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Chapter 19: Problem 24

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.

Short Answer

Expert verified
The entropy change of vaporizing water at 25掳C (0.1369 kJ/mol K) is greater than the entropy change of vaporizing water at 100掳C (0.1093 kJ/mol K). This result is derived from the formula \(\Delta S = \frac{\Delta H}{T}\), using a constant enthalpy of vaporization for water and the given temperatures in Kelvin. Since the initial and final states of the system remain the same, the entropy change does not depend on whether the process is carried out reversibly or not.

Step by step solution

01

Enthalpy of Vaporization

First, let's recall the relationship between enthalpy change (鈭咹) and entropy change (鈭哠) for a phase transition at constant temperature, T: \[\Delta G = \Delta H - T \Delta S\] Where 鈭咷 is the Gibbs free energy change. For a phase transition, such as vaporization, 鈭咷 = 0. Therefore, we can rearrange the equation to solve for 鈭哠: \[\Delta S = \frac{\Delta H}{T}\] Since we are given the assumption that the enthalpy of vaporization does not change significantly between the two temperatures, we can use a constant value for 鈭咹. The enthalpy of vaporization for water at 100掳C is approximately 40.79 kJ/mol. We can then convert the given temperatures to Kelvin by adding 273.15: T1 = 298.15 K and T2 = 373.15 K.
02

Calculate Entropy Change for Both Processes

Now, we'll use the enthalpy of vaporization and the temperatures to calculate the entropy change for each process: Process 1: Vaporizing water at 25掳C (298.15 K) \[\Delta S_1 = \frac{\Delta H}{T_1} = \frac{40.79 \, \text{kJ/mol}}{298.15 \, \text{K}} = 0.1369 \, \text{kJ/mol K}\] Process 2: Vaporizing water at 100掳C (373.15 K) \[\Delta S_2 = \frac{\Delta H}{T_2} = \frac{40.79 \, \text{kJ/mol}}{373.15 \, \text{K}} = 0.1093 \, \text{kJ/mol K}\]
03

Compare Entropy Changes

Now that we have calculated the entropy change for both processes, we can directly compare them: \[\Delta S_1 = 0.1369 \, \text{kJ/mol K} > \Delta S_2 = 0.1093 \, \text{kJ/mol K}\] The entropy change of vaporizing water at 25掳C is greater than the entropy change of vaporizing water at 100掳C.
04

Discuss Reversibility and Entropy Change

Entropy change depends on the heat transfer between the system and its surroundings. In a reversible process, the heat exchange is carried out infinitely slowly, allowing for the system to always be in equilibrium with the surroundings. In an irreversible process, the heat exchange occurs at a finite rate, and the system may not always be in equilibrium. However, in both processes (reversible and irreversible), the initial and final states of the system are the same, which means the values of the initial and final entropies do not depend on the path (reversible or irreversible) taken between them. Thus, the entropy change in either process does not depend on whether we carry out the process reversibly or not.

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Key Concepts

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

Enthalpy of Vaporization
The enthalpy of vaporization is the amount of energy required to change a substance from a liquid to a gas at its boiling point. It's an integral part of the vaporization process. This energy is needed to overcome the intermolecular forces keeping the water molecules together as a liquid. For water, the enthalpy of vaporization is relatively high at 40.79 kJ/mol. This is because of the strong hydrogen bonds present between water molecules.

The presence of strong hydrogen bonds means that a substantial amount of energy is necessary to vaporize water. Given that the problem states that the enthalpy of vaporization doesn鈥檛 change much between 25掳C and 100掳C, we can simplify calculations by assuming a constant value. Remember, this constant value is used to calculate changes in entropy, which is critical for understanding the difference in entropy change at various temperatures.
Vaporization Process
Vaporization is a phase transition in which a liquid is transformed into a gas. This process can occur in two ways: boiling or evaporation. During vaporization, molecules within the liquid gain sufficient energy to escape as gas into the atmosphere. In our exercise, we focus on the vaporization of water at two different temperatures: 25掳C and 100掳C.

Despite the same enthalpy change, the vaporization process involves different changes in entropy based on temperature. Entropy, a measure of disorder or randomness, increases as a liquid turns into a gas. Obtaining the entropy change during vaporization is essential. We calculate it using the formula \( \Delta S = \frac{\Delta H}{T} \). The results show that vaporizing water at 25掳C yields a larger entropy change than at 100掳C. This is because at lower temperatures, the molecules are less energetic, providing a stark contrast when they transition to a gaseous state.
Reversible and Irreversible Processes
In thermodynamics, any process can occur reversibly or irreversibly. A reversible process is an ideal concept where the system changes states in infinite slow steps, maintaining equilibrium with its surroundings. Irreversible processes occur more quickly, with the system not in constant equilibrium.

Despite this difference in path, entropy hinges only on the initial and final states of the system. This means that whether a vaporization process occurs reversibly or irreversibly doesn't affect the overall entropy change. For our example with water, both vaporization at 25掳C and at 100掳C result in the same final and initial states, meaning the entropy changes remain unaffected by process reversibility. This principle underscores the path-independence property of entropy in thermodynamics.

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Most popular questions from this chapter

Acetylene gas, \(\mathrm{C}_{2} \mathrm{H}_{2}(g),\) is used in welding. (a) Write a balanced equation for the combustion of acetylene gas to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\). (b) How much heat is produced in burning \(1 \mathrm{~mol}\) of \(\mathrm{C}_{2} \mathrm{H}_{2}\) under standard conditions if both reactants and products are brought to \(298 \mathrm{~K}\) ? (c) What is the maximum amount of useful work that can be accomplished under standard conditions by this reaction?

(a) If you are told that the entropy of a certain system is zero, what do you know about the system and the temperature? (b) The energy of a gas is increased by heating it. Using \(\mathrm{CO}_{2}\) as an example, illustrate the different ways in which additional energy can be distributed among the molecules of the gas. (c) \(\mathrm{CO}_{2}(g)\) and \(\mathrm{Ar}(g)\) have nearly the same molar mass. At a given temperature, will they have the same number of microstates? Explain.

Use data from Appendix \(\mathrm{C}\) to calculate the equilibrium constant, \(K,\) at \(298 \mathrm{~K}\) for each of the following reactions: (a) \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{Hl}(g)\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{6}(g)\)

For the isothermal expansion of a gas into a vacuum, \(\Delta E=0, q=0\), and \(w=0\). (a) Is this a spontaneous process? (b) Explain why no work is done by the system during this process. (c) In thermodynamics, what is the "driving force" for the expansion of the gas?

The normal boiling point of \(\mathrm{Br}_{2}(l)\) is \(58.8{ }^{\circ} \mathrm{C},\) and its molar enthalpy of vaporization is \(\Delta H_{\text {vap }}=29.6 \mathrm{~kJ} /\) mol. (a) When \(\mathrm{Br}_{2}(l)\) boils at its normal boiling point, does its entropy increase or decrease? (b) Calculate the value of \(\Delta S\) when \(1.00 \mathrm{~mol}\) of \(\mathrm{Br}_{2}(l)\) is vaporized at \(58.8{ }^{\circ} \mathrm{C}\).
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