Chemistry - Thermodynamics - I Question with Solution | TestHub
ChemistryThermodynamics - IAdiabatic Process/PolytropicMedium2 minQB
ChemistryMediumnumerical
The number of degrees of freedom of molecules in a gas whose molar heat capacity is 29 J/mol-K in the process: PT = constant is
Answer:
3.00
Solution:
PT = const with PV = RT (1 mole) gives T²/V = const, i.e. V ∝ T², so dV = 2cT dT. Work per mole: W = ∫P dV = ∫(RT/V)(2cT) dT = 2RΔT. Hence C = ΔU/ΔT + W/ΔT = Cv + 2R ⇒ Cv = 29 − 2 × 8.314 = 12.37 J K⁻¹ mol⁻¹. Degrees of freedom: f = 2Cv/R = 2 × 12.37/8.314 ≈ 3.
Stream:JEESubject:ChemistryTopic:Thermodynamics - ISubtopic:Adiabatic Process/Polytropic
⏱ 2mℹ️ Source: QB
Doubts & Discussion
Loading discussions...
