Problem 39
Question
For each of the following pairs, predict which substance has the higher entropy per mole at a given temperature: (a) \(\operatorname{Ar}(l)\) or \(\operatorname{Ar}(g),(\mathbf{b}) \operatorname{He}(g)\) at 3 atm pressure or \(\operatorname{He}(g)\) at 1.5 atm pressure, (c) 1 mol of \(\mathrm{Ne}(g)\) in 15.0 \(\mathrm{L}\) or 1 \(\mathrm{mol}\) of \(\mathrm{Ne}(g)\) in \(1.50 \mathrm{L},(\mathbf{d}) \mathrm{CO}_{2}(g)\) or \(\mathrm{CO}_{2}(s) .\)
Step-by-Step Solution
Verified Answer
(a) Ar(g) has higher entropy than Ar(l). (b) He(g) at 1.5 atm has higher entropy than He(g) at 3 atm. (c) 1 mol of Ne(g) in 15.0 L has higher entropy than 1 mol of Ne(g) in 1.50 L. (d) CO2(g) has higher entropy than CO2(s).
1Step 1: (a) Ar(l) or Ar(g)
When comparing the entropy of a substance in the liquid state (Ar(l)) and the gas state (Ar(g)), the gas state will always have higher entropy. This is because gas particles have more randomness and disorder than particles in the liquid state. Therefore, Ar(g) has higher entropy per mole at a given temperature than Ar(l).
2Step 2: (b) He(g) at 3 atm or He(g) at 1.5 atm
In this case, we are comparing the entropy of helium gas (He(g)) at two different pressures: 3 atm and 1.5 atm. At a lower pressure, a gas will have a higher entropy. This is because lower pressure allows the gas particles to have more freedom of movement, resulting in greater disorder. Therefore, He(g) at 1.5 atm has higher entropy per mole at a given temperature than He(g) at 3 atm.
3Step 3: (c) Ne(g) in 15.0 L or Ne(g) in 1.50 L
In this scenario, we are comparing the entropy of neon gas (Ne(g)) in two different volumes: 15.0 L and 1.50 L. Entropy increases with an increase in volume, as the gas particles have more space to move around and create disorder. Thus, 1 mol of Ne(g) in 15.0 L has higher entropy per mole at a given temperature than 1 mol of Ne(g) in 1.50 L.
4Step 4: (d) CO2(g) or CO2(s)
When comparing the entropy of a substance in the solid state (CO2(s)) and the gas state (CO2(g)), the gas state will always have higher entropy. As discussed earlier, gas particles have more randomness and disorder than particles in the solid state. Therefore, CO2(g) has higher entropy per mole at a given temperature than CO2(s).
Key Concepts
Thermodynamic EntropyStates of MatterGas Laws
Thermodynamic Entropy
Entropy is a fundamental concept in thermodynamics, often associated with the disorder or randomness of particles within a system. It's a measure of the number of possible arrangements the atoms or molecules of a system can have, and it increases when the system becomes more disordered. In a broader sense, it represents the energy not available for doing work and is a key ingredient in the second law of thermodynamics, which states that the total entropy of an isolated system can never decrease over time.
In the context of our textbook exercise, the concept of entropy helps predict which substance in a pair will have higher entropy per mole. Generally, the state of matter with the particles more spread out and less ordered - like gases compared to liquids or solids - will have higher entropy. Moreover, factors such as pressure and volume also affect entropy. A lower pressure or a greater volume allows for more molecular movement and, consequently, a higher entropy. This principle is clearly seen in the solutions provided, showing that gases have higher entropy than liquids or solids and that gas entropy increases with volume and decreases with pressure.
In the context of our textbook exercise, the concept of entropy helps predict which substance in a pair will have higher entropy per mole. Generally, the state of matter with the particles more spread out and less ordered - like gases compared to liquids or solids - will have higher entropy. Moreover, factors such as pressure and volume also affect entropy. A lower pressure or a greater volume allows for more molecular movement and, consequently, a higher entropy. This principle is clearly seen in the solutions provided, showing that gases have higher entropy than liquids or solids and that gas entropy increases with volume and decreases with pressure.
States of Matter
The states of matter—solid, liquid, and gas—are distinguished by changes in the properties of matter related to external factors such as pressure and temperature. Solids are characterized by a fixed volume and shape with particles vibrating in place. Liquids have a definite volume but no fixed shape and the particles are more mobile than in solids. Gases, on the other hand, have neither a fixed volume nor shape, with particles that move freely and are far apart. This freedom of movement in gases results in high entropy because there is a greater number of possible ways in which the particles can be arranged.
This is exemplified in the problem statements where the gas states of substances such as Argon (Ar) and Carbon Dioxide (CO2) have higher entropy than their liquid or solid counterparts. The transformations between these states, known as phase transitions, involve changes in energy and entropy, which are critical to understanding thermal phenomena and thermodynamic processes.
This is exemplified in the problem statements where the gas states of substances such as Argon (Ar) and Carbon Dioxide (CO2) have higher entropy than their liquid or solid counterparts. The transformations between these states, known as phase transitions, involve changes in energy and entropy, which are critical to understanding thermal phenomena and thermodynamic processes.
Gas Laws
Gas laws describe how various properties of gases (pressure, volume, and temperature) are related and change under different conditions. These laws are grounded in simple mathematical relationships and are critical in understanding and predicting the behavior of gases. The most fundamental laws include Boyle's Law, which describes the inverse relationship between pressure and volume, Charles's Law, which describes the direct relationship between volume and temperature, and Avogadro's Law, that states that equal volumes of all gases at the same temperature and pressure contain the same number of molecules.
The solutions to our textbook problems involve concepts from these gas laws. For instance, lower pressure as per Boyle's Law allows for higher entropy, explained by the increased volume for the gas particles to move, hence more randomness (in the case of He(g) at different pressures). Similarly, an increase in volume would also mean greater entropy because the gas particles would be more spread out and have more potential arrangements (as seen with Ne(g) in different volumes). Understanding these laws not only helps students with textbook exercises but also provides foundational knowledge for grasping complex thermodynamic principles.
The solutions to our textbook problems involve concepts from these gas laws. For instance, lower pressure as per Boyle's Law allows for higher entropy, explained by the increased volume for the gas particles to move, hence more randomness (in the case of He(g) at different pressures). Similarly, an increase in volume would also mean greater entropy because the gas particles would be more spread out and have more potential arrangements (as seen with Ne(g) in different volumes). Understanding these laws not only helps students with textbook exercises but also provides foundational knowledge for grasping complex thermodynamic principles.
Other exercises in this chapter
Problem 37
Indicate whether each statement is true or false. (a) The third law of thermodynamics says the entropy of a perfect, pure crystal at absolute zero increases wit
View solution Problem 38
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)
View solution Problem 40
For each of the following pairs, predict which substance possesses the larger entropy per mole: (a) 1 1 mol of \(\mathrm{O}_{2}(g)\) at \(300^{\circ} \mathrm{C}
View solution Problem 41
Predict the sign of the entropy change of the system for each of the following reactions: $$\begin{array}{l}{\text { (a) } \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)
View solution