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Comment on the statement: "Just talking about entropy increases its value in the universe."

Short Answer

Expert verified
No, just talking about entropy does not increase its value in the universe. The statement is a metaphor indicating that discussing such complex topics could lead to confusion or misunderstanding, metaphorically 'increasing entropy' in the sense of information disorder.

Step by step solution

01

Understand the Concept of Entropy

The entropy in thermodynamics is a physical quantity that represents the degree of randomness or disorder of a system. It's often denoted by the letter 'S'. According to the second law of thermodynamics, in any energy exchange, if no energy enters or leaves the system, the potential energy of the state will always be less than that of the initial state, which implies that the entropy of an isolated system always increases or remains constant.
02

Interpret the Phrase

The statement 'Just talking about entropy increases its value in the universe' is more of a metaphoric representation rather than a scientific fact. The act of talking about entropy or any other topic doesn't physically contribute to the total entropy of the universe. This statement may refer metaphorically to the idea that discussing complex or abstract topics, such as entropy, can lead to a state of 'disorder' or 'confusion', especially for those not familiar with such subjects. The 'increased entropy' is, therefore, on the level of information understood or misunderstood by the listener.
03

Conclude the Discussion

Based on the analysis and interpretation, we conclude that speaking about entropy, or anything else, doesn't impact the total entropy of the universe in the context of the second law of thermodynamics. The statement should be understood as a metaphor about communication and comprehension rather than a factual scientific assertion.

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Most popular questions from this chapter

In the Mond process for the purification of nickel, carbon monoxide is reacted with heated nickel to produce \(\mathrm{Ni}(\mathrm{CO})_{4},\) which is a gas and can therefore be separated from solid impurities: $$ \mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) $$ Given that the standard free energies of formation of \(\mathrm{CO}(g)\) and \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) are \(-137.3 \mathrm{~kJ} / \mathrm{mol}\) and \(-587.4 \mathrm{~kJ} / \mathrm{mol}\), respectively, calculate the equilibrium constant of the reaction at \(80^{\circ} \mathrm{C}\). Assume that \(\Delta G_{f}^{\circ}\) is temperature independent.

Arrange the following substances ( 1 mole each) in order of increasing entropy at \(25^{\circ} \mathrm{C}:\) (a) \(\mathrm{Ne}(g)\), (b) \(\mathrm{SO}_{2}(g),\) (c) \(\mathrm{Na}(s)\) (d) \(\mathrm{NaCl}(s)\) (e) \(\mathrm{H}_{2}(g)\). Give the reasons for your arrangement.

A carbon monoxide (CO) crystal is found to have entropy greater than zero at absolute zero of temperature. Give two possible explanations for this observation.

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

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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