Chemistry - Thermodynamics - II Question with Solution | TestHub
ChemistryThermodynamics - IIGibb's Free EnergyEasy2 minQB
ChemistryEasymultiple choice
The normal boiling point of a liquid is 400 K and Δ Hvap is 40 kJ/mol. Assume that Δ Hvap is independent of temperature and pressure. The correct statement(s) is/are
Options:(select one or more)
Answer:
A, D
Solution:
At 1 atm and 400 K, Δ S = 40000/400 = 100 J/K-mol and Δ G = 0. At pressure P: Δ S = 100 - Rln P and Δ G = RTln P. At 0.25 atm, Δ S = 100 + 11.5 = 111.5 > 100 (a true) and Δ G = 8.314 × 400 × ln 0.25 = -4610 J < 0 (b false). At 4 atm, Δ S = 88.5 < 100 (c false) and Δ G = +4610 J > 0 (d true). Ans: (a), (d).
Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Gibb's Free Energy
⏱ 2mℹ️ Source: QB
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