Problem 24
Question
Suppose we vaporize a mole of liquid water at \(25^{\circ} \mathrm{C}\) and another mole of water at \(100^{\circ} \mathrm{C}\). (a) Assuming that the enthalpy of vaporization of water does not change much between \(25^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C}\), which process involves the larger change in entropy? (b) Does the entropy change in either process depend on whether we carry out the process reversibly or not? Explain.
Step-by-Step Solution
Verified Answer
The process of vaporizing water at \(25^\circ \mathrm{C}\) involves a larger change in entropy than at \(100^\circ \mathrm{C}\) because the temperature is lower. Entropy change in either process does not depend on whether we carry out the process reversibly or not, as entropy is a state function and depends only on the initial and final states of the system.
1Step 1: Calculate \(\Delta S_1\) for vaporizing water at \(25^\circ \mathrm{C}\)
Assuming that the enthalpy of vaporization (\(\Delta H\)) doesn't change much between \(25^\circ \mathrm{C}\) and \(100^\circ \mathrm{C}\), the formula to calculate the change in entropy (\(\Delta S_1\)) for vaporizing water at \(25^\circ \mathrm{C}\) is:
\[
\Delta S_1 = \frac{\Delta H}{T_1}
\]
Where \(T_1 = 25^\circ \mathrm{C} + 273.15 = 298.15 \mathrm{K}\) is the temperature in Kelvin.
2Step 2: Calculate \(\Delta S_2\) for vaporizing water at \(100^\circ \mathrm{C}\)
Similarly, the formula to calculate the change in entropy (\(\Delta S_2\)) for vaporizing water at \(100^\circ \mathrm{C}\) is:
\[
\Delta S_2 = \frac{\Delta H}{T_2}
\]
Where \(T_2 = 100^\circ \mathrm{C} + 273.15 = 373.15 \mathrm{K}\) is the temperature in Kelvin.
3Step 3: Compare the change in entropy for the two processes
As we need to compare (\(\Delta S_1\)) and (\(\Delta S_2\)), let's put both into one equation:
\[
\Delta S_1 = \frac{\Delta H}{T_1} \quad \text{and} \quad \Delta S_2 = \frac{\Delta H}{T_2}
\]
Since the enthalpy of vaporization (\(\Delta H\)) is assumed constant for both processes, we can now compare the change in entropy by comparing their denominators (\(T_1\) and \(T_2\)). Since \(T_2 > T_1\), it's clear that:
\[
\Delta S_1 > \Delta S_2
\]
This means that the process of vaporizing water at \(25^\circ \mathrm{C}\) involves a larger change in entropy than at \(100^\circ \mathrm{C}\).
4Step 4: Discuss the dependence of entropy change on the reversibility of the process
Entropy is a state function, meaning its change depends only on the initial and final states of the system and not on the path or reversibility of the process. Therefore, the entropy change in either process does not depend on whether we carry out the process reversibly or not. The comparison we made in Step 3 will be the same whether the process is reversible or not.
Key Concepts
Enthalpy of VaporizationTemperature and Entropy RelationshipState Function Property of Entropy
Enthalpy of Vaporization
The concept of enthalpy of vaporization, denoted as \(\Delta H_{vap}\), refers to the amount of heat energy required for the conversion of one mole of a substance from its liquid phase to its gaseous phase at constant pressure and temperature. This value is critical in understanding the energy dynamics during phase transitions.
Enthalpy of vaporization is effectively the energy needed to break the intermolecular forces that hold the liquid molecules together. In the context of the exercise, assuming that the \(\Delta H_{vap}\) of water remains nearly constant between temperatures of \(25^\circ \mathrm{C}\) and \(100^\circ \mathrm{C}\), we utilize this assumption to compare the entropy changes during vaporization at these two distinct temperatures.
Since \(\Delta H_{vap}\) is constant, the entropy change directly corresponds to the temperature at which vaporization occurs. The higher the temperature, the lower the entropy change for a given amount of heat due to the relationship between temperature and entropy, which we'll explore in the following section.
Enthalpy of vaporization is effectively the energy needed to break the intermolecular forces that hold the liquid molecules together. In the context of the exercise, assuming that the \(\Delta H_{vap}\) of water remains nearly constant between temperatures of \(25^\circ \mathrm{C}\) and \(100^\circ \mathrm{C}\), we utilize this assumption to compare the entropy changes during vaporization at these two distinct temperatures.
