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Chapter 19: Problem 108

When most elastomeric polymers (e.g., a rubber band) are stretched, the molecules become more ordered, as illustrated here:Suppose you stretch a rubber band. (a) Do you expect the entropy of the system to increase or decrease? (b) If the rubber band were stretched isothermally, would heat need to be absorbed or emitted to maintain constant temperature?

Short Answer

Expert verified
(a) When a rubber band is stretched, its molecules become more ordered, causing the entropy of the system to decrease. (b) In an isothermal stretching process, heat would need to be emitted (released to the surroundings) to maintain constant temperature, as this increases the entropy of the surroundings to ensure a non-negative total entropy change.

Step by step solution

01

Understanding entropy increase or decrease

When a rubber band is stretched, its molecules become more ordered. Entropy is a measure of the disorder of the system. As the system becomes more ordered, the entropy decreases. Therefore, when the rubber band is stretched, the entropy of the system decreases.
02

Heat absorbed or emitted in isothermal stretching

In an isothermal process, the temperature remains constant. According to the second law of thermodynamics, the total entropy change in an isothermal process is given by: \(\Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings} \geq 0\). Since we found in the previous step that when the rubber band is stretched, its entropy decreases \((\Delta S_{system} < 0)\), the entropy of the surroundings \((\Delta S_{surroundings})\) must increase to ensure that the total entropy change is non-negative. This increase in the entropy of the surroundings occurs when heat is transferred from the system to the surroundings. Therefore, if the rubber band were stretched isothermally, heat would need to be emitted (released to the surroundings) to maintain constant temperature.

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Key Concepts

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

Isothermal Process
An isothermal process is a concept where a system undergoes changes while maintaining a constant temperature throughout. This is important when discussing thermodynamic processes, as temperature directly affects the energy and dynamics of molecules. During an isothermal process:
  • The system absorbs or releases heat energy, making sure that its temperature stays the same.
  • Though temperature remains constant, other properties of the system, such as pressure or volume, can change.
  • For a perfect gas, the amount of heat involved is exactly equal to the work done by or on the system.
In the specific case of stretching a rubber band isothermally, adjustments in energy must keep the temperature unchanged, leading to a transfer of heat outward as detailed in the exercise above.
Second Law of Thermodynamics
The second law of thermodynamics is a fundamental principle that governs the directional flow of energy and the nature of various processes. Essentially, it states that the total entropy of an isolated system can never decrease over time. This law highlights:
  • The tendency of energy spread and disorder within a closed system.
  • The idea that no process is perfectly efficient, as some energy is always lost to the surroundings.
  • Entropy always increases or stays the same; it does not spontaneously decrease, ensuring natural processes are irreversible.
In practice, when stretching a rubber band, the second law ensures that any decrease in the system's entropy (due to the increased orderliness of rubber molecules) is balanced by an increase in the surroundings' entropy when heat is emitted.
Entropy Decrease
Entropy, fundamentally, is a measure of disorder within a system. When a system becomes more ordered, its entropy decreases, which is significant in the context of thermodynamics and energy transfer. Key points about entropy decrease include:
  • Increased order of molecules corresponds to a decline in entropy.
  • A decrease in entropy in one part of a system must be compensated by a corresponding increase elsewhere to satisfy the second law of thermodynamics.
  • Decreased entropy scenarios, like stretching rubber bands, involve energy change and transfer to maintain overall balance.
This illustrates that though a stretched rubber band becomes more organized, the emitted heat ensures the surrounding environment's entropy compensates, maintaining the second law's stipulations.

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Most popular questions from this chapter

A particular reaction is spontaneous at \(450 \mathrm{~K}\). The enthalpy change for the reaction is \(+34.5 \mathrm{~kJ} .\) What can you conclude about the sign and magnitude of \(\Delta S\) for the reaction?

A certain reaction has \(\Delta H^{\circ}=-19.5 \mathrm{~kJ}\) and \(\Delta S^{\circ}=\) \(+42.7 \mathrm{~J} / \mathrm{K} .\) (a) Is the reaction exothermic or endothermic? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (c) Calculate \(\Delta G^{\circ}\) for the reaction at \(298 \mathrm{~K} .\) (d) Is the reaction spontaneous at \(298 \mathrm{~K}\) under standard conditions?

One way to derive Equation \(19.3\) depends on the observation that at constant \(T\) the number of ways, \(W\), of arranging \(m\) ideal-gas particles in a volume \(V\) is proportional to the volume raised to the \(m\) power: \(W \propto V^{m}\) Use this relationship and Boltzmann's relationship between entropy and number of arrangements (Equation 19.5) to derive the equation for the entropy change for the isothermal expansion or compression of \(n\) moles of an ideal gas.

(a) 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. (b) You are told that the number of microstates for a system increases. What does this tell you about the entropy of the system?

Calculate \(\Delta S^{\circ}\) values for the following reactions by using tabulated \(S^{\circ}\) values from Appendix \(C\). In each case explain the sign of \(\Delta S^{\circ}\). (a) \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) \(\mathrm{K}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{KO}_{2}(s)\) (c) \(\mathrm{Mg}(\mathrm{OH})_{2}(s)+2 \mathrm{HCl}(\mathrm{g}) \longrightarrow \mathrm{MgCl}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\) (d) \(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g)\)
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