Chemistry - Thermodynamics - II Question with Solution | TestHub
ChemistryThermodynamics - IICalculation of DSMedium2 minQB
ChemistryMediumsingle choice
The entropy change of 2.0 moles of an ideal gas with adiabatic exponent γ = 5/3, when its volume increases 4.0 times while its pressure drops 8.0 times, is (ln 2 = 0.7, R = 8.314 J/K-mol)
Options:
Answer:
B
Solution:
CV = R/(γ - 1) = 1.5R, CP = 2.5R. Δ S = n[CVln(P2/P1) + CPln(V2/V1)] = 2R[1.5ln(1/8) + 2.5ln 4] = 2R[-4.5ln 2 + 5ln 2] = 2R × 0.5ln 2 = 8.314 × 0.7 = +5.82 J/K. Matches (b).
Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Calculation of DS
⏱ 2mℹ️ Source: QB
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