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Chapter 20: Problem 18

Vaporization at the normal boiling point \(\left(T_{\text {vap }}\right)\) of a substance (the boiling point at one atm) can be regarded as a reversible process because if the temperature is decreased infinitesimally below \(T_{\mathrm{vap}},\) all the vapor will condense to liquid, whereas if it is increased infinitesimally above \(T_{\mathrm{vap}},\) all the liquid will vaporize. Calculate the entropy change when two moles of water vaporize at \(100.0^{\circ} \mathrm{C}\). The value of \(\Delta_{\mathrm{vap}} \bar{H}\) is \(40.65 \mathrm{kJ} \cdot \mathrm{mol}^{-1} .\) Comment on the sign of \(\Delta_{\text {vap }} S\)

Short Answer

Expert verified
The entropy change for vaporizing two moles of water at 100掳C is positive, indicating increased disorder.

Step by step solution

01

Understanding the given data

We are given that 2 moles of water vaporize at the normal boiling point of 100掳C, where \( \Delta_{\text{vap}} \bar{H} \) is 40.65 kJ/mol. We need to calculate the entropy change \( \Delta_{\text{vap}} S \) for this process.
02

Recall the formula for entropy change

The formula to calculate the entropy change for vaporization at the boiling point is \( \Delta_{\text{vap}} S = \frac{\Delta_{\text{vap}} \bar{H}}{T_{\text{vap}}} \). Since \( \Delta_{\text{vap}} \bar{H} \) is given per mole, we must take into account the number of moles.
03

Convert the temperature to Kelvin

Since the enthalpy and entropy are calculated using absolute temperature, we need to convert the boiling point from Celsius to Kelvin. Use the formula: \[ T_{\text{vap}} = 100掳C + 273.15 = 373.15 \, K \]
04

Calculate the entropy change for one mole

Using the formula and the data provided:\[ \Delta_{\text{vap}} S = \frac{\Delta_{\text{vap}} \bar{H}}{T_{\text{vap}}} = \frac{40.65 \, \text{kJ/mol}}{373.15 \, \text{K}} \]. First, calculate this value for one mole of water.
05

Calculate the total entropy change for two moles

Since we have 2 moles, we multiply the result from Step 4 by 2:\[ \Delta_{\text{vap}} S_{\text{total}} = 2 \times \frac{40.65}{373.15} \]Evaluate this to get the total entropy change in J/K.
06

Evaluate the sign of \( \Delta_{\text{vap}} S \)

Since vaporization involves a transition from an ordered (liquid) to a less ordered state (vapor), the entropy change \( \Delta_{\text{vap}} S \) should be positive, indicating an increase in disorder of the system.

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Key Concepts

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

Reversible Process
Imagine a scenario during the phase change when water is transitioning from liquid to vapor at its boiling point. This action happens at the normal boiling point, where the atmospheric pressure is 1 atm. Here, vaporization can be considered as a reversible process. But what is a reversible process in thermodynamics?
A reversible process is one where the system and its surroundings can be returned to their initial states by an infinitesimal change. In simpler terms, it means that the transition, either to vapor or liquid, can be undone by just a tiny change in temperature.
  • If we slightly decrease the temperature below the boiling point, vapor will condense back into liquid.
  • If we slightly increase the temperature above the boiling point, the liquid will convert into vapor.
For the vaporization of water at its boiling point, it operates so closely to equilibrium that the process can be continually adjusted, qualifying it as a reversible process. This concept is essential because it means maximum possible work can be extracted, contributing to efficient energy transformations.
Entropy Calculation
Entropy is a measure of disorder or randomness in a system, and during vaporization, when water turns from liquid into vapor, we observe a change in entropy. To understand how to calculate entropy change for such a process, let's dive deeper into the considerations.
The formula used to calculate entropy change during vaporization is:\[ \Delta_{\text{vap}} S = \frac{\Delta_{\text{vap}} \bar{H}}{T_{\text{vap}}} \]This equation stems from the Second Law of Thermodynamics, which relates entropy change to heat and temperature. Here:
  • \( \Delta_{\text{vap}} \bar{H} \) is the molar enthalpy of vaporization, which is the energy required to vaporize one mole of a substance.
  • \( T_{\text{vap}} \) is the absolute temperature in Kelvin.
In the given exercise, the enthalpy change is \( 40.65 \text{ kJ/mol} \), and the temperature at the boiling point is \( 373.15 \text{ K} \). Substituting these values into the formula allows us to determine the entropy change for vaporizing water. Since the problem pertains to 2 moles, we need to multiply the resulting entropy change by 2, emphasizing that we're considering 2 moles of water. The positive result highlights an increase in disorder, which matches our expectations for a liquid-to-gas transition.
Boiling Point
The boiling point is a critical temperature point where a liquid turns into vapor, and understanding it is key to grasping how vaporization processes work. For water specifically, the normal boiling point is \( 100^{\circ} C \), equivalent to \( 373.15 \text{ K} \) in absolute terms because thermodynamic processes are usually measured in Kelvin.
The boiling point is defined under a pressure of 1 atm, which is the standard atmospheric pressure at sea level. This condition is important as boiling points can vary with pressure changes. For example, water will boil at a lower temperature in a high-altitude location where the atmospheric pressure is lower.
Understanding the boiling point helps you appreciate why water readily turns to steam at this precise temperature under normal conditions. During vaporization, the temperature doesn't fall or rise but remains constant as the energy input goes into breaking intermolecular forces rather than increasing temperature. This constancy at the boiling point defines why it鈥檚 pivotal in calculations relating to reversible processes and entropy changes.
Recognizing the importance of the boiling point provides us the perfect reference to predict and control the behavior of substances undergoing phase changes across various conditions.

