Chemistry - Thermodynamics - I Question with Solution | TestHub
ChemistryThermodynamics - IAdiabatic Process/PolytropicMedium2 minQB
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A real gas is subjected to an adiabatic process from (2 bar, 40 L, 300 K) to (4 bar, 30 L, 300 K) against a constant pressure of 4 bar. The enthalpy change for the process is
Options:
Answer:
C
Solution:
Adiabatic, so q = 0 and Δ U = w = -PextΔ V = -4 × (30 - 40) = +40 bar-L = 4000 J (1 bar-L = 100 J). Δ (PV) = 4 × 30 - 2 × 40 = 40 bar-L = 4000 J. Δ H = Δ U + Δ(PV) = 4000 + 4000 = 8000 J. Matches (c).
Stream:JEESubject:ChemistryTopic:Thermodynamics - ISubtopic:Adiabatic Process/Polytropic
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