For a spontaneous process, ΔSsys = –300 J/K. What does this tell us about ΔSsurr?

07B

Question

For a spontaneous process, ΔSsys = –300 J/K. What does this tell us about ΔSsurr?

07B

BlackTom AI · Accepted Answer

First, restating the question and options: The problem asks, for a spontaneous process, given ΔSsys = –300 J/K, what does this tell us about ΔSsurr? The answer choices are: - ΔSsurr < –300 J/K - ΔSsurr = 300 J/K - ΔSsurr = –300 J/K - ΔSsurr > 300 J/K Now, evaluate each option through the lens of the second law of thermodynamics. A process is spontaneous when the total entropy change ΔStotal = ΔSsys + ΔSsurr is greater than zero. Option 1: ΔSsurr < –300 J/K. If the surroundings’ entropy change were less than –300 J/K, the sum ΔSsys + ΔSsurr would be less than (–300) + (–300) = –600 J/K, which is negative. A negative total entropy change would imply a non-spontaneous or impossible process under the given conditions, so this option cannot describe a spontaneous process. Option 2: ΔSsurr = 300 J/K. If ΔSsurr were exactly +300 J/K, then ΔStotal = (–300) + (300) = 0. A zero total entropy change corresponds to thermodynamic equilibrium, not a spontaneous process. Therefore this choice does not describe spontaneity. Option 3: ΔSsurr = –300 J/K. Here the total would be ΔStotal = (–300) + (–300) = –600 J/K, a clearly negative total. This would indicate a strongly non-spontaneous process, contradicting the requirement for spontaneity. So this is incorrect. Option 4: ΔSsurr > 300 J/K. In this case, ΔStotal = (–300) + ΔSsurr > (–300) + 300 = 0, i.e., the total entropy increases despite the system losing entropy. Any ΔSsurr greater than 300 J/K yields a positive total entropy change, which aligns with the condition for spontaneity. In summary, to satisfy spontaneity with ΔSsys = –300 J/K, the surroundings must contribute more than +300 J/K to the entropy, making ΔSsurr > 300 J/K. This eliminates the first three options and confirms the fourth as the correct condition for spontaneity. Note on potential confusions: Some students might think only the system’s entropy change matters, but the second law requires considering the universe (system plus surroundings). A negative ΔSsys can still lead to a spontaneous process if the surroundings’ entropy increase is sufficiently large to overcome the decrease in the system’s entropy. Conversely, achieving exactly zero total entropy (ΔStotal = 0) corresponds to equilibrium, not spontaneity, which is why ΔSsurr = 300 J/K is not acceptable here.

For a spontaneous process, ΔSsys = –300 J/K. What does this tell us about ΔSsurr? 07B单项选择题

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