One mole of an ideal monoatomic gas is heated in a process PV 2 = constant. By what amount heat is absorbed in the process in a 40°C rise in temperature? (R = 8.314 J/K-mol)

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One mole of an ideal monoatomic gas is heated in a process PV² = constant. By what amount heat is absorbed in the process in a 40°C rise in temperature? (R = 8.314 J/K-mol)

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Answer:

Solution:

For PV^x = constant, C = C_V + R/(1-x). With x = 2: C = 1.5R - R = 0.5R. Q = 0.5 × 8.314 × 40 = 166 J. Matches (a).

Stream:JEESubject:ChemistryTopic:Thermodynamics - ISubtopic:Adiabatic Process/Polytropic

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