Chemistry - Thermodynamics - I Question with Solution | TestHub
ChemistryThermodynamics - IAdiabatic Process/PolytropicHard2 minQB
ChemistryHardsingle choice
One mole of nitrogen at 300 K undergoes a polytropic process in which its volume increases 2 times while its pressure decreases 4 times. Treating nitrogen as an ideal diatomic gas with Cv = 5/2R and R = 8.3 J/K-mol, the heat exchanged by the gas in this process is
Options:
Answer:
A
Solution:
T2 = T1×(P2V2)/(P1V1) = 300 × 2/4 = 150 K, so Δ T = -150 K. PVx = const: 2x = 4, x = 2. C = Cv + R/(1 - x) = 2.5R - R = 1.5R = 12.45 J/K-mol. q = 12.45 × (-150) = -1867.5 J. Check: Δ U = 2.5(8.3)(-150) = -3112.5 J and wby gas = (nR(T1 - T2))/(x - 1) = +1245 J, so q = Δ U + wby = -1867.5 J. Heat is released even though the gas expands. Matches (a).
Stream:JEESubject:ChemistryTopic:Thermodynamics - ISubtopic:Adiabatic Process/Polytropic
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