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Chapter 18: Problem 96

Comment on the correctness of the analogy sometimes used to relate a student's dormitory room becoming untidy to an increase in entropy.

Short Answer

Expert verified
The analogy of comparing a student's dormitory room becoming untidy to an increase in entropy partially holds up, because both cases involve an increase in disorder or randomness. However, it's an oversimplification because in thermodynamics, entropy also involves the dispersal of energy, and a system naturally tends towards a state of higher entropy, neither of which are captured in the analogy.

Step by step solution

01

Understanding the concept of entropy

Entropy in thermodynamics refers to the degree of disorder or randomness in a system. It is a state function, meaning its value is determined by the current state of the system, rather than how the system got to that state.
02

Relating entropy to the dormitory room

In the analogy, a tidy room can be considered a state of low entropy (organized) and an untidy room can be considered a state of high entropy (disorganized). However, it's important to note that in the context of thermodynamics, 'order' and 'disorder' are about the arrangements of particles at a microscopic level, not necessarily about the 'neatness' of the objects in a room.
03

Assessing the limitations of the analogy

While the dormitory room analogy can help in understanding the basic idea of entropy, it may oversimplify the concept and ignore some key aspects of it. In thermodynamics, increasing entropy is related to the dispersal or spreading out of energy, not just the disorder in arrangement of items. Additionally, in thermodynamics, a process naturally tends towards a state of higher entropy, but this might not necessarily be true for a dorm room; a student might regularly tidy up the room, which would decrease its 'entropy'.

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Most popular questions from this chapter

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?

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?

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

Find the temperatures at which reactions with the following \(\Delta H\) and \(\Delta S\) values would become spontaneous: (a) \(\Delta H=-126 \mathrm{~kJ} / \mathrm{mol}, \Delta S=84 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\) (b) \(\Delta H=-11.7 \mathrm{~kJ} / \mathrm{mol}, \Delta S=-105 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\).

The equilibrium constant \(\left(K_{P}\right)\) for the reaction $$ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) $$ is 4.40 at \(2000 \mathrm{~K}\). (a) Calculate \(\Delta G^{\circ}\) for the reaction. (b) Calculate \(\Delta G\) for the reaction when the partial pressures are \(P_{\mathrm{H}_{2}}=0.25 \mathrm{~atm}, P_{\mathrm{CO}_{2}}=0.78 \mathrm{~atm}\) \(P_{\mathrm{H}_{2} \mathrm{O}}=0.66 \mathrm{~atm},\) and \(P_{\mathrm{CO}}=1.20 \mathrm{~atm}\)
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