Chemistry - Thermodynamics - I Question with Solution | TestHub

ChemistryThermodynamics - IFirst law of ThermodynamicsHard2 minPYQ_2023
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1mole of ideal gas is allowed to expand reversibly and adiabatically from a temperature of27°C. The work done is3 kJ mol-1. The final temperature of the gas is _____K(Nearest integer). GivenCv=20 J mol-1 K-1

Answer:
150
Solution:

From first law of thermodynamics:
∆U = q + W
We know for adiabatic process,
∴q=0

∴ΔU=w⇒m × Cv × ∆T = w

⇒1×20×T2-300=-3000

⇒T2-300=-150

∴T2=150 K

Stream:JEESubject:ChemistryTopic:Thermodynamics - ISubtopic:First law of Thermodynamics
⏱ 2mℹ️ Source: PYQ_2023

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