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Chapter 18: Problem 7

State the second law of thermodynamics in words and express it mathematically.

Short Answer

Expert verified
The second law states that entropy in a closed system always increases. Mathematically: \(\Delta S \geq 0\).

Step by step solution

01

Understanding the Second Law of Thermodynamics

The second law of thermodynamics is a fundamental principle of physics that describes the direction of natural processes. In simple words, it states that in any energy transfer or transformation, the total entropy of a closed system will always increase over time, approaching a maximum value.
02

Expressing the Law Mathematically

Mathematically, the second law of thermodynamics can be expressed in terms of entropy change: \[ \Delta S = S_{final} - S_{initial} \geq 0 \]This inequality indicates that the change in entropy (\(\Delta S\)) of an isolated system is never negative, implying that entropy either increases or remains constant.

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Key Concepts

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

Entropy
Entropy is a key concept in the second law of thermodynamics. It measures the level of disorder or randomness in a system. Imagine it like the messiness of a room; the more disordered the room, the higher its entropy. There are a few important points to understand about entropy:
  • When energy is transferred or transformed, some of it inevitably becomes dispersed or spread out, increasing entropy.
  • The second law of thermodynamics states that the total entropy of a closed system will always increase over time. This means processes naturally progress towards a state of greater disorder.
  • Entropy can only remain constant or increase; it never decreases. The system grows more disorderly unless energy is added to organize it.
Understanding entropy helps explain why certain processes are irreversible in nature, like how heat flows from a hot object to a cold one. It pushes systems toward a state of equilibrium.
Energy Transfer
Energy transfer is a fundamental concept in understanding how systems evolve over time according to the second law of thermodynamics. Whenever there is energy transfer or transformation in a closed system:
  • Some energy becomes less useful, often converting into heat or other forms which are less orderly.
  • This transformation increases the entropy of the system, as energy spreads out and chaos increases.
  • The system tends towards equilibrium, where energy is more evenly distributed and no further work is possible without external input.
Energy transfer is crucial for understanding natural processes, such as the work done by engines or biological systems. It explains why perpetual motion machines are impossible, as energy will always become more evenly spread and less available for doing work without adding new energy.
Closed System
A closed system is one where neither matter nor energy is allowed to enter or exit. It is a key setting for applying the second law of thermodynamics.
  • Within a closed system, any changes must occur using the internal energy and matter already present.
  • Because of this isolation, the entropy of a closed system can only remain constant or increase, dictated by the transformations within.
  • No energy input means that the system will naturally progress toward greater disorder, aligning with the second law.
Understanding closed systems helps explain environments where energy transformations occur without external factors, like a sealed thermos. The concept underscores the inevitable increase in entropy over time in such systems.

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Most popular questions from this chapter

Calculate the pressure of \(\mathrm{O}_{2}\) (in atm) over a sample of \(\mathrm{NiO}\) at \(25^{\circ} \mathrm{C}\) if \(\Delta G^{\circ}=212 \mathrm{~kJ} / \mathrm{mol}\) for the reaction $$ \mathrm{NiO}(s) \rightleftharpoons \mathrm{Ni}(s)+\frac{1}{2} \mathrm{O}_{2}(g) $$

A \(74.6-\mathrm{g}\) ice cube floats in the Arctic Sea. The temperature and pressure of the system and surroundings are at 1 atm and \(0^{\circ} \mathrm{C}\). Calculate \(\Delta S_{\mathrm{sys}}, \Delta S_{\text {surr }},\) and \(\Delta S_{\text {univ }}\) for the melting of the ice cube. What can you conclude about the nature of the process from the value of \(\Delta S_{\text {univ }}\) ? (The molar heat of fusion of water is \(6.01 \mathrm{~kJ} / \mathrm{mol} .)\)

Why is it more convenient to predict the direction of a reaction in terms of \(\Delta G_{\text {sys }}\) instead of \(\Delta S_{\text {univ }}\) ? Under what conditions can \(\Delta G_{\text {sys }}\) be used to predict the spontaneity of a reaction?

At \(25^{\circ} \mathrm{C}, \Delta G^{\circ}\) for the process $$ \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g) $$ is \(8.6 \mathrm{~kJ} / \mathrm{mol}\). Calculate the vapor pressure of water at this temperature.

Calculate \(K_{P}\) for the following reaction at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) \quad \Delta G^{\circ}=2.60 \mathrm{~kJ} / \mathrm{mol} $$
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