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Chapter 18: Problem 85

Entropy has sometimes been described as "time's arrow" because it is the property that determines the forward direction of time. Explain.

Short Answer

Expert verified
The entropy of an isolated system either remains constant or increases due to all spontaneous processes, but never decreases. This continual increase gives a direction to time, equating to the perception of time's arrow always pointing from the past to the future. Therefore, entropy is often described as 'time's arrow'.

Step by step solution

01

Understanding of Entropy

Entropy is a concept in the field of thermodynamics that measures the number of specific ways in which a system may be arranged, also referred to as its randomness or disorder. The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time, and is constant if and only if all processes are reversible. Isolated systems spontaneously evolve towards thermodynamic equilibrium, the state with maximum entropy.
02

Introducing the Concept of 'Time's Arrow'

The term 'time's arrow' is a concept that is used to refer to the one-directional or asymmetric nature of time. It's usually associated with the idea of time progressing from the past to the future, and it doesn't, in normal conditions, 'turn back'.
03

Link between Entropy and 'Time's Arrow'

Because the entropy of an isolated system either remains constant or increases in all spontaneous processes, and never decreases, entropy can be considered a measure of the passage of time. The continuing increase of entropy seems to give a direction to time, leading to our perception of time's arrow always pointing from past to future.

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Most popular questions from this chapter

Ammonium nitrate \(\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right)\) dissolves spontaneously and endothermically in water. What can you deduce about the sign of \(\Delta S\) for the solution process?

Explain the difference between \(\Delta G\) and \(\Delta G^{\circ}\).

For the autoionization of water at \(25^{\circ} \mathrm{C}\), $$ \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ \(K_{\mathrm{w}}\) is \(1.0 \times 10^{-14}\). What is \(\Delta G^{\circ}\) for the process?

For reactions carried out under standard-state conditions, Equation (18.10) takes the form \(\Delta G^{\circ}=\Delta H^{\circ}-\) \(T \Delta S^{\circ} .\) (a) Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are independent of temperature, derive the equation $$\ln \frac{K_{2}}{K_{1}}=\frac{\Delta H^{\circ}}{R}\left(\frac{T_{2}-T_{1}}{T_{1} T_{2}}\right)$$ where \(K_{1}\) and \(K_{2}\) are the equilibrium constants at \(T_{1}\) and \(T_{2}\), respectively. (b) Given that at \(25^{\circ} \mathrm{C} K_{\mathrm{c}}\) is \(4.63 \times 10^{-3}\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) \quad \Delta H^{\circ}=58.0 \mathrm{~kJ} / \mathrm{mol} $$ calculate the equilibrium constant at \(65^{\circ} \mathrm{C}\)

For each pair of substances listed here, choose the one having the larger standard entropy value at \(25^{\circ} \mathrm{C}\). The same molar amount is used in the comparison. Explain the basis for your choice. (a) \(\operatorname{Li}(s)\) or \(\operatorname{Li}(l)\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) or \(\mathrm{CH}_{3} \mathrm{OCH}_{3}(l)\) (c) \(\operatorname{Ar}(g)\) or \(\operatorname{Xe}(g)\) (d) \(\mathrm{CO}(g)\) or \(\mathrm{CO}_{2}(g)\) (e) \(\mathrm{O}_{2}(g)\) or \(\mathrm{O}_{3}(g)\) (f) \(\mathrm{NO}_{2}(g)\) or \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\)
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