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Chapter 17: Problem 9

At 1 atm, liquid water is heated above \(100^{\circ} \mathrm{C}\). For this process, which of the following choices (i-iv) is correct for \(\Delta S_{\text {surr }}\) ? \(\Delta S\) ? \(\Delta S_{\text {univ }} ?\) Explain each answer. i. greater than zero ii. less than zero iii. equal to zero iv. cannot be determined

Short Answer

Expert verified
For the process of heating liquid water above \(100^{\circ}\mathrm{C}\) at 1 atm pressure, the change in entropy (∆S) of the system is greater than zero (choice i), as the water gains energy and becomes more disordered. The change in entropy of the surroundings (∆S_{surr}) is less than zero (choice ii), as the surroundings lose heat to the system. The change in entropy of the universe (∆S_{univ}) cannot be determined (choice iv), as more information is needed to determine the overall change in entropy of the universe. There is no choice that corresponds to a change in entropy equal to zero (choice iii).

Step by step solution

01

Analyze the heating process of liquid water

In this process, liquid water is being heated above its boiling point at 1 atm pressure, meaning it will turn into steam. We should know that when a substance changes from a liquid to a gas, its entropy increases as the molecules become more disordered and occupy a much larger volume. Therefore, the enthalpy change (∆H) for the system is positive, as energy is supplied to the water.
02

Determine the change in entropy of the system (∆S)

Since the water is being heated, it's gaining energy and turning into steam. This results in an increase in disorder for the water, which means that the change in entropy of the system, ∆S, is greater than zero. This corresponds to choice i.
03

Determine the change in entropy of the surroundings (∆S_surr)

Given that the system is absorbing heat, the surroundings are transferring heat to the system. The relationship between heat transfer (q) and entropy change for the surroundings can be described by the equation: ∆S_surr = -q/T Since heat is being transferred from the surroundings, q is negative, meaning that the change in entropy of the surroundings, ∆S_surr, is less than zero. This corresponds to choice ii.
04

Determine the change in entropy of the universe (∆S_univ)

Considering both the system and the surroundings, we can calculate the change in entropy of the universe (∆S_univ) as: ∆S_univ = ∆S + ∆S_surr Since ∆S > 0 and ∆S_surr < 0, it cannot be determined whether the overall entropy change of the universe is greater than zero, less than zero, or equal to zero without more information. Therefore, the answer for ∆S_univ is choice iv.
05

Explaining each choice

For the given process of heating liquid water above 100°C at 1 atm pressure, we determined the following changes in entropy: - ∆S (system): greater than zero, as the water gains energy and becomes more disordered (choice i). - ∆S_surr (surroundings): less than zero, as the surroundings lose heat to the system (choice ii). - ∆S_univ (universe): cannot be determined, as more information is needed to determine the overall change in entropy of the universe (choice iv). There is no choice that corresponds to a change in entropy equal to zero (choice iii).

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Most popular questions from this chapter

If wet silver carbonate is dried in a stream of hot air, the air must have a certain concentration level of carbon dioxide to prevent silver carbonate from decomposing by the reaction $$ \mathrm{Ag}_{2} \mathrm{CO}_{3}(s) \rightleftharpoons \mathrm{Ag}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g) $$ \(\Delta H^{\circ}\) for this reaction is \(79.14 \mathrm{~kJ} / \mathrm{mol}\) in the temperature range of 25 to \(125^{\circ} \mathrm{C}\). Given that the partial pressure of carbon dioxide in equilibrium with pure solid silver carbonate is \(6.23 \times 10^{-3}\) torr at \(25^{\circ} \mathrm{C}\), calculate the partial pressure of \(\mathrm{CO}_{2}\) necessary to prevent decomposition of \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) at \(110 .{ }^{\circ} \mathrm{C}\). (Hint: Manipulate the equation in Exercise 79.)

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