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

Short Answer

Expert verified
The second law states entropy increases; mathematically: \( \Delta S \geq 0 \).

Step by step solution

01

Understand the Second Law in Words

The second law of thermodynamics states that in any cyclic process, the entropy will either increase or remain the same. It signifies that natural processes have a preferred direction of progression, meaning that the universe tends towards a state of maximum entropy.
02

Expressing the Second Law Mathematically

The second law of thermodynamics can be mathematically represented by the inequality: \[ \Delta S \geq 0 \] where \( \Delta S \) is the change in entropy. For reversible processes, the change in entropy \( \Delta S = 0 \), whereas for irreversible processes, \( \Delta S > 0 \).
03

Explanation of Mathematical Expression

The inequality \( \Delta S \geq 0 \) implies that the total entropy of an isolated system cannot decrease over time. Entropy remains constant for reversible processes and increases for irreversible processes.

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Key Concepts

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

Entropy
Entropy is a measure of disorder or randomness within a system. It reflects how energy is distributed among the particles in a system.
According to the second law of thermodynamics, the entropy of an isolated system will tend to increase over time, meaning systems naturally progress towards more disordered states.
Imagine a tidy room that left uncared for, becomes messy. Similarly, in physical systems, the tendency is for energy to spread out as evenly as possible, increasing entropy.
This concept of entropy relates to the second law, as it emphasizes how natural processes have a direction, always moving towards higher entropy. This foundational idea helps us understand why certain processes, like mixing substances, are irreversible. Understanding entropy is crucial in many fields, from physics to information theory.
Reversible Processes
Reversible processes are idealized processes that can be reversed without leaving any change in the system or the surroundings. Imagine a swing that always returns to the same height with every push—it perfectly swings back and forth.
In thermodynamics, if a process is reversible, the change in entropy (\( \Delta S \)) is zero. This means the system can go back and forth without losing energy to its surroundings.
While truly reversible processes don't exist in the real world, they are a valuable theoretical tool, offering insights into how real processes deviate from perfect reversibility.
  • Key Feature: No net change in entropy
  • Significance: Ideal standard to measure real processes
Irreversible Processes
Irreversible processes are the opposite of reversible processes. They are processes that cannot completely return to their original state without leaving changes in the environment. For instance, think of a cracked egg.
Once you break it, you cannot unbreak it perfectly. In such processes, entropy increases (\( \Delta S > 0 \)), showing the system's progression towards greater disorder.
Practically, all real-world processes are irreversible, like friction, chemical reactions, or heat transfer between different temperatures.
Recognizing the irreversibility of processes helps us understand the natural direction of time, emphasizing the flow from order to chaos.
  • Key Feature: Increase in entropy
  • Examples: Heat flowing from hot to cold, mixing of gases
Cyclic Process
A cyclic process is one where a system undergoes a series of changes and finally returns to its initial state. Imagine a cycle in a heat engine, which eventually comes back to its starting point.
In terms of entropy, if a system returns to its initial state, that means any changes in entropy over one complete cycle sum to zero. However, in practice, no process is perfectly cyclic due to real-world inefficiencies.
These processes are significant in engines and refrigerators, where cycles allow the system to perform work or transfer heat repeatedly. Even though each cycle is a return journey, some loss or gain always occurs if the process is not perfectly reversible.
Mathematical Expression of Second Law
The mathematical expression of the second law of thermodynamics is given by the inequality:\[ \Delta S \geq 0 \]Here, \( \Delta S \) represents the change in entropy of a system. This formula encapsulates the law's essence, indicating that the entropy of an isolated system never decreases, it either increases or remains constant.
For reversible processes, entropy does not change (\( \Delta S = 0 \)). However, for irreversible processes, entropy increases (\( \Delta S > 0 \)).
This inequality solidifies the idea that energy disperses and spreads out, influencing process directions and outcomes. The concept is crucial for predicting and understanding energy transformation and transfer in various physical scenarios and systems.

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Most popular questions from this chapter

Predict whether the entropy change is positive or negative for each of the following reactions. Give reasons for your predictions. (a) \(2 \mathrm{KClO}_{4}(s) \longrightarrow 2 \mathrm{KClO}_{3}(s)+\mathrm{O}_{2}(g)\) (b) \(\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) (c) \(2 \mathrm{Na}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g)\) (d) \(\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{~N}(g)\)

Consider two carboxylic acids (acids that contain the \(-\) COOH group \(): \mathrm{CH}_{3} \mathrm{COOH}\) (acetic acid, \(\left.K_{\mathrm{a}}=1.8 \times 10^{-5}\right)\) and \(\mathrm{CH}_{2} \mathrm{ClCOOH}\) (chloroacetic acid, \(K_{\mathrm{a}}=1.4 \times 10^{-3}\) ). (a) Calculate \(\Delta G^{\circ}\) for the ionization of these acids at \(25^{\circ} \mathrm{C}\). (b) From the equation \(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ},\) we see that the contributions to the \(\Delta G^{\circ}\) term are an enthalpy term \(\left(\Delta H^{\circ}\right)\) and a temperature times entropy term \(\left(T \Delta S^{\circ}\right)\). These contributions are listed here for the two acids: \begin{tabular}{lcc} & \(\Delta \boldsymbol{H}^{\circ}(\mathbf{k} \mathbf{J} / \mathbf{m o l})\) & \(T \Delta S^{\circ}(\mathbf{k} \mathbf{J} / \mathbf{m o l})\) \\ \hline \(\mathrm{CH}_{3} \mathrm{COOH}\) & -0.57 & -27.6 \\\ \(\mathrm{CH}_{2} \mathrm{ClCOOH}\) & -4.7 & -21.1 \end{tabular} Which is the dominant term in determining the value of \(\Delta G^{\circ}\) (and hence \(K_{\mathrm{a}}\) of the acid)? (c) What processes contribute to \(\Delta H^{\circ} ?\) (Consider the ionization of the acids as a Bronsted acid-base reaction.) (d) Explain why the \(T \Delta S^{\circ}\) term is more negative for \(\mathrm{CH}_{3} \mathrm{COOH} .\)

In the metabolism of glucose, the first step is the conversion of glucose to glucose 6 -phosphate: glucose \(+\mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow\) glucose 6 -phosphate \(+\mathrm{H}_{2} \mathrm{O}\) $$ \Delta G^{\circ}=13.4 \mathrm{~kJ} / \mathrm{mol} $$ Because \(\Delta G^{\circ}\) is positive, this reaction does not favor the formation of products. Show how this reaction can be made to proceed by coupling it with the hydrolysis of ATP. Write an equation for the coupled reaction, and estimate the equilibrium constant for the coupled process.

State whether the sign of the entropy change expected for each of the following processes will be positive or negative, and explain your predictions. (a) \(\mathrm{PCl}_{3}(l)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(g)\) (b) \(2 \mathrm{HgO}(s) \longrightarrow 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}(g)\) (d) \(\mathrm{U}(s)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{UF}_{6}(s)\)

Heating copper(II) oxide at \(400^{\circ} \mathrm{C}\) does not produce any appreciable amount of Cu: $$ \mathrm{CuO}(s) \rightleftharpoons \mathrm{Cu}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \quad \Delta G^{\circ}=127.2 \mathrm{~kJ} / \mathrm{mol} $$ However, if this reaction is coupled to the conversion of graphite to carbon monoxide, it becomes spontaneous. Write an equation for the coupled process, and calculate the equilibrium constant for the coupled reaction.

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