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 whose adiabatic exponent γ = 1.50, if as a result of a certain process, the gas volume increased 2.0 times while the pressure dropped 4.0 times, is (ln 2 = 0.7)
Options:
Answer:
A
Solution:
CV = R/(γ - 1) = 2R, CP = γ R/(γ - 1) = 3R. Δ S = n[CVln(P2/P1) + CPln(V2/V1)] = 2R[2ln(1/4) + 3ln 2] = 2R[-4ln 2 + 3ln 2] = -2Rln 2 = -2 × 8.314 × 0.7 = -11.64 J/K. Matches (a).
Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Calculation of DS
⏱ 2mℹ️ Source: QB
Doubts & Discussion
Loading discussions...
