Chemistry - Thermodynamics - II Question with Solution | TestHub
ChemistryThermodynamics - IICalculation of DSMedium2 minQB
ChemistryMediumsingle choice
A quantity of 1.6 g helium gas is expanded adiabatically 3.0 times and then compressed isobarically to the initial volume. Assume ideal behaviour of gas and both the processes reversible. The entropy change of the gas in this process is (ln 3 = 1.1)
Options:
Answer:
C
Solution:
n = 1.6/4 = 0.4 mol. The reversible adiabatic step has Δ S = 0. The isobaric step goes from 3V back to V: Δ S = nCPln(V/3V) = 0.4 × (5R/2) × (-ln 3). With R = 2 cal/K-mol, CP = 5: Δ S = 0.4 × 5 × (-1.1) = -2.2 cal/K. Matches (c).
Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Calculation of DS
⏱ 2mℹ️ Source: QB
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