Since \(\Delta H_{vap}\) is constant, the entropy change directly corresponds to the temperature at which vaporization occurs. The higher the temperature, the lower the entropy change for a given amount of heat due to the relationship between temperature and entropy, which we'll explore in the following section.
Temperature and Entropy Relationship
Entropy, symbolized by \(S\), is a measure of the disorder or randomness in a system. The relationship between temperature and entropy is intrinsic to thermodynamic processes. During the phase transition of vaporization, increasing temperature typically corresponds to an increase in entropy because higher temperatures allow particles more freedom to move and spread out, thereby increasing disorder.
However, when delving into the calculation of entropy change \(\Delta S\) during vaporization, as shown in the problem's solution, we see the inverse relationship: \(\Delta S = \frac{\Delta H_{vap}}{T}\). In this equation, the higher temperature (in Kelvin), serving as the denominator, results in a smaller entropy change for the same amount of enthalpy absorbed by the system.
The provider of the step by step solution correctly explains this concept by illustrating that vaporizing water at a cooler temperature (\(25^\circ \mathrm{C}\) or \(298.15 \mathrm{K}\)) leads to a larger entropy change than doing so at a higher temperature (\(100^\circ \mathrm{C}\) or \(373.15 \mathrm{K}\)). This is essential for students to note, as it is a common point of confusion.
However, when delving into the calculation of entropy change \(\Delta S\) during vaporization, as shown in the problem's solution, we see the inverse relationship: \(\Delta S = \frac{\Delta H_{vap}}{T}\). In this equation, the higher temperature (in Kelvin), serving as the denominator, results in a smaller entropy change for the same amount of enthalpy absorbed by the system.
The provider of the step by step solution correctly explains this concept by illustrating that vaporizing water at a cooler temperature (\(25^\circ \mathrm{C}\) or \(298.15 \mathrm{K}\)) leads to a larger entropy change than doing so at a higher temperature (\(100^\circ \mathrm{C}\) or \(373.15 \mathrm{K}\)). This is essential for students to note, as it is a common point of confusion.
State Function Property of Entropy
A state function is a property of a system that depends only on the current state, and not on the path taken to reach that state. Entropy is a state function, meaning the change in entropy \(\Delta S\) from one state to another is the same no matter the process path. This concept is crucial in thermodynamics because it allows one to determine the entropy difference between two states without needing to know the details of the process.
In the exercise, this property lets us conclude that the entropy change for vaporization of water does not depend on whether the process is carried out reversibly or irreversibly. This is an important lesson, as it reinforces the foundational thermodynamic principle that focuses on initial and final states rather than process pathways.
The exercise solution emphasizes this point by stating that whether the vaporization is reversible or not, the entropy change \(\Delta S\) for a given enthalpy of vaporization will remain consistent. This understanding aids students in appreciating the universality of entropy changes in isolated scenarios, decoupling it from the complexities of real-world processes which may vary in their reversibility.
In the exercise, this property lets us conclude that the entropy change for vaporization of water does not depend on whether the process is carried out reversibly or irreversibly. This is an important lesson, as it reinforces the foundational thermodynamic principle that focuses on initial and final states rather than process pathways.
The exercise solution emphasizes this point by stating that whether the vaporization is reversible or not, the entropy change \(\Delta S\) for a given enthalpy of vaporization will remain consistent. This understanding aids students in appreciating the universality of entropy changes in isolated scenarios, decoupling it from the complexities of real-world processes which may vary in their reversibility.
Other exercises in this chapter
Problem 21
Consider a system consisting of an ice cube. (a) Under what conditions can the ice cube melt reversibly? (b) If the ice cube melts reversibly, is \(\Delta E\) z
View solution Problem 23
Indicate whether each statement is true or false. (a) \(\Delta S\) for an isothermal process depends on both the temperature and the amount of heat reversibly t
View solution Problem 25
The normal boiling point of \(\mathrm{Br}_{2}(l)\) is \(58.8^{\circ} \mathrm{C}\), and its molar enthalpy of vaporization is \(\Delta H_{\text {vap }}=29.6 \mat
View solution Problem 27
Indicate whether cach statement is true or false. (a) The second law of thermodynamics says that entropy is conserved. (b) If the entropy of the system increase
View solution