of solid and gaseous Carbon Dioxide

Lastly, we need to compare the entropy of solid and gaseous Carbon Dioxide. Similar to the case of Argon, the entropy increases from solid to liquid to gas. The gaseous state of Carbon Dioxide has more disorder and randomness than the solid state, so the entropy of \(\mathrm{CO}_{2}(g)\) is higher than the entropy of \(\mathrm{CO}_{2}(s)\).

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Key Concepts

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

States of Matter
Matter exists primarily in three states: solid, liquid, and gas. Each state varies in terms of particle arrangement and movement.
In solids, particles are tightly packed in a fixed, orderly arrangement, resulting in low entropy, the measure of disorder within a system.
In liquids, particles are more loosely arranged and can move past one another, increasing the system's entropy compared to solids.
Gases, however, have particles in random, irregular motion, filling any available space and causing the highest entropy among the three states.
Generally, when transitioning from a solid to a liquid, and then to a gas, the entropy increases as the molecular order decreases.Understand that more disorder and randomness equate to higher entropy.
  • Solids have the least entropy.
  • Liquids have moderate entropy.
  • Gases have the highest entropy.
This is why gaseous Argon \(\operatorname{Ar}(g)\) has more entropy than liquid Argon \(\operatorname{Ar}(l)\).
Pressure Effects on Gases
Pressure significantly influences the entropy of a gas. When the pressure increases, the gas molecules are forced closer together,
resulting in a more ordered state and hence, lower entropy. Conversely, decreasing pressure allows the particles more space to move, leading to higher entropy levels.
This principle is crucial in understanding the behavior of gases under compression and expansion.In the case of gaseous Helium, the sample at lower pressure \(1.5 \text{ atm}\) has higher entropy than the same gas at \(3 \text{ atm}\).
The reduced pressure gives molecules more freedom and volume to roam around, increasing disorder and entropy.
  • High pressure brings more order and lowers entropy.
  • Low pressure introduces more disorder and increases entropy.
Recognize these pressure outcomes, essential for gas-related calculations.
Volume Effects on Gases
Volume changes directly affect gas entropy. When the volume of a gas increases, there is more space available,
which allows for increased molecular movement and higher entropy. A larger volume means more disorder,
as gas molecules are not confined to a smaller space and can roam more freely.If we look at the example with Neon gas,Neon in a larger volume \(15 \text{ L}\) will have higher entropy than when confined to \(1.5 \text{ L}\).
This concept is key for understanding gas behavior in different conditions, such as during expansion or contraction.
  • Larger volume equals more disorder and higher entropy.
  • Smaller volume equals less disorder and lower entropy.
It is crucial to keep in mind the relationship between volume and entropy in gas calculations.
Disorder and Randomness in Molecules
Entropy fundamentally captures the degree of disorder and randomness in a system. Increased molecular disorder translates to higher entropy.
In gases, particles are free to move randomly, creating significant disorder and high entropy.
For instance, Carbon Dioxide in its gaseous state \(\mathrm{CO}_2(g)\) has a much more disordered arrangement of molecules compared to its solid state \(\mathrm{CO}_2(s)\).
Hence, the entropy in the gaseous state is substantially higher.Understanding this concept is vital in comprehending why gases typically possess greater entropy than liquids and solids.
  • Gases exhibit high molecular randomness and entropy.
  • Solids display a structured order with lower entropy.
Using these principles, you can predict and compare entropies across different phases of matter.

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Most popular questions from this chapter

About \(86 \%\) of the world's electrical energy is produced by using steam turbines, a form of heat engine. In his analysis of an ideal heat engine, Sadi Carnot concluded that the maximum possible efficiency is defined by the total work that could be done by the engine, divided by the quantity of heat available to do the work (for example from hot steam produced by combustion of a fuel such as coal or methane). This efficiency is given by the ratio \(\left(T_{\text {high }}-T_{\text {low }}\right) / T_{\text {high }}\), where \(T_{\text {high }}\) is the temperature of the heat going into the engine and \(T_{\text {low }}\) is that of the heat leaving the engine. (a) What is the maximum possible efficiency of a heat engine operating between an input temperature of \(700 \mathrm{~K}\) and an exit temperature of \(288 \mathrm{~K} ?\) (b) Why is it important that electrical power plants be located near bodies of relatively cool water? (c) Under what conditions could a heat engine operate at or near \(100 \%\) efficiency? (d) It is often said that if the energy of combustion of a fuel such as methane were captured in an electrical fuel cell instead of by burning the fuel in a heat engine, a greater fraction of the energy could be put to useful work. Make a qualitative drawing like that in Figure \(5.10\) that illustrates the fact that in principle the fuel cell route will produce more useful work than the heat engine route from combustion of methane.

The oxidation of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) in body tissue produces \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} .\) In contrast, anaerobic decomposition, which occurs during fermentation, produces ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and \(\mathrm{CO}_{2} .\) (a) Using data given in Appendix \(\mathrm{C}\), compare the equilibrium constants for the following reactions: $$ \begin{aligned} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) & \rightleftharpoons 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) & \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+2 \mathrm{CO}_{2}(g) \end{aligned} $$ (b) Compare the maximum work that can be obtained from these processes under standard conditions.

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 \(10 \mathrm{~atm}\) 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 Châtelier's principle. \(\infty\) (Section 15.7)

(a) Express the second law of thermodynamics as a mathematical equation. (b) In a particular spontaneous process the entropy of the system decreases. What can you conclude about the sign and magnitude of \(\Delta S_{\text {surr }} ?\) (c) During a certain reversible process, the surroundings undergo an entropy change, \(\Delta S_{\text {surr }}=-78 \mathrm{~J} / \mathrm{K}\). What is the entropy change of the system for this process?

Consider the vaporization of liquid water to steam at a pressure of 1 atm. (a) Is this process endothermic or exothermic? (b) In what temperature range is it a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) At what temperature are the two phases in equilibrium?

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