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Most popular questions from this chapter

Show that if a process involves only an isothermal transfer of energy as heat (pure heat transfer), then $$ d S_{\mathrm{sys}}=\frac{d q}{T} \quad \text { (pure heat transfer) } $$

In this problem, we will illustrate the condition \(d S_{\mathrm{prod}} \geq 0\) with a concrete example. Consider the two-component system shown in Figure \(20.8\). Each compartment is in equilibrium with a heat reservoir at different temperatures \(T_{1}\) and \(T_{2}\), and the two compartments are separated by a rigid heat-conducting wall. The total change of energy as heat of compartment 1 is $$ d q_{1}=d_{\mathrm{e}} q_{1}+d_{\mathrm{i}} q_{1} $$ where \(d_{\mathrm{e}} q_{1}\) is the energy as heat exchanged with the reservoir and \(d_{\mathrm{i}} q_{1}\) is the energy as heat exchanged with compartment \(2 .\) Similarly, $$ d q_{2}=d_{\mathrm{e}} q_{2}+d_{\mathrm{i}} q_{2} $$ Clearly, $$ d_{i} q_{1}=-d_{i} q_{2} $$ Show that the entropy change for the two-compartment system is given by $$ \begin{aligned} d S &=\frac{d_{\mathrm{e}} q_{1}}{T_{1}}+\frac{d_{\mathrm{e}} q_{2}}{T_{2}}+d_{\mathrm{i}} q_{1}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) \\\ &=d S_{\text {exchange }}+d S_{\text {prod }} \end{aligned} $$ $$ d S_{\text {exchange }}=\frac{d_{\mathrm{e}} q_{1}}{T_{1}}+\frac{d_{\mathrm{e}} q_{2}}{T_{2}} $$ is the entropy exchanged with the reservoirs (surroundings) and $$ d S_{\mathrm{prod}}=d_{\mathrm{i}} q_{1}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) $$ is the entropy produced within the two-compartment system. Now show that the condition \(d S_{\text {prod }} \geq 0\) implies that energy as heat flows spontaneously from a higher temperature to a lower temperature. The value of \(d S_{\text {exchange }}\), however, has no restriction and can be positive, negative, or zero.

Calculate the value of \(\Delta S\) if one mole of an ideal gas is expanded reversibly and isothermally from 1.00 bar to 0.100 bar. Explain the sign of \(\Delta S\).

In this problem, we will consider a two-compartment system like that in Figure 20.4 except that the two subsystems have the same temperature but different pressures and the wall that separates them is flexible rather than rigid. Show that in this case, $$d S=\frac{d V_{\mathrm{B}}}{T}\left(P_{\mathrm{B}}-P_{\mathrm{A}}\right)$$ Interpret this result with regard to the sign of \(d V_{\mathrm{B}}\) when \(P_{\mathrm{B}}>P_{\mathrm{A}}\) and when \(P_{\mathrm{B}}

Show that $$\Delta S \geq \frac{q}{T}$$ for an isothermal process. What does this equation say about the sign of \(\Delta S ?\) Can \(\Delta S\) decrease in a reversible isothermal process? Calculate the entropy change when one mole of an ideal gas is compressed reversibly and isothermally from a volume of \(100 \mathrm{dm}^{3}\) to \(50.0 \mathrm{dm}^{3}\) at \(300 \mathrm{K}\)